XI Mathematics IIT JEE Advanced Study Package 2014 159

BRILLIANT PUBLIC SCHOOL, SITAMARHI

(Affiliated up to +2 level to C.B.S.E., New Delhi)

Class-XI IIT-JEE Advanced Mathematics Study Package Session: 2014-15 Office: Rajopatti, Dumra Road, Sitamarhi (Bihar), Pin-843301 Ph.06226-252314 , Mobile:9431636758, 9931610902 Website: www.brilliantpublicschool.com; E-mail: [email protected]

STUDY PACKAGE Target: IIT-JEE (Advanced) SUBJECT: MATHEMATICS-XI Chapter 1 Trigonometric Ratio and Identity 2 Trigonometric Equations 3 Properties of Triangle 4 Functions 5 Complex Numbers 6 Quadratic Equations 7 Permutations and Combinations 8 Binomial Theorem 9 Probability 10 Progressions 11 Straight Lines 12 Circles 13 Parabola, Ellipse and Hyperbola 14 Highlights on Conic Sections 15 Vector Algebra and 3-D Geometry 16 Limits 17 Differentiation

Pages 19 14 24 40 37 23 19 24 36 25 21 23 68 25 62 18 17

Exercises 5 3 5 5 5 6 5 8 5 5 5 5 15 8 5 2

STUDY PACKAGE Target: IIT-JEE (Advanced) SUBJECT: MATHEMATICS TOPIC: 1 XI M 1. Trigonometric Ratio and Identity Index: 1. Key Concepts 2. Exercise I to V 3. Answer Key 4. Assertion and Reasons 5. 34 Yrs. Que. from IIT-JEE 6. 10 Yrs. Que. from AIEEE

1

Trigonometric Ratios & Identities 1.

Basic Trigonometric Identities: (a) sin² θ + cos² θ = 1; −1 ≤ sin θ ≤ 1; −1 ≤ cos θ ≤ 1 ∀ θ ∈ R π   (b) sec² θ − tan² θ = 1 ; sec θ ≥ 1 ∀ θ ∈ R – (2n + 1) , n ∈ Ι  2  

(c) cosec² θ − cot² θ = 1 ; cosec θ ≥ 1 ∀ θ ∈ R – {nπ , n ∈ Ι} Solved Example # 1 Prove that (i) cos4A – sin4A + 1 = 2 cos2A (ii) Solution (i)

(ii)

tan A + sec A − 1 1+ sin A = tan A − sec A + 1 cos A

cos4A – sin4A + 1 = (cos2A – sin2A) (cos2A + sin2A) + 1 = cos2A – sin2A + 1 [∴ cos2A + sin2A = 1] 2 = 2 cos A tan A + sec A − 1 tan A − sec A + 1

=

tan A + sec A − (sec 2 A − tan 2 A ) tan A − sec A + 1

=

(tan A + sec A )(1 − sec A + tan A ) tan A − sec A + 1

= tan A + sec A =

1+ sin A cos A

Solved Example # 2 If sin x + sin2x = 1, then find the value of cos12x + 3 cos10x + 3 cos8x + cos6x – 1 Solution cos12x + 3 cos10x + 3 cos8x + cos6x – 1 = (cos4x + cos2x)3 – 1 = (sin2x + sinx)3 – 1 [∵ cos2x = sin x] =1–1=0 Solved Example # 3 If tan θ = m –

1 1 , then show that sec θ – tan θ = – 2m or 4m 2m

Solution Depending on quadrant in which θ falls, sec θ can be ±

So, if sec θ =

1 4m2 + 1 =m+ 4m 4m

2

4m 2 + 1 4m

1   1  and if sec θ = –  m + 4m   2m

⇒

sec θ – tan θ =

⇒

sec θ – tan θ = – 2m

Self Practice Problem 1.

Prove the followings : (i) cos6A + sin6A + 3 sin2A cos2A = 1 (ii) sec 2A + cosec2A = (tan A + cot A)2 (iii) sec 2A cosec2A = tan2A + cot 2A + 2 (iv) (tan α + cosec β)2 – (cot β – sec α)2 = 2 tan α cot β (cosec α + sec β) 1 1   1 − sin 2 α cos 2 α +   cos2α sin2α = 2 2 2 2  sec α − cos α cos ec α − sin α  2 + sin 2 α cos 2 α

(v)

m 2 + 2mn

2.

If sin θ =

2.

C irc ul ar sin θ =

m 2 + 2mn + 2n 2

PM OP

, then prove that tan θ =

Defi nit i o n cos θ =

Of

m 2 + 2mn 2mn + 2n 2

T rig o no met ri c

Func t i o ns:

OM OP

sin θ tan θ = cos θ , cos θ ≠ 0 cos θ cot θ = sin θ , sin θ ≠ 0

sec θ =

3.

1 , cos θ ≠ 0 cos θ

T ri g o no met ri c

cosec θ =

Fu nc t io ns

If θ is any angle, then − θ, 90 (a) sin (− θ) = − sin θ (b) sin (90° − θ) = cos θ (c) sin (90° + θ) = cos θ (d) sin (180° − θ) = sin θ (e) sin (180° + θ) = − sin θ (f) sin (270° − θ) = − cos θ (g) sin (270° + θ) = − cos θ (h) tan (90° − θ) = cot θ Solved Example # 4

1 , sin θ ≠ 0 sin θ

Of

A ll i ed

A ngl es:

± θ, 180 ± θ, 270 ± θ, 360 ± θ etc. are called ALLIED ANGLES. ; cos (− θ) = cos θ ; cos (90°− θ) = sin θ ; cos (90° + θ) = − sin θ ; cos (180° − θ) = − cos θ ; cos (180° + θ) = − cos θ ; cos (270° − θ) = − sin θ ; cos (270° + θ) = sin θ ; cot (90° − θ) = tan θ

Prove that cot A + tan (180º + A) + tan (90º + A) + tan (360º – A) = 0 (i) (ii) sec (270º – A) sec (90º – A) – tan (270º – A) tan (90º + A) + 1 = 0 Solution (i)

cot A + tan (180º + A) + tan (90º + A) + tan (360º – A) = cot A + tan A – cot A – tan A = 0 (ii) sec (270º – A) sec (90º – A) – tan (270º – A) tan (90º + A) + 1 = – cosec 2A + cot 2A + 1 = 0 Self Practice Problem 3. Prove that (i) sin 420º cos 390º + cos (–300º) sin (–330º) = 1 (ii) tan 225º cot 405º + tan 765º cot 675º = 0 3

4.

Graphs of Trigonometric functions: (a) y = sin x

x ∈ R; y ∈ [–1, 1]

(b) y = cos x x ∈ R; y ∈ [ – 1, 1]

(c) y = tan x x ∈ R – (2n + 1) π/2, n ∈ Ι ; y ∈ R

(d) y = cot x

x ∈ R – nπ , n ∈ Ι; y ∈ R

(e) y = cosec x

x ∈ R – nπ , n ∈ Ι ; y ∈ (− − ∞, − 1] ∪ [1, ∞ )

(f) y = sec x

x ∈ R – (2n + 1) π/2, n ∈ Ι ; y ∈ (− ∞, − 1] ∪ [1, ∞) 4

Solved Example # 5

Find number of solutions of the equation cos x = |x| Solution

Clearly graph of cos x & |x| intersect at two points. Hence no. of solutions is 2 Solved Example # 6

Find range of y = sin2x + 2 sin x + 3 ∀ x ∈ R Solution We know – 1 ≤ sin x ≤ 1 ⇒ 0 ≤ sin x +1 ≤ 2 ⇒ 2 ≤ (sin x +1)2 + 2 ≤ 6 Hence range is y ∈ [2, 6] Self Practice Problem 4 xy

4.

Show that the equation sec2θ =

5.

Find range of the followings. (i) y = 2 sin2x + 5 sin x +1∀ x ∈ R

Answer

[–2, 8]

y = cos2x – cos x + 1 ∀ x ∈ R

Answer

3   4 , 3  

(ii)

( x + y )2

is only possible when x = y ≠ 0

 3 − 1,  2  

6.

 2π  2π  Find range of y = sin x, x ∈  3 

5.

Trigonometric Functions of Sum or Difference of Two Angles:

Answer

(a)

sin (A ± B) = sinA cosB ± cosA sinB

(b) (c) (d)

cos (A ± B) = cosA cosB ∓ sinA sinB sin²A − sin²B = cos²B − cos²A = sin (A+B). sin (A− B) cos²A − sin²B = cos²B − sin²A = cos (A+B). cos (A − B)

(e)

tan A ± tan B tan (A ± B) = 1 ∓ tan A tan B

(f)

cot A cot B ∓ 1 cot (A ± B) = cot B ± cot A

5

(g)

tan A + tan B + tanC−tan A tan B tan C tan (A + B + C) = 1 − tan A tan B − tan B tan C− tan C tan A .

Solved Example # 7

Prove that (i) sin (45º + A) cos (45º – B) + cos (45º + A) sin (45º – B) = cos (A – B) (ii)

π   3π  + θ  = –1 tan  + θ  tan  4   4 

Solution

(i)

(ii)

Clearly sin (45º + A) cos (45º – B) + cos (45º + A) sin (45º – B) = sin (45º + A + 45º – B) = sin (90º + A – B) = cos (A – B) π   3π  + θ tan  + θ  × tan  4   4 

=

1 + tan θ −1 + tan θ × =–1 1 − tan θ 1 + tan θ

Self Practice Problem 3 5 , cos β = , then find sin (α + β) 5 13

33 63 , 65 65

7.

If sin α =

8.

Find the value of sin 105º

9.

Prove that 1 + tan A tan

6.

Fa c t o ris at i o n o f t he Su m o r D i fferenc e o f T w o Si nes o r Cosines:

Answer

–

Answer

2 2

A A = tan A cot – 1 = sec A 2 2

C+D C−D cos 2 2

(a)

sinC + sinD = 2 sin

(c)

cosC + cosD = 2 cos

C+D C−D cos 2 2

sinC − sinD = 2 cos

(d)

cosC − cosD = − 2 sin

Prove that sin 5A + sin 3A = 2sin 4A cos A Solution L.H.S. sin 5A + sin 3A = 2sin 4A cos A

C+D C−D sin 2 2

(b)

Solved Example # 8

[ ∵ sin C + sin D = 2 sin

3 +1

= R.H.S.

C+D C −D cos ] 2 2

Solved Example # 9

Find the value of 2 sin 3θ cos θ – sin 4θ – sin 2θ Solution 2 sin 3θ cos θ – sin 4θ – sin 2θ = 2 sin 3θ cos θ – [2 sin 3θ cos θ ] = 0 Self Practice Problem 6

C+D C−D sin 2 2

10.

7.

Proved that 13 x 3x sin 2 2

(i)

cos 8x – cos 5x = – 2 sin

(iii)

sin A + sin 3 A + sin 5 A + sin 7 A = tan 4A cos A + cos 3 A + cos 5 A + cos 7 A

(iv)

sin A + 2 sin 3 A + sin 5 A sin 3 A = sin 3 A + 2 sin 5 A + sin 7 A sin 5 A

(v)

sin A − sin 5 A + sin 9 A − sin 13 A = cot 4A cos A − cos 5 A − cos 9 A + cos 13 A

(ii)

sin A + sin 2A A = cot cos A − cos 2 A 2

Transformat io n of Prod uc ts into Sum or Dif ference of S ines & Cosines: (a)

2 sinA cosB = sin(A+B) + sin(A−B)

(b)

2 cosA sinB = sin(A+B) − sin(A−B)

(c)

2 cosA cosB = cos(A+B) + cos(A−B)

(d)

2 sinA sinB = cos(A−B) − cos(A+B)

Solved Example # 10

Prove that (i)

sin 8θ cos θ − sin 6θ cos 3θ = tan 2θ cos 2θ cos θ − sin 3θ sin 4θ

(ii)

tan 5θ + tan 3θ = 4 cos 2θ cos 4θ tan 5θ − tan 3θ

Solution

(i)

2 sin 8θ cos θ − 2 sin 6θ cos 3θ 2 cos 2θ cos θ − 2 sin 3θ sin 4θ

=

(ii)

sin 9θ + sin 7θ − sin 9θ − sin 3θ 2 sin 2θ cos 5θ = = tan 2θ cos 3θ + cos θ − cos θ + cos 7θ 2 cos 5θ cos 2θ

tan 5θ + tan 3θ sin 5θ cos 3θ + sin 3θ cos 5θ sin 8θ = = = 4 cos2θ cos 4θ tan 5θ − tan 3θ sin 5θ cos 3θ − sin 3θ cos 5θ sin 2θ

Self Practice Problem θ 7θ 3θ 11θ sin + sin sin = sin 2θ sin 5θ 2 2 2 2

11.

Prove that sin

12.

Prove that cos A sin (B – C) + cos B sin (C – A) + cos C sin (A – B) = 0

13.

Prove that 2 cos

8.

Multiple and Sub-multiple Angles :

π 9π 3π 5π cos + cos + cos =0 13 13 13 13

θ θ cos 2 2

(a)

sin 2A = 2 sinA cosA ; sin θ = 2 sin

(b)

cos 2A = cos²A − sin²A = 2cos²A − 1 = 1 − 2 sin²A; 2 cos²

(c)

tan 2A =

(d)

sin 2A =

2 tan A 1 − tan A 2

2 tan A 1 + tan A 2

; tan θ =

2 tan 2θ 1 − tan 2 2θ

, cos 2A =

1−tan 2 A 1+ tan 2 A

7

θ θ = 1 + cos θ, 2 sin² = 1 − cos θ. 2 2

(e)

sin 3A = 3 sinA − 4 sin3A

(g)

tan 3A =

cos 3A = 4 cos3A − 3 cosA

(f)

3 tan A − tan3 A 1 − 3 tan 2 A

Solved Example # 11

Prove that (i)

sin 2A = tan A 1 + cos 2A

(ii)

tan A + cot A = 2 cosec 2 A

(iii)

1 − cos A + cos B − cos( A + B) A B = tan cot 1 + cos A − cos B − cos( A + B) 2 2

Solution

2 sin A cos A sin 2A = = tan A 1 + cos 2 A 2 cos 2 A

(i)

L.H.S.

(ii)

 1 + tan2 A  2 1 + tan 2 A   = 2  2 tan A  = L.H.S. tan A + cot A = = 2 cosec 2 A sin 2 A tan A  

(iii)

L.H.S.

1 − cos A + cos B − cos( A + B) 1 + cos A − cos B − cos( A + B)

A A A  + 2 sin sin + B  2 2 2   = A A  2 A 2 cos − 2 cos cos + B  2 2 2  2 sin 2

 A +B  B  A A  sin + sin + B    2 sin 2 cos 2   A A  2 2       = tan = tan  2  2 + A B B  A A   2 sin 2 sin 2    cos 2 − cos 2 + B        

= tan

A B cot 2 2

Self Practice Problem sin θ + sin 2θ = tan θ 1 + cos θ + cos 2θ

14.

Prove that

15.

Prove that sin 20º sin 40º sin 60º sin 80º =

16.

Prove that tan 3A tan 2A tan A = tan 3A – tan 2A – tan A

17.

A  Prove that tan  45 º +  = sec A + tan A 2 

9.

Important Trigonometric Ratios: (a)

sin n π = 0

;

3 16

cos n π = (−1)n ; tan n π = 0, 8

where n ∈ Ι

(b)

sin 15° or sin

3 −1 5π π = = cos 75° or cos 12 12 2 2

cos 15° or cos

tan 15° =

(c)

sin

3 −1 3 +1

3+1 5π π = sin 75° or sin = 12 12 2 2 = 2− 3 = cot 75° ; tan 75° =

3 +1 3 −1

π π 5−1 or sin 18° = & cos 36° or cos = 10 5 4

1 0 . C o nd it i o na l

;

;

= 2+ 3 = cot 15° 5 +1 4

Ident it ies :

If A + B + C = π then : (i)

sin2A + sin2B + sin2C = 4 sinA sinB sinC

(ii)

sinA + sinB + sinC = 4 cos

(iii)

cos 2 A + cos 2 B + cos 2 C = − 1 − 4 cos A cos B cos C

(iv)

cos A + cos B + cos C = 1 + 4 sin

(v)

tanA + tanB + tanC = tanA tanB tanC

(vi)

tan

A B C cos cos 2 2 2

A B C sin sin 2 2 2

A B B C C A tan + tan tan + tan tan =1 2 2 2 2 2 2

(viii)

A B C A B C + cot + cot = cot . cot . cot 2 2 2 2 2 2 cot A cot B + cot B cot C + cot C cot A = 1

(ix)

A+B+C=

(vii)

cot

π 2

then tan A tan B + tan B tan C + tan C tan A = 1

Solved Example # 12

If A + B + C = 180°, Prove that, sin2A + sin2B + sin2C = 2 + 2cosA cosB cosC. Solution. Let S = sin2A + sin2B + sin2C so that 2S = 2sin2A + 1 – cos2B +1 – cos2C = 2 sin2A + 2 – 2cos(B + C) cos(B – C) = 2 – 2 cos2A + 2 – 2cos(B + C) cos(B – C) ∴ S = 2 + cosA [cos(B – C) + cos(B+ C)] since cosA = – cos(B+C) ∴ S = 2 + 2 cos A cos B cos C Solved Example # 13

If x + y + z = xyz, Prove that

2x 1− x2

+

2y 1− y2

+

2z 1− z2

Solution. Put x = tanA, y = tanB and z = tanC, so that we have tanA + tanB + tanC = tanA tanB tanC ⇒ Hence L.H.S.

=

2x 1− x

2

.

2y 1− y

2

.

2z 1 − z2

.

A + B + C = nπ, where n ∈ Ι

9

∴

2y

2x 1− x

+

2

+

1− y 2

2z 1− z2

=

2 tan A 1 − tan 2 A

= tan2A + tan2B + tan2C = tan2A tan2B tan2C =

2y

2x 1− x

.

2

1− y2

.

[

+

2 tan B 1 − tan2 B

+

2 tan C 1 − tan2 C

.

∵ A + B + C = nπ ]

2z 1 − z2

Self Practice Problem 18.

19.

If A + B + C = 180°, prove that (i)

sin(B + 2C) + sin(C + 2A) + sin(A + 2B) = 4sin

(ii)

sin 2 A + sin 2B + sin 2C A B C = 8 sin sin sin . sin A + sin B + sin C 2 2 2

B−C C−A A −B sin sin 2 2 2

If A + B + C = 2S, prove that (i) sin(S – A) sin(S – B) + sinS sin (S – C) = sinA sinB. (ii)

sin(S – A) + sin (S – B) + sin(S – C) – sin S = 4sin

1 1 . Range

of

Trigonometric

Expression: Expression :

E = a sin θ + b cos θ E = a 2 + b 2 sin (θ + α), where tan α =

b a

= a 2 + b 2 cos (θ − β), where tan β =

a b

Hence for any real value of θ, − a 2 + b 2 ≤ E ≤

a2 + b2

Solved Example # 14

Find maximum and minimum values of following : (i) 3sinx + 4cosx (ii) 1 + 2sinx + 3cos2x Solution. (i) We know – (ii)

3 2 + 4 2 ≤ 3sinx + 4cosx ≤

32 + 42

– 5 ≤ 3sinx + 4cosx ≤ 5 1+ 2sinx + 3cos2x = – 3sin2x + 2sinx + 4 2 sin x   2  +4 = – 3  sin x − 3   2

1 13  = – 3  sin x −  + 3 3   2

Now

1  16 0 ≤  sin x −  ≤ 3 9 

⇒

–

A B C sin sin . 2 2 2

2

1  16 ≤ – 3  sin x −  ≤ 0 3 3  10

2

1  13 13 – 1 ≤ – 3  sin x −  + ≤ 3 3 3  

Self Practice Problem 20. Find maximum and minimum values of following

(i) (ii)

3 + (sinx – 2) 2 10cos2x – 6sinx cosx + 2sin2x

Answer Answer

max = 12, min = 4. max = 11, min = 1.

(iii)

π  cosθ + 3 2 sin  θ +  + 6 4  

Answer

max = 11, min = 1

1 2 . Sine a nd Cosine

Series: Series :

(

)

sin α + sin (α + β) + sin (α + 2β ) +...... + sin α + n− 1β =

nβ

sin 2 n −1   β α + β sin  2  sin 2 nβ

sin 2 n −1   β α + cos α + cos (α + β) + cos (α + 2β ) +...... + cos α + n − 1β = β cos  2  sin 2

(

)

Solved Example # 15

Find the summation of the following (i)

cos

2π 4π 6π + cos + cos 7 7 7

(ii)

cos

π 2π 3π 4π 5π 6π + cos + cos + cos + cos + cos 7 7 7 7 7 7

(iii)

cos

π 3π 5π 7π 9π + cos + cos + cos + cos 11 11 11 11 11

Solution.

(i)

 2π 6 π  +   3π 7 7  cos  sin 2π 4π 6π 2 7 + cos + cos = cos π 7 7 7 sin 7

cos

=

4π 3π sin 7 7 π sin 7

3π 3π sin 7 7 π sin 7

− cos

=

6π 7 =– 1 =– π 2 2 sin 7 sin

(ii)

cos

π 2π 3π 4π 5π 6π + cos + cos + cos + cos + cos 7 7 7 7 7 7

11

 π 6π   +  6π cos  7 7  sin 14  2  π 6π   cos sin   2 14 = = =0 π π sin sin 14 14

(iii)

cos

π 3π 5π 7π 9π + cos + cos + cos + cos 11 11 11 11 11

cos

=

10 π 5π sin 22 11 π sin 11

10π 11 = 1 π 2 2 sin 11

sin

=

Self Practice Problem

Find sum of the following series :

21.

cos

π 3π 5π + cos + cos + ...... + to n terms. 2n + 1 2n + 1 2n + 1

Answer

1 2

22.

sin2α + sin3α + sin4α + ..... + sin nα, where (n + 2)α = 2π

Answer

0.

12

SHORT REVISION Trigonometric Ratios & Identities 1.

2.

3.

4.

BASIC TRIGONOMETRIC IDENTITIES : (a)sin2θ + cos2θ = 1 ; −1 ≤ sin θ ≤ 1 ; −1 ≤ cos θ ≤ 1 ∀ θ ∈ R 2 (b)sec θ − tan2θ = 1 ; sec θ ≥ 1 ∀ θ ∈ R 2 2 (c)cosec θ − cot θ = 1 ; cosec θ ≥ 1 ∀ θ ∈ R IMPORTANT T′ RATIOS: cos n π = (-1)n ; tan n π = 0 where n ∈ I (a)sin n π = 0 ; ( 2 n + 1 ) π ( 2n + 1)π = (−1)n &cos = 0 where n ∈ I (b)sin 2 2 5π π 3−1 (c)sin 15° or sin = = cos 75° or cos ; 12 12 2 2 3+1 π 5π cos 15° or cos = = sin 75° or sin ; 12 12 2 2 3 +1 3 −1 tan 15° = = 2 − 3 = cot 75° ; tan 75° = = 2 + 3 = cot 15° 3 +1 3 −1 3π π π π 2+ 2 2− 2 ; tan = 2−1 ; tan = 2+1 (d)sin = ; cos = 8 8 8 8 2 2 π π 5+1 5−1 (e) sin or sin 18° = & cos 36° or cos = 10 5 4 4 TRIGONOMETRIC FUNCTIONS OF ALLIED ANGLES : If θ is any angle, then − θ, 90 ± θ, 180 ± θ, 270 ± θ, 360 ± θ etc. are called ALLIED ANGLES. (a) sin (− θ) = − sin θ ; cos (− θ) = cos θ (b) sin (90°- θ) = cos θ ; cos (90° − θ) = sin θ (c) sin (90°+ θ) = cos θ ; cos (90°+ θ) = − sin θ (d)sin (180°− θ) = sin θ; cos (180°− θ) = − cos θ (e) sin (180°+ θ) = − sin θ ; cos (180°+ θ) = − cos θ (f) sin (270°− θ) = − cos θ ; cos (270°− θ) = − sin θ (g) sin (270°+ θ) = − cos θ ; cos (270°+ θ) = sin θ

TRIGONOMETRIC FUNCTIONS OF SUM OR DIFFERENCE OF TWO ANGLES : (a) sin (A ± B) = sinA cosB ± cosA sinB (b) cos (A ± B) = cosA cosB ∓ sinA sinB (c) sin²A − sin²B = cos²B − cos²A = sin (A+B) . sin (A− B) (d) cos²A − sin²B = cos²B − sin²A = cos (A+B) . cos (A − B) (e) tan (A ± B) = tan A ± tan B (f) cot (A ± B) = cot B ± cot A 1 ∓ tan A tan B FACTORISATION OF THE SUM OR DIFFERENCE OF TWO SINES OR COSINES : C−D C+ D C− D C+ D cos (b) sinC − sinD = 2 cos sin (a) sinC + sinD = 2 sin 2 2 2 2 C+ D C− D C+ D C−D (c) cosC + cosD = 2 cos cos (d) cosC − cosD = − 2 sin sin 2 2 2 2 TRANSFORMATION OF PRODUCTS INTO SUM OR DIFFERENCE OF SINES & COSINES : (a) 2 sinA cosB = sin(A+B) + sin(A−B) (b) 2 cosA sinB = sin(A+B) − sin(A−B) (c) 2 cosA cosB = cos(A+B) + cos(A−B) (d) 2 sinA sinB = cos(A−B) − cos(A+B) MULTIPLE ANGLES AND HALF ANGLES : cot A cot B ∓ 1

5.

6. 7.

(a)

θ 2

θ 2

sin 2A = 2 sinA cosA ; sin θ = 2 sin cos 13

(b)

cos2A = cos2A − sin2A = 2cos2A − 1 = 1 − 2 sin2A ; cos θ = cos2

8.

9.

10.

θ 2

− sin²

θ 2

= 2cos2

θ 2

θ 2

− 1 = 1 − 2sin2 .

1 − cos 2A 2 cos2A = 1 + cos 2A , 2sin2A = 1 − cos 2A ; tan2A = 1 + cos 2A θ θ 2 2 2 cos = 1 + cos θ , 2 sin = 1 − cos θ. 2 2 2tan(θ 2) 2tanA (c) tan 2A = ; tan θ = 2 1−tan 2 (θ 2) 1−tan A 2tanA 1−tan 2 A (d) sin 2A = , cos 2A = (e) sin 3A = 3 sinA − 4 sin3A 2 1+ tan 2 A 1+ tan A 3tanA− tan 3 A (f) cos 3A = 4 cos3A − 3 cosA (g) tan 3A = 1−3tan 2 A THREE ANGLES : tan A + tan B+ tanC− tan A tan BtanC (a) tan (A+B+C) = 1− tan A tan B− tan BtanC− tanCtan A NOTE IF : (i) A+B+C = π then tanA + tanB + tanC = tanA tanB tanC π (ii) A+B+C = then tanA tanB + tanB tanC + tanC tanA = 1 2 (b) If A + B + C = π then : (i) sin2A + sin2B + sin2C = 4 sinA sinB sinC C A B (ii) sinA + sinB + sinC = 4 cos cos cos 2 2 2 MAXIMUM & MINIMUM VALUES OF TRIGONOMETRIC FUNCTIONS: (a) Min. value of a2tan2θ + b2cot2θ = 2ab where θ ∈ R (b) Max. and Min. value of acosθ + bsinθ are a 2 + b 2 and – a 2 + b 2 (c) If f(θ) = acos(α + θ) + bcos(β + θ) where a, b, α and β are known quantities then – a 2 + b 2 + 2ab cos(α − β) < f(θ) < a 2 + b 2 + 2ab cos(α − β)  π (d) If α,β ∈  0,  and α + β = σ (constant) then the maximum values of the expression 2 cosα cosβ, cosα + cosβ, sinα + sinβ and sinα sinβ occurs when α = β = σ/2. (e) If α,β ∈  0, π  and α + β = σ(constant) then the minimum values of the expression  2 secα + secβ, tanα + tanβ, cosecα + cosecβ occurs when α = β = σ/2. If A, B, C are the angles of a triangle then maximum value of (f) sinA + sinB + sinC and sinA sinB sinC occurs when A = B = C = 600 (g) In case a quadratic in sinθ or cosθ is given then the maximum or minimum values can be interpreted by making a perfect square. Sum of sines or cosines of n angles, nβ sin 2  n−1  β sin α + sin (α + β) + sin (α + 2β ) + ...... + sin α + n − 1 β = β sin  α + 2  sin 2  nβ sin 2  n−1  α + n − 1 β cos α + cos (α + β) + cos (α + 2β ) + ...... + cos = cos  α + β  β 2   sin

(

)

(

)

EXERCISE–I

2

that cos²α + cos² (α + β) − 2cos α cos β cos (α + β) = sin²β that cos 2α = 2 sin²β + 4cos (α + β) sin α sin β + cos 2(α + β) that , tan α + 2 tan 2α + 4 tan 4α + 8 cot 8 α = cot α . that : (a) tan 20° . tan 40° . tan 60° . tan 80° = 3 3π 5π 7π 3 4 π + sin 4 + sin 4 + sin 4 = (b) tan 9° − tan 27° − tan 63° + tan 81° = 4 . (c) sin 16 16 16 16 2 Q.5 Calculate without using trigonometric tables : 2 cos 40° − cos20° (b) 4 cos 20° − 3 cot 20° (c) (a) cosec 10° − 3 sec 10° sin 20° 3π 5π 7π π  sec5° cos40°  + −2sin35° (d) 2 2 sin10° (e) cos6 + cos6 + cos6 + cos6 16 16 16 16 sin5°  2  (f) tan 10° − tan 50° + tan 70° 7π  7π   π 3π  π 3π       Q.6(a) If X = sin  θ +  + sin  θ −  + sin  θ +  , Y = cos  θ +  + cos  θ −  + cos  θ +  14 12 12    12  12    12    12  

Q.1 Q.2 Q.3 Q.4

Prove Prove Prove Prove

X Y = 2 tan2θ. then prove that − Y X (b) Prove that sin²12° + sin² 21° + sin² 39° + sin² 48° = 1+ sin² 9° + sin² 18° . Q.7

Q.8 Q.9 Q.10 Q.11 Q.12

Show that :

1° 2

(a)

cot 7

(b)

tan 142

or tan 82 1° 2

1° = 2

(

3+ 2

)(

)

2 +1 or

2 + 3+ 4 + 6

=2+ 2 − 3 − 6 .

m+ n . 2( m −n ) sin y 3 + sin 2 x π y π x If tan  +  = tan3  +  , prove that = . sin x 1 + 3 sin 2 x 4 2 4 2 π 4 5 If cos (α + β) = ; sin (α - β) = & α , β lie between 0 & , then find the value of tan 2α. 5 13 4 tanβ 1+ tanα 1−tanα tanβ sinβ n = ( m > n ) then . Prove that if the angles α & β satisfy the relation = sin(2α +β ) m m+ n m −n (a) If y = 10 cos²x − 6 sin x cos x + 2 sin²x , then find the greatest & least value of y . (b) If y = 1 + 2 sin x + 3 cos2 x , find the maximum & minimum values of y ∀ x ∈ R . (c) If y = 9 sec2x + 16 cosec2x, find the minimum value of y ∀ x ∈ R.

If m tan (θ - 30°) = n tan (θ + 120°), show that cos 2 θ =

π  (d) Prove that 3 cos  θ +  + 5 cos θ + 3 lies from - 4 & 10 . 

(

3

)

(

)

(

)

(e) Prove that 2 3 + 4 sin θ + 4 cos θ lies between − 2 2+ 5 & 2 2+ 5 .  tan A    = ∑ (tan A) − 2 ∑ (cot A).  tan B.tanC  If α + β = c where α, β > 0 each lying between 0 and π/2 and c is a constant, find the maximum or minimum value of (a) sin α + sin β (b) sin α sin β (c) tan α + tan β (d) cosec α + cosec β Let A1 , A2 , ...... , An be the vertices of an n-sided regular polygon such that ; 1 1 1 . Find the value of n. = + A1 A 2 A1 A 3 A1 A 4 Prove that : cosec θ + cosec 2 θ + cosec 22 θ + ...... + cosec 2 n − 1 θ = cot (θ/2) − cot 2 n - 1θ For all values of α , β , γ prove that; β+ γ α +β γ +α cos α + cos β + cos γ + cos (α + β + γ) = 4 cos .cos . cos . 2 2 2 1 + sin A cos B 2 sin A − 2 sin B + = . Show that cos A 1 − sin B sin(A − B) + cos A − cos B

Q.13 If A + B + C = π, prove that Q.14

Q.15

Q.16 Q.17

Q.18

Q.19 If tan β =

tan α + tan γ 1 + tan α . tan γ

∑

, prove that sin 2β =

sin 2 α + sin 2 γ 1 + sin 2 α . sin 2 γ

.

Q.20 If α + β = γ , prove that cos² α + cos² β + cos² γ = 1 + 2 cos α cos β cos γ . (1 − tan α2 ) 1 − tan β2 1 − tan 2γ sin α + sin β + sin γ − 1 π Q.21 If α + β + γ = , show that = . 2 (1 + tan α2 ) 1 + tan β2 1 + tan 2γ cos α + cos β + cos γ Q.22 If A + B + C = π and cot θ = cot A + cot B + cot C, show that , sin (A − θ) . sin (B − θ) . sin (C − θ) = sin3 θ . 3π 5π 17 π π + cos + ......... + cos Q.23 If P = cos + cos and 19 19 19 19 2π 4π 6π 20π + cos + cos + ......... + cos Q = cos , then find P – Q. 21 21 21 21 Q.24 If A, B, C denote the angles of a triangle ABC then prove that the triangle is right angled if and only if sin4A + sin4B + sin4C = 0. Q.25 Given that (1 + tan 1°)(1 + tan 2°)......(1 + tan 45°) = 2n, find n.

( (

)( )(

) )

EXERCISE–II

Q.1 Q.2

If tan α = p/q where α = 6β, α being an acute angle, prove that; 1 (p cosec 2 β − q sec 2 β) = p 2 +q 2 . 2 Let A1 , A2 , A3 ............ An are the vertices of a regular n sided polygon inscribed in a circle of radius R. If (A1 A2)2 + (A1 A3)2 + ......... + (A1 An)2 = 14 R2 ,15find the number of sides in the polygon.

Q.3 Q.4 Q.5

cos 3θ + cos 3φ = (cosθ + cosφ) cos(θ + φ) – (sinθ + sinφ) sin(θ + φ) 2 cos(θ − φ) − 1 Without using the surd value for sin 180 or cos 360 , prove that 4 sin 360 cos 180 = 5 sin x sin3x sin9x 1 + + = (tan27x − tanx) Show that , cos3x cos9x cos27x 2

Prove that:

5

Q.6

Let x1 =

rπ

∏ cos 11 r =1

5

and x2 =

rπ

∑ cos 11 , then show that r =1

Q.25

1  π  x1 · x2 =  cos ec − 1 , where Π denotes the continued product. 64  22  2π If θ = , prove that tan θ . tan 2 θ + tan 2 θ . tan 4 θ + tan 4 θ . tan θ = − 7. 7 cosx π prove that , > 8. For 0 < x < 2 sin x(cosx −sinx ) 4 3π 2π π 2π 7 7 prove that, sin α + sin 2α + sin 4α = (b) sin . sin . sin = (a) If α = 7 7 7 7 2 8 88 1 cos k Let k = 1°, then prove that ∑ = sin 2 k n =0 cos nk · cos(n + 1)k 3 Prove that the value of cos A + cos B + cos C lies between 1 & where A + B + C = π. 2 If cosA = tanB, cosB = tanC and cosC = tanA , then prove that sinA = sinB = sinC = 2 sin18°. 3 + cos x Show that ∀ x ∈ R can not have any value between − 2 2 and 2 2 . What inference sin x sin x ? can you draw about the values of 3 + cos x 5 If (1 + sin t)(1 + cos t) = . Find the value of (1 – sin t)(1 – cos t). 4 sin 8 α cos8 α 1 sin 4 α cos4 α 1 + 3 = + = Prove that from the equality follows the relation ; 3 a b a b a +b (a +b )3 . Prove that the triangle ABC is equilateral iff , cot A + cot B + cot C = 3 . Prove that the average of the numbers n sin n°, n = 2, 4, 6, ......., 180, is cot 1°. Prove that : 4 sin 27° = 5+ 5 1 / 2 − 3− 5 1 / 2 . A C B If A+B+C = π; prove that tan2 + tan2 + tan2 ≥ 1. 2 2 2 A B C 1 If A+B+C = π (A , B , C > 0) , prove that sin . sin . sin ≤ . 2 2 2 8 Show that elliminating x & y from the equations , sin x + sin y = a ; 8ab cos x + cos y = b & tan x + tan y = c gives 2 2 2 = c. a +b −4a 2 Determine the smallest positive value of x (in degrees) for which tan(x + 100°) = tan(x + 50°) tan x tan (x – 50°). x tan n n 2 Evaluate : ∑ x n − 1 n =1 2 cos n −1 2 β+ γ−α  γ +α −β  α+β−γ  If α + β + γ = π & tan   · tan   · tan   = 1, then prove that; 4 4 4       1 + cos α + cos β + cos γ = 0. ∀ x ∈ R, find the range of the function, f (x) = cos x (sin x + sin 2 x + sin 2 α ) ; α ∈ [0, π]

Q.1

sec2θ =

Q.7 Q.8 Q.9 Q.10 Q.11 Q.12 Q.13

Q.14 Q.15 Q.16 Q.17 Q.18 Q.19 Q.20 Q.21

(

) (

)

(

Q.22

Q.23

Q.24

Q.2

)

EXERCISE–III

4xy is true if and only if : ( x + y) 2 (A) x + y ≠ 0 (B) x = y , x ≠ 0 (a)

Let n be an odd integer. If sin nθ = (A) b0 = 1, b1 = 3

[JEE ’96, 1] n

(C) x = y

(D) x ≠ 0 , y ≠ 0

∑ br sinr θ, for every value of θ, then :

r=0

16

(B) b0 = 0, b1 = n

(b)

(C) b0 = − 1, b1 = n (D) b0 = 0, b1 = n2 − 3n + 3 Let A0 A1 A2 A3 A4 A5 be a regular hexagon inscribed in a circle of unit radius . Then the product of the lengths of the line segments A0 A1, A0 A2 & A0 A4 is : 3

Q.3

3 3

(B) 3 3 (C) 3 (D) (A) 4 2 (c) Which of the following number(s) is/are rational ? [ JEE '98, 2 + 2 + 2 = 6 out of 200 ] (A) sin 15º (B) cos 15º (C) sin 15º cos 15º (D) sin 15º cos 75º θ   For a positive integer n, let fn (θ) =  tan  (1+ sec θ) (1+ sec 2θ) (1+ sec 4θ) .... (1 + sec2nθ) Then 2    π

 π

 π

 π 

 =1 (A) f2   = 1 (B) f3   = 1 (C) f4   = 1 (D) f 5  16 32 64 128  Q.4(a) Let f (θ) = sin θ (sin θ + sin 3 θ) . Then f (θ) : [ JEE 2000 Screening. 1 out of 35 ] (A) ≥ 0 only when θ ≥ 0 (B) ≤ 0 for all real θ (C) ≥ 0 for all real θ (D) ≤ 0 only when θ ≤ 0 .

(b) In any triangle ABC, prove that, cot

[JEE '99,3]

A B A B C C + cot + cot = cot cot cot . [JEE 2000] 2 2 2 2 2 2

Q.5(a) Find the maximum and minimum values of 27cos 2x · 81sin 2x. π (b) Find the smallest positive values of x & y satisfying, x − y = , cot x + cot y = 2. [REE 2000, 3] 4 π and β + γ = α then tanα equals [ JEE 2001 (Screening), 1 out of 35 ] Q.6 If α + β = 2 (A) 2(tanβ + tanγ) (B) tanβ + tanγ (C) tanβ + 2tanγ (D) 2tanβ + tanγ 1 1 Q.7 If θ and φ are acute angles satisfying sinθ = , cos φ = , then θ + φ ∈ [JEE 2004 (Screening)] 2 3  π π  π 2π   2π 5π   5π  (A)  ,  (D)  , π  (B)  ,  (C)  ,  2 3    6   3 2  3 6  Q.8 In an equilateral triangle, 3 coins of radii 1 unit each are kept so that they touch each other and also the sides of the triangle. Area of the triangle is (A) 4 + 2 3 (B) 6 + 4 3 (C) 12 +

Q.9

7 3 4

(D) 3 +

7 3 4

[JEE 2005 (Screening)]

 π Let θ ∈  0,  and t1 = (tanθ)tanθ, t2 = (tanθ)cotθ, t3 = (cotθ)tanθ , t4 = (cotθ)cotθ, then  4 (A) t1 > t2 > t3 > t4 (B) t4 > t3 > t1 > t2 (C) t3 > t1 > t2 > t4 (D) t2 > t3 > t1 > t4 [JEE 2006, 3]

ANSWER SHEET (EXERCISE–I)

Q 5. (a) 4

(b) −1

(c) 3

(d) 4

(e) 13

5 4

(f) 3

Q 10.

56 33

Q 12. (a) ymax = 11 ; ymin = 1 (b) ymax = ; ymin = − 1, (c) 49 3 2 Q14. (a) max = 2 sin (c/2), (b) max. = sin (c/2), (c) min = 2 tan (c/2), (d) min = 2 cosec (c/2) Q 15. n = 7 Q23. 1 Q.25 n = 23

EXERCISE –II

1 1   − 2 2 , 2 2    2 1 − Q 23. Q.25 – sin 2 x 2 n −1 sin x 2 n −1

Q.2

n = 7 Q.13

Q.14

13 − 10 4

Q.22

x = 30°

1 + sin 2 α ≤ y ≤ 1 + sin 2 α

EXERCISE–III

Q.1 Q.5 Q.8

Q.2 (a) B, (b) C, (c) C

B

Q.3 A, B, C, D

π 5π (a) max. = 35 & min. = 3–5 ; (b) x = ;y= 12 6

B

Q.9

Q.6 C

B

17

Q.7 B

Q.4 (a) C

EXERCISE–IV (Objective) Part : (A) Only one correct option

(

(32π + x ) −sin3 (72π − x ) when simplified reduces to: cos(x − 2π ).tan (32π + x ) )

tan x − 2π .cos 1 .

(C) − sin x cos x (D) sin2x  4  3π   6 π    4 6 The expression 3 sin  2 − α  + sin (3π + α) – 2 sin  2 + α  + sin (5π + α ) is equal to         (A) 0 (B) 1 (C) 3 (D) sin 4α + sin 6α If tan A & tan B are the roots of the quadratic equation x 2 − ax + b = 0, then the value of sin2 (A + B). (A) sin x cos x

2. 3.

5. 6. 7. 8. 9. 10.

11.

13. 14. 15.

16.

17.

18.

a2

a2

The greatest and least value of log (A) 2 & 1 (B) 5 & 3

2 3 3

3π < α < π, then 4 (A) 1 + cot α

2

(sin x − cos x + 3 2 ) are respectively: (C) 7 & 5

(D) 9 & 7

(B)

4 3 3

(C)

(D) none

3

1 is equal to sin2 α (B) – 1 – cot α (C) 1 – cot α (D) – 1 + cot α π x   3π   If x ∈  π,  then 4 cos2  −  + 4 sin 4 x + sin 2 2 x is always equal to 2  4 2  (A) 1 (B) 2 (C) – 2 (D) none of these If 2 cos x + sin x = 1, then value of 7 cos x + 6 sin x is equal to (A) 2 or 6 (B) 1 or 3 (C) 2 or 3 (D) none of these 11 If cosec A + cot A = , then tan A is 2 15 117 21 44 (A) (B) (C) (D) 16 43 22 117 1 If cot α + tan α = m and – cos α = n, then cos α 2 1/3 2 1/3 (A) m (mn ) – n(nm ) = 1 (B) m(m 2n)1/3 – n(nm 2)1/3 = 1 (C) n(mn2)1/3 – m(nm 2)1/3 = 1 (D) n(m 2n)1/3 m(mn2)1/3 = 1 cos 6 x + 6 cos 4 x + 15 cos 2x + 10 The expression is equal to cos 5 x + 5 cos 3 x + 10 cos x (B) 2 cos x (C) cos2 x (D) 1 + cos x (A) cos 2x sin A cos A 3 5 If = and = , 0 < A, B < π/2, then tan A + tan B is equal to sin B cos B 2 2

If

(A)

3/ 5

2 cot α +

(B)

If sin 2θ = k, then the value of (A)

1− k2 k

(B)

tan θ 1 + tan θ

2 − k2 k

(D) ( 5 + 3 ) / 5

(C) 1

5/ 3 3

19.

a2

In a right angled triangle the hypotenuse is 2 2 times the perpendicular drawn from the opposite vertex. Then the other acute angles of the triangle are π π π 3π π 3π π π (A) & (B) & (C) & (D) & 3 6 8 8 5 10 4 4 1 1 cos290 ° + 3 sin250 ° = (A)

12.

a2

(C) (D) 2 (B) 2 2 a 2 +(1−b)2 a +b (b+c )2 b (1−a)2 2 2 The value of log2 [cos (α + β) + cos (α − β) − cos 2α. cos 2β] : (A) depends on α & β both (B) depends on α but not on β (C) depends on β but not on α (D) independent of both α & β. cos20°+8sin70°sin50 °sin10° is equal to: sin 2 80° (A) 1 (B) 2 (C) 3/4 (D) none If cos A = 3/4, then the value of 16cos2 (A/2) – 32 sin (A/2) sin (5A/2) is (A) – 4 (B) – 3 (C) 3 (D) 4 If y = cos2 (45º + x) + (sin x − cos x)2 then the maximum & minimum values of y are: (A) 2 & 0 (B) 3 & 0 (C) 3 & 1 (D) none π 3π 5π 17π The value of cos + cos + cos +...... + cos is equal to: 19 19 19 19 (A) 1/2 (B) 0 (C) 1 (D) none (A)

4.

(B) − sin2 x

2

cot θ 3

+

1 + cot 2 θ

is equal to

(C) k182 + 1

(D) 2 – k 2

Part : (B) May have more than one options correct 20. Which of the following is correct ? (A) sin 1° > sin 1 (B) sin 1° < sin 1 (C) cos 1° > cos 1 (D) cos 1° < cos 1 21. If 3 sin β = sin (2α + β), then tan (α + β) – 2 tan α is (A) independent of α (B) independent of β (C) dependent of both α and β (D) independent of α but dependent of β 22.

4 It is known that sin β = & 0 < β < π then the value of 5

(A) independent of α for all β in (0, π)

(B)

5 3

3 sin(α + β ) −

2 cos (α + β) cos 6π

sinα

is:

for tan β > 0

3 (7 + 24 cot α ) for tan β < 0 (D) none 15 If the sides of a right angled triangle are {cos2α + cos2β + 2cos(α + β)} and {sin2α + sin2β + 2sin(α + β)}, then the length of the hypotenuse is: α−β α +β (A) 2[1+cos(α − β)] (B) 2[1 − cos(α + β)] (C) 4 cos2 (D) 4sin2 2 2 If x = sec φ − tan φ & y = cosec φ + cot φ then: y+1 1+ x y −1 (A) x = y − 1 (B) y = 1 − x (C) x = y + 1 (D) xy + x − y + 1 = 0 (a + 2) sin α + (2a – 1) cos α = (2a + 1) if tan α = 2a 2a 4 3 (A) (B) (C) 2 (D) 2 3 4 a +1 a −1 2b If tan x = , (a ≠ c) a−c y = a cos2x + 2b sin x cos x + c sin2x z = a sin2x – 2b sin x cos x + c cos2x, then (A) y = z (B) y + z = a + c (C) y – z = a – c (D) y – z = (a – c)2 + 4b2

(C)

23.

24.

25.

26.

n

27.

28.

n

 cos A + cos B   sin A + sinB    +    sin A − sinB   cos A − cos B  A −B A −B (B) 2 cotn : n is even (C) 0 : n is odd (A) 2 tann (D) none 2 2 6 6 2 The equation sin x + cos x = a has real solution if 1   1 1 1   (D) a ∈  , 1 (C) a ∈  − (A) a ∈ (–1, 1) (B) a ∈  − 1, −  2   2 2 2 

EXERCISE–IV (Subjective) 1. 2. 3.

4. 5. 6.

7. 8. 9. 10. 11.

The minute hand of a watch is 1.5 cm long. How far does its tip move in 50 minutes? (Use π = 3.14). If the arcs of the same length in two circles subtend angles 75° and 120°at the centre, find the ratio of their radii. Sketch the following graphs : x (i) y = 3 sin 2x (ii) y = 2 tan x (iii) y = sin 2   3π    3π  − θ  + cot (2π + θ) = 1. + θ  cos (2π + θ) cot  Prove that cos    2    2  θ 9θ 5θ – cos 3 θ cos = sin 5 θ sin . 2 2 2 3 3π x x If tan x = , π < x < , find the value of sin and cos . 4 2 2 2   2  α −π   1 − cot  4   9α    + cos α cot 4α  prove that  = cosec 4α.  sec 2 π α − 2   1 + cot 2       4  Prove that, sin 3 x. sin3 x + cos 3 x. cos3 x = cos3 2 x. p 1 If tan α = where α = 6 β, α being an acute angle, prove that; (p cosec 2 β − q sec 2 β) = p 2 + q 2 . q 2 tan α + tan γ sin 2α + sin 2γ If tan β = 1 + tan α. tan γ , prove that sin 2β = 1 + sin 2α. sin 2γ .

Prove that cos 2 θ cos

(i)

(ii)

1° = 2 + 2− 3− 6 . 2

tan 142

cot 7

1° 1° or tan 82 = 2 2

Show that:

( 3 + 2 )( 2 +1) 19

(iii)

or

2+ 3+ 4+ 6

(

4 sin 27° = 5 + 5

) − (3 − 5 ) 1/ 2

1/ 2

12. 13.

14. 15.

16. 17.

18. 19.

20. 21. 22.

Prove that, tan α + 2 tan 2α + 4 tan 4α + 8 cot 8 α = cot α. −3 If cos (β − γ) + cos (γ − α) + cos (α − β) = , prove that 2 cos α + cos β + cos γ = 0, sin α + sin β + sin γ = 0. 1 sin8 α cos 8 α sin4 α cos 4 α 1 + = + = follows the relation a3 b3 a b a+b (a + b)3 Prove that: cosec θ + cosec 2 θ + cosec 22 θ +... + cosec 2 n − 1θ = cot (θ/2) − cot 2n − 1 θ. Hence or 4π 8π 16π 32π otherwise prove that cosec + cosec + cosec + cosec =0 15 15 15 15 1 1 1 Let A1, A2,......, An be the vertices of an n−sided regular polygon such that; A A = A A + A A . 1 2 1 3 1 4 Find the value of n. If A + B + C = π, then prove that 1 A B C A B C ≥1 (ii) sin . sin . sin ≤ . (i) tan² + tan² + tan² 8 2 2 2 2 2 2 3 (iii) cos A + cos B + cos C ≤ 2 ax sin θ by cos θ ax by + = a2 – b2, – = 0. Show that (ax)2/3 + (by)2/3 = (a2 – b2) 2/3 If 2 cos θ sin θ cos θ sin2 θ n n n n If Pn = cos θ + sin θ and Q n = cos θ – sin θ, then show that Pn – Pn – 2 = – sin2θ cos2θ Pn – 4 Q n – Q n – 2 = – sin2θ cos2θ Q n – 4 and hence show that P4 = 1 – 2 sin2θ cos2θ , Q 4 = cos2θ – sin2θ If sin (θ + α) = a & sin (θ + β) = b (0 < α, β, θ < π/2) then find the value of cos2 (α − β) − 4 ab cos(α − β) If A + B + C = π, prove that tan B tan C + tan C tan A + tan A tan B = 1 + sec A. sec B. sec C. If tan2α + 2tanα. tan2β = tan2β + 2tanβ. tan2α, then prove that each side is equal to 1 or tan α = ± tan β.

Prove that from the equality

EXERCISE–IV

EXERCISE–V

1. D

2. B

3. A

4. D

5. B

6. C

7. B

8. A

9. B

10. B

11. B

12. B

13. B

14. A

15. C

16. A

17. B

18. D

19. B

20. BC

21. AB 22. BC 23. AC 24. BCD

1. 7.85 cm

6. sin

x = 2

25. BD 26. BC 16. n = 7

27. BC 28. BD

20

2. r 1 : r2 = 8 : 5 3 10

and cos

x =– 2

20. 1 − 2a2 − 2b2

1 10

STUDY PACKAGE Target: IIT-JEE (Advanced) SUBJECT: MATHEMATICS TOPIC: 2 XI M 2. Trigonometric Equations Index: 1. Key Concepts 2. Exercise I to III 3. Answer Key 4. Assertion and Reasons 5. 34 Yrs. Que. from IIT-JEE 6. 10 Yrs. Que. from AIEEE

1

1. 2.

Trigonometric Equation : An equation involving one or more trigonometric ratios of an unknown angle is called a trigonometric equation.

Solution of Trigonometric Equation :

A solution of trigonometric equation is the value of the unknown angle that satisfies the equation. π 3π 9π 11π 1 e.g. if sinθ = , , , ........... ⇒ θ= , 4 4 4 4 2 Thus, the trigonometric equation may have infinite number of solutions (because of their periodic nature) and can be classified as : (i) Principal solution (ii) General solution. Principal solutions: 2 .1 The solutions of a trigonom etric equation which lie in the interv al [0, 2π) are called Principal solutions. 1 e.g Find the Principal solutions of the equation sinx = . 2 Solution.

1 2 ∵ there exists two values π 5π 1 i.e. and which lie in [0, 2π) and whose sine is 6 6 2 π 1 ∴ Principal solutions of the equation sinx = are , 6 2 General Solution :

∵

2 .2

sinx =

5π Ans. 6

The expression involving an integer 'n' which gives all solutions of a trigonometric equation is called General solution. General solution of some standard trigonometric equations are given below.

3.

General Solution of Some Standard Trigonometric Equations : (i)

If sin θ = sin α

⇒ θ = n π + (−1)n α

(ii)

If cos θ = cos α

⇒ θ = 2nπ ± α

(iii)

If tan θ = tan α

⇒ θ = nπ + α

(iv)

If sin² θ = sin² α

⇒ θ = n π ± α, n ∈ Ι.

(v)

If cos² θ = cos² α

⇒ θ = n π ± α, n ∈ Ι.

(vi) If tan² θ = tan² α Some Important deductions : ⇒ (i) sinθ = 0 (ii)

sinθ = 1

⇒

(iii)

sinθ = – 1

⇒

(iv)

cosθ = 0

⇒

(v) (vi) (vii)

cosθ = 1 cosθ = – 1 tanθ = 0

⇒ ⇒ ⇒

⇒ θ = n π ± α, n ∈ Ι. θ = nπ,

n∈Ι π θ = (4n + 1) , n ∈ Ι 2 π θ = (4n – 1) , n ∈ Ι 2 π θ = (2n + 1) , n ∈ Ι 2 n∈Ι θ = 2nπ, θ = (2n + 1)π, n ∈ Ι θ = nπ, n∈Ι

Solved Example # 1 Solve sin θ =

Solution.

3 . 2 2

 π π where α ∈ − ,  ,  2 2 where α ∈ [0, π],  π π where α ∈  − ,  ,  2 2

n ∈ Ι. n ∈ Ι. n ∈ Ι.

[ Note: α is called the principal angle ]

Page : 2 of 15 TRIG. EQUATIONS

Trigonometric Equation

⇒ ∴

3 2 π sinθ = sin 3

Page : 3 of 15 TRIG. EQUATIONS

∵

sin θ =

θ = n π + (– 1)n

π ,n∈Ι 3

Ans.

Solved Example # 2 2

Solve sec 2θ = –

3

Solution. ∵ ⇒

2

sec 2θ = – cos2θ = –

3

3 2

⇒

5π ,n∈Ι 6 5π ,n∈Ι θ = nπ ± 12

cos2θ = cos

5π 6

2θ = 2nπ ±

⇒ ⇒

Ans.

Solved Example # 3 Solve tan θ = 2 Solution. ∵ tanθ = 2 ............(i) Let 2 = tanα ⇒ tanθ = tanα ⇒ θ = n π + α, where α = tan–1(2), n ∈ Ι Self Practice Problems: 1.

Solve

cot θ = – 1

2.

Solve

cos3θ = –

Ans.

(1)

1 2

θ = nπ –

π , n∈Ι 4

(2)

2nπ 2π ± ,n∈Ι 3 9

Solved Example # 4 Solve cos2θ = Solution. ∵

cos2θ =

1 2 1 2 2

 1   ⇒ cos θ =   2 π ⇒ cos2θ = cos2 4 π ⇒ θ = n π ± , n ∈ Ι Ans. 4 Solved Example # 5 2

Solve 4 tan 2θ = 3sec2θ Solution. ∵

4 tan2θ = 3sec2θ

.............(i) π For equation (i) to be defined θ ≠ (2n + 1) , n ∈ Ι 2 ∵ equation (i) can be written as: 4 sin 2 θ cos θ 2

=

3 cos 2 θ

⇒

4 sin2θ = 3

⇒

 3  sin2θ =   2  

∵

θ ≠ (2n + 1)

∴ cos2θ ≠ 0

π ,n∈Ι 2

2

3

π 3 π π ± , n ∈ Ι Ans. θ = nπ 3

sin2θ = sin2

⇒

Self Practice Problems : 1.

Solve

7cos2θ + 3 sin2θ = 4.

2.

Solve

2 sin2x + sin22x = 2 π (1) nπ ± , n ∈ Ι 3

Ans.

(2)

(2n + 1)

π ,n∈Ι 2

or

nπ ±

π ,n∈Ι 4

Types of Trigonometric Equations : Type -1

Trigonometric equations which can be solved by use of factorization. Solved Example # 6 Solve (2sinx – cosx) (1 + cosx) = sin2x. Solution. ∵ ⇒ ⇒ ⇒ ⇒

+ cosx) = sin2x + cosx) – sin2x = 0 + cosx) – (1 – cosx) (1 + cosx) = 0 – 1) = 0 or 2sinx – 1 = 0 1 ⇒ cosx = – 1 or sinx = 2 π ⇒ x = (2n + 1)π, n ∈ Ι or sin x = sin 6 ∴ Solution of given equation is π (2n + 1)π π, n ∈ Ι or nπ π + (–1)n ,n∈Ι 6 Self Practice Problems : (2sinx – cosx) (1 (2sinx – cosx) (1 (2sinx – cosx) (1 (1 + cosx) (2sinx 1 + cosx = 0

Solve

cos3x + cos2x – 4cos2

2.

Solve

cot 2θ + 3cosecθ + 3 = 0

Ans.

(1)

Type - 2

π , n∈Ι 6

Ans.

x =0 2

1.

(2)

⇒ x = nπ + (– 1) n

(2n + 1)π, n ∈ Ι π 2nπ – , n ∈ Ι or 2

nπ + (–1)n + 1

π ,n∈Ι 6

Trigonometric equations which can be solved by reducing them in quadratic equations. Solved Example # 7 Solve Solution. ∵ ⇒ ⇒ ⇒ ∵

∴ ∴

2 cos2x + 4cosx = 3sin2x 2cos2x + 4cosx – 3sin2x = 0 2cos2x + 4cosx – 3(1– cos2x) = 0 5cos2x + 4cosx – 3 = 0   − 2 + 19      cos x −  − 2 − 19  cos x −      = 0 5 5       cosx ∈ [– 1, 1] ∀ x ∈ R

........(ii)

− 2 − 19 5 equation (ii) will be true if

cosx ≠

cosx =

− 2 + 19 5 − 2 + 19 5  − 2 + 19  , n ∈ Ι α = cos–1   5 4  

⇒

cosx = cosα,

where cosα =

⇒

x = 2nπ π±α

where

Ans.

Page : 4 of 15 TRIG. EQUATIONS

⇒

2.

Solve Ans.

1.

Solve

4cosθ – 3secθ = tanθ π (1) 2nπ ± , n ∈ Ι 3 (2)

 1   = 0 cos2θ – ( 2 + 1)  cos θ − 2 

Page : 5 of 15 TRIG. EQUATIONS

Self Practice Problems :

π ,n∈Ι 4  − 1 − 17  , n∈Ι nπ + (– 1)n α where α = sin–1   8    − 1 + 17  , n ∈Ι or nπ + (–1)n β where β = sin–1   8  

or

2nπ ±

Type - 3

Trigonometric equations which can be solved by transforming a sum or difference of trigonometric ratios into their product. Solved Example # 8 Solve cos3x + sin2x – sin4x = 0 Solution. ⇒ ⇒ ⇒ ⇒

∴

cos3x + sin2x – sin4x = 0 cos3x – 2cos3x.sinx = 0 cos3x = 0 π 3x = (2n + 1) , n ∈ Ι 2 π x = (2n + 1) , n ∈ Ι 6 solution of given equation is π ,n∈Ι or (2n + 1) 6

⇒ ⇒ or

cos3x + 2cos3x.sin(– x) = 0 cos3x (1 – 2sinx) = 0 1 – 2sinx = 0 1 sinx = 2 π x = nπ + (–1) n , n ∈ Ι 6

or or nπ π + (–1)n

π ,n∈Ι 6

Ans.

Self Practice Problems : 1.

Solve

sin7θ = sin3θ + sinθ

2.

Solve

5sinx + 6sin2x +5sin3x + sin4x = 0

3.

Solve

cosθ – sin3θ = cos2θ nπ ,n∈Ι (1) 3 nπ (2) ,n∈Ι 2 2nπ (3) ,n∈Ι 3

Ans.

or or or

nπ π ± ,n∈Ι 2 12 2π 2nπ ± ,n∈Ι 3 π 2nπ – , n ∈ Ι 2

or

nπ +

π ,n∈Ι 4

Type - 4 Trigonometric equations which can be solved by transforming a product of trigonometric ratios into their sum or difference. Solved Example # 9 Solve Solution. ∵ ⇒ ⇒ ⇒ ⇒ ⇒

∴

Type - 5

sin5x.cos3x = sin6x.cos2x sin5x.cos3x = sin6x.cos2x ⇒ sin8x + sin2x = sin8x + sin4x ⇒ 2sin2x.cos2x – sin2x = 0 ⇒ sin2x = 0 or 2cos2x – 1 = 0 1 2x = nπ, n ∈ Ι or cos2x = 2 π nπ x= , n ∈ Ι or 2x = 2nπ ± , n ∈ Ι 3 2 π ⇒ x = nπ ± , n ∈ Ι 6 Solution of given equation is π nπ ,n∈Ι or nπ π± ,n∈Ι 6 2

2sin5x.cos3x = 2sin6x.cos2x sin4x – sin2x = 0 sin2x (2cos2x – 1) = 0

Ans.

Trigonometric Equations of the form a sinx + b cosx = c, where a, b, c ∈ R, can be solved by dividing both sides of the equation by

a2 + b2 .

5

Solve sinx + cosx =

2

Solution. 2

∵ Here

sinx + cosx = a = 1, b = 1.

∴

divide both sides of equation (i) by 1 1 sinx . + cosx. =1 2 2 π π sinx.sin + cosx.cos = 1 4 4 π  cos  x −  = 1 4 

⇒ ⇒ ⇒ ⇒

∴

..........(i) 2 , we get

π = 2nπ, n ∈ Ι 4 π ,n∈Ι x = 2nπ + 4 Solution of given equation is π 2nπ π+ ,n∈Ι Ans. 4

x–

Note : Trigonometric equation of the form a sinx + b cosx = c can also be solved by changing sinx and cosx into their corresponding tangent of half the angle. Solved Example # 11 Solve 3cosx + 4sinx = 5 Solution. ∵

∵

∴ ⇒

Let ∴

⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒

3cosx + 4sinx = 5 .........(i) x 2 x 2 tan 1 − tan 2 2 cosx = & sinx = x x 1 + tan 2 1 + tan 2 2 2 equation (i) becomes x    2 x   2 tan   1 − tan  2 2   =5  +4 3  ........(ii)  2 x   2 x   1 + tan   1 + tan  2  2  x tan =t 2 equation (ii) becomes  1− t2  2t   + 4   =5 3  2 2   1+ t   1+ t  4t2 – 4t + 1 = 0 (2t – 1)2 = 0 1 x t= ∵ t = tan 2 2 x 1 tan = 2 2 x 1 tan = tanα, where tanα = 2 2 x = nπ + α 2  1 α where α = tan –1   , n ∈ Ι x = 2nπ π + 2α 2

Self Practice Problems : 1.

Solve

3 cosx + sinx = 2 6

Ans.

Page : 6 of 15 TRIG. EQUATIONS

Solved Example # 10

x =0 2 π 2nπ + , n ∈ Ι 6

Solve

sinx + tan

Ans.

(1)

(2)

x = 2nπ, n ∈ Ι

Type - 6

Trigonometric equations of the form P(sinx ± cosx, sinx cosx) = 0, where p(y, z) is a polynomial, can be solved by using the substitution sinx ± cosx = t. Solved Example # 12 Solve sinx + cosx = 1 + sinx.cosx Solution. ∵ Let ⇒

sinx + cosx = 1 + sinx.cosx sinx + cosx = t sin2x + cos2x + 2 sinx.cosx = t2

........(i)

t2 − 1 2

⇒

sinx.cosx =

Now

put

⇒ ⇒ ⇒

t2 − 1 2 t2 – 2t + 1 = 0 t=1 sinx + cosx = 1

sinx + cosx = t

t2 − 1 in (i), we get 2

and sinx.cosx =

t=1+

∵

t = sinx + cosx .........(ii)

divide both sides of equation (ii) by 2 , we get 1 1 1 ⇒ sinx. + cosx. = 2 2 2 ⇒ ⇒

(i)

(ii)

π π  cos  x −  = cos 4 4  π π x– = 2nπ ± 4 4 if we take positive sign, we get π ,n∈Ι Ans. x = 2nπ π+ 2 if we take negative sign, we get x = 2nπ Ans. π, n ∈ Ι

Self Practice Problems: 1.

Solve

sin2x + 5sinx + 1 + 5cosx = 0

2.

Solve

3cosx + 3sinx + sin3x – cos3x = 0

3.

Solve

(1 – sin2x) (cosx – sinx) = 1 – 2sin2x. π π (1) nπ – , n ∈ Ι (2) nπ – , n ∈ Ι 4 4 π (3) 2nπ + , n ∈ Ι or 2nπ, n ∈ Ι or 2

Ans.

Type - 7

nπ +

π ,n∈Ι 4

Trigonometric equations which can be solved by the use of boundness of the trigonometric ratios sinx and cosx. Solved Example # 13 x  x    Solve sinx  cos − 2 sin x  +  1 + sin − 2 cos x  cos x = 0 4 4     Solution. x   x   .......(i) ∵ sinx  cos − 2 sin x  + 1 + sin − 2 cos x  cos x = 0 4   4   ⇒ ⇒ ⇒

x x – 2sin2x + cosx + sin .cosx – 2cos2x = 0 4 4 x x    sin x. cos + sin . cos x  – 2 (sin2x + cos2x) + cosx = 0 4 4  

sinx.cos

sin

5x + cosx = 2 4

7

........(ii)

Page : 7 of 15 TRIG. EQUATIONS

2.

and

cosx = 1

and

x = 2m π, m ∈ Ι

and

x = 2m π, m ∈ Ι

........(iv)

– 4, p ∈ Ι

∴

general solution of given equation can be obtained by substituting either m = 4p – 3 in equation (iv) or n = 5p – 4 in equation (iii)

∴

general solution of equation (i) is (8p – 6)π π, p ∈ Ι Ans.

Self Practice Problems : 1.

Solve

sin3x + cos2x = – 2

2.

Solve

3 sin 5 x − cos 2 x − 3 = 1 – sinx π (1) (4p – 3) , p ∈ Ι (2) 2

Ans.

2m π +

π , m ∈Ι 2

SHORT REVISION TRIGONOMETRIC EQUATIONS & INEQUATIONS THINGS TO REMEMBER : π π

2.

If sin θ = sin α ⇒ θ = n π + (−1)n α where α ∈ − ,  , n ∈ I .  2 2 If cos θ = cos α ⇒ θ = 2 n π ± α where α ∈ [0 , π] , n ∈ I .

3.

If tan θ = tan α ⇒ θ = n π + α where α ∈  − π , π  , n ∈ I .

4.

If sin² θ = sin² α ⇒ θ = n π ± α.

5.

cos² θ = cos² α ⇒ θ = n π ± α.

6. 7.

tan² θ = tan² α ⇒ θ = n π ± α. [ Note : α is called the principal angle ] TYPES OF TRIGONOMETRIC EQUATIONS :

1.

 2

2

(a)

Solutions of equations by factorising . Consider the equation ; (2 sin x − cos x) (1 + cos x) = sin² x ; cotx – cosx = 1 – cotx cosx

(b)

Solutions of equations reducible to quadratic equations. Consider the equation :

(c)

3 cos² x − 10 cos x + 3 = 0 and 2 sin2x + 3 sinx + 1 = 0 Solving equations by introducing an Auxilliary argument . Consider the equation :

(d)

sin x + cos x = 2 ; 3 cos x + sin x = 2 ; secx – 1 = ( 2 – 1) tanx Solving equations by Transforming a sum of Trigonometric functions into a product. Consider the example : cos 3 x + sin 2 x − sin 4 x = 0 ; sin2x + sin22x + sin23x + sin24x = 2 ; sinx + sin5x = sin2x + sin4x

(e)

Solving equations by transforming a product of trigonometric functions into a sum. 8

Page : 8 of 15 TRIG. EQUATIONS

Now equation (ii) will be true if 5x =1 sin 4 π 5x ⇒ = 2nπ + , n ∈ Ι 2 4 (8n + 2)π ⇒ x = ,n∈Ι ........(iii) 5 Now to find general solution of equation (i) (8n + 2)π = 2m π 5 ⇒ 8n + 2 = 10m 5m − 1 ⇒ n= 4 if m=1 then n=1 if m=5 then n=6 ......... ......... ......... ......... ......... ......... if m = 4p – 3, p ∈ Ι then n = 5p

sin 5 x . cos 3 x = sin 6 x .cos 2 x ; 8 cosx cos2x cos4x = (f)

sin 6 x ; sin3θ = 4sinθ sin2θ sin4θ sin x

Solving equations by a change of variable : (i) Equations of the form of a . sin x + b . cos x + d = 0 , where a , b & d are real numbers & a , b ≠ 0 can be solved by changing sin x & cos x into their corresponding tangent of half the angle. Consider the equation 3 cos x + 4 sin x = 5. (ii)

Many equations can be solved by introducing a new variable . eg. the equation sin4 2 x + cos4 2 x = sin 2 x . cos 2 x changes to 

1 2

2 (y + 1)  y −  = 0 by substituting , sin 2 x . cos 2 x = y.. Solving equations with the use of the Boundness of the functions sin x & cos x or by making two perfect squares. Consider the equations : x x     sin x  cos − 2 sin x +  1+ sin − 2cos x  . cos x = 0 ;   4 4   4 11 sin2x + 2tan2x + tanx – sinx + =0 3 12 TRIGONOMETRIC INEQUALITIES : There is no general rule to solve a Trigonometric inequations and the same rules of algebra are valid except the domain and range of trigonometric functions should be kept in mind. (g)

8.

x 1   Consider the examples : log 2  sin  < – 1 ; sin x  cos x +  < 0 ; 5 − 2 sin 2 x ≥ 6 sin x − 1 2 2  

EXERCISE–I 1 52

1 1 + log15 cos x + log 5 (sin x ) 2 = 15 2 +5

Q.1

Solve the equation for x,

Q.2

Find all the values of θ satisfying the equation; sin θ + sin 5 θ = sin 3 θ such that 0 ≤ θ ≤ π.

Q.3

Find all value of θ, between 0 & π, which satisfy the equation; cos θ . cos 2 θ . cos 3 θ = 1/4.

Q.4

Solve for x , the equation

Q.5

Determine the smallest positive value of x which satisfy the equation,

Q.6

π  2 sin  3 x +  =  4

Q.7

Find the general solution of the trigonometric equation 3

Q.8

Find all values of θ between 0° & 180° satisfying the equation; cos 6 θ + cos 4 θ + cos 2 θ + 1 = 0 .

Q.9

Find the solution set of the equation, log −x 2 −6x (sin 3x + sin x) = log −x 2 −6x (sin 2x).

13 − 18 tanx = 6 tan x – 3, where – 2π < x < 2π.

1 + 8 sin 2 x . cos2 2 x

1 + sin 2 x − 2 cos 3 x = 0 .

1   + log 3 (cos x + sin x )  2 

−2

log 2 (cos x − sin x )

= 2.

10

10

Q.10 Find the value of θ, which satisfy 3 − 2 cosθ − 4 sinθ − cos 2θ + sin 2θ = 0. Q.11

Find the general solution of the equation, sin πx + cos πx = 0. Also find the sum of all solutions in [0, 100].

Q.12 Find the least positive angle measured in degrees satisfying the equation sin3x + sin32x + sin33x = (sinx + sin2x + sin3x)3. 9

Page : 9 of 15 TRIG. EQUATIONS

Consider the equation :

Q.14 Prove that the equations (a) have no solution. Q.15 (a) (b) (c)

sin x · sin 2x · sin 3x = 1

(b)

sin x · cos 4x · sin 5x = – 1/2

Let f (x) = sin6x + cos6x + k(sin4x + cos4x) for some real number k. Determine all real numbers k for which f (x) is constant for all values of x. all real numbers k for which there exists a real number 'c' such that f (c) = 0. If k = – 0.7, determine all solutions to the equation f (x) = 0.

Q.16 If α and β are the roots of the equation, a cos θ + b sin θ = c then match the entries of column-I with the entries of column-II. Column-I Column-II 2b (P) (A) sin α + sin β a+c c−a (B) sin α . sin β (Q) c+ a 2bc α β (C) tan + tan (R) 2 2 2 a +b 2 (D)

tan

α 2

. tan

β 2

=

(S)

c 2 −a 2 a 2 +b 2

Q.17 Find all the solutions of, 4 cos2x sin x − 2 sin2x = 3 sin x. Q.18 Solve for x, (− π ≤ x ≤ π) the equation; 2 (cos x + cos 2 x) + sin 2 x (1 + 2 cos x) = 2 sin x. Q.19 Solve the inequality sin2x >

2 sin2x + (2 –

2 )cos2x.

Q.20 Find the set of values of 'a' for which the equation, sin4 x + cos4 x + sin 2x + a = 0 possesses solutions. Also find the general solution for these values of 'a'. Q.21 Solve: tan22x + cot22x + 2 tan 2x + 2 cot 2x = 6. Q.22 Solve: tan2x . tan23x . tan 4x = tan2x − tan23x + tan 4x. Q.23 Find the set of values of x satisfying the equality 2 cos 7 x π 3π    > 2cos 2 x . sin  x −  – cos  x +  = 1 and the inequality cos 3 + sin 3 4 4   

Q.24 Let S be the set of all those solutions of the equation, (1 + k)cos x cos (2x − α) = (1 + k cos 2x) cos(x − α) which are independent of k & α. Let H be the set of all such solutions which are dependent on k & α. Find the condition on k & α such that H is a non-empty set, state S. If a subset of H is (0, π) in which k = 0, then find all the permissible values of α. Q.25 Solve for x & y,

x cos 3 y + 3x cos y sin 2 y = 14 x sin 3 y + 3x cos 2 y sin y = 13

Q.26 Find the value of α for which the three elements set S = {sin α, sin 2α, sin 3α} is equal to the three element set T = {cos α, cos 2α, cos 3α}. Q.27 Find all values of 'a' for which every root of the equation, a cos 2x + a cos 4x + cos 6x = 1 10

is also a root of the equation, sin x cos 2 x = sin 2x cos 3x −

1 sin 5x , and conversely, every root 2

Page : 10 of 15 TRIG. EQUATIONS

Q.13 Find the general values of θ for which the quadratic function cos θ + sin θ is the square of a linear function. (sinθ) x2 + (2cosθ)x + 2

Q.28 Solve the equations for 'x' given in column-I and match with the entries of column-II. Column-I Column-II (A)

cos 3x . cos3 x + sin 3x . sin3 x = 0

(P)

nπ ±

π 3

(B)

sin 3α = 4 sin α sin(x + α) sin(x − α)

(Q)

nπ +

π , n∈I 4

(R)

nπ π , n∈I + 4 8

(S)

nπ 2

where α is a constant ≠ nπ. (C)

| 2 tan x – 1 | + | 2 cot x – 1 | = 2.

(D)

sin10x + cos10x =

29 cos42x. 16

±

π 4

EXERCISE–II Q.1 Q.2

Solve the following system of equations for x and y [REE ’2001(mains), 3] (cos ec 2 x − 3 sec 2 y) ( 2 cos ecx + 3 |sec y |) 5 = 1 and 2 = 64. The number of integral values of k for which the equation 7cosx + 5sinx = 2k + 1 has a solution is (A) 4 (B) 8 (C) 10 (D) 12 [JEE 2002 (Screening), 3]

Q.3

cos(α – β) = 1 and cos(α + β) = 1/e, where α, β ∈ [– π, π], numbers of pairs of α, β which satisfy both the equations is (A) 0 (B) 1 (C) 2 (D) 4 [JEE 2005 (Screening)]

Q.4

If 0 < θ < 2π, then the intervals of values of θ for which 2sin2θ – 5sinθ + 2 > 0, is

Q.5

 π   5π   π 5π   π   π 5π   41π  , π  [JEE 2006, 3] (A)  0,  ∪  , 2π  (B)  ,  (C)  0,  ∪  ,  (D)  8 6   6  6   8 6 6   48  The number of solutions of the pair of equations 2 sin2θ – cos2θ = 0 2 cos2θ – 3 sin θ = 0 in the interval [0, 2π] is (A) zero (B) one (C) two (D) four [JEE 2007, 3]

ANSWER Q.1

x = 2nπ +

π , n∈I 6

Q.2

0,

π 6

,

π 3

,

2π 3

Q.4 α − 2 π ; α − π , α , α + π , where tan α = Q.6

x = 2 nπ +

Q.9

x=−

Q.13 2nπ +

EXERCISE–I

17 π π or 2nπ + ; n ∈ I Q.7 12 12

,

5π 6

&π

2 3

x = 2nπ +

Q.3

π π 3π 5π 2π 7π , , , , , 8 3 8 8 3 8

Q.5

x = π/16

π Q.8 12

30° , 45° , 90° , 135° , 150°

5π 1 π Q.10 θ = 2 n π or 2 n π + ; n ∈ I Q.11 x = n – , n ∈ I; sum = 5025Q.12 2 3 4 π or (2n+1)π – tan–12 , n ∈ I 4

Q.16 (A) R; (B) S; (C) P; (D) Q Q.17 Q.18

±π −π , ,± π 3 2

Q.19

nπ +

Q.15 (a) –

3 ; (b) k ∈ 2

nπ ; n π + (−1)n

π π < x < nπ + 8 4 11

π 10

1  nπ π − 1, − 2  ; (c) x = 2 ± 6

or n π + (−1)n

 3π  −   10 

72°

Page : 11 of 15 TRIG. EQUATIONS

of the second equation is also a root of the first equation.

[

Q.21 x = Q.22

(

1 n π + (− 1) n sin −1 1 − 2 a + 3 2 nπ 4

+ (−1)n

π 8

or

nπ 4

)]

 3 1

where n ∈ I and a ∈ − ,  2 2 



π 24

+ (−1)n+1

(2 n + 1) π , k π , where n , k ∈ I 4

3π , n ∈I 4 (i) k sin α ≤ 1 (ii) S = n π , n ∈ I (iii) α ∈ (− m π , 2 π − m π) m ∈ I

Q.23 x = 2nπ + Q.24

Q.25 x = ± 5 5 & y = n π + tan−1

1 2

Q.26

Q.27 a = 0 or a < − 1

Q.28

nπ π + 2 8 (A) S; (B) P; (C) Q; (D) R

EXERCISE–II Q.1 Q.2

π π x = nπ + (–1)n and y = mπ + where m & n are integers. 6 6 B Q.3 D Q.4 A Q.5 C

Part : (A) Only one correct option 1.

The solution set of the equation 4sinθ .cosθ – 2cosθ – 2 3 sinθ + 3 = 0 in the interval (0, 2π) is  3 π 7π   (A)  , 4 4

2.

2π , n∈Ι 3

(C) nπ or m π ±

5.

π where n, m ∈ Ι 3

If 20 sin2 θ + 21 cos θ − 24 = 0 &

(A) 3 4.

π 5π   3π  (C)  , π, , 4 3 3  

 π 5 π 11π  ,  (D)  , 6  6 6

All solutions of the equation, 2 sinθ + tanθ = 0 are obtained by taking all integral values of m and n in: (A) 2nπ +

3.

 π 5π   (B)  , 3 3 

(B)

(B) nπ or 2m π ±

2π where n, m ∈ Ι 3

(D) nπ or 2m π ±

π where n, m ∈ Ι 3

7π θ < θ < 2π then the values of cot is: 4 2

15 3

(C) −

15 3

The general solution of sinx + sin5x = sin2x + sin4x is: (A) 2 nπ ; n ∈ Ι (B) nπ ; n ∈ Ι (C) nπ/3 ; n ∈ Ι A triangle ABC is such that sin(2A + B) =

(D) − 3 (D) 2 nπ/3 ; n ∈ Ι

1 . If A, B, C are in A.P. then the angle A, B, C are 2

respectively. (A) 6.

7.

5π π π , , 12 4 3

(B)

π π 5π , , 4 3 12

The maximum value of 3sinx + 4cosx is (A) 3 (B) 4

(C)

π π 5π , , 3 4 12

(C) 5

If sin θ + 7 cos θ = 5, then tan (θ/2) is a root of the equation (A) x 2 − 6x + 1 = 0 (B) 6x 2 − x − 1 = 0 (C) 6x 2 + x + 1 = 0 12

(D)

π 5π π , , 3 12 4

(D) 7 (D) x 2 − x + 6 = 0

Page : 12 of 15 TRIG. EQUATIONS

Q.20

sin 3 θ − cos 3 θ cos θ − − 2 tan θ cot θ = − 1 if: sin θ − cos θ 1 + cot 2 θ

 

(A) θ ∈  0 ,

π  2

π  , π 2 

 

(C) θ ∈  π ,

(B) θ ∈ 

3π   2

 3π  , 2π  2 

(D) θ ∈ 

9.

The number of integral values of a for which the equation cos 2x + a sin x = 2a − 7 possesses a solution is (A) 2 (B) 3 (C) 4 (D) 5

10.

The principal solution set of the equation, 2 cos x = 2 + 2 sin 2 x is

 π 13 π  (A)  ,  8 8  11.

 π 13 π   (B)  , 4 8 

 π 13 π   (C)  ,  4 10 

 π 13 π  (D)  8 , 10   

The number of all possible triplets (a1, a2, a3) such that : a1 + a2 cos 2x + a3 sin2x = 0 for all x is (A) 0 (B) 1 (C) 2 (D) infinite

12.

 nπ  , n ∈ N, then greatest value of n is If 2tan2x – 5 secx – 1 = 0 has 7 different roots in 0, 2   (A) 8 (B) 10 (C) 13 (D) 15

13.

The solution of |cosx| = cosx – 2sinx is (A) x = nπ, n ∈ Ι (C) x = nπ + (–1) n

(B) x = nπ + π , n ∈Ι 4

π ,n∈Ι 4

(D) (2n + 1)π +

π ,n∈Ι 4

14.

The arithmetic mean of the roots of the equation 4cos3x – 4cos2x – cos(π + x) – 1 = 0 in the interval [0, 315] is equal to (A) 49π (B) 50π (C) 51π (D) 100π

15.

Number of solutions of the equation cos 6x + tan2 x + cos 6x . tan2 x = 1 in the interval [0, 2π] is : (B) 5 (C) 6 (D) 7 (A) 4

Part : (B) May have more than one options correct 16.

sinx − cos2x − 1 assumes the least value for the set of values of x given by: (A) x = nπ + (−1) n+1 (π/6) , n ∈ Ι (B) x = nπ + (−1)n (π/6) , n ∈ Ι n (C) x = nπ + (−1) (π/3), n ∈ Ι (D) x = nπ − (−1) n ( π/6) , n ∈ Ι

17.

cos4x cos8x − cos5x cos9x = 0 if (A) cos12x = cos 14 x (C) sinx = 0

18.

The equation 2sin (A) sin2x = 1

19.

20.

(B) sin13 x = 0 (D) cosx = 0

x x . cos2x + sin2x = 2 sin . sin2x + cos2x has a root for which 2 2 1 1 (B) sin2x = – 1 (C) cosx = (D) cos2x = – 2 2

sin2x + 2 sin x cos x − 3cos2x = 0 if (A) tan x = 3 (C) x = nπ + π/4, n ∈ Ι

(B) tanx = − 1 (D) x = nπ + tan−1 (−3), n ∈ Ι

sin2x − cos 2x = 2 − sin 2x if (A) x = nπ/2, n ∈ Ι (C) x = (2n + 1) π/2, n ∈ Ι

(B) x = nπ − π/2, n ∈ Ι (D) x = nπ + ( −1)n sin−1 (2/3), n ∈ Ι

13

Page : 13 of 15 TRIG. EQUATIONS

8.

Page : 14 of 15 TRIG. EQUATIONS

1.

Solve

cot θ = tan8θ

2.

Solve

x x cot   – cosec   = cotx 2   2

3.

Solve

 1   cotθ + 1 = 0. cot 2θ +  3 + 3  

4.

Solve

cos2θ + 3 cosθ = 0.

5.

Solve the equation: sin 6x = sin 4x − sin 2x .

6.

Solve: cos θ + sin θ = cos 2 θ + sin 2 θ .

7.

Solve

4 sin x . sin 2x . sin 4x = sin 3x .

8.

Solve

sin2nθ – sin2(n – 1)θ = sin2θ, where n is constant and n ≠ 0, 1

9.

Solve

tanθ + tan2θ +

10.

Solve: sin3 x cos 3 x + cos3 x sin 3 x + 0.375 = 0

11.

Solve the equation,

12.

Solve the equation: sin 5x = 16 sin5 x .

13.

If tan θ + sin φ =

14.

Solve for x , the equation

15.

Find the general solution of sec 4 θ − sec 2 θ = 2 .

16.

Solve the equation

17.

Solve for x: 2 sin  3 x +

18.

Solve the equation for 0 ≤ θ ≤ 2 π; sin 2θ + 3 cos2θ

19.

Solve: tan2 x . tan2 3 x . tan 4 x = tan2 x − tan2 3 x + tan 4 x .

20.

Find the values of x, between 0 & 2 π, satisfying the equation; cos 3x + cos 2x = sin

3 tanθ tan2θ =

3.

sin 3 x − cos 3 x cos x 2 2 = . 3 2 + sin x

7 3 & tan² θ + cos² φ = then find the general value of θ & φ . 4 2

 

13 − 18 tan x = 6 tan x − 3, where − 2 π < x < 2 π .

3 sin x − cos x = cos² x . 2

π  = 4

1 + 8 sin 2 x . cos 2 2 x .

(

)

2

14

π  − 5 = cos  − 2θ  . 6 

3x x + sin . 2 2

Solve: cos

22.

Solve the equation, sin2 4 x + cos2 x = 2 sin 4 x cos4 x .

EXERCISE # 1

1 π  n +  , n∈Ι 3 3 

9. 1. D

2. B

3. D

4. C

5. B

6. C

7. B

8. B

9. D

10. A

11. D

12. D

13. D

14. C

15. D

16. AD 17. ABC

10. x =

nπ π + ( − 1)n + 1 , n∈Ι 4 24

18. ABCD 19. CD 11. x = (4 n + 1)

20. BC

π ,n∈Ι 2

EXERCISE # 2 12. x = n π ; x = n π ± 1.

1 π  n +  , n ∈ Ι 2 9 

13. θ = n π + 2. x = 4nπ ±

3. θ = nπ –

2π ,n∈Ι 3

π ,n∈Ι 3

or nπ –

π ,n∈Ι 6

 17 − 3  , n ∈Ι 4. 2nπ ± α where α = cos–1    4   nπ π , n ∈ Ι or n π ± , n ∈ Ι 4 6

2nπ π 6. 2 n π, n ∈ Ι or + , n∈Ι 3 6

17. (24 + 1)

18. θ =

π , n∈ Ι 3

π π ,  ∈ Ι or x = (24k – 7) , k∈Ι 12 12

7 π 19 π , 12 12

π 5π 9 π 13 π , ,π, , 7 7 7 7

21. φ

22. x = (2 n + 1)

15

or 2 n π ±

(2 n + 1) π , k π, where n, k ∈ Ι 4

nπ π ± , n ∈Ι 3 9

1 π  mπ , m ∈ Ι or  m +  ,m ∈Ι 2   n n −1

2 3

2nπ π π ± or 2nπ ± , n ∈ Ι 5 10 2

20.

8. m π, m ∈ Ι or

π π , φ = n π + (−1)n , n ∈ I 4 6

16. x = (2 n + 1)π, , n ∈ Ι

19. 7. x = n π, n ∈ Ι or

π ,n∈Ι 6

14. α − 2 π; α − π, α, α + π, where tan α =

15.

5.

Page : 15 of 15 TRIG. EQUATIONS

2x cos 6 x = − 1 . 3

21.

π , n∈I 2

STUDY PACKAGE Target: IIT-JEE (Advanced) SUBJECT: MATHEMATICS TOPIC: 3 XI M 3. Properties of Triangle Index: 1. Key Concepts 2. Exercise I to V 3. Answer Key 4. Assertion and Reasons 5. 34 Yrs. Que. from IIT-JEE 6. 10 Yrs. Que. from AIEEE

1

Properties & Solution of Triangle 1. Sine Rule: In any triangle ABC, the sines of the angles are proportional to the opposite sides i.e. a b c = = . sin A sin B sin C  A −B  cos    2  . C sin 2  A −B cos   a+b  2  = . We have to prove C c sin 2 From sine rule, we know that a b c = = = k (let) sin A sin B sin C a = k sinA, b = k sinB and c = k sinC a+b L.H.S. = c

Example :

a+b In any ∆ABC, prove that = c

Solution.

∵ ∵ ⇒ ∵

k (sin A + sin B ) = k sin C

C  A −B cos   2  2  = C C sin cos 2 2 = R.H.S. Hence L.H.S. = R.H.S. Proved In any ∆ABC, prove that (b2 – c2) cot A + (c 2 – a2) cot B + (a2 – b2) cot C = 0 ∵ We have to prove that (b2 – c 2) cot A + (c2 – a2) cot B + (a2 – b2) cot C = 0 ∵ from sine rule, we know that a = k sinA, b = k sinB and c = k sinC ∴ (b2 – c 2) cot A = k 2 (sin2B – sin2C) cot A ∵ sin2B – sin2C = sin (B + C) sin (B – C) ∴ (b2 – c2) cot A = k2 sin (B + C) sin (B – C) cotA cos A ∴ (b2 – c 2) cot A = k2 sin A sin (B – C) sin A = – k2 sin (B – C) cos (B + C) cos

Example : Solution.

k2 [2sin (B – C) cos (B + C)] 2 k2 ⇒ (b2 – c2) cot A = – [sin 2B – sin 2C] 2 k2 Similarly (c2 – a2) cot B = – [sin 2C – sin 2A] 2 k2 and (a2 – b2) cot C = – [sin 2A – sin 2B] 2 adding equations (i), (ii) and (iii), we get (b2 – c 2) cot A + (c2 – a2) cot B + (a2 – b2) cot C = 0 Self Practice Problems In any ∆ABC, prove that A  A 1. a sin  + B  = (b + c) sin   . 2   2

 A +B  A −B sin   cos    2   2  = C C sin cos 2 2  A −B cos    2  = C sin 2

∵

B+C=π–A

∵

cosA = – cos(B + C)

=–

2.

a 2 sin(B − C) b 2 sin(C − A ) c 2 sin( A − B) + + =0 sin B + sin C sin C + sin A sin A + sin B

3. 2

..........(i) ..........(ii) ..........(iii) Hence Proved

A B tan + tan c 2 2 . = A B a−b tan − tan 2 2

2. Cosine Formula: (i) cos A =

b 2 + c2 − a 2 2b c

or a² = b² + c² − 2bc cos A = b2 + c2 + 2bc cos (B + C)

c2 + a 2 − b 2 a 2 + b 2 − c2 (iii) cos C = 2 ca 2a b In a triangle ABC if a = 13, b = 8 and c = 7, then find sin A.

(ii) cos B = Example : Solution.

*Example : Solution.

64 + 49 − 169 b2 + c 2 − a2 = 2 .8 .7 2bc 2π 1 ⇒ cosA = – ⇒ A= 3 2 2π 3 = Ans. ∴ sinA = sin 3 2 In a ∆ ABC, prove that a(b cos C – c cos B) = b2 – c 2 ∵ We have to prove a (b cosC – c cosB) = b2 – c 2. ∵ from cosine rule we know that

∵

cosA =

a 2 + b2 − c 2 & 2ab   a 2 + b 2 − c 2  L.H.S. = a  b  2ab  

cosC ∴

=

a2 + c 2 − b 2 2ac 2 2    2  − c  a + c − b     2ac   

cos B =

a2 + b2 − c 2 (a 2 + c 2 − b 2 ) – 2 2 = (b2 – c2) Hence L.H.S. = R.H.S. Proved =

Example : Solution.

= R.H.S.

 a b  c a If in a ∆ABC, ∠A = 60° then find the value of 1 + +  1 + −  .  c c  b b ∵ ∠A = 60°  a b  c a c +a+b b+c −a  1 + +  1 + −  =     ∵ c c  b b c b     

=

(b + c )2 − a 2 bc

=

(b 2 + c 2 − a 2 ) + 2bc bc

=

b2 + c 2 − a2 +2 bc

 b 2 + c 2 − a2    =2   +2 2bc   ∵ ∠A = 60°

= 2cosA + 2

⇒

cos A =

1 2

a b  c a  1 + +  1 + −  = 3 Ans. c c b b   Self Practice Problems :

∴

a 2 + ab + b 2 , then prove that the greatest angle is 120°. A a(cosB + cosC) = 2(b + c) sin2 . 2

1.

The sides of a triangle ABC are a, b,

2.

In a triangle ABC prove that

3.

Projection Formula:

(i) a = b cosC + c cosB (ii) b = c cosA + a cosC (iii) c = a cosB + b cosA Example : In a triangle ABC prove that a(b cosC – c cosB) = b2 – c2 Solution. ∵ L.H.S. = a (b cosC – c cosB) = b (a cosC) – c (a cosB) ............(i) ∵ From projection rule, we know that b = a cosC + c cosA ⇒ a cosC = b – c cosA ⇒ a cosB = c – b cosA & c = a cosB + b cosA Put values of a cosC and a cosB in equation (i), we get L.H.S. = b (b – ccos A) – c(c – b cos A) = b2 – bc cos A – c2 + bc cos A = b2 – c 2 = R.H.S. Hence L.H.S. = R.H.S. Proved Note: We have also proved a (b cosC – ccosB) = b2 – c 2 by using cosine – rule in solved *Example. Example : In a ∆ABC prove that (b + c) cos A + (c + a) cos B + (a + b) cos C = a + b + c. 3

Solution.

∵

L.H.S. = (b + c) cos A (c + a) cos B + (a + B) cos C

= = = = Hence L.H.S. =

b cos A + c cos A + c cos B + a cos B + a cos C + b cos C (b cos A + a cos B) + (c cos A + a cos C) + (c cos B + b cos C) a+b+c R.H.S. R.H.S. Proved

Self Practice Problems

1.

In a ∆ABC, prove that B  2 C + c cos 2  = a + b + c. 2  b cos 2 2 

2.

cos B c − b cos A = . cos C b − c cos A

3.

cos A cos B cos C a2 + b2 + c 2 + + = . c cos B + b cos C a cos C + c cos A a cos B + b cos A 2abc

4. Napier’s Analogy - tangent rule: B−C = 2 A−B (iii) tan = 2 Example : Find the

(i) tan

Solution.

c −a B A C−A b−c cot (ii) tan = cot c +a 2 2 2 b+c a −b C cot a +b 2 unknown elements of the ∆ABC in which a = 3 + 1, b = 3 – 1, C = 60°.

∵

a = 3 + 1, b = 3 – 1, C = 60° A + B + C = 180°

∴

A + B = 120°

∵

From law of tangent, we know that

∵

.......(i)

a−b C  A −B tan   = cot a + b 2 2  

= =

( 3 + 1) − ( 3 − 1)

cot 30°

( 3 + 1) + ( 3 − 1) 2 2 3

cot 30°

 A −B  =1 tan   2  π A −B ∴ = 45° = 4 2 ⇒ A – B = 90° From equation (i) and (ii), we get A = 105° and B = 15° Now,

⇒

∵

From sine-rule, we know that

∴

c=

.......(ii)

a b c = = sin A sin B sin C

a sin C ( 3 + 1) sin 60° = sin A sin105°

3 2 3 +1

( 3 + 1)

=

∵

sin105° =

3 +1 2 2

2 2

⇒

c=

∴ c= Self Practice Problem 1.

6 6 , A = 105°, B = 15°

In a ∆ABC if b = 3, c = 5 and cos (B – C) = Ans.

Ans. 7 A , then find the value of tan . 25 2

1 3 4

2.

A B C B −C C−A   A −B If in a ∆ABC, we define x = tan   tan , y = tan   tan and z = tan   tan 2 2 2  2   2   2  then show that x + y + z = – xyz.

5. Trigonometric Functions of Half Angles: (i)

sin

(s − c) (s − a ) (s − b) (s − c) A B C ; sin = ; sin = = ca b c 2 2 2

(ii)

cos

s (s − b) s (s − a ) A B C = ; cos = ; cos = ca bc 2 2 2

(iii)

tan

A = 2

(iv)

sin A =

(s − a ) (s − b) ab

s (s − c) ab

(s − b) (s − c) ∆ a+b+c = where s = is semi perimetre of triangle. s (s − a ) s (s − a ) 2

2 bc

s(s − a )(s − b)(s − c) =

2∆ bc

6. Area of Triangle (∆) ∆=

1 1 1 ab sin C = bc sin A = ca sin B = s (s − a ) (s − b) (s − c) 2 2 2

Example :

In a ∆ABC if a, b, c are in A.P. then find the value of tan

Solution.

∵

∴ ∴ ∵ ⇒ ∵

∴ ∴ ⇒

tan

∆ A = s(s − a) 2

and tan

A C . tan . 2 2

∆ C = s(s − c ) 2

∆2 A C . tan = 2 s ( s − a)(s − c ) 2 2 s −b b A C tan . tan = =1– s s 2 2 it is given that a, b, c are in A.P. 2b = a + c a+b+c 3b s= = 2 2 b 2 = put in equation (i) s 3 2 A C tan . tan =1– 3 2 2 1 A C . tan = Ans. tan 3 2 2 tan

∵

∆ 2 = s (s – a) (s – b) (s – c)

........(i)

Example :

In a ∆ABC if b sinC(b cosC + c cosB) = 42, then find the area of the ∆ABC.

Solution.

∵ ∵

∵

∴ Example : Solution.

b sinC (b cosC + c cosB) = 42 From projection rule, we know that a = b cosC + c cosB put in (i), we get ab sinC = 42 1 ∆= ab sinC 2 ∆ = 21 sq. unit Ans.

........(i) given ........(ii)

C A B  In any ∆ABC prove that (a + b + c)  tan + tan  = 2c cot . 2 2 2  B A  ∵ L.H.S. = (a + b + c)  tan + tan  2 2  A 2

=

(s − b)(s − c ) s(s − a)

and tan

B = 2

(s − a)(s − c ) s(s − b)

∵

tan

∴

 (s − b)(s − c ) (s − a)(s − c )  +  L.H.S. = (a + b + c)  s(s − a) s(s − b)  

= 2s

s−c s

 s−b s−a +   s − a s − b   5

=2

=2 = 2c

 s−b+s−a   s(s − c )   (s − a)(s − b ) 

∵

2s= a + b + c

∴

2s – b – a = c

∵

cot

  c  s(s − c )   (s − a )(s − b ) 

s(s − c ) (s − a)(s − b)

= 2c cot = R.H.S. Hence L.H.S. = R.H.S.

C = 2

s(s − c ) (s − a)(s − b)

C 2

Proved

7. m - n Rule:

(m + n) cot θ = m cot α − n cot β = n cot B − m cot C Example :

If the median AD of a triangle ABC is perpendicular to AB, prove that tan A + 2tan B = 0.

Solution.

From the figure, we see that θ = 90° + B (as θ is external angle of ∆ABD)

Now if we apply m-n rule in ∆ABC, we get (1 + 1) cot (90 + B) = 1. cot 90° – 1.cot (A – 90°) ⇒ – 2 tan B = cot (90° – A) ⇒ – 2 tan B = tan A ⇒ tan A + 2 tan B = 0 Hence proved. Example :

The base of a triangle is divided into three equal parts. If t 1, t2, t3 be the tangents of the angles subtended by these parts at the opposite vertex, prove that  1 1  1 1 1 4 1 + 2  =  +   +  . t2   t1 t 2   t 2 t 3  

Solution.

Let point D and E divides the base BC into three equal parts i.e. BD = DE = DC = d (Let) and let α, β and γ be the angles subtended by BD, DE and EC respectively at their opposite vertex. ⇒ t 1 = tanα, t2 = tanβ and t3 = tanγ Now in ∆ABC ∵ BE : EC = 2d : d = 2 : 1 ∴ from m-n rule, we get (2 + 1) cotθ = 2 cot (α + β) – cotγ ⇒ 3cotθ = 2 cot (α + β) – cotγ .........(i) again ∵ in ∆ADC ∵ DE : EC = x : x = 1 : 1 ∴ if we apply m-n rule in ∆ADC, we get (1 + 1) cotθ = 1. cotβ – 1 cotγ 2cotθ = cotβ – cotγ .........(ii) from (i) and (ii), we get 2 cot(α + β) − cot γ 3 cot θ = cot β − cot γ 2 cot θ ⇒ 3cotβ – 3cotγ = 4cot (α + β) – 2 cotγ ⇒ 3cotβ – cotγ = 4 cot (α + β)  cot α. cot β − 1  ⇒ 3cotβ – cotγ = 4   cot β + cot α  2 ⇒ 3cot β + 3cotα cotβ – cotβ cotγ – cotα cotγ = 4 cotα cotβ – 4 ⇒ 4 + 3cot2β = cotα cotβ + cotβ cotγ + cotα cotγ ⇒ 4 + 4cot2β = cotα cotβ + cotα cotγ + cotβ cotγ + cot 2β ⇒ 4(1 + cot 2β) = (cotα + cotβ) (cotβ + cotγ)   1 1   1 1  1     6 +  + ⇒ 4 1 + 2  =  tan α tan β tan β tan γ  tan β     

 1 1 1 1 1 4 1 + 2  =  +   +   t1 t 2   t 2 t 3   t2  Self Practice Problems : ⇒

1.

Hence proved

1

In a ∆ABC, the median to the side BC is of length

11 − 6 3 30° and 45°. Prove that the side BC is of length 2 units.

and it divides angle A into the angles of

8. Radius of Circumcirlce : R=

c a b a bc = = = 2 sinA 2 sinB 2 sinC 4∆ s R

Example :

In a ∆ABC prove that sinA + sinB + sinC =

Solution.

In a ∆ABC, we know that a b c = = = 2R sin A sin B sin C a b c ∴ sin A = , sinB = and sinC = . 2R 2R 2R a+b+c ∴ sinA + sinB + sinC = ∵ a + b + c = 2s 2R 2s s = ⇒ sinA + sinB + sinC = . 2R R In a ∆ABC if a = 13 cm, b = 14 cm and c = 15 cm, then find its circumradius. abc .......(i) ∵ R= 4∆ ∵ ∆ = s(s − a )(s − b)(s − c )

Example : Solution.

a+b+c = 21 cm 2 ∴ ∆ = 21.8.7.6 = 7 2.4 2.3 2 ⇒ ∆ = 84 cm 2 13 .14.15 65 ∴ R= = cm 4.84 8 65 cm. ∴ R= 8 A B C In a ∆ABC prove that s = 4R cos . cos . cos . 2 2 2 In a ∆ABC,

∵

Example : Solution.

s(s − a ) s(s − b) B C = = , cos and cos bc ca 2 2 A B C ∵ R.H.S. = 4R cos . cos . cos . 2 2 2 s( s − a)(s − b)(s − c ) abc = .s ∵ (abc )2 ∆ = s = L.H.S. Hence R.H.L = L.H.S. proved 1 1 1 1 4R In a ∆ABC, prove that + + – = . s−a s −b s−c s ∆ 4R 1 1 1 1 + + – = s−a s −b s−c s ∆ 1  1  1  1 + −   +  ∵ L.H.S. =  s−a s−b s−c s

∵

Example : Solution.

s=

cos

A = 2

2s − a − b (s − s + c ) + ( s − a)(s − b) s( s − c ) c c + = ( s − a)(s − b) s(s − c ) =

∵

s( s − c ) abc and R = ab 4∆

∆=

s(s − a)(s − b)(s − c )

2s = a + b + c

 2s2 − s(a + b + c ) + ab   s(s − c ) + ( s − a )(s − b)   =c   =c  ∆2   s(s − a)(s − b)(s − c )   7

∴

 2s 2 − s(2s) + ab  abc 4R∆ 4R L.H.S. = c  =  = 2 = 2 2 ∆ ∆ ∆ ∆  

abc 4∆ abc = 4R∆

R=

∵ ⇒

4R ∴ L.H.S. = ∆ Self Practice Problems :

In a ∆ABC, prove the followings : 1.

a cot A + b cotB + cos C = 2(R + r).

2.

s  s  s  r 4  − 1  − 1  − 1 = . R a  b  c 

3.

If α, β, γ are the distances of the vertices of a triangle from the corresponding points of contact with the αβ y incircle, then prove that = r2 α+β+y

9. Radius of The Incircle : ∆ s a sin B2 sin C2 (iii) r = cos A2

A B C = (s − b) tan = (s − c) tan 2 2 2 A B C (iv) r = 4R sin sin sin 2 2 2

(ii) r = (s − a) tan

(i) r =

& so on

10. Radius of The Ex- Circles : A B C ∆ ; ∆ ; ∆ r2 = r3 = (ii) r1 = s tan ; r2 = s tan ; r 3 = s tan 2 2 2 s−a s−b s−c a cos B2 cos C2 A B C (iii) r1 = & so on (iv) r 1 = 4 R sin . cos . cos 2 2 2 cos A2 Example : In a ∆ABC, prove that r1 + r2 + r3 – r = 4R = 2a cosecA

(i) r1 =

Solution.

∵

L.H.S

= r1 + r2 + r 3 – r ∆ ∆ ∆ ∆ = + + – s−a s −b s−c s 1 1 1 1    + −   +∆  =∆  s−a s−b s−c s  s − b + s − a   s − s + c  = ∆  (s − a)(s − b)  +  s(s − c )       c c  + =∆    (s − a)(s − b) s(s − c )   s(s − c ) + (s − a)(s − b)  = c∆    s(s − a)(s − b )(s − c )   2s 2 − s(a + b + c ) + ab   = c∆  ∆2   abc = ∆

= 4R = 2acosecA

∵

a + b + c = 2s

∵ ∵

R=

abc 4∆

a = 2R = acosecA sin A

Example :

= R.H.S. Hence L.H.S. = R.H.S. proved If the area of a ∆ABC is 96 sq. unit and the radius of the escribed circles are respectively 8, 12 and 24. Find the perimeter of ∆ABC.

Solution.

∵ ∵ ∵ ∵

∴

∆ = 96 sq. unit r1 = 8, r2 = 12 and r3 = 24 ∆ ⇒ s – a = 12 r1 = s−a ∆ r2 = ⇒ s–b=8 s−b ∆ r3 = ⇒ s–c=4 s−c 8 adding equations (i), (ii) & (iii), we get

.........(i) .........(ii) .........(iii)

∴

3s – (a + b + c) = 24 s = 24 perimeter of ∆ABC = 2s = 48 unit.

Self Practice Problems

In a ∆ABC prove that 1.

r 1r2 + r2r3 + r3r1 = s2

2.

rr1 + rr 2 + rr 3 = ab + bc + ca – s2

3.

If A, A1, A2 and A3 are the areas of the inscribed and escribed circles respectively of a ∆ABC, then prove 1 1 1 1 = that + + . A A1 A2 A3

4.

c r1 − r r2 − r + = r . a b 3

11. Length of Angle Bisectors, Medians & Altitudes :

(i) Length of an angle bisector from the angle A = β a =

2 bc cos A 2 b+c

;

1 2 b2 + 2 c2 − a 2 2 2∆ & (iii) Length of altitude from the angle A = Aa = a 3 2 2 2 NOTE : ma + m b + m c = (a2 + b2 + c2) 4 (ii) Length of median from the angle A = m a =

Example :

AD is a median of the ∆ABC. If AE and AF are medians of the triangles ABD and ADC respectively, and AD = m 1, AE = m 2 , AF = m 3 , then prove that m 22 + m 32 – 2m 12 =

Solution.

In ∆ABC 1 (2b2 + 2c2 – a2) = m 12 AD2 = 4 1 a2 (2c2 + 2AD2 – ) ∵ In ∆ABD, AE2 = m 22 = 4 4 2 1  2AD2 + 2b 2 − a  Similarly in ∆ADC, AF 2 = m 32 = 4  4  by adding equations (ii) and (iii), we get

a2 . 8

∵

∵

.........(i) .........(ii) ........(iii)

2    4 AD2 + 2b 2 + 2c 2 − a   2   2 1  2b 2 + 2c 2 − a  2 = AD + 2  4  2 1  2b 2 + 2c 2 − a 2 + a  2 = AD +  2  4 

m 22 + m 3 2 =

1 4

1 a2 (2b2 + 2c2 – a2) + 4 8 2 a = AD2 + AD2 + 8 2 a ∵ = 2AD2 + 8 a2 = 2m 12 + 8

= AD2 +

∴

m 22 + m 32 – 2m 12 =

a2 8

AD2 = m 12

Hence Proved 9

Self Practice Problem : 3.

In a ∆ABC a = 5, b = 4, c = 3. ‘G’ is the centroid of triangle, then find circumradius of ∆GAB. 5 Ans. 13 12

12. The Distances of The Special Points from Vertices and Sides of Triangle: (i)

Circumcentre (O)

:

OA = R & Oa = R cos A

(ii)

Incentre (I)

:

IA = r cosec

(iii)

Excentre (I1)

:

(iv)

Orthocentre (H)

:

HA = 2R cos A & Ha = 2R cos B cos C

(v)

Centroid (G)

:

GA =

Example :

Solution.

A & Ia = r 2 A & I 1a = r1 I1 A = r1 cosec 2

1 2∆ 2b2 +2c2 −a 2 & Ga = 3 3a

If x, y and z are respectively the distances of the vertices of the ∆ABC from its orthocentre, then prove that abc a c b (i) + + = (ii) x y + z = 2(R + r) xyz x z y ∵ x = 2R cosA, y = 2R cosB, z = 2R cosC and and a = 2R sinA, b = 2R sinB, c = 2R sinC a c b ∴ + + = tanA + tan B + tan C .........(i) x z y & ∵ ∴

∵ ∵

∴

∴

abc ........(ii) xyz = tanA. tanB. tanC We know that in a ∆ABC Σ tanA = Π tanA From equations (i) and (ii), we get abc a c b + + = xyz x z y x + y + z = 2R (cosA + cosB + cosC) A B C in a ∆ABC cosA + cosB + cosC = 1 + 4sin sin sin 2 2 2 A B C  x + y + z = 2R 1 + 4 sin . sin . sin  2 2 2  A B C  = 2  R + 4R sin . sin . sin  2 2 2  x + y + z = 2(R + r)

r = 4R sin

∵

B C A sin sin 2 2 2

Self Practice Problems A B C tan tan . 2 2 2

1.

If Ι be the incentre of ∆ABC, then prove that ΙA . ΙB . ΙC = abc tan

2.

If x, y, z are respectively be the perpendiculars from the circumcentre to the sides of ∆ABC, then prove abc a c b that + + = . 4 xyz x z y

13. Orthocentre and Pedal Triangle: The triangle KLM which is formed by joining the feet of the altitudes is called the Pedal Triangle. (i) Its angles are π − 2A, π − 2B and π − 2C. (ii) Its sides are a cosA = R sin 2A, b cosB = R sin 2B and c cosC = R sin 2C (iii) Circumradii of the triangles PBC, PCA, PAB and ABC are equal.

14. Excentral Triangle: The triangle formed by joining the three excentres Ι1, Ι2 and Ι 3 of ∆ ABC is called the excentral or excentric triangle. (i) ∆ ABC is the pedal triangle of the ∆ Ι1 Ι 2 Ι 3. (ii) Its angles are

π C π A π B − , − & − . 2 2 2 2 2 2

10

(iii)

(iv)

(v)

A , 2 B C & 4 R cos . 4 R cos 2 2 A Ι Ι1 = 4 R sin ; 2 B C Ι Ι2 = 4 R sin ; Ι Ι 3 = 4 R sin . 2 2 Its sides are 4 R cos

Incentre Ι of ∆ ABC is the orthocentre of the excentral ∆ Ι 1 Ι 2 Ι 3.

15. Distance Between Special Points : (i) Distance between circumcentre and orthocentre OH2 = R2 (1 – 8 cosA cos B cos C) (ii) Distance between circumcentre and incentre A B C OΙ2 = R2 (1 – 8 sin sin sin ) = R2 – 2Rr 2 2 2 (iii) Distance between circumcentre and centroid 1 OG2 = R2 – (a2 + b2 + c2) 9 In Ι is the incentre and Ι1, Ι 2, Ι3 are the centres of escribed circles of the ∆ABC, prove that Example : 2 (ii) ΙΙ12 + Ι 2Ι32 = ΙΙ22 + Ι3Ι 12 = ΙΙ 32 + Ι1Ι 22 (i) ΙΙ 1. ΙΙ2 . ΙΙ3 = 16R r Solution. (i) ∵ We know that A B C ΙΙ1 = a sec , ΙΙ2 = b sec and ΙΙ3 = c sec 2 2 2 C A B ∵ Ι 1Ι2 = c. cosec , Ι2 Ι3 = a cosec and Ι 3Ι1 = b cosec 2 2 2 A B C ∵ ΙΙ1 . ΙΙ 2 . ΙΙ3 = abc sec sec .sec ........(i) 2 2 2 ∵ a = 2R sin A, b = 2R sinB and c = 2R sinC ∴ equation (i) becomes A B C ∵ ΙΙ1. ΙΙ 2 . ΙΙ3 = (2R sin A) (2R sin B) (2R sinC) sec sec sec 2 2 2

∴

A A B B  C C   2 sin cos   2 sin cos   2 sin cos  2 2 2 2 2 2  = 8R3 . C A B cos . cos . cos 2 2 2 A B C A B C = 64R3 sin sin sin ∵ r = 4R sin sin sin 2 22 2 2 2 2 ΙΙ1 . ΙΙ2 . ΙΙ 3 = 16R r Hence Proved

ΙΙ1 + Ι 2Ι3 = ΙΙ2 + Ι3Ι1 = ΙΙ3 + Ι1Ι2 2

(ii)

2

2

2

2

2

a2 A A + a2 cosec2 = A A 2 2 sin2 cos 2 2 2 A A 16 R 2 sin2 . cos2 A A 2 2 2 2 2 ∵ a = 2 R sinA = 4R sin cos ∴ ΙΙ 1 + Ι 2Ι3 = = 16R 2 A 2 A 2 2 sin . cos 2 2 2 Similarly we can prove ΙΙ22 + Ι3Ι 12 = ΙΙ32 + Ι1Ι22 = 16R Hence ΙΙ12 + Ι 2Ι32 = ΙΙ22 + Ι3Ι12 = ΙΙ32 + Ι1Ι22 Self Practice Problem : π 1. In a ∆ ABC, if b = 2 cm, c = 3 cm and ∠A = , then find distance between its circumcentre and 6 incentre.

∵

Ans.

ΙΙ1 + Ι2Ι3 = a2 sec2 2

2

2 − 3 cm

11

SHORT REVISION SOLUTIONS OF TRIANGLE I.

SINE FORMULA :

In any triangle ABC ,

II.

COSINE FORMULA :

(i) cos A =

a b c . = = sin A sin B sin C

b 2 +c 2 −a 2 2bc

or a² = b² + c² − 2bc. cos A

c 2 +a 2 −b 2 2ca (i) a = b cos C + c cos B

a 2 +b 2 −c 2 2ab (ii) b = c cos A + a cos C

(iii) cos C =

(ii) cos B = III.

IV.

V.

VI.

VII.

PROJECTION FORMULA :

(iii) c = a cos B + b cos A A B−C b−c NAPIER’S ANALOGY − TANGENT RULE : (i) tan = cot b+ c 2 2 C− A c−a B C A− B a −b (ii) tan = cot (iii) tan = cot a +b 2 2 c+a 2 2 TRIGONOMETRIC FUNCTIONS OF HALF ANGLES :

(s−c)(s−a ) C ; sin = ca 2

(s−a )(s−b) ab

A = 2

(s−b)(s−c) B ; sin = bc 2

cos

A = 2

s(s−a ) B ; cos = bc 2

(iii)

tan

A = 2

∆ (s−b)(s−c) a + b+c = where s = & ∆ = area of triangle. s(s−a ) s(s−a ) 2

(iv)

Area of triangle = s(s−a )(s−b)(s−c) .

(i)

sin

(ii)

s(s−b) C ; cos = ca 2

s(s−c) ab

M −N

RULE : In any triangle , (m + n) cot θ = m cot α − n cot β = n cot B − m cot C 1 2

ab sin C =

1 2

bc sin A =

1 2

ca sin B = area of triangle ABC.

a b c = = = 2R sin A sin B sin C a bc Note that R = 4 ∆ ; Where R is the radius of

circumcircle & ∆ is area of triangle

VIII. Radius of the incircle ‘r’ is given by: (a) r = (c) r =

∆ a +b+c where s = 2 s a sin B2 sin C2 & so on cos A2

(b) r = (s − a) tan (d) r = 4R sin

Radius of the Ex− circles r1 , r2 & r3 are given by :

(a)

r1 = (c)

r1 =

a cos B2 cos C2 cos A2

r2 = 4 R sin X.

B 2

(b)

. cos

A 2

. cos

r1 = s tan

(d)

& so on C 2

= (s − b) tan

= (s − c) tan

A 2

. cos

C 2

. cos

r3 = 4 R sin

B 2

. cos

A 2

C 2

. cos

LENGTH OF ANGLE BISECTOR & MEDIANS : If ma and βa are the lengths of a median and an angle bisector from the angle A then, 12

C 2

C A B ; r2 = s tan ; r3 = s tan 2 2 2

r1 = 4 R sin

;

B 2

A B C sin sin 2 2 2

IX.

∆ ∆ ∆ ; r2 = ; r3 = s−c s−b s−a

A 2

B 2

;

ma =

1 2

2 b 2 + 2 c 2 − a 2 and βa =

Note that m2a + m2b + m2c = XI.

− − − − XII

− − −

2 bc cos A 2 b+c

3 2 (a + b2 + c2) 4

ORTHOCENTRE AND PEDAL TRIANGLE : The triangle KLM which is formed by joining the feet of the altitudes is called the pedal triangle. the distances of the orthocentre from the angular points of the ∆ ABC are 2 R cosA , 2 R cosB and 2 R cosC the distances of P from sides are 2 R cosB cosC, 2 R cosC cosA and 2 R cosA cosB the sides of the pedal triangle are a cosA (= R sin 2A), b cosB (= R sin 2B) and c cosC (= R sin 2C) and its angles are π − 2A, π − 2B and π − 2C. circumradii of the triangles PBC, PCA, PAB and ABC are equal . EXCENTRAL TRIANGLE : The triangle formed by joining the three excentres I1, I2 and I3 of ∆ ABC is called the excentral or excentric triangle. Note that : Incentre I of ∆ ABC is the orthocentre of the excentral ∆ I1I2I3 . ∆ ABC is the pedal triangle of the ∆ I1I2I3 . the sides of the excentral triangle are A C B , 4 R cos and 4 R cos 2 2 2 π A π B and its angles are , and π − C . − − 2 2 2 2 2 2 C B A I I1 = 4 R sin ; I I2 = 4 R sin ; I I3 = 4 R sin . 2 2 2

4 R cos

−

XIII. THE DISTANCES BETWEEN THE SPECIAL POINTS : The distance between circumcentre and orthocentre is = R . 1 − 8 cos A cos B cos C (a) (b)

The distance between circumcentre and incentre is = R 2 − 2 R r

(c) XIV.

The distance between incentre and orthocentre is 2 r 2 − 4 R 2 cos A cos B cos C Perimeter (P) and area (A) of a regular polygon of n sides inscribed in a circle of radius r are given by π 1 2π P = 2nr sin and A = nr2 sin 2 n n Perimeter and area of a regular polygon of n sides circumscribed about a given circle of radius r is given by π π P = 2nr tan and A = nr2 tan n n

EXERCISE–I Q.1

With usual notations, prove that in a triangle ABC: b−c c−a a −b + + =0 r3 r1 r2

Q.2

a cot A + b cot B + c cot C = 2(R + r)

Q.3

r3 r1 r2 3 + + = (s − b) (s − c) (s − c) (s − a ) (s − a ) (s − b) r

Q.4

r1 − r r2 − r c + = a b r3

Q.5

abc A B C cos cos cos = ∆ s 2 2 2

Q.6

(r1 + r2)tan

Q.7

(r1− r) (r2− r)(r3− r) = 4 R r2

Q.8 (r + r1)tan 13

C C = (r3 − r) cot = c 2 2

B−C C−A A−B +(r + r2)tan +(r + r3) tan =0 2 2 2

Q.9

1 1 1 1 a 2 + b2 + c2 + + + = r 2 r12 r2 2 r32 ∆2

Q.10 (r3+ r1) (r3+ r2) sin C = 2 r3 r2 r3 + r3r1 + r1r2

Q.11

1 1 1 1 + + = bc ca ab 2Rr

Q.12

Q.13

bc − r2 r3 ca − r3r1 ab − r1r2 = = =r r3 r1 r2

 1 1   1 1   1 1  4R  −  −  − = 2 2      r r1   r r2   r r3  r s 2

Q.14

Q.15 Rr (sin A + sin B + sin C) = ∆

Q.16

1 1 1 1  41 1 1   + + +  =  + +  r r r r  r  r1 r2 r3  1 2 3  2R cos A = 2R + r – r1

A B C s2 a 2 + b2 + c2 + cot + cot = Q.18 cot A + cot B + cot C = 2 2 2 ∆ 4∆ Given a triangle ABC with sides a = 7, b = 8 and c = 5. If the value of the expression

Q.17 cot Q.19

(∑ sin A ) ∑ cot A  can be expressed in the form qp where p, q ∈ N and qp is in its lowest form find 2  the value of (p + q). Q.20 If r1 = r + r2 + r3 then prove that the triangle is a right angled triangle. Q.21 If two times the square of the diameter of the circumcircle of a triangle is equal to the sum of the squares of its sides then prove that the triangle is right angled. Q.22 In acute angled triangle ABC, a semicircle with radius ra is constructed with its base on BC and tangent to the other two sides. rb and rc are defined similarly. If r is the radius of the incircle of triangle ABC then 1 1 1 2 prove that, = + + . ra rb rc r Q.23 Given a right triangle with ∠A = 90°. Let M be the mid-point of BC. If the inradii of the triangle ABM and ACM are r1 and r2 then find the range of r1 r2 . Q.24 If the length of the perpendiculars from the vertices of a triangle A, B, C on the opposite sides are p1, p2, p3 then prove that

1 1 1 1 1 1 1 + + = = + + . p1 p2 p3 r1 r r2 r3

 a b   b c   c a   bc ca ab Q.25 Prove that in a triangle r + r + r = 2R  b + a  +  c + b  +  a + c  − 3 .        1 2 3

EXERCISE–II Q.1 Q.2

b+c c+a a+b = = ; then prove that, cos A = cos B = cos C . 11 12 13 7 19 25 A b−c For any triangle ABC , if B = 3C, show that cos C = b + c & sin = . 2 2c 4c

With usual notation, if in a ∆ ABC,

π 3 · l (AB) and ∠ DBC = . Determine the ∠ABC. 2 4

Q.3

In a triangle ABC, BD is a median. If l (BD) =

Q.4

ABCD is a trapezium such that AB , DC are parallel & BC is perpendicular to them. If angle ADB = θ , BC = p & CD = q , show that AB =

Q.5

(p 2 + q 2 ) sinθ . p cos θ + q sin θ

If sides a, b, c of the triangle ABC are in A.P., then prove that A B C cosec 2A; sin2 cosec 2B; sin2 cosec 2C are in H.P.. sin2 2 2 2 14

Q.6

Find the angles of a triangle in which the altitude and a median drawn from the same vertex divide the angle at that vertex into 3 equal parts.

Q.7

In a triangle ABC, if tan

Q.8

ABCD is a rhombus. The circumradii of ∆ ABD and ∆ ACD are 12.5 and 25 respectively. Find the area of rhombus. cot C In a triangle ABC if a2 + b2 = 101c2 then find the value of . cot A + cot B

Q.9

A B C , tan , tan are in AP. Show that cos A, cos B, cos C are in AP. 2 2 2

Q.10 The two adjacent sides of a cyclic quadrilateral are 2 & 5 and the angle between them is 60°. If the area of the quadrilateral is 4 3 , find the remaining two sides. Q.11

If I be the in−centre of the triangle ABC and x, y, z be the circum radii of the triangles IBC, ICA & IAB, show that 4R3 − R (x2 + y2 + z2) − xyz = 0.

Q.12 Sides a, b, c of the triangle ABC are in H.P. , then prove that cosec A (cosec A + cot A) ; cosec B (cosec B + cot B) & cosec C (cosec C + cot C) are in A.P. Q.13 In a ∆ ABC, (i) (iii) tan2

a b = cos A cos B

(ii) 2 sin A cos B = sin C

A A C + 2 tan tan − 1 = 0, prove that (i) ⇒ (ii) ⇒ (iii) ⇒ (i). 2 2 2

Q.14 The sequence a1, a2, a3, ........ is a geometric sequence. The sequence b1, b2, b3, ........ is a geometric sequence. b1 = 1;

b2 =

4

7 − 28 + 1; 4

a1 =

4

28 and

∞

1

∑a n =1

n

∞

= ∑ bn n =1

If the area of the triangle with sides lengths a1, a2 and a3 can be expressed in the form of p q where p and q are relatively prime, find (p + q). Q.15 If p1 , p2 , p3 are the altitudes of a triangle from the vertices A , B , C & ∆ denotes the area of the 1 1 1 2ab 2 C triangle , prove that p + p − p = (a + b + c)∆ cos 2 . 1 2 3 Q.16 The triangle ABC (with side lengths a, b, c as usual) satisfies log a2 = log b2 + log c2 – log (2bc cosA). What can you say about this triangle? Q.17 With reference to a given circle, A1 and B1 are the areas of the inscribed and circumscribed regular polygons of n sides, A2 and B2 are corresponding quantities for regular polygons of 2n sides. Prove that (1) A2 is a geometric mean between A1 and B1. (2) B2 is a harmonic mean between A2 and B1. Q.18 The sides of a triangle are consecutive integers n, n + 1 and n + 2 and the largest angle is twice the smallest angle. Find n. Q.19 The triangle ABC is a right angled triangle, right angle at A. The ratio of the radius of the circle circumscribed

(

)

to the radius of the circle escribed to the hypotenuse is, 2 : 3 + 2 . Find the acute angles B & C. Also find the ratio of the two sides of the triangle other than the hypotenuse.

15

Q.20 ABC is a triangle. Circles with radii as shown are drawn inside the triangle each touching two sides and the incircle. Find the radius of the incircle of the ∆ABC. Q.21 Line l is a tangent to a unit circle S at a point P. Point A and the circle S are on the same side of l, and the distance from A to l is 3. Two tangents from point A intersect line l at the point B and C respectively. Find the value of (PB)(PC). Q.22 Let ABC be an acute triangle with orthocenter H. D, E, F are the feet of the perpendiculars from A, B, and C on the opposite sides. Also R is the circumradius of the triangle ABC. Given (AH)(BH)(CH) = 3 and (AH)2 + (BH)2 + (CH)2 = 7. Find (a) the ratio

∏ cos A , ∑ cos 2 A

(b) the product (HD)(HE)(HF)

(c) the value of R.

EXERCISE–III Q.1

The radii r1, r2, r3 of escribed circles of a triangle ABC are in harmonic progression. If its area is 24 sq. cm and its perimeter is 24 cm, find the lengths of its sides. [REE '99, 6]

Q.2(a) In a triangle ABC , Let ∠ C = 2(r + R) is equal to: (A) a + b (b)

Q.3

Q.4

Q.5

Q.6

(B) b + c

In a triangle ABC , 2 a c sin (A) a2 + b2 − c2

1 (A − B + C) = 2

(B) c2 + a2 − b2

(C) c + a

(D) a + b + c

(C) b2 − c2 − a2

(D) c2 − a2 − b2 [JEE '2000 (Screening) 1 + 1]

Let ABC be a triangle with incentre ' I ' and inradius ' r ' . Let D, E, F be the feet of the perpendiculars from I to the sides BC, CA & AB respectively . If r1 , r2 & r3 are the radii of circles inscribed in the quadrilaterals AFIE , BDIF & CEID respectively, prove that r r1 r r1 r2 r3 + 2 + 3 = . [JEE '2000, 7] r − r1 r − r2 r − r3 (r − r1 )(r − r2 )(r − r3 ) 1 If ∆ is the area of a triangle with side lengths a, b, c, then show that: ∆ < (a + b + c)abc 4 Also show that equality occurs in the above inequality if and only if a = b = c. [JEE ' 2001] Which of the following pieces of data does NOT uniquely determine an acute–angled triangle ABC (R being the radius of the circumcircle)? (A) a, sinA, sinB (B) a, b, c (C) a, sinB, R (D) a, sinA, R [ JEE ' 2002 (Scr), 3 ] If In is the area of n sided regular polygon inscribed in a circle of unit radius and On be the area of the polygon circumscribing the given circle, prove that On In = 2

Q.7

π . If ' r ' is the inradius and ' R ' is the circumradius of the triangle, then 2

2  1 + 1 −  2 I n       n    

[JEE 2003, Mains, 4 out of 60]

The ratio of the sides of a triangle ABC is 1 : 3 : 2. The ratio A : B : C is (A) 3 : 5 : 2

(B) 1 : 3 : 2

(D) 1 : 2 : 3 [JEE 2004 (Screening)] Q.8(a) In ∆ABC, a, b, c are the lengths of its sides and A, B, C are the angles of triangle ABC. The correct relation is  B−C A  = a cos  (A) ( b − c) sin   2  2

(C) 3 : 2 : 1

A B−C ( b − c) cos  = a sin   (B) 16 2  2 

B+C A  = a cos  (C) ( b + c) sin   2  2

A  B+C  (D) (b − c) cos  = 2a sin  2  2  [JEE 2005 (Screening)] (b) Circles with radii 3, 4 and 5 touch each other externally if P is the point of intersection of tangents to these circles at their points of contact. Find the distance of P from the points of contact. [JEE 2005 (Mains), 2]

Q.9(a) Given an isosceles triangle, whose one angle is 120° and radius of its incircle is 3 . Then the area of triangle in sq. units is (A) 7 + 12 3

(C) 12 + 7 3

(B) 12 – 7 3

(D) 4π

[JEE 2006, 3] (b) Internal bisector of ∠A of a triangle ABC meets side BC at D. A line drawn through D perpendicular to AD intersects the side AC at E and the side AB at F. If a, b, c represent sides of ∆ABC then (A) AE is HM of b and c (C) EF =

(B) AD =

4bc A sin b+c 2

2bc A cos b+c 2

(D) the triangle AEF is isosceles

[JEE 2006, 5]

Q.10 Let ABC and ABC′ be two non-congruent triangles with sides AB = 4, AC = AC′ = 2 2 and angle B = 30°. The absolute value of the difference between the areas of these triangles is [JEE 2009, 5]

EXERCISE–I Q.19 107

Q.23

1   , 2 2 

Q.3

Q.6

π/6, π/3, π/2

EXERCISE–II 120°

Q.14 9 Q.20 r = 11

Q.8

Q.16 triangle is isosceles Q.21 3

Q.9

400

Q.18 4

Q.22 (a)

50 Q.10

Q.19 B =

5π 12

;C=

3 cms & 2 cms π 12

;

b = 2+ 3 c

9 3 3 , (b) 3 , (c) 14R 2 8R

EXERCISE–III Q.1 6, 8, 10 cms Q.2 (a) A, (b) B Q.9 (a) C, (b) A, B, C, D Q.10 4

Q.5 D

Q.7

D

Q.8

(a) B; (b) 5

P. T. O.

17

Part : (A) Only one correct option 1.

2.

In a triangle ABC, (a + b + c) (b + c − a) = k. b c, if : (B) k > 6 (C) 0 < k < 4 (A) k < 0 In a ∆ABC, A =

2π 9 3 , b – c = 3 3 cm and ar ( ∆ABC) = cm 2. Then a is 3 2

(A) 6 3 cm

3.

(D) k > 4

(B) 9 cm

If R denotes circumradius, then in ∆ ABC,

(A) cos (B – C)

(B) sin (B – C)

(C) 18 cm

b2 − c 2 is equal to 2a R (C) cos B – cos C

(D) none of these

(D) none of these

4.

If the radius of the circumcircle of an isosceles triangle PQR is equal to PQ (= PR), then the angle P is π π 2π π (D) (A) (B) (C) 6 3 3 2

5.

In a ∆ ABC, the value of (A)

6.

r R

acosA + bcosB + ccosC is equal to: a+b+c

(B)

R 2r

In a right angled triangle R is equal to s+r s−r (A) (B) 2 2

(C)

R r

(C) s – r

(D)

2r R

(D)

s+r a

7.

In a ∆ABC, the inradius and three exradii are r, r1, r2 and r3 respectively. In usual notations the value of r. r1. r2. r3 is equal to abc (A) 2∆ (D) none of these (B) ∆ 2 (C) 4R

8.

In a triangle if r1 > r2 > r3, then (B) a < b < c (A) a > b > c

9.

1 1 With usual notation in a ∆ ABC  r + r   1 2 where 'K' has the value equal to: (A) 1 (B) 16

(C) a > b and b < c 1 1  +  r   2 r3 

(D) a < b and b > c

 1 1 KR 3  +  = , r  2 a b2c 2  3 r1 

(C) 64

(D) 128

10.

The product of the arithmetic mean of the lengths of the sides of a triangle and harmonic mean of the lengths of the altitudes of the triangle is equal to: (A) ∆ (B) 2 ∆ (C) 3 ∆ (D) 4 ∆

11.

In a triangle ABC, right angled at B, the inradius is: AB + BC − AC AB + AC − BC AB + BC + AC (A) (B) (C) (D) None 2 2 2 The distance between the middle point of BC and the foot of the perpendicular from A is :

12.

(A)

13.

− a2 + b2 + c 2 2a

(B)

b2 − c 2 2a

(C)

b2 + c 2 bc

(D) none of these

In a triangle ABC, B = 60° and C = 45°. Let D divides BC internally in the ratio 1 : 3, then, (A)

2 3

(B)

1 3

(C)

1 6

(D)

sin ∠BAD = sin ∠CAD

1 3

14.

Let f, g, h be the lengths of the perpendiculars from the circumcentre of the ∆ ABC on the sides a, b and a b c abc c respectively. If + + = λ then the value of λ is: f g h f gh (A) 1/4 (B) 1/2 (C) 1 (D) 2

15.

A triangle is inscribed in a circle. The vertices of the triangle divide the circle into three arcs of length 3, 4 and 5 units. Then area of the triangle is equal to: 18

(A)

9 3 (1 + 3 ) π

2

(B)

9 3 ( 3 − 1) π

(C)

2

9 3 (1 + 3 ) 2π

2

(D)

9 3 ( 3 − 1) 2π 2

16.

If in a triangle ABC, the line joining the circumcentre and incentre is parallel to BC, then cos B + cos C is equal to: (A) 0 (B) 1 (C) 2 (D) none of these

17.

If the incircle of the ∆ ABC touches its sides respectively at L, M and N and if x, y, z be the circumradii of the triangles MIN, NIL and LIM where I is the incentre then the product xyz is equal to: (A) R r2

18.

19.

(B) r R2

(C)

1 R r2 2

(D)

1 r R2 2

r 1 A  tan B + tan C    is equal to : If in a ∆ABC, r = , then the value of tan 2 2 2 2  1 1 (B) (C) 1 (D) None of these (A) 2 2

In any ∆ABC, then minimum value of (A) 3

(B) 9

r1 r2 r3 r3

is equal to (C) 27

(D) None of these

20.

In a acute angled triangle ABC, AP is the altitude. Circle drawn with AP as its diameter cuts the sides AB and AC at D and E respectively, then length DE is equal to ∆ ∆ ∆ ∆ (B) (C) (D) (A) 3R 2R 4R R

21. 22.

AA1, BB1 and CC1 are the medians of triangle ABC whose centroid is G. If the concyclic, then points A, C1, G and B1 are (A) 2b2 = a2 + c2 (B) 2c2 = a2 + b2 (C) 2a2 = b2 + c2 (D) None of these In a ∆ABC, a, b, A are given and c1, c2 are two values of the third side c. The sum of the areas of two triangles with sides a, b, c1 and a, b, c2 is 1 1 (A) b2 sin 2A (B) a2 sin 2A (C) b2 sin 2A (D) none of these 2 2

23.

In a triangle ABC, let ∠C = is equal to (A) a + b – c

π . If r is the inradius and R is the circumradius of the triangle, then 2(r + R) 2 [IIT - 2000] (B) b + c (C) c + a (D) a + b + c

24.

Which of the following pieces of data does NOT uniquely determine an acute - angled triangle ABC (R being the radius of the circumcircle )? [IIT - 2002] (A) a , sin A, sin B (B) a, b, c (C) a, sin B, R (D) a, sin A, R

25.

If the angles of a triangle are in the ratio 4 : 1 : 1, then the ratio of the longest side to the perimeter is (A) 3 : (2 + 3 ) (B) 1 : 6 (C) 1 : 2 + 3 (D) 2 : 3 [IIT - 2003]

26.

The sides of a triangle are in the ratio 1 : (A) 1 : 3 : 5

27.

In an equilateral triangle, 3 coincs of radii 1 unit each are kept so that they touche each other and also the sides of the triangle. Area of the triangle is [IIT - 2005]

(A) 4 + 2 3

28.

(B) 6 + 4

3

(C) 12 +

If P is a point on C1 and Q is a point on C2, then (A) 1/2

29.

(B) 2 : 3 : 4

3 : 2, then the angle of the triangle are in the ratio [IIT - 2004] (C) 3 : 2 : 1 (D) 1 : 2 : 3

(B) 3/4

7 3 4

(D) 3 +

7 3 4

PA 2 + PB 2 + PC 2 + PD 2

equals QA 2 + QB 2 + QC 2 + QD 2 (C) 5/6 (D) 7/8

A circle C touches a line L and circle C1 externally. If C and C1 are on the same side of the line L, then locus of the centre of circle C is (A) an ellipse (B) a circle (C) a parabola (D) a hyperbola 19

30.

Let
be a line through A and parallel to BD. A point S moves such that its distance from the line BD and the vertex A are equal. If the locus of S meets AC in A1, and
in A2 and A3, then area of ∆A1 A2A3 is (A) 0.5 (unit)2 (B) 0.75 (unit)2 (C) 1 (unit)2 (D) (2/3) (unit)2

Part : (B) May have more than one options correct 31.

In a ∆ABC, following relations hold good. In which case(s) the triangle is a right angled triangle? (A) r2 + r3 = r1 − r (B) a2 + b2 + c2 = 8 R2 (C) r1 = s (D) 2 R = r1 − r

32.

In a triangle ABC, with usual notations the length of the bisector of angle A is : A abc cos ec 2 bc cos A 2 bc sin A 2∆ . A 2 2 2 (A) (B) (C) (D) b + c cos ec 2 b+c b+c 2R (b + c ) AD, BE and CF are the perpendiculars from the angular points of a ∆ ABC upon the opposite sides, then : Perimeter of ∆DEF r (B) Area of ∆DEF = 2 ∆ cosA cosB cosC (A) = Perimeter of ∆ABC R R (C) Area of ∆AEF = ∆ cos2A (D) Circum radius of ∆DEF = 2

33.

34.

The product of the distances of the incentre from the angular points of a ∆ ABC is: (abc ) R (abc ) r (A) 4 R2 r (B) 4 Rr 2 (C) (D) s s

35.

In a triangle ABC, points D and E are taken on side BC such that BD = DE = EC. If angle ADE = angle AED = θ, then: (A) tanθ = 3 tan B (B) 3 tanθ = tanC 6 tan θ = tan A (D) angle B = angle C (C) tan2 θ − 9

36.

With usual notation, in a ∆ ABC the value of Π (r 1 − r) can be simplified as:

A (A) abc Π tan 2

1.

(B) 4 r R

If in a triangle ABC, angled.

2

(C)

(a b c)2 2 R (a + b + c)

(D) 4 R r2

cos A + 2 cos C sin B = , prove that the triangle ABC is either isosceles or right cos A + 2 cos B sin C

 A + B  , prove that triangle is isosceles.  2 

2.

In a triangle ABC, if a tan A + b tan B = (a + b) tan 

3.

 r  r  If  1 − 1   1 − 1  = 2 then prove that the triangle is the right triangle. r2   r3  

4. 5. 6.

In a ∆ ABC, ∠ C = 60° & ∠ A = 75°. If D is a point on AC such that the area of the ∆ BAD is 3 times the area of the ∆ BCD, find the ∠ ABD. The radii r1, r 2, r3 of escribed circles of a triangle ABC are in harmonic progression. If its area is 24 sq. cm and its perimeter is 24 cm, find the lengths of its sides. ABC is a triangle. D is the middle point of BC. If AD is perpendicular to AC, then prove that

7.

8.

9.

(

)

2 c 2 −a 2 . 3ac Two circles, of radii a and b, cut each other at an angle θ. Prove that the length of the common chord is 2ab sin θ . 2 a + b 2 + 2ab cos θ In the triangle ABC, lines OA, OB and OC are drawn so that the angles OAB, OBC and OCA are each equal to ω, prove that (i) cot ω = cot A + cot B + cot C cosec2 ω = cosec2 A + cosec2 B + cosec2 C (ii) In a plane of the given triangle ABC with sides a, b, c the points A′, B′, C′ are taken so that the ∆ A′ BC, ∆ AB′C and ∆ ABC′ are equilateral triangles with their circum radii Ra, Rb, Rc ; in−radii ra, rb, r c & ex − radii ra′, rb′ & rc ′ respectively. Prove that; cos A. cos C =

[∑ (3R +6r +2r′ )] Πtan A 3

(i) 10.

Π r a: Π Ra: Π r a′ = 1: 8: 27

(ii)

r 1 r2 r 3 =

a

a

a

2 648 3 The triangle ABC is a right angled triangle, right angle at A. The ratio of the radius of the circle 20

(

)

(

)

11.

circumscribed to the radius of the circle escribed to the hypotenuse is, 2 : 3 + 2 . Find the acute angles B & C. Also find the ratio of the two sides of the triangle other than the hypotenuse. The triangle ABC is a right angled triangle, right angle at A. The ratio of the radius of the circle

12.

circumscribed to the radius of the circle escribed to the hypotenuse is, 2 : 3 + 2 . Find the acute angles B & C. Also find the ratio of the two sides of the triangle other than the hypotenuse. If the circumcentre of the ∆ ABC lies on its incircle then prove that,

13.

cosA + cosB + cosC = 2 Three circles, whose radii area a, b and c, touch one another externally and the tangents at their points of contact meet in a point; prove that the distance of this point from either of their points of contacts 1

 abc  2 is   . a+b+c 

EXERCISE # 2

EXERCISE # 1 1. C

2. B

3. B

4. D

5. A

6. B

7. B

8. A

9. C

10. B

11. A

12. B

13. C

14. A

15. A

16. B

17. C

18. B

19. C

20. D

21. C

22. A

23. A

24. D

25. A

26. A

27. B

28. B

29. C

30. C

31. ABCD 32. ACD

34. BD 35. ACD

4. ∠ ABD = 30°

5. 6, 8, 10 cms

10. B =

5π π b ,C= , = 2+ 3 12 12 c

11. B =

5π π b ,C= , = 2+ 3 12 12 c

33. ABCD

36. ACD

21

Some questions (Assertion–Reason type) are given below. Each question contains Statement – 1 (Assertion) and Statement – 2 (Reason). Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct. So select the correct choice : (A) Statement – 1 is True, Statement – 2 is True; Statement – 2 is a correct explanation for Statement – 1. (B) Statement – 1 is True, Statement – 2 is True; Statement – 2 is NOT a correct explanation for Statement – 1. (C) Statement – 1 is True, Statement – 2 is False.(D) Statement – 1 is False, Statement – 2 is True. 576.

−1 −1 Statement-1: The value of tan 2 + tan 3 =

3π 4

 x+y  .  1 − xy 

Statement-2: If x > 0, y > 0, xy > 1, then tan–1x + tan–1y = π + tan −1  577. 578. 579. 580.

7π  7π −1  –1 is the principal value of cos  cos  Statement-2: cos (cos x) = x if x∈[0, π] 6 6   3 π −1 −1 Statement-1: The value of cot–1(–1) is Statement-2: cot (− x) = π − cot x, x ∈ R 4 1 π Statement-1: If x + = 2 then the principal value of sin–1x is x 2 Statement-1:

Statement-2: sin–1(sin x) = x ∀ x∈R. Statement-1: If A, B, C are the angles of a triangle such that angle A is obtuse then tan B tan C > 1. Statement-2: In any triangle tan A =

581.

582. 583. 584. 585.

tan B + tan C . tan B tan C − 1

Let f(θ) = sinθ.sin (π/3 + θ) . sin (π/3 – θ) Statement-1: f(θ) ≤ 1/4 Statement-2: f(θ) = 1/4 sin2 Statement–1 : Number of ordered pairs (θ, x) satisfying 2sinθ = ex + e–x, θ∈[0, 3π] is 2. Statement–2 : Number of values of x for which sin2x + cos4x = 2 is zero. Statement–1 : The number of values of x∈ [0, 4π] satisfying | 3 cosx – sinx| ≥ 2 is 2. Statement–2 : |cos (x + π/6)| = 1 ⇒ number of solutions of | 3 cosx – sinx| ≥ 2 is 4 Statement–1 : Number of solutions of sin–1 (sinx) = 2π – x; x∈[3π/2, 5π/2] is 1 Statement–2 : sin–1 (sinx) = x, x∈ [–π/2, π/2] Statement–1 : Number of ordered pairs (x, y) satisfying sin–1x = π – sin–1y and cos–1x + cos–1y = 0 simultaneously is 1 Statement–2 : Ordered pairs (x, y) satisfying sin–1x = π – sin–1y and cos–1x + cos–1y = 0 will lie on x2 + y2 = 2.

586.

Statement–1

: The equation k cos x – 3 sin x = k + 1 is solvable only if k belongs to the interval ( −∞, 4

587.

Statement–2 Statement–1 Statement–2

: − a + b ≤ a sin x ± b cos x ≤ a + b . : The equation 2 sec2x – 3 sec x + 1 = 0 has no solution in the interval (0, 2π) : sec x ≤ – 1 as sec x ≥ 1. 2

2

2

2

588.

Statement–1

: The number of solution of the equation sin x =| x | is only one.

589.

Statement–2 Statement–1 Statement–2

: The number of point of intersection of the two curves y = |sin x| and y = |x| is three. : The equation sin x = 1 has infinite number of solution. : The domain of f(x) = sin x is (– ∞, ∞).

590.

Statement–1

: There is no solution of the equation | sin x | + | cos x |= tan x + cot x .

591.

592.

2

 

Statement–1 :If sin θ = a for exactly one value of θ ∈  0,

594. 595.

Statement–2 : – 1 ≤ sin θ ≤ 1. Statement–1 : tan 5° is an irrational number. Let θ be an acute angle Statement–1 : sin6 θ + cos6 θ ≤ 1.

596.

2

Statement–2 : 0 ≤ | sin x | + | cos x |≤ 2 and tan2 x + cot2x ≥ 2. Statement–1 : The equation sin2x + cos2 y = 2 sec2 z is only solvable cos y = 1 an sec z = 1 where x, y , z ∈ R. Statement–2 : Maximum value of sin x and cos y is 1 and minimum value of sec z is 1. Statement–1 : If cot–1x < n, n∈ R then x < cot (n) Statement–2 : cot–1 (x) is an decreasing function.

593.

S–1 : sin

π is a root of 8x3 – 6x + 1 = 0. S–2 18

]

when

sin

x

7π  , then a can take infinite value in the interval [– 1, 1]. 3 

Statement–2

:

tan 15° is an irrational number.

Statement–2

:

sin θ + cos θ ≤ 1

: For any θ ∈ R, sin 3θ = 3 sin θ – 4 sin3 θ. 1

=

22

1,

597. 598.

Let f be any one of the six trigonometric functions. Let A, B ∈ R satisfying f(2A) = f(2B). Statement–1 : A = nπ + B, for some n ∈ I. Statement–2 : 2π is one of the period of f. Let x ∈ [-1, 1] Statement–1

(

: 2 sin-1 x = sin -1 2x 1 − x

599.

Let f(x) = cos–1 x Statement–1 : f is a decreasing function. Statement–2 : f(– x) = π – f(x).

600.

Statement–1

601.

602.

603.

2

).

Statement–2

: - 1 ≤ 2x

1 − x 2 ≤ 1.

: The total number of 2 real roots of the equation x2 tan x = 1 lies between the interval (0, 2π).

Statement–2

: The total number of solution of equation cos x − sin x = 2 cos x in [0, 2π] is 3.

Statement–1

: The number of real solutions of equation sin ex cos ex = 2x – 2 + 2- x – 2 is 0.

Statement–2

: The number of solutions of the eqution 1 + sin x sin2

Statement–1

x = 0 n [- π, π] is 0. 2 2 2 −1  −1  4  −1  : Equation tan  x +  − tan   − tan  x −  = 0 has 3 real roots. x x  x 

Statement–2

: the number of real solution of

Statement–1

: If

1 + cos 2x = 2 sin −1 ( sin x ) ; x ∈ [ −π, π] is 2.

1 1 1 1 n + tan −1 + tan −1 ... + tan −1 tan-1 = tan -1 θ, then θ = . 1+ 2 1 + 2.3 1 + 3.4 1 + n ( n + 1) n +1 Statement–2

: The sum of series cos-1 2 + cot-1 8 + cot-1 18 + . . . is

604.

Statement-1: If tanθ + secθ = 3 , 0 < θ < π, then θ = π/6 Statement-2: General solution of cosθ = cosα is θ = α, if 0 < α < π/2

605.

Statement-1: If x < 0, tan-1x + tan -1

π . 4

1 = π/2 x

Statement-2: tan-1x + cos-1x = π/2,∀x∈R 606.

Statement-1: sin-1 (sin10) = 10 Statement-2: For principal value sin -1 (sinx) = x

607.

Statement-1: cos

π 2π 4π 1 cos cos = − 7 7 7 8

Statement-2: cosθ cos2θ cos23θ .... cos2n-1θ = -

1 π if θ = n , n ∈ N, n ≥ 2. n 2 2 −1

TRI 608.

Statement-1: sin3 < sin1 < sin2 is true Statement-2: sinx is positive in first and second quadrants.

609.

Statement-1: The equation 2sin2x – (P + 3) sinx + (2P – 2) = 0 possesses a real solution if P∈[-1, 3] Statement-2 : -1 ≤ sinx ≤ 1

610.

 

Statement-1: The maximum value of 3sinθ + 4cos  θ + Statement-2:: - a + b 2

2

≤ asinθ + bcos θ ≤

π  is 5 here θ∈R. 4

a 2 + b2

2

23

611.

Statement-1: If A + B + C = π, cosA + cosB + cosC ≤ 3/2 Statement-2:: If A + B + C = π, sin

612.

A B C 1 sin sin ≤ 2 2 2 8

Statement-1: The maximum & minimum values of the function f(x) =

1 does not exists. 6sin x − 8cos x + 5

Statement-2: The given function is an unbounded function. 613.

1  = π/2 x

Statement-1: If x < 0 tan-1x +tan-1 

Statement-2: tan-1x + cot-1x = π/2 ∀ x∈R. 614.



Statement-1: In any triangle square of the length of the bisector AD is bc  1 −



Statement-2: In any triangle length of bisector AD =

a2   (b + c) 2 

bc A cos b+c 2

615.

Statement-1: If in a triangle ABC, C = 2acosB, then the triangle is isosceles. Statement-2: Triangle ABC, the two sides are equal i.e. a = b.

616.

Statement-1: If the radius of the circumcircle of an isosceles triangle pqR is equal to pq = PR then the angle p = 2π/3. Statement-2: OPQ and oPR will be equilateral i.e., ∠OPq = 60°, ∠OPR = 60°

617.

Statement-1: The minimum value of the expression sinα + sinβ + sinγ is negative, where α, β, γ are real numbers such that α + β + γ = π. Statement-2: If α, β, γ are angle of a triangle then sinα + sinβ + sinγ = 4cos

α β γ cos cos . 2 2 2

618.

Statement-1: If in a triangle sin2A + sin2B + sin2C = 2 then one of the angles must be 90°. Statement-2: In any triangle sin2A + sin2B + sin2C = 2 + 2cosA cosB cosC.

619.

Statement-1: If in a ∆ABC a → 2c and b → 3c then cosB must tend to –1. Statement-1: In a ∆ABC cosB =

c2 + a 2 − b2 . 2ac

620.

Statement-1: cos(45 − A) cos(45 − B) − sin(45 − A) × sin (45 − B) = sin(A + B). Statement-2: cos(90 − θ) = − sin θ.

621.

Statement-1: The maximum and minimum values of 7cosθ + 24sinθ are 25 and − 25 respectively. Statement-2:

622.

a 2 + b 2 ≤ a cos θ ± b sin θ ≤ a 2 + b 2 for all θ.

Statement-1: If sin

−1

x + sin −1 (1 − x) = sin −1 1 − x 2 then x = 0,

−1

Statement-2: sin sin x = x ∀x ∈ R

1 2

TE 623.

Statement-1: The numbers sin 18° and –sin54° are roots of same quadratic equation with integer coefficients. Statement-2: If x = 18°, then 5x = 90°, if y = -54°, then 5y = -270°.

Inverse Trigonometry 624.

(

) (

)

Statement-1: The number of solution of the equation cos( π x − 4 cos π x = 1 is one. Statement-2: cosx = cosα ⇒ x = 2nπ ± α n∈I 3

24

Inverse Trigonometric Function

 π 3π 

627.

Statement-1: The range of sin-1x + cos-1x + tan −1x is  ,  4 4  -1 -1 Statement-2: sin x + cos x = π/2 for every x∈R.

628.

Statement-1: sin-1 (sin10) = 10 Statement-2: sin-1 (sinx) = x for - π/2 ≤ x ≤ π/2

629.

Statement-1: If sin–1x + sin–1y =

2π , the value of cos–1x + cos–1y is π/3. 3

Statement-2: sin–1x + cos–1x = π/2 ∀x∈ [–1, 1]. 630.

 

Statement-1: 7π/6 is the principal value of cos–1  cos

7π   6 

Statement-2: cos–1 (cosx) = x, if x∈ [0, π]. 631.

Statement-1:

3 cos θ + sin θ = 5 has no solution.

Statement-2: a cos θ + b sin θ = c has solution if | c |≤

632.

a 2 + b2  3 1 ,  2 2 

Statement-1: The equation sin4x + cos4x + sin2x + a = 0 is valid if a ∈  −

Statement-2: If discriminant of a quadratic equation is −ve. Then its roots are real. 633.

Statement-1: In a ∆ABC cosAcosB + sinAsinBsinC = 1 then ∆ABC must be isosceles as well as right angled triangle. Statement-2: In a ∆ABC if A =

634.

π tanA tanB = k. then k must satisfy k2 − 6k + 1 ≥ 0 4

Statement-1: If r1, r2, r 3 in a ∆ABC are in H.P. then sides a, b, c are in A.P. Statement-2:: r1 =

∆ ∆ ∆ . , r2 = , r3 = s−a s−b s−c

Answer Key 576. A

577. D

578. A

579. C

580. D

581.C

582.B

583. D

584. B

585. B

586. A

587. B

588. C

589.A

590. C

591. A

592. D

593. D

594. A

595. C

596. A

597. A

598. D

599. B

600. C

601. B

602. D

603. D

604. A

605. D

606. D

607. D

608. B

609. A

610. D

611. B

612. A

613. D

614. C

615. A

616. A

617. B

618. A

619. A

620. C

621. A

622. C

623. A

624. B

627. C

628. D

629. A

630. D

631. A

4

25

STUDY PACKAGE Target: IIT-JEE (Advanced) SUBJECT: MATHEMATICS TOPIC: 4 XI M 4. Functions Index: 1. Key Concepts 2. Exercise I to V 3. Answer Key 4. Assertion and Reasons 5. 34 Yrs. Que. from IIT-JEE 6. 10 Yrs. Que. from AIEEE

1

A.

Definition :

Functions

Function is a special case of relation, from a non empty set A to a non empty set B, that associates each member of A to a unique member of B. Symbolically, we write f: A → B. We read it as "f is a function from A to B". Set 'A' is called domain of f and set 'B' is called co-domain of f. For example, let A ≡ {–1, 0, 1} and B ≡ {0, 1, 2}. Then A × B ≡ {(–1, 0), (–1, 1), (–1, 2), (0, 0), (0, 1), (0, 2), (1, 0), (1, 1), (1, 2)} Now, " f : A → B defined by f(x) = x 2 " is the function such that f ≡ {(–1, 1), (0, 0), (1, 1)} f can also be show diagramatically by following picture.

Every function say f : A → B satisfies the following conditions: (a) f ⊆ A x B, (b) ∀ a ∈ A ⇒ (a, f(a)) ∈ f and (c) (a, b) ∈ f & (a, c) ∈ f ⇒ b = c Illustration # 1: (i) Which of the following correspondences can be called a function ? (A) f(x) = x 3 ; {–1, 0, 1} → {0, 1, 2, 3} (B) f(x) = ± x ; {0, 1, 4} → {–2, –1, 0, 1, 2} ; {0, 1, 4} → {–2, –1, 0, 1, 2} (C) f(x) = x ; {0, 1, 4} → {–2, –1, 0, 1, 2} (D) f(x) = – x Solution: f(x) in (C) & (D) are functions as definition of function is satisfied. while in case of (A) the given relation is not a function, as f(–1) ∉ codomain. Hence definition of function is not satisfied. While in case of (B), the given relation is not a function, as f(1) = ± 1 and f(4) = ± 2 i.e. element 1 as well as 4 in domain is related with two elements of codomain. Hence definition of function is not satisfied. (ii) Which of the following pictorial diagrams represent the function (A)

(B)

(C) (D) Solution: B & D. In (A) one element of domain has no image, while in (C) one element of domain has two images in codomain Assignment: 1. Let g(x) be a function defined on [−1, 1]. If the area of the equilateral triangle with two of its vertices at (0,0) & (x,g(x)) is 3 / 4 sq. units, then the function g(x) may be. 2.

(B*) g(x) = (1 − x 2 ) (C*) g(x) = − (1 − x 2 ) (A) g(x)= ± (1 − x 2 ) Represent all possible functions defined from {α, β} to {1, 2} Answer (1) B

B.

Domain, Co-domain & Range of a Function :

(2)

(i)

(ii)

(iii)

(D) g(x) =

(1 + x 2 )

(iv)

Let f: A → B, then the set A is known as the domain of f & the set B is known as co−domain of f. If a member 'a' of A is associ at ed to t he member 'b' of B, t hen ' b' i s cal led the f -image of 'a' and we writ e b = f (a). Further 'a' is called a pre-image of 'b'. The set {f(a): ∀ a ∈A} is called the range of f and is denoted by f(A). Clearly f(A) ⊂ B. Sometimes if only definition of f (x) is given (domain and codomain are not mentioned), then domain is set of those values of ' x' for which f (x) is defined, while codomain is considered to be (– ∞, ∞) A function whose domain and range both are sets of real numbers is called a real function. Conventionally the word "FUNCTION” is used only as the meaning of real function. Illustration # 2 : Find the domain of following functions : (i)

Solution :(i)

(ii)

sin–1 (2x – 1)

f(x) =

x2 − 5

f(x) =

2 x 2 − 5 is real iff x – 5 ≥ 0

(ii)

⇒

|x| ≥

5

⇒

x ≤ – 5 or x ≥

5

∴ the domain of f is (–∞ , – 5 ] ∪ [ 5 , ∞) –1 ≤ 2x – 1 ≤ + 1 ∴ domain is x ∈ [0, 1] Algebraic Operations on Functions : If f & g are real valued functions of x with domain set A and B respectively, then both f & g are defined in A ∩ B. Now we define f + g, f − g, (f . g) & (f /g) as follows:

f f( x )   (x) = g g ( x ) domain is {x  x ∈ A ∩ B such that g(x) ≠ 0}.   Note :  For domain of φ(x) = {f(x)}g(x) , conventionally, the conditions are f(x) > 0 and g(x) must be defined.  For domain of φ(x) = f(x) Cg(x) or φ(x) = f(x)Pg(x) conditions of domain are f(x) ≥ g(x) and f(x) ∈ N and g(x) ∈ W Illustration # 3: Find the domain of following functions :

(iii)

2

(i) Solution:

f(x) =

sin x − 16 − x 2

(ii)

f(x) =

3

log(x 3 − x)

4−x sin x is real iff sin x ≥ 0 ⇔ x∈[2nπ, 2nπ + π], n∈I.

(i)

2

(iii)

f(x) = x cos

−1

x

2 16 − x is real iff 16 − x ≥ 0 ⇔ − 4 ≤ x ≤ 4. Thus the domain of the given function is {x : x∈[2nπ, 2nπ + π], n∈I }∩[−4, 4] = [−4, −π] ∪ [0, π]. (ii) Domain of 4 − x 2 is [−2, 2] but 4 − x 2 = 0 for x = ± 2 ⇒ x ∈ (–2, 2) log(x 3 − x) is defined for x 3 − x > 0 i.e. x(x − 1)(x + 1) > 0. ∴ domain of log(x 3 − x) is (−1, 0 ) ∪ (1, ∞). Hence the domain of the given function is {(−1, 0 ) ∪ (1, ∞)}∩ (−2, 2) = (−1, 0 ) ∪ (1, 2). (iii) x > 0 and –1 ≤ x ≤ 1 ∴ domain is (0, 1] Assignment : 3. Find the domain of following functions. 1 –1 2 x − 1 f(x) = log( 2 − x ) + x + 1 (ii) f(x) = 1 − x – sin (i) 3 Ans. (i) [–1, 1) ∪ (1, 2) (ii) [–1, 1] Methods of determining range : (i) Representing x in terms of y Definition of the function is usually represented as y (i.e. f(x) which is dependent variable) in terms of an expression of x (which is independent variable). To find range rewrite given definition so as to represent x in terms of an expression of y and thus obtain range (possible values of y). If y = f(x) ⇔ x = g(y), then domain of g(y) represents possible values of y, i.e. range of f(x).

2

Find the range of f(x) =

Illustration # 4: x + x +1

x2 + x + 1 x2 + x − 1

2

Solution

x2 + x + 1

Illustration # 5: Solution

∴

(iii)

{x 2 + x + 1 and x 2 + x – 1 have no common factor}

x2 + x − 1

⇒ yx 2 + yx – y = x 2 + x + 1 x2 + x − 1 2 ⇒ (y – 1) x + (y – 1) x – y – 1 = 0 If y = 1, then the above equation reduces to –2 = 0. Which is not true. Further if y ≠ 1, then (y – 1) x 2 + (y – 1) x – y – 1 = 0 is a quadratic and has real roots if (y – 1)2 – 4 (y – 1) (–y – 1) ≥ 0 i.e. if y ≤ –3/5 or y ≥ 1 but y ≠ 1 Thus the range is (–∞, –3/5] ∪ (1, ∞) Graphical Method : Values covered on y-axis by the graph of function is range

y=

(ii)

f(x) =

f(x) =

Find the range of f(x) = x2 − 4 x−2

x2 − 4 x−2

= x + 2; x ≠ 2

graph of f(x) would be

Thus the range of f(x) is R – {4} Using Monotonocity/Maxima-Minima (a) Continuous function: If y = f(x) is continuous in its domain then range of f(x) is y ∈ [min f(x), max. f(x)] (b) Sectionally continuous function: In case of sectionally continuous functions, range will be union of [min f(x), max. f(x)] over all those intervals where f(x) is continuous, as shown by following example. Let graph of function y = f(x) is

Then range of above sectionally continuous function is [y 2, y 3] ∪ (y4, y5] ∪ (y 6, y7] Note :  In case of monotonic functions minimum and maximum values lie at end points of interval. Illustration # 6 : Find the range of following functions : (i) y =
n (2x – x 2) (ii) y = sec–1 (x 2 + 3x + 1) Solution : (i) Step – 1 Using maxima-minima, we have 2x – x 2 ∈ (–∞, 1] Step – 2 For log to be defined accepted values are 2x – x 2 ∈ (0, 1] {i.e. domain (0, 1]} Now, using monotonocity
n (2x – x 2) ∈ (–∞, 0] ∴ range is (– ∞, 0] Ans. 3

(ii)

y = sec –1 (x 2 + 3x + 1) Let t = x 2 + 3x + 1 for x ∈ R  5   5  but y = sec –1 (t) ⇒ t ∈ − , − 1 ∪ [1, ∞) then t ∈ − , ∞  4 4    

  π −1 5   from graph range is y ∈ 0,  ∪ sec  − 4 , π     2 

Assignment: Find domain and range of following functions. 4. x 2 − 2x + 5 (i) y = x3 (ii) y= 2 x + 2x + 5 Answer

C.

(i)

1

domain R; range R

(iii)

y=

(iv)

y = cot–1 (2x – x 2)

(v)

3  2 y =
n sin–1  x + x +  Answer 4 

x −x 2

Answer Answer

(ii)

3 − 5 3 + 5  domain R ; range  2 , 2   

domain R – [0, 1] ; range (0, ∞) π  domain R ; range  , π  4  − 2 − 5 − 2 + 5  π  π , domain x ∈   ; range 
n 6 ,
n 2    4 4  

Classification of Functions : Functions can be classified as : (i) One − One Function (Injective Mapping) and Many − One Function: One − One Function : A function f : A → B is said to be a one-one function or injective mapping if different elements of A have different f images in B. Thus for x 1, x 2 ∈ A & f(x 1), f(x 2) ∈ B, f(x 1) = f(x 2) ⇔ x 1 = x 2 or x 1 ≠ x 2 ⇔ f(x 1) ≠ f(x 2). Diagrammatically an injective mapping can be shown as

OR Many − One function : A function f : A → B is said to be a many one function if two or more elements of A have the same f image in B. Thus f : A → B is m any one iff there exi sts atleast two elem ents x 1 , x 2 ∈ A, such that f(x 1) = f(x 2) but x 1 ≠ x 2. Diagrammatically a many one mapping can be shown as

OR Note :  If a function is one−one, it cannot be many−one and vice versa. Methods of determining whether function is ONE-ONE or MANY-ONE : (a) If x 1, x 2 ∈ A & f(x 1), f(x 2) ∈ B, f(x 1) = f(x 2) ⇔ x 1 = x 2 or x 1 ≠ x 2 ⇔ f(x 1) ≠ f(x 2), then function is ONE-ONE otherwise MANY-ONE. (b) If there exists a straight line parallel to x-axis, which cuts the graph of the function atleast at two points, then the function is MANY-ONE, otherwise ONE-ONE. (c) If either f′(x) ≥ 0, ∀ x ∈ complete domain or f′(x) ≤ 0 ∀ x ∈ complete domain, where equality can hold at discrete point(s) only, then function is ONE-ONE, otherwise MANY-ONE. (ii) Onto function (Surjective mapping) and Into function : Onto function : If the function f : A → B is such that each element in B (co−domain) must have atleast one pre−image in A, then we say that f is a function of A 'onto' B. Thus f : A → B is surjective iff ∀ b ∈ B, there exists some a ∈ A such that f (a) = b. Diagrammatically surjective mapping can be shown as

OR Method of determining whether function is ONTO or INTO : Find the range of given function. If range ≡ co−domain, then f(x) is onto, otherwise into Into function : If f : A → B is such that there exists atleast one element in co−domain which is not the image of any element in domain, then f(x) is into. 4

Diagrammatically into function can be shown as

OR Note :  If a function is onto, it cannot be into and vice versa. Thus a function can be one of these four types:

(a)

one−one onto (injective & surjective)

(b)

one−one into (injective but not surjective)

(c)

many−one onto (surjective but not injective)

(d)

many−one into (neither surjective nor injective)

Note : 

If f is both injective & surjective, then it is called a bijective mapping. The bijective functions are also named as invertible, non singular or biuniform functions.  If a set A contains 'n' distinct elements then the number of different functions defined from A → A is nn and out of which n! are one one. Illustration # 7 (i) Find whether f(x) = x + cos x is one-one. Solution The domain of f(x) is R. f′ (x) = 1 − sin x. ∴ f′ (x) ≥ 0 ∀ x ∈ complete domain and equality holds at discrete points only ∴ f(x) is strictly increasing on R. Hence f(x) is 2one-one. 3 (ii) Identify whether the function f(x) = –x + 3x – 2x + 4 ; R → R is ONTO or INTO Solution As codomain ≡ range, therefore given function is ONTO (iii) f(x) = x 2 – 2x + 3; [0, 3] → A. Find whether f(x) is injective or not. Also find the set A, if f(x) is surjective. Solution f′(x) = 2(x – 1); 0 ≤ x ≤ 3 − ve ; 0 ≤ x < 1 f′(x) =  + ve ; 1 < x < 3 ∴ f(x) is not monotonic. Hence it is not injective. For f(x) to be surjective, A should be equal to its range. By graph range is [2, 6] ∴ A ≡ [2, 6]

∴

Assignment: 5. For each of the following functions find whether it is one-one or many-one and also into or onto (i) f(x) = 2 tan x; (π/2, 3π/2) → R

one-one onto

Answer

(ii) (iii)

D.

1

; (–∞, 0) → R 1+ x2 Answer one-one into f(x) = x 2 +
n x f(x) =

Answer

one-one onto

Various Types of Functions :

Polynomial Function : If a function f is defined by f (x) = a0 x n + a1 x n−1 + a2 x n−2 +... + an−1 x + an where n is a non negative integer and a0, a1, a2,........., an are real numbers and a0 ≠ 0, then f is called a polynomial function of degree n. Note :  There are two polynomial functions, satisfying the relation; f(x).f(1/x) = f(x) + f(1/x), which are f(x) = 1 ± x n (ii) Algebraic Function : y is an algebraic function of x, if it is a function that satisfies an algebraic equation of the form, P0 (x) yn + P1 (x) yn−1 +....... + Pn−1 (x) y + Pn (x) = 0 where n is a positive integer and P0 (x), P1 (x)....... are polynomials in x. e.g. y = x is an algebraic function, since it satisfies the equation y² − x² = 0. Note :  All polynomial functions are algebraic but not the converse.  A function that is not algebraic is called Transcendental Function . g( x ) (iii) Fractional / Rational Function : A rational function is a function of the form, y = f (x) = , where g (x) h( x ) & h (x) are polynomials and h (x) ≡/ 0. (iv) Exponential Function : A function f(x) = ax = ex In a (a > 0, a ≠ 1, x ∈ R) is called an exponential function. Graph of exponential function can be as follows : Case - Ι Case - ΙΙ For a > 1 For 0 < a < 1 (i)

5

(v)

Logarithmic Function : f(x) = logax is called logarithmic function where a > 0 and a ≠ 1 and x > 0. Its graph can be as follows Case- Ι For a > 1

(vi)

Case- ΙΙ For 0 < a < 1

Absolute Value Function / Modulus Function :  x if The symbol of modulus function is f (x) = x and is defined as: y = x=  − x if

(vi)

Signum Function :

x≥0 . x 0  f (x) = sgn (x) =  0 for x = 0 − 1 for x < 0 

| x |  ; x≠0 It is also written as sgn x =  x  0 ; x = 0 | f ( x ) | ; f ( x) ≠ 0  Note : sgn f(x) =  f ( x )  0 ; f (x) = 0

(vii)

Greatest Integer Function or Step Up Function : The function y = f (x) = [x] is called the greatest integer function where [x] equals to the greatest integer less than or equal to x. For example : for − 1 ≤ x < 0 ; [x] = − 1 ; for 0 ≤ x < 1 ; [x] = 0 for 1 ≤ x < 2 ; [x] = 1 ; for 2 ≤ x < 3 ; [x] = 2 and so on. Alternate Definition : The greatest integer occur on real number line while moving L.H.S. of x (starting from x) is [x]

(a) (c) (viii)

Properties of greatest integer function : x − 1 < [x] ≤ x (b) [x ± m] = [x] ± m iff m is an integer.  0 ; if x is an int eger [x] + [y] ≤ [x + y] ≤ [x] + [y] + 1 (d) [x] + [− x] =   − 1 otherwise Fractional Part Function: It is defined as, y = {x} = x − [x]. e.g. the fractional part of the number 2.1 is 2.1 − 2 = 0.1 and the fractional part of − 3.7 is 0.3. The period of this function is 1 and graph of this function is as shown. 6

Identity function : The function f : A → A defined by, f(x) = x ∀ x ∈ A is called the identity function on A and is denoted by ΙA. It is easy to observe that identity function is a bijection. (x) Constant function : A function f : A → B is said to be a constant function, if every element of A has the same f image in B. Thus f : A → B; f(x) = c, ∀ x ∈ A, c ∈ B is a constant function. Illustration # 8 (i) Let {x} & [x] denote the fractional and integral part of a real number x respectively. Solve 4{x} = x + [x] Solution As x = [x] + {x} 2 [ x] 4{x} = [x] + {x} + [x] ⇒ {x} = ∴ Given equation ⇒ 3 As [x] is always an integer and {x} ∈ [0, 1), possible values are [x] {x} x = [x] + {x} 0 0 0 2 5 1 3 3 5 ∴ There are two solution of given equation x = 0 and x = 3 (ii) Draw graph of f(x) = sgn (
n x) (ix)

Solution

Assignment: 6. If f : R → R satisfying the conditions f(0) = 1, f(1) = 2 and f(x + 2) = 2f (x) + f(x + 1), then find f (6). Answer 64 7. Draw the graph of following functions where [.] denotes greatest integer function (i) y=[2x]+ 1 (ii) y = x [x], 1 ≤ x ≤ 3 (iii) y = sgn (x 2 – x)

Answer (i)

E.

(ii)

(iii)

Odd & Even Functions :

(i) If f (−x) = f (x) for all x in the domain of ‘f’ then f is said to be an even function. If f (x) − f (−x) = 0 ⇒ f (x) is even. e.g. f (x) = cos x; g (x) = x² + 3. (ii) If f (−x) = −f (x) for all x in the domain of ‘f’ then f is said to be an odd function. If f (x) + f (−x) = 0 ⇒ f (x) is odd. e.g. f (x) = sin x; g (x) = x 3 + x. Note :  A function may neither be odd nor even. (e.g. f(x) = ex , cos–1x)  If an odd function is defined at x = 0, then f(0) = 0 Properties of Even/Odd Function (a) Every even function is symmetric about the y−axis & every odd function is symmetric about the origin. For example graph of y = x 2 is symmetric about y-axis, while graph of y = x 3 is symmetric about origin

(b)

All functions (whose domain is symmetrical about origin) can be expressed as the sum of an even & an odd function, as follows f(x) =

(c) The only function which is defined on the entire number line and is even & odd at the same time is f(x) = 0. 7

(d)

If f and g both are even or both are odd then the function f.g will be even but if any one of them is odd then f.g will be odd. (e) If f(x) is even then f′(x) is odd but converse need not be true.  x + x 2 + 1  is an odd function. Illustration # 9: Show that log    2  x + x 2 + 1   . Solution Let f(x) = log  Then f(–x) = log  − x + ( − x ) + 1       x 2 + 1 − x  x 2 + 1 + x     1 2      = log = log – log  x + x + 1  = –f(x) 2   2 x + 1 + x x +1 + x Hence f(x) is an odd function. Illustration # 10 Show that ax +a–x is an even function. Let f(x) = ax + a–x Then f(–x) = a–x + a–(–x) = a–x +ax = f(x). Solution Hence f(x) is an even function Illustration # 11 Show that cos–1 x is neither odd nor even. Let f(x) = cos–1x. Then f(–x) = cos–1 (–x) = π – cos–1 x which is neither equal to f(x) nor equal to f(–x). Solution Hence cos–1 x is neither odd nor even Assignment: 8. Determine whether following functions are even or odd? e x + e −x (i) Answer Odd e x − e−x  2  (ii) log  x + 1 − x  Answer Odd   2   (iii) x log  x + x + 1  Answer Even   Answer Odd (iv) sin–1 2x 1− x 2 Even extension / Odd extension : Let the defincition of the function f(x) is given only for x ≥ 0. Even extension of this function implies to define the function for x < 0 assuming it to be even. In order to get even extension replace x by –x in the given defincition Similarly, odd extension implies to define the function for x < 0 assuming it to be odd. In order to get odd extension, multiply the definition of even extension by –1 Illustration # 12 What is even and odd extension of f(x) = x 3 – 6x 2 + 5x – 11 ; x > 0 Solution Even extension f(x) = –x 3 – 6x 2 + 5x – 11 ;x 0) then k =

(1 + y ′ ) 2

3

(B) dependent of x but independent of m (D) independent of m & x . where k in terms of R alone is equal to

1 2 1 2 (B) – (C) (D) – 2 2 R R R R Q.35 If f & g are differentiable functions such that g ′ (a) = 2 & g(a) = b and if fog is an identity function then f ′ (b) has the value equal to : (A) 2/3 (B) 1 (C) 0 (D) 1/2 (A) –

x3

Q.36 Given f(x) = − + x2 sin 1.5 a − x sin a . sin 2a − 5 arc sin (a2 − 8a + 17) then : 3 (A) f(x) is not defined at x = sin 8 (B) f ′ (sin 8) > 0 12

(

Q.38 Given : f(x) = 4x3 − 6x2 cos 2a + 3x sin 2a . sin 6a +
n 2 a − a 2 (B) f ′ (1/2) < 0 (A) f(x) is not defined at x = 1/2 (C) f ′ (x) is not defined at x = 1/2 (D) f ′ (1/2) > 0 emx + (m −

1)−2 ex

)

then :

d 2y dy then 2 − 2m + m2y is equal to : dx dx

Q.39 If y = (A + Bx) (A) ex (B) emx (C) e−mx (D) e(1 − m) x ax bx Q.40 Suppose f (x) = e + e , where a ≠ b, and that f '' (x) – 2 f ' (x) – 15 f (x) = 0 for all x. Then the product ab is equal to (A) 25 (B) 9 (C) – 15 (D) – 9 Q.41 Let h (x) be differentiable for all x and let f (x) = (kx + ex) h(x) where k is some constant. If h (0) = 5, h ' (0) = – 2 and f ' (0) = 18 then the value of k is equal to (A) 5 (B) 4 (C) 3 (D) 2.2 Q.42 Let ef(x) = ln x . If g(x) is the inverse function of f(x) then g ′ (x) equals to : (A) ex (B) ex + x (C) e ( x + e x ) (D) e(x + ln x) dy Q.43 The equation y2e xy = 9e –3 · x2 defines y as a differentiable function of x. The value of for dx x = – 1 and y = 3 is 15 9 (A) – (B) – (C) 3 (D) 15 2 5

Q.44 Let f(x) = xx and g(x) = x ( ) then : (A) f ′ (1) = 1 and g ′ (1) = 2 (B) f ′ (1) = 2 and g ′ (1) = 1 (C) f ′ (1) = 1 and g ′ (1) = 0 (D) f ′ (1) = 1 and g ′ (1) = 1 Q.45 The function f(x) = ex + x, being differentiable and one to one, has a differentiable inverse f–1(x). The value of d –1 (f ) at the point f(l n2) is dx 1 1 1 (A) (B) (C) (D) none
n2 3 4 3 π log sin|x| cos x for |x| < x≠0 Q.46 If f (x) = 3 3 x  log sin|3 x| cos   2 =4 for x = 0  π π then, the number of points of discontinuity of f in  − ,  is 3 3 (A) 0 (B) 3 (C) 2 (D) 4

( )

Q.47 If y = (A)

xx

x

(a − x) a − x − ( b − x) x − b a −x + x−b x + (a + b) (a − x) (x − b)

(B)

then

dy wherever it is defined is equal to : dx

2 x − (a + b) 2 (a − x) (x − b)

(C) −

(a + b ) 2 (a − x) (x − b)

(D)

2 x + (a + b ) 2 (a − x) (x − b)

2

Q.48 If y is a function of x then (A)

dy d y = 0 . If x is a function of y then the equation becomes : 2 +y dx dx

dx d2 x d2 x = 0 (B) + x +y dy d y2 d y2

3

 dx   =0  dy

(C)

d2 x −y d y2

2

 dx   =0  dy

(D)

d2 x −x d y2

2

 dx   =0  dy

Q.49 A function f (x) satisfies the condition, f (x) = f ′ (x) + f ′′ (x) + f ′′′ (x) + ...... ∞ where f (x) is a differentiable function indefinitely and dash denotes the order of derivative . If f (0) = 1, then f (x) is : (A) ex/2 (B) ex (C) e2x (D) e4x cos 6x + 6 cos 4 x + 15 cos 2 x + 10 dy , then = Q.50 If y = cos 5x + 5 cos 3x + 10 cos x dx (A) 2 sinx + cosx (B) –2sinx (C) cos2x (D) sin2x 3

d 2 x  dy  d2y  + 2 = K then the value of K is equal to Q.51 If 2  dy  dx  dx (A) 1 (B) –1 (C) 2 13

(D) 0

DIFFRENTIATION / Page 13 of 18

(C) f ′ (x) is not defined at x = sin 8 (D) f ′ (sin 8) < 0 Q.37 A function f, defined for all positive real numbers, satisfies the equation f(x2) = x3 for every x > 0 . Then the value of f ′ (4) = (B) 3 (C) 3/2 (D) cannot be determined (A) 12

)



1 2 2 x (1− x)

2

(A)

x (1 − x)

−

 e y = f(x) =    0

Q.53 Let

(C) −

(B) zero

(D) π

1

if x ≠ 0

x2

if x = 0

Then which of the following can best represent the graph of y = f(x) ? (B)

(A)

(C) 1

(D)

1

1


+ m  n −
 m+ n 
− m  n +
 m − n m− n  .  x n−
 . x
− m Q.54 Diffrential coefficient of  x           

w.r.t. x is

(C) – 1 (D) x
mn x + h f ( x) − 2 h f ( h) Q.55 Let f (x) be diffrentiable at x = h then Lim is equal to x→ h x−h (A) f(h) + 2hf '(h) (B) 2 f(h) + hf '(h) (C) hf(h) + 2f '(h) (D) hf(h) – 2f '(h) (A) 1

(B) 0

b g

d 3y

Q.56 If y = at2 + 2bt + c and t = ax2 + 2bx + c, then equals dx 3 2 2 (A) 24 a (at + b) (B) 24 a (ax + b) (C) 24 a (at + b)2 Q.57

Limit+ x→0

(A)

(D) 24 a2 (ax + b)

 x x  a arc tan  has the value equal to − b arc tan a b  x x  1

a−b 3

(B) 0

(C)

(a 2 − b 2 ) 6a 2 b 2

(D)

a 2 − b2 3a 2 b2

x Q.58 Let f (x) be defined for all x > 0 & be continuous. Let f(x) satisfy f   = f ( x ) − f ( y) for all x, y & f(e) = 1. y Then :

(A) f(x) is bounded

 1 (B) f   → 0 as x → 0  x

(C) x.f(x)→1 as x→ 0 (D) f(x) = ln x

Q.59 Suppose the function f (x) – f (2x) has the derivative 5 at x = 1 and derivative 7 at x = 2. The derivative of the function f (x) – f (4x) at x = 1, has the value equal to (A) 19 (B) 9 (C) 17 (D) 14 4 2 x − x +1 dy Q.60 If y = 2 and = ax + b then the value of a + b is equal to dx x + 3x + 1 5π 5π 5π 5π (A) cot (B) cot (C) tan (D) tan 8 12 12 8 Q.61 Suppose that h (x) = f (x)· g(x) and F(x) = f (g ( x ) ) , where f (2) = 3 ; g(2) = 5 ; g '(2) = 4 ; f '(2) = –2 and f '(5) = 11, then (A) F'(2) = 11 h'(2) (B) F'(2) = 22h'(2) (C) F'(2) = 44 h'(2) (D) none Q.62 Let f (x) = x3 + 8x + 3 which one of the properties of the derivative enables you to conclude that f (x) has an inverse? (A) f ' (x) is a polynomial of even degree. (B) f ' (x) is self inverse. (C) domain of f ' (x) is the range of f ' (x). (D) f ' (x) is always positive. Q.63 Which one of the following statements is NOT CORRECT ? (A) The derivative of a diffrentiable periodic function is a periodic function with the same period. (B) If f (x) and g (x) both are defined on the entire number line and are aperiodic then the function F(x) = f (x) . g (x) can not be periodic. (C) Derivative of an even differentiable function is an odd function and derivative of an odd differentiable function is an even function. (D) Every function f (x) can be represented as the sum of an even and an odd function Select the correct alternatives : (More than one are correct) Q.64 If y = tan x tan 2x tan 3x then

dy has the value equal to : 14 dx

DIFFRENTIATION / Page 14 of 18

(

−1 −1 Q.52 If f(x) = 2 sin 1 − x + sin 2 x (1− x) where x ∈  0 ,  then f ' (x) has the value equal to

2 2 2 dy = (A) 2 ln x . x x (B) (2 ln x + 1). x x (C) (2 ln x + 1). x x dx dy = Let y = x + x + x + ...... ∞ then dx 1 1 x y (A) (B) (C) (D) 2y − 1 x + 2y 2x + y 1 + 4x

2

Q.66 If y = xx then Q.67

Q.68 If 2x + 2y = 2x + y then (A) −

2y 2x

dy has the value equal to : dx 1 (C) 1 − 2y (B) 1 − 2x

(D)

dv du −u = u2 + v2 dx dx

Q.70 Let f (x) =

x − 2 x −1 x −1 −1

d2u = 2v dx2

(B)

(C)

(D) x x

2 +1

. ln ex2

) − 1)

2x 1 − 2y 2

y

Q.69 The functions u = ex sin x ; v = ex cos x satisfy the equation : (A) v

( (2

+1

x

d 2v = − 2u dx 2

(D) none of these

. x then :

(B) f ′ (3/2) = − 1 (C) domain of f (x) is x ≥ 1 (D) none (A) f ′ (10) = 1 Q.71 Two functions f & g have first & second derivatives at x = 0 & satisfy the relations, 2 , f ′ (0) = 2 g ′ (0) = 4g (0) , g ′′ (0) = 5 f ′′ (0) = 6 f(0) = 3 then : g(0) 15 f (x) (A) if h(x) = then h ′ (0) = (B) if k(x) = f(x) . g(x) sin x then k ′ (0) = 2 4 g(x) 1 g ′ (x ) (D) none = (C) Limit x→0 2 f ′ (x)

f(0) =

Q.72 If y = x (
nx )

(


n (
n x )

dy is equal to : dx + 2
n x
n (
n x )

, then

y
n x
nx − 1 x y (C) ((ln x)2 + 2 ln (ln x)) x
n x

(A)

)

y (ln x)ln (ln x) (2 ln (ln x) + 1) x y
n y (D) (2 ln (ln x) + 1) x
n x

(B)

ANSWER KEY Q.1 Q.6 Q.11 Q.16 Q.21 Q.26 Q.31 Q.36 Q.41 Q.46 Q.51 Q.56 Q.61 Q.64 Q.68

A B B D A C D D C C D D B A,B,C A,B,C,D

Q.2 Q.7 Q.12 Q.17 Q.22 Q.27 Q.32 Q.37 Q.42 Q.47 Q.52 Q.57 Q.62 Q.65 Q.69

C B D A B C D B C B B D D A,C A,B,C

Q.3 Q.8 Q.13 Q.18 Q.23 Q.28 Q.33 Q.38 Q.43 Q.48 Q.53 Q.58 Q.63 Q.66 Q.70

B C B B B D D D D C C D B C,D A,B

Q.4 Q.9 Q.14 Q.19 Q.24 Q.29 Q.34 Q.39 Q.44 Q.49 Q.54 Q.59

D A D C B D B A D A B A

Q.67 A,C,D Q.71 A,B,C 15

Q.5 Q.10 Q.15 Q.20 Q.25 Q.30 Q.35 Q.40 Q.45 Q.50 Q.55 Q.60

B C C C C C D C B B A B

Q.72 B,D

DIFFRENTIATION / Page 15 of 18

(A) 3 sec2 3x tan x tan 2x + sec2 x tan 2x tan 3x + 2 sec2 2x tan 3x tan x (B) 2y (cosec 2x + 2 cosec 4x + 3 cosec 6x) (C) 3 sec2 3x − 2 sec2 2x − sec2 x (D) sec2 x + 2 sec2 2x + 3 sec2 3x dy x − x then equals Q.65 If y = e + e dx 1 1 e x − e− x e x − e− x y2 − 4 y2 + 4 (A) (B) (C) (D) 2 x 2 x 2 x 2x

1.

dy 2 2x 2 − 1 and y = f(x ) then dx at x = 1 is (B) 1 (C) – 2

If f′(x) = (A) 2

2.

If y = x

x2

(A) 2
n x. 3.

4.

If f(x) = e 1 (A) 2

xx

(B) (2
n x + 1).

x

2

1 2

then

x b+

xx

(C) (2
n x + 1). x

(C) 1

7.

9.

10.

a ab + 2 ay

(B)

12.

13.

14.

.
n ex 2

dy = dx

b ab + 2 by

(C)

a ab + 2 by

Let f(x) = sin x; g(x) = x 2 & h(x) = loge x & F(x) = h[g(f(x))] then

(D)

b ab + 2 ay

d2 F is equal to: dx 2

(C) 2x cot x 2 (D) − 2 cosec2 x dy If y = (1 + x) (1 + x 2) (1 + x 4) .....(1 + x 2n ), then at x = 0 is dx (A) –1 (B) 1 (C) 0 (D) 2n dy 1 If y = sin−1  x 1 − x + x 1 − x 2  and = + p, then p =   dx 2 x (1 − x) 1

(B)

1

(C) sin−1 x

1− x

 y  If x 2 + y 2 = et where t = sin–1   x 2 +y 2 

(D)

x−y x+y

(B)

(A)

x 4 x −1

(B)

x2 x4 − 1

The differential coefficient of sin−1

x−y (D) 2x + y

(D) 1

(C) 0

t 1+ t

1− x 2

 dy  :  then dx 

2x + y x+y (C) x−y x−y 2 2 dy x −1 x +1 + sec−1 2 , x  > 1 then is equal to: If y = sin−1 2 dx x +1 x −1 (A)

(A) 1

11.

2 +1

x a +...............

(A) 0 8.

(D) x x

(D) –1

(A) 2 cosec3 x (B) 2 cot (x 2) − 4x 2 cosec2 (x 2)

6.

x2 + 1

, then f ′(0). (B) –

a+

5.

2

x  tan−1  sin  2 

If y =

(A)

(D) –1

dy then = dx

2

w.r.t. cos−1

(B) t

(C)

1 1 + t2 1

is: (D) none

1 + t2

 tan − 1 x  −1  − 1  w.r.t. tan x is: 1 + tan x  

Differentiation of 

−1 1 1    (A)  (B) − 1 (C) (D) −1 2 (1 + tan −1 x )2  1 + tan x  1 + tan −1 x Let f(x) be a polynomial in x. Then the second derivative of f(ex), is: (B) f ′′ (ex). e2x + f ′ (ex). e2x (C) f ′′ (ex) e2x (D) f ′′ (ex). e2x + f ′ (ex). ex (A) f ′′ (ex). ex + f ′ (ex) f g h f ′ g′ h′ If f(x), g(x), h(x) are polynomials in x of degree 2 and F(x) = , then F′(x) is equal to f ′′ g′′ h′′ (A) 1 (B) 0 (C) –1 (D) f(x) . g(x) . h(x) y y1 y 2 If y = sin mx then the value of y 3 y 4 y 5 (where settings of y shows the order of derivative) is: 16 y6 y7 y8

(

)

DIFFRENTIATION / Page 16 of 18

EXERCISE -1 Part : (A) Only one correct option

(B) dependent of x but independent of m (D) independent of m & x.

f (5 + t ) − f (5 − t ) = 2t

15.

If f ′ (5) = 7 then Limit t →0

16.

(A) 0 (B) 3.5 (C) 7 (D) 14 Let ef(x) = ln x. If g(x) is the inverse function of f(x) then g′ (x) equals to: (A) ex

(C) ex+e

(B) ex + x

x

(D) ex + ln x

n

17.

If u = ax + b then (A)

18.

dn [f(u)] du n

(C) an

dy  2x − 1 =  & f ′ (x) = sin x then 2 dx  x + 1

dn [f(u)] du n

1 + x − x2

(1 + x ) 2

(C)

2

1 − x + x2

(1 + x ) 2

2

 2x − 1 sin  2   x + 1

dn [f(u)] dx n

(B)

( ) sin  2x − 1  x + 1 (1 + x )

2 1 + x − x2 2

 2x − 1   x 2 + 1

sin 

2

2

(D) none

 d   3 d2y    y . 2  equals:  dx   dx 

If y2 = P(x), is a polynomial of degree 3, then 2 

(A) P ′′′ (x) + P ′ (x) (B) P ′′ (x). P ′′′ (x) (C) P (x). P ′′′ (x) Part : (B) May have more than one options correct 20.

(D) a−n

If y = f  (A)

19.

d [f(ax + b)] is equal to: dx n dn (B) a n [f(u)] du

(D) a constant

Two functions f & g have first & second derivatives at x = 0 & satisfy the relations,

2 , f ′ (0) = 2 g ′ (0) = 4g (0), g ′′ (0) = 5 f ′′ (0) = 6 f(0) = 3 then: g(0) 15 f (x) (A) if h(x) = then h ′ (0) = (B) if k(x) = f(x). g(x) sin x then k ′ (0) = 2 4 g(x) g′ (x) 1 = (D) none (C) Limit x →0 f ′ (x) 2 f(0) =

fn − 1 ( x )

for all n ∈ N and f o (x) = x, then

d {f (x)} is equal to: dx n

21.

If f n (x) = e

22.

d {f (x)} (B) f n (x). f n − 1 (x) dx n − 1 (C) f n (x). f n − 1 (x)........ f 2 (x). f 1 (x) (D) none of these If f is twice differentiable such that f ′′(x) = –f(x) and f ′(x) = g(x). If h(x) is a twice differentiable function such that h′(x) = [f(x)]2 + [g(x)] 2 . If h(0) = 2, h(1) = 4, then the equation y = h(x) represents: (A) a curve of degree 2 (B) a curve passing through the origin (C) a straight line with slope 2 (D) a straight line with y intercept equal to 2.

(A) f n (x).

23.

24. 25.

x3 + x 2 sin 1.5 a − x sin a. sin 2a − 5 sin–1 (a2 − 8a + 17) then: 3 2 (A) f ′(x) = – x + 2x sin6 – sin4 sin8 (B) f ′ (sin 8) > 0 (C) f ′ (x) is not defined at x = sin 8 (D) f ′ (sin 8) < 0 If f(x) = x 3 + x 2f ′(1) + xf ′′(3) for all x ∈ R then

Given f(x) = −

(A) f(0) + f(2) = f(1) (B) f(0) + f(3) = 0 (C) f(1) + f(3) = f(2) (D) none of these If f(x) = (ax + b) sin x + (cx + d) cos x, then the values of a, b, c and d such that f ′(x) = x cos x for all x are (A) a = d = 1 (B) b = 0 (C) c = 0 (D) b = c

EXERCISE -2 d2 y

dy +2k + n2 y = 0, where n2 = p2 + k2. dt dt 2 Evaluate the following limits using L ′ hospitale rule as otherwise 2. Limit log tan2 x (tan2 2 x)

1.

If y = A e − kt cos (p t + c) then prove that

x →0

3.

( x−a) 4 ( x−b) 4 If f (x) = ( x−c ) 4

( x−a)3 1 ( x−a) 4 ( x−b)3 1 ( x−b) 4 then f ′ (x) = λ. 3 ( x−c ) 1 ( x−c ) 4

( x−a)2 1 ( x−b)2 1 . Find the value of λ. ( x−c )2 1 17

DIFFRENTIATION / Page 17 of 18

(A) independent of x but dependent on m (C) dependent on both m & x

If x = a t3 & y = b t2, where t is a parameter, then prove that

d3 y dx

3

8.b

=

27a 3 .t 7

sina

5. 6. 7.

If sin y = x sin (a + y), show that

dy = 2 . dx 1 − 2xcosa+x

F" f " g" 2c F′′′ f ′′′ g′′′ If F(x) = f(x). g(x) & f ′ (x). g ′ (x) = c, prove that F = f + g + f g & F = f + g . If α be a repeated root of a quadratic equation f(x) = 0 & A(x), B(x), C(x) be the polynomials of degree 3, 4 & 5 A( x ) B( x ) C( x ) A(α ) B(α ) C(α ) respectively, then show that is divisible by f(x), where dash denotes the derivative. A ' (α ) B' (α) C' (α) 2  1 +  dy     dx  

3/2

can be reduced to the form R2/3 =

1

+

8.

Show that R =

9.

 d2 y   dx 2  dx 2    Also show that, if x = a sin 2θ (1 + cos 2θ) & y = a cos 2θ (1 − cos 4 a cos 3θ. Differentiate the following functions with respect to x.

(i)

x 2.
n x. ex

d2 y

(i)

sinx − xcos x xsinx + cos x

2 3

1

2 .  d2 x  3    dy 2    2θ) then the value of R equals to

(iii)

  −1 1 − cos x  tan  tan 1 + cos x  

9. 20.

C ABC

(iii)

1 x sec2 2 2

Exercise # 1 1. A 2. C 3. A 4. D 5. D 12. D 13. B 14. D 15. C 16. C 21. AC 22. CD 23. AD 24. ABC

6. B 7. D 17. C 18. B 25. ABC

8. 19.

B C

10.

Exercise # 2 2. 1

3. 3

9. (i) ex x (2
n x + 1 + x
n x)

(ii)

x2

( xsinx + cos x)

2

18

A

11.

C

DIFFRENTIATION / Page 18 of 18

4.