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CONTENT

MATHEMATICS CLASS : XII Preface

1.

Functions Exercise

2.

79 - 100

Application of derivatives Exercise

6.

54 - 78

Method of differentiation Exercise

5.

29 - 53

Continuity and derivability Exercise

4.

01 - 28

Limits Exercise

3.

Page No.

101 - 126

Solution of triangle Exercise

127 - 149

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FUNCTIONS EXERCISE # 1 PART - I Section (A) : A–3.

(i)

f(x) =

f(x) =

x 3 5x 3 x2 1 x 3 5x 3 ( x 1)( x 1)

Division by zero is undefined Domain x R – {1, –1}

x±1 x (–, –1) (–1, 1) (1, )

(ii)

f(x) =

sin1 x x For sin–1x, x [–1, 1] and division by zero is undefined x 0

Domain x [–1, 0) (0, 1]

1 (iii)

(iv)

(v)

f(x) =

x | x |

for function to be defined x + |x| > 0 for x > 0, x + |x| = 2x > 0 for x 0, x + |x| = 0 Domain is x (0, ) f(x) = ex + sin x Domain x R as there is no restriction for exponent of e.

1 f(x) = log (1 x ) + 10

x2

1 – x > 0 and x + 2 0 x (– , 1) – {0} and x – 2

(vi)

f(x) =

and

1–x1 x [–2, 0) (0, 1)

x

3x 1 –1 1 2x + 3 sin 2

1 – 2x 0 and – 1

3x 1 1 2

1 2

and –

1 x1 3

Taking intersection

1 1 Domain x , 3 2

(vii)

f(x) = 2 sin

1

x

x

1

1

x

x (0, 1) (1, ) and 0 < x –

1 f(x) = logx log 2 x 1/ 2 In case of composite function in log. We start with outer log. x > 0, x 1 and

1/2

–

x2 – 1 x 1 and x > 2 (viii)

–1/3

1

1 >1 x 2 1 3 x (0, ) – {1} and 0 Range 0 < sin–1 x

(iv)

1

>1+

x 1

Range y (0, 1]

x (0, 1]

2

– < n (sin–1x) n 2 Inequality doesn't change as n is increasing function f(x) = 2 – 3x – 5x2 Domain x R Method 1 y = – 5x2 – 3x + 2 opening downward parabola 0

D Range y , 4a

49 y , 20 Method 2 5x2 + 3x + (y – 2) = 0

D0 (v)

– D/4a 2

9 – 20 (y – 2) 0

20y – 49 0

y

49 20

f(x) = 3 |sin x| – 4|cos x| f(x) is a periodic function with period . So analysis is limited in [0, ]

, |sin x| = 1, |cos x| = 0 2 fmin = 3.0 – 4.1 = – 1 at x = 0, |sin x| = 0, |cos x| = 1 fmax = 3.1 – 4.0 = + 3 at x =

(vi)

f(x) =

sin x

Range y [–4, 3]

cos x

1 tan x 1 cot 2 x f(x) = sin x |cos x| + cos x |sin x| periodic period = 2 sin 2x 0 f(x) = sin 2x 0

+

2

x 0, 2 , x , 2 , x , 2 3 , x , 2 2 ,

Range y [–1, 1]

RESONANCE

S OLUTIONS (XII) # 2

Section (B) : B–1.

(ii)

2 x 2 and g(x) = ( x ) Domain x R, Domain x [0, ) non-identical functions f(x) = sec(sec –1x) and g(x) = cosec (cosec –1x) Domain x (–, –1] [1, ), Domain x (–, –1] [1, ) f(x) = x g(x) = x Identical functions

(iii)

f(x) =

(i)

(iv)

B–5.

(i)

(ii)

(iii)

(iv)

f(x) =

1 cos x and g(x) = cos x 2 f(x) = |cos x| non-identical function f(x) = x and g(x) = enx, Domain x R+ Domain x R non-identical function f(x) = ex and g(x) = n x fog(x) = en x = x, x > 0 gof(x) = n ex = x, x R f(x) = |x| and g(x) = sin x fog(x) = f(sin x) = |sin x| gof (x) = g(|x|) = sin |x| f(x) = sin–1x and g(x) = x2 fog(x) = sin–1(g(x)) = sin–1x2 gof(x) = (f2(x)) = (sin–1x)2 f(x) = x2 + 2, g(x) =

fog(x) = g2(x) + 2 =

gof(x) =

B–6.

x x 1

x2 ( x 1)2

+2=

3x 2 4x 2 ( x 1)2

f (x) x2 2 = 2 f (x) 1 x 1

1 x 2 , x 1 f(x) = 1 x , 1 x 2 g(x) = 1 – x, – 2 x 1

1 g2 , g( x ) 1 x [0,1] fog(x) = 1 g( x ) , 1 g( x ) 2 x [–1,0)

1 (1 – x )2 fog (x) = 1 (1 – x )

,

x [0,1]

, x [–1, 0)

2 – 2x x 2 , x [0,1] fog (x) = 2 – x , x [–1, 0)

Section (C) : C–1.

(i)

y = |(x + 2) (x + 3)| many - one function

(ii)

y = |nx| many - one function

RESONANCE

S OLUTIONS (XII) # 3

(iii)

f(x) =sin 4x, x – , 8 8

2 one-one function period =

1 1 , x (0, ) x x many one function

(iv)

f(x) = x +

(v)

f(x) =

1 –1

1 – e x 1 –1 .

e x f =

2 1– e

1 x2 0

1 –1 x

increasing function

Hence one - one

C–5.

3x 2 – cos( x ) 4 Hence many - one

(vi)

f(x) =

(vii)

f(x) = sin–1 x – cos–1 x = 2 sin–1 x –

(i) (ii) (iii)

f(x) = sin (x2 +1) f(x) = x + x2 f(x) = x – x3

f(– x) = f (x) = even function f(– x) = x2 – x f (x) or – f(x) Neither even nor odd function f(–x) = – x + x3 = – f(x) odd function

(iv)

a x – 1 f(x) = x x a 1

ax – 1 f(–x) = – x – x a 1

even function

monotonically increasing. 2

a x – 1 = f(x) even function f(–x) = x x a 1

(v)

f(x) = log (x +

(vi)

2 2 f(x) + f(– x) = log ( x x 1)(– x x 1) = log [(x2 + 1) – x2] = 0 hence odd function f(x) = sin x + cos x f(– x) = – sin x + cos x f(x) or – f(x) Neither even nor odd. f(x) = (x2 – 1) |x| f(–x) = f(x) even function.

(vii)

(viii)

x2 1 )

f(–x) = log (–x +

x2 1 )

| tan(tan 1 x ) | x 2 x [ 2 x ] 1 x 1 f(x) = sec(sec 1 x ) x x | x | x x0 f(x) = 3 0 x 1 x x

RESONANCE

x x x x0 f(x) = 3 0 x 1 x x

S OLUTIONS (XII) # 4

C–6.

(i)

x 2 – sin x , – 1 x 0 even extension of f(x) = f(–x) = – x e x , x –1

(ii)

– x 2 sin x , – 1 x 0 odd extension of f(x) = – f(– x) = x – e x , x –1

Section (D) : D–2.

(i)

f(x) = 2 + 3 cos (x – 2)

(ii)

f(x) = sin 3x + cos2 x + |tan x|

fundamental period = 2

2 period of f(x) = L.C.M. , , = 2 3 f(x + ) = – sin x + cos2 x + |tan x| f(x)

f(x) = sin

(iv)

f(x) = cos

for fundamental period

fundament period = 2

3x sin 2x – 5 7

period

10 ,7 3

10 , 7 = 70 period of f(x) = L.C.M. 3

Fundament period = 70

f(x) = [sin 3x] – |cos 6x|

period

2 2 period of f(x) = L.C.M. , = 3 3 3

Fundamental period =

(vi)

f(x)=

(vii)

f(x) =

2 3

3 2 3

1 fundamental period = 2 1 cos x sin12x

period of f(x) = L.C.M. , = 6 3 3

2

1 cos 6 x for fundamental period

fx = 6

(viii)

2 , , 3

x x + sin 4 3 period 8, 6 period of f(x) = L.C.M. (8, 6) = 24 fundamental period = 24

(iii)

(v)

period

sin12 x 6 1 cos 2 6 x 6

= f(x)

f(x) = sec3 x + cosec3 x period 2 Fundamental period = L.C.M. (2, 2) = 2

Fundament period =

6

2

Section (E) : E–1.

(i)

f:DR f(x) = 1 – 2–x

f (x) = 2– x n2 > 0 increasing function one one function

D : [x R), Range : (–, 1) codomain

function is not bijective

RESONANCE

f –1 does not exist S OLUTIONS (XII) # 5

(ii)

f(x) = (4 – (x – 7)3)1/5 f (x) =

1 (4 – (x – 7)3) 5

– 4/5

. (– 3 (x – 7)2) 0 decreasing function one one function

Lim f ( x ) – x

Lim f ( x )

x –

D:R

Range : R = codomain

onto function

function is bijective (invertible)

y = (4 – (x – 7)3)1/5 4 – y5 = (x – 7)3 x = 7 + (4 – y5)1/3 (iii)

f –1 (x) =

E–6.

f –1(x) = 7 + (4 – x 5)1/3

or

x=

f(x) = 1 ± xn

or

f(x) = 1 – x3 f(1) = – 3

f(x) = n x 1 x 2 D : x R, Range : R y = n x 1 x 2

E–5.

or

e y ey 2

e x e x 2

1 1 f(x) . f = f(x) + f x x f(3) = – 26 f(x) = – 3x2

f(x + y) = f(x) . f(y) and f(1) = 2 10

f (n) = f(1) + f(2) + ........... + f(10)

n 1

210 1 = 21 + 22 + 23 + ....... + 210 = 2 2 1 = 2046

PART - II Section (A) : A–2.

For domain – log0.3(x – 1) 0 log0.3(x – 1) 0 (x – 1) 1 x2 Taking intersection x [2, )

A–3.

f(x) = cot–1

x( x 3 ) + cos –1

for domain x(x + 3) 0 x (–, –3] [0, ) Taking intersection x {–3, 0}

RESONANCE

and and and

x2 + 2x + 8 > 0 (x + 1)2 + 7 > 0 xR

x 2 3x 1

and and

0 x 2 + 3x + 1 1 x 2 + 3x + 1 0 and

x 2 + 3x 0 x [–3, 0]

S OLUTIONS (XII) # 6

A–5.

f(x) = 4x + 2x + 1 Let 2x = t > 0, x R

f(x) = g(t) = t2 + t + 1,

t>0

2

1 3 g(t) = t + 2 4 2

1 1 t > 2 2

1 1 t > 2 4

2

1 3 t + >1 2 4

Range is (1, )

Section (B) : B–2.*

–1

f(x) = en(sec x ) = sec–1x, x (–, – 1] (1, ) –1 g(x) = sec x, x (–, – 1] [1, ) non-identical functions f(x) = tan (tan–1 x) = x, x R g(x) = cot (cot–1 x) = x, x R identical functions

(A)

(B)

1 x 0 f(x) = sgn (x) = 0 x 0 – 1 x 0

(C)

1 x 0 g(x) = sgn(sgn x) = 0 x 0 – 1 x 0

Identical functions f(x) = cot2 x . cos2 x, x R – {n }, g(x) = cot2 x – cos2 x = cot2 x (1 – sin2 x) = cot2 x. cos2 x Identical functions

(D)

B–4.

B–6.*

Domain of f(g(x)) Range of g(x) Domain of f(x) – 5 |2x + 5| 7 – 12 2x 2 f(x) =

1– x , 1 x fog(x) =

0x1

n I x R – {n },

n I

0 |2x + 5| 7 –6x1

g(x) = 4x (1 – x),0 x 1

–7 2x + 5 7

1 – g( x ) 1 – 4 x(1 – x ) 1 – 4 x 4x 2 = = 1 g( x ) 1 4 x(1 – x ) 1 4x – 4x 2

1– x 1– x gof(x) = 4f(x) . (1 – f(x)) = 4 1 x 1 – 1 x

=

8 x(1 – x ) (1 x )2

Section (C) : C–2.

One One / Many One f(x) =

f(x) =

f(x) =

2x 2 x 5 7 x 2 2x 10

, Domain x R

( 4x 1)(7 x 2 2x 10) (14 x 2)(2x 2 x 5) (7 x 2 2x 10)2 11x 2 30x 20 2

(7 x 2x 10)

2

30 > 0 x (– , 0) , 11 30 f (x) < 0 x 0, 11

f(x) = 0

RESONANCE

x = 0,

30 11 S OLUTIONS (XII) # 7

Function is increasing and decreasing in different intervals, so non monotonic Many one function. Onto / Into f(x) =

2x 2 x 5

7 x 2 2x 10 2x 2 – x + 5 > 0, x R and 7x 2 + 2x + 10 > 0 x R a = 2 > 0 and a = 7 and D = 4 – 280 < 0 D = 1 – 40 = – 39 < 0 f(x) > 0 x R Also f(x) never tends to ± as 7x 2 + 2x + 10 has no real roots, Range Codomain so into function.

C–3.

f(x) = x 3 + x 2 + 3x + sin x, f(x) = 3x 2 + 2x + 3 + cos x

3x 2 + 2x + 3

–1 cos x 1

lim f(x) = +

xR

32 as a = 3 > 0 and D < 0 12 so f(x) > 0 x R

lim f(x) = –

x

x

Hence f(x) is one-one and onto function (as f(x) is continuous function) C–6.

f(g(x1)) = f(g(x2)) as f is one - one function hence f(g(x1)) = f(g(x2)) x1 = x2

g(x1) = g(x2) x1 = x2

f(g(x)) is one - one function

as

g is one - one function

Section (D) : 2 D–2.

f(x) = sin

[a] x .

[a] = 4 D–3.

Period =

[ a]

=

a [4, 5)

f(x) = x + a – [x + b] + sin x + cos 2x + sin (3x) + cos (4x) + ........ + sin (2n – 1) + cos (2px) f(x) = {x + b} + a – b + sin (x) + cos (2x) + sin (3x) + cos (4x) + .... + sin (2n – 1) + cos (2nx) Period of f(x) = L.C.M (1, 2,

2 2 2 2 , , ........., , )=2 3 4 2n 1 2n

period of f(x) = 2 since f(1 + x) f(x) , hence fundamental period is 2 D–7.*

(A) (B)

f(x) = cos (cos–1 x) = x, x [–1, 1] odd function f (x + ) = cos (sin (x +)) + cos (cos (x + )) f (x + ) = cos (sin x) + cos (cos x) = f(x)

f x = cos sin x + cos cos x 2 2 2 = cos (cos x) + cos (sin x) = f(x) fundamental period = (C)

2

f(x) = cos (3 sin x), x [–1, 1] – 3 sin1 3 sin x 3 sin 1 cos (3 sin 1) cos (3 sin x) 1

Range is [cos (3 sin1), 1]

1 x f–1(x) = n 1 x

Section (E) : E–1.

ex ex y = x 1 e e x By compnendo and dividendo

1 y 2e x = 1 y 2

RESONANCE

1 y x = n 1 y

S OLUTIONS (XII) # 8

E–7.

E–8.

f(1) = 1 = 2 – 1 f(n + 1) = 2f(n) + 1 f(3) = 7 = 23 – 1 f(4) = 15 = 24 – 1 Similarly f(n) = 2n – 1

f(2) = 2f(1) + 1 = 2. 1 + 1 = 3 = 22 – 1

Method 1 : (usual but lengthy) x2 f(x) + f(1 – x) = 2x – x4 .....(1) replace x by (1 – x) in equation (1) (1 – x)2 f(1 – x)+ f(x) = 2 (1– x) – (1 – x)4 .....(2) eliminate f(1 – x) by equation (1) and (2) we get f(x) = 1 – x2 Method 2 : Since R.H.S. is polynomial of 4th degree and also by options consider f(x) = ax2 + bx + c x2 f(x) + f(1 – x) = 2x – x4 x 2 (ax2 + bx + c) + a (1 – x)2 + b (1 – x) + c = 2x – x4 by comparing coefficients a=–1 b=0 c=1 f(x) = – x2 + 1

EXERCISE # 2 PART - I 1.

(i)

f(x) =

or

3 2 x 2 .2 x 3 – 2x – 2 .2–x 0 (2x – 1) (2x – 2) 0

(ii)

f(x) =

(iii)

1 – 1 x 2 0 f(x) = (x2 + x + 1)–3/2 D:xR

(iv)

f(x) =

or

(2x)2 – 3.2x + 2 0 2x [1, 2] x [0, 1]

1 1 x2

x2 + x2

1 x 2 1

0 1 – x2 1

or

x

or

x 2n

or

x (0, 1] [4, 5)

x [– 1, 1]

1 x 1 x

x2 1 x 0 and 0 x2 1 x x (– , –2) [2, ) and x (–1, 1] D: (v)

f(x) =

tan x tan 2 x

tan x – tan2x 0

or

0 tan x 1

n

n , n 4

1

(vi)

(vii)

f(x) = 2 sin x 2

f(x) =

sin

x 0 2

5x x 2 log1 / 4 4

5x x 2 1 4

RESONANCE

and

5x – x2 > 0

S OLUTIONS (XII) # 9

(viii) or 2.

(i)

f(x) = log10 (1 – log10(x2 – 5x + 16)) 1 – log10 (x2 – 5x + 16) > 0 x (2, 3) f(x) = 1 – |x – 2| |x – 2| [0, )

or

x2 – 5x + 6 < 0

f(x) (– , 1]

1 f(x) , 1 3

1 (ii)

(iii)

f(x) =

x5 D : x (5, ) R : f(x) (0, ) 1 f(x) = 2 cos 3 x range of cos 3x is [–1, 1] cos 3x [–1, 1]

(iv)

x2

f(x) =

=y x 8x 4 x + 2 = yx2 – 8yx – 4y for x to be real D 0 (8y + 1)2 + 4y (4y + 2) 0 64y2 + 16y + 1 + 16y2 + 8y 0 2

80y2 + 24y + 1 0 (v)

f(x) =

(viii) (ix)

f(x) = 3 sin

(i)

1 1 , y , 4 20

or

1 y , 3 3

2 x2 16

D : x , 4 4

f(x) = x3 – 2x2 + 5 = (x2 – 1)2 + 4 R : [4, ) f(x) = x3 – 12x , x [–3, 1] = x (x2 – 12) f(x) = 3x2 – 12 = 0 or f(x) = sin2x + cos4x = sin2x + 1 + sin4x – 2 sin2x = sin 4x – sin2x + 1 1 2 3 = sin x + 2 4

7.

or

=y x 2 2x 4 x2 – 2x + 4 = yx2 + 2xy + 4y x2 (1 – y) – 2x(1 + y) + 4(1 – y) = 0 D0

2 x 2 0 , 4 16

(vii)

yx2 – x (8y + 1) – (4y + 2) = 0

x 2 2x 4

4(1 + y)2 – 16(1 – y)2 0

(vi)

or

f(–x) =

3 f(x) 0 , 2

x=±2

R : [–11, 16]

3 R : , 1 . 4

(2 x 1)7 (2 x )6

neither even non add (ii) (iii)

sec x x 2 9 = f(x) x sin x f(–x) = – f(x) odd f(–x) =

RESONANCE

even

S OLUTIONS (XII) # 10

(iv)

x 1 x2 2 [ x ] [ x ] 1 x 1 , even by graph of function f(x) = 2 x x 1

(v)

f(x) =

2x(sin x tan x ) x 2 1

if x = n, f(n) = 0 if x n

8.

x x = – 1

f(–x) = – f(x) odd function

(i)

f(x) = 1 –

cos2 x sin2 x – fx = 1 – f(x) 2 1 – tan x 1 – cot x fundamental period =

(ii)

(iii)

(iv)

sin2 x cos2 x – 1 cot x 1 tan x period of f(x) = L.C.M. (, ) = For fundamental period

f(x) = tan [ x ] : [x] 2n + 1 2 f (x) = 0 By graph fundamental period = 2 f(x) = log (2 + cos 3x) fundamental period of f(x) = fundamental period of (2 + cos 3x) (as log is a monotonic function) f(x) = en sin x + tan3x – cos (3x – 5) f(x) = sin x + tan3x – cos (3x – 5), sin x > 0

period 2 ; ,

2 3

2 = 2 Period of f(x) = L.C.M. 2 , , 3

(v)

x x x x x x x f(x) = sin x sin 2 sin 4 .... sin n –1 tan tan 3 tan 5 ... tan n 2 2 2 2 2 2 2

Period of f(x) = L.C.M. of 2, 23 , 25 ,......2n , 2, 23 .....2n = 2 n

(vi)

f(x) =

sin x sin 3 x cos x cos 3 x

2 2 period of f(x) = L.C.M. 2. , 2, = 2 3 3 For fundamental period

f(x + ) = 11.

f(x) =

sin ( x ) sin(3 x 3 ) – sin x – sin 3 x = cos( x ) cos(3 x 3) – cos x – cos 3 x

1 x

1 x2

f(x) = 0 at x = 1 ±

Fundamental period =

2

for x 2 1, 1 2 f is bijective function hence f is invertible.

RESONANCE

S OLUTIONS (XII) # 11

1 x 1 x2 or

or

=y x2y + x + (y – 1) = 0 x=

1 1 4 y( y 1) 2y

=

1 4y 4 y 2 1

1 4x 4x 2 1 , f–1(x) = 2x , 1

2y

x0 x0

as f (1) 0

n

f (a k ) = 16 (2

n

13.

k 1

or

or Now or 15.

(i)

(ii)

– 1)

f(a + 1) + f(a + 2) + ......... + f(a + n) = 16 (2n – 1) f(x + y) = f(x) . f(y) f(0) = 1, f(1) = 2 f(x) = 2x f(a + 1) + f(a + 2) + ........ + f(a + n) = 2a [2 + 4 + ......... + 2n] = 2a . 2(2n – 1) 16 = 2a + 1 or a=3 f(x) = Ax2 + Bx + C x and f(x) at x = 0, f(0) = C at x = 1, f(1) = A + B + C C is integer at x = –1, f(–1) = A – B + C f(1) + f(–1) = 2A + 2C 2A is also integer f(x) = A x(x – 1) + (A + B) x + C f(x) = 2A

C is integer A + B is also integer C is integer

x( x 1) + (A + B)x + C 2

x( x 1) is also an integer and 2A, (A+ B), C 2 f(x) is also an integer.

If x is an integer then

PART - II 2.

f(x) = here

1

x

1 cos 1 (2 x 1) tan 3 x

– 1 2x + 1 < 1

– 2 2x < 0

–1x 3x > –

2

or

x , 0 6

Domain : , 0 (–1, 0) , 0 6 6

RESONANCE

S OLUTIONS (XII) # 12

3.

1 x3 f(x) = sin 3 / 2 + sin(sin x ) + log(3{x} + 1) (x 2 + 1) 2x Domain : 3{x} + 1 1 or 0 x –1

and

6.

1 x 3

1 2x 3 / 2 – 2x 3/2 1 + x 3 2x 3/2 1 + x 3 + 2x 3/2 0 (1 + x 3/2)2 0 xR 1 + x 3 – 2x 3/2 0 or (1 – x 3/2)2 0 3/2 or 1–x =0 or x=1 Hence domain x f(x) = (sin–1x + cos –1x)3 – 3 sin–1x cos –1x (sin–1x + cos –1x) –1

=

3 3 1 3 2 – 3 sin–1x cos x = – sin–1 x + 3 (sin–1x)2 2 2 8 8 2 4

3 3 = + 8 2

2 2 2 1 1 3 3 3 3 1 sin x (sin x ) sin x – = + 2 16 4 32 32 2

maximum value of f(x) at x = – 1 8.

f maximum =

9 3 7 3 3 3 + × = 32 2 16 8

Here (2 – log2 (16 sin2x + 1) > 0

0 < 16 sin2x + 1 < 4

1 16 sin2x + 1 4

3 16 0 log2 (16 sin2x + 1) < 2

log

2

2 2 – log2 (16 sin x + 1) > 0

0 sin2x <

2

2 log

2

(2 – log2 (16 sin2x + 1)) > –

2y>– Hence range is y (– 2] 11.

(A)

f(x) = e1/2 n x = g(x) =

12.

D:x>0

x,D:x0

D : x ± (2n +1)

2

(B)

tan–1 (tan x) = x

(C)

cot–1 (cot x) = x D : x ± n f(x) = cos 2x + sin4x = cos 2x + (1 – cos 2x)2 = 1 – cos 2x + cos 4x = sin2x + cos 4x g(x) = sin2x + cos 4x

(D)

f(x) =

|x| , D:x0 x g(x) = sgn (x), D : x R

f(6{x}2 – 5{x} + 1) (3{x} – 1) (2{x} – 1) 0

13.

x,

f((3{x} – 1) (2 {x} – 1)) or

1 1 {x} , 3 2

x

n

1 1 n 3 , n 2

R+ 0 , 2 g(x) = 2x – x2 R R f(g(x)) = cot–1 (2x – x2), where x (0, 1]

f(x) = cot–1x

hence f(g(x)) , 4 2

RESONANCE

S OLUTIONS (XII) # 13

20.

21.

25.

1 1 2 1 f x = x + x + [x + 1] – 3 x + 15 3 3 3 3 1 2 = x + x + [x] – 3x + 15 = f(x) 3 3 f(x) = |x – 1| f : R+ R x g(x) = e , g : [–1, ) R fog(x) = f[g(x)] = |ex – 1| D : [–1, ) R : [0, )

f(x) = (A) (B) (C)

(D)

fundamental period is 1/3

sin( [ x ]) =0,x { x} By graph fundamental period is one f(–x) = 0 = f(x) even function Range y {0} { x } y = sgn sgn – 1, x { x } y = sgn (1) – 1 y=1–1 y = 0, x Identical to f(x)

28.

f(x) = sin x + tan x + sgn (x2 – 6x + 10) f(x) = sinx + tan x + sgn ((x – 3)2 +1) f(x) = sin x + tan x + 1 period = L.C.M. (2, ) = 2 fundamental period = 2

29.

f:NI

n – 1 , n odd f(n) = 2 n – , n even 2 For For

n odd numbers f(n) 0, 1, 2, 3, ...... n even numbers f(n) –1, –2, –3, ...... range I

f(n) is one -one onto function.

EXERCISE # 3 2.

x = 2

(A)

sin–1 x + cos–1

(B)

2 sin–1 x + cos–1 1 x = 0

(C)

x [0, 1]

x [–1, 0]

x [0, )

1 x tan–1 x + tan–1= tan–1 1– x

1 – x2 g 2 = 2h(x) 1 x

(D)

cos–1 1 x 2 = – sin–1(x)

cos

–1

1 – x2 –1 1 x 2 = 2 tan x

1 x h(x) + h (1) = h 1– x x (–, 1)

RESONANCE

S OLUTIONS (XII) # 14

Comprehension # 2 (6, 7, 8) Period of e

x tan 4

is 4

(1 2 [ x]) =0 2

cos

xR

[x]

Period of sin 2 is 4 then y =

8 2[ x ] [ x ] 2

p=4

[x]2 – 2[x] – 8 0 – 2 [x] 4 q=–2 , r=5 r–q–1=5+2–1=6 x 2 , f2 (x) = 2 x ,

Period of f(x) is 4

– [x]2 + 2 [x] + 8 0

i.e.,

([x] – 4) ([x] + 2) 0 –2 xx–1 f(x + 1) < f(x – 1) as f(x) is M.D. g(f(x + 1) < g(f(x – 1)) as g(x) is M..

Let

f(x) =

, x

f(x) = Let

f(x) 0

.

g(x) = x sec2x – tan x g(x) = 2x sec2x tanx > 0 x>0 g(x) > g(0), g(x) > 0 x1 < x2 f(x1) < f(x2) <

RESONANCE

f(x) > 0

f(x) is M.I.

< S OLUTIONS (XII) # 102

Section (D) : D-2.

f(x) = 3x 3 f(x) = 0 x=0 x = – 2, f(–2) = – 8 x = 0, f(0) = 0 x = 2, f(2) = 8 Minimum = – 8, maximum = 8 (ii) f(x) = cos x – sin x (i)

f(x) = 0

x=

x = 0, x=

f(0) = 1 ,

x = ,

f

=

f() = – 1

Minimum = –1, Maximum = (iii) f(x) = 4 – x f(x) = 0 x=4 x = – 2,f(–2) = – 10 x = 4, f(4) = 8 x=

,f

=

Minimum = –10, (iv)

Maximum = 8

f(x) = cos x – sin 2x f(x) = 0

x = 0, f(0) = Minimum = D-6_.

cos x = 0, sin x =

x=

, x=

x=

f

=

x=

, f

=

, Maximum =

Let No. of children of john & anglina = y x + (x + 1) + y = 24 y = 23 – 2x Number of fights F = x(x + 1) + x(23 – 2x) + (x + 1) (23 – 2x) F = – 3x2 + 45x + 23 But 'x' wil be integral.

D-8.

,

2

2

= 0 – 6x + 45 = 0

x = 7.5

check x = 6 or x = 7, F = 191

2

R +r =h R2 = h2 – r2 volume of cylinder , V = R2 (2h) = (2h) (

)2

= 2 (r2 – h2) + 2h(–2h) = 0

r2 = 3h2

h=

< 0 at h =

Vmax = 2

RESONANCE

= S OLUTIONS (XII) # 103

D-10.

2 + 2r = 440 A = 2r = – 2r2 + 440r = – 4r + 440 = 0

D-12.

at

r=

Let Let

x, y be dimensions of rectangle. 2x + 2y = 36. V be volume sweeped V = x 2y V = x 2(18 – x) = x.3.(12 – x)

At

x = 12, V has maximum value

y=6

Section (E) : E-2.

Let (h, k) be point of inflection h sin h = k y = sin x + x cos x y = cos x + cos x – x sin x y = 0 2 cos h – h sin h = 0 2 cos h = k sin2 h + cos 2 h = 1 = 1 4k 2 + h2k 2 = 4h2

+

...(1)

...(2)

locus y2 (4 + x 2) = 4x 2

Section (F) : F-2.

F-4.

Let

f(x) = 3x2 + px –1

f(x) = x3 +

f(x) satisfies conditions in Rolle's theorem 3x2 + px –1 = 0 has atleast one root in (–1,1).

–x+c

f(–1) =

+ c = f(1)

f(c) = 0 for atleast one c (–1,1)

Let h(x) = f(x) g(x) h(a) = 0 = h(b) By Rolle’s theorem on [a, b] h(x) = 0, for at least one c (a, b). f (c) g(c) + f(c) g(c) = 0

PART - II Section (A) : A-1.

V=

V=

=

77 × 103 =

× 70 × 70 ×

( 1 litre = 103 c.c.)

= 20 cm/min.

RESONANCE

S OLUTIONS (XII) # 104

A-3_.

= Let x = 25 and x = 0.2 such that f(x) = f (x) =

f(x + x) = f(x) + f(x). x

=

+

.x

= A-4_.

+

× 0.2 =

=5+

= 5 + 0.02 = 5.02

V = x3 x = (3x 2) x = (3x 2) (0.04x) = 0.12x 3m 3

V =

Section (B) : B-5*.

2y3 = ax 2 + x 3 6y2

= 2ax + 3x 2

=

=

Tangent at (a, a) is 5x – 6y = – a =

,

2 + 2 = 61

= = 61

a2 = 25.36 a = ± 30 B-8.

P1 : y2 = 8x C1 : x 2 + (y + 6)2 = 1

Equation of normal of parabola y = mx – 2am – am 3 if passes through (0,–6) –6 = – 2am – am 3 a=2 3 = 2m + m 3 3 m + 2m – 3 = 0 m = 1. Point on parabola (am 2 , – 2am) (2, –4).

Section (C) : C-2.

5x 4 – 3

f(x) =

It is sufficient to solve for p, the condition f(x) 0

xR

5x 4 – 3 0 x R

RESONANCE

S OLUTIONS (XII) # 105

Case -

1–p1 Inequality holds true. 1–p>0 p 0 and for x cos x – x < 0,

RESONANCE

f(x) has maxima at x 0

S OLUTIONS (XII) # 106

D-8*.

f(x) =

,x>0

=

x > 0.

f(x) is decreasing On

x > 0.

, greatest value is

f(

)=

–

n

f

=

n

and least value is

.

D-10. S = 2rh = 2H

= 2H

Maximum at r =

D-13.

Let d be distance between (k, 0) and any point (x, y) on curve. d= ( y2 = 2x – 2x 2).

d=

Maximum d =

Maximum d =

Section (E) : E-3.

x=1

a=–

E-4.

3=a+b

=0

,

b=

3a + b = 0

f(x) = n(x – 2) –

f(x) =

=

=

=

.

As n(x – 2) is defined when x > 2 f(x) is M.. for x (2, )

f –1(x) is M.. wherever defined

Also

f(x) is always concave downward

f(x) =

RESONANCE

0

If x > 0, (x) > 0 Hence (x) is increasing As we know ex x + 1 (ex) (x + 1) ex + 13.

x+1+

Let x > –1 Consider f(x) = (1 + x)n(1 + x) – tan–1 x f(x) = n(1 + x) + 1 – f(x) =

>0

f(x) is increasing f(x) < 0

For x < 0, f(x) < f(0) f(x) decreasing f(x) > f(0)

(1 + x)n(1 + x) – tan–1 x > 0 For x > 0,

f(x) > f(0) f(x) > 0

f(x) > 0

n(1 + x) > f(x) is increasing

f(x) > f(0)

f(x) > 0

n(1 + x) > Hence larger of these is n(1 + x). 15.

f(x) = a0 + a1x + a2x 2 + a3x 3 + a4x 4 + a5x 5 + a6x 6 a0 = a1 = a2 = a3 = 0 = e2

a4 = 2

f(x) = 2x 4 + a5x 5 + a6x 6 f(x) = x 3 ( 8 + 5a5x + 6a6x 2) f(1) = 0, f(2) = 0 , 17.

f(x) = 2x 4

xy = 18 Area of printed space

=

=

Maximum when x=

RESONANCE

y= S OLUTIONS (XII) # 109

21.

f(x) = 0 sin x=

=0

= n

,nN

x = .......,

,

, .......

,

Consider interval

, 1.

=0=

By Rolle’s theorem f(x) vanishes at least once in Infinite number of such intervals are there. Hence f(x) vanishes at infinite number of points in (0, 1) 24.

Let g(x) =

, x [a, b]

By Rolle’s theorem, g’(x0) = 0 =0

25.

f(x0) =

Let f(x) = (x + a) – (x) f(x) = (x + a) – (x) + k f(0) = (a) – (0) + k f(2a) = (3a) – (2a) + k f(0) = f(2a) By Rolle's theorem on [0, 2a], f(c) = 0 for at least one c (0, 2a) (x + a) = (x) has at least one root in (0, 2a)

PART - II 3.

f(x) =

>0

f(x) increasing hence g(x) is also increasing function 7.

Let

be quantity y=

y2 = y

=

=

=a 9.

= = a

f(x) = ; g(x) = for a > 1, a 1 and x R n a h(x) = n f(x) + ng(x)

(n a) h(x) =

h(x) =

na +

n a

+ |x|

h(x) = a sgn x Now h(–x) = a|–x| sgn (–x) = –h(x) h(x) is an odd function Also graph of h(x) is It is clear from the graph that h(x) is an increasing function

RESONANCE

S OLUTIONS (XII) # 110

12.

f(x) =

f(x) = For 13.

16.

0 < x < 1,

tan

f(1–) f(1) and f(1+) f(1) – 2 + log2 (b2 – 2) 5

f (x) > 0

0 < b2 – 2 128

f(x) is increasing.

2 < b2 130

2 = h2 + x 2 Area of base (triangle) is

3x =

a

Volume V = h

. 4 . 3.x 2 = 3

=h

=3 18.

a2

(2 – 3h2)

h (2 – h2)

V is maximum when h =

Maximum of f(x) is Given expression is f(x 1) + f(x 2)

20.

.

f(x 1)

f(x 2)

f(x 1) + f(x 2)

f (x) = 1 +1– 2– At x = 0, f(x) is least. Least value = f(0) = 1

23.

f(x) = m – n is odd. f (x) < 0

25.

>0

x (–, 0)

f (x) > 0

x (0, )

(x) = (3(f(x))2 – 6(f(x)) + 4)f(x) + 5 + 3 cos x – 4 sin x 5–

5 + 3 cosx – 4 sin x 5 +

adding (3(f(x))2 – 6(f(x)) + 4)f(x) (3(f(x))2 – 6(f(x)) + 4)f(x) (x) (3(f(x))2 – 6(f(x)) + 4)f(x) + 10 3(f(x))2 – 6f(x) + 4 = 3 (f(x) – 1)2 + 1 > 0 2 (3(f(x)) – 6(f(x)) + 4)f(x) 0 when ever f(x) is increasing. (x) 0 (x) is increasing, when ever f(x) is increasing. If f(x) = – 11 then (3(f(x))2 – 6f(x) + 4) f(x) + 10 = – 33 (f(x) – 1)2 – 1 < 0 (x) < 0 (x) is decreasing.

RESONANCE

S OLUTIONS (XII) # 111

28.

f(x) = f(x) = 0 at x = 0, x = – 3, x = 1 so at x = 0, f(x) has local minima. and at x = –3, x = 1 ; f(x) has local maxima f(1) =

,

f(– 3) =

. f(–3) < 0, f(1) > 0 and f(x) 0

f(x) is undefined at point(s) in (–3, 1). Hence f(x) has no absolute maxima.

EXERCISE # 3 PART - I 2.

(A) (B)

f(x) is continuous and differentiable f(0) = f() Hence condition in Rolle’s theorem and LMVT are satisfied. f(1–) = –1, f(1) = 0, f(1+) = 1 f(x) is not continuous at x = 1, belonging to Hence, atleast one condition in LMVT and Rolle’s theorem is not satisfied

(C)

f’(x) =

(x – 1)–3/5, x 1

(D)

At x = 1, f(x) is not differentiable. Hence at least one condition in LMVT and Rolle’s theorem is not satisfied. At x = 0

L.H.D. =

3.

=

= –1

R.H.D. = 1 At x = 0, f(x) is not differentiable Hence at least one condition in LMVT and Rolle’s theorem is not satisfied. (A) Let PQ = x Then BP =

PS =

tan60º =

area A of rectangle =

=

=–

(4 – x) x

(4 – 2x) = 0

x=2

0

g(x) = –

× f(x).g(x)

g(x) > 0 g(x) = f –1 (x) concave upward

15.

f(x) =

17.

(1 – nx) f(x) 0, when x e f(x) is decreasing function, when x e >e f() < f(e) 1/ < e1/e e > e Statement-1 is True, Statement-2 is False

f ’ (x) = 50x49 – 20x19 = 10x19(5x30 – 2) x = 0 is stationary point. Statement-2 is ture. f(0) = 0 =

–

0 increasing

gmax = g(1) = e +

h(x) = x2

+

+

hmax = h(1) = e +

RESONANCE

h(x) = 2x ,

+ 2x3 so

– 2x

= 2x

>0

a=b=c S OLUTIONS (XII) # 120

26.

(A)

Re

= Re

= –1 y 1 =

1 or

= Re

= Re (–1/y) =

– 1

Alternate

Re

= Re

= Re

= Re

= Re

= Re as –1 sin 1 (– , 0 ) (0, ) (B)

(C) (D)

–1

1

–1

0

1

–1

0

0

0

1

– 1 0

t (– , –9] [–1 , 1] [9, ) x (– , 0) [2 , ) f() = 2 sec2 f() 2 f(x) = x3/2 (3x – 10)

f ’(x) = x3/2 3 +

0

f() [2, )

x1/2 (3x –10)

asf ’(x) 0 0

27.

– 15 0

x 2

3x +

– 15 0

x [2, )

f(x) = x4 – 4x3 + 12x2 + x – 1 f(x) = 4x3 – 12x2 + 24x + 1 f(x) = 12x2 – 24x + 24 = 12 (x2 – 2x + 2) > 0 xR f(x) is S.I. function Let is a real root of the eqution f(x) = 0 f(x) is MD for x (– , ) and M.I. for x (, ) where < 0 f(0) = – 1 and < 0 f() is also negative f(x) = 0 has two real & distinct roots.

RESONANCE

S OLUTIONS (XII) # 121

28.

p = (x – 1) (x – 3) = (x2 – 4x + 3) p(x) = (x3/3 –2x2 + 3x) + p(1) = 6 6 = (1/3 – 2 + 3) + 6 = (1/3 + 1) + 18 = 4 + 3 ...(i) p(3) = 2 2 = (27/3 – 2 × 9 + 9) + 2= =2=3 p(x) = 3(x – 1) (x – 3) p(0) = 3(–1)(–3) =9

29.

f(x) = |x| + |x2 – 1|

f(x) =

f(x) =

PART - II 1.

Let

f(x) = x +

f(x) = Minimum occures at x = 1. 2.

3.

f(x) = 6x2 – 18ax + 12a2 f(x) = 0 x = a, 2a f(x) = 12 x – 18a f(a) = – 6a < 0 and f (2a) = 6a > 0 p = a, q = 2a p2 = q a = 2 (a > 0) Let

f (x) = ax2 + bx + c

f(x) =

a2 – 2a = 0

a = 0, 2

f(0) = d = f(1) (2a + 3b + 6c = 0) By Rolle's theorem, at least one root of f(x) = 0 lies in (0, 1) 4.

y2 = 18x Differentiating w.r.t. t

...... (1)

From equation (1), x =

RESONANCE

2y.2 = 18

y=

Required point is S OLUTIONS (XII) # 122

5.

x = a (1 + cos ), y = a sin

Equation of normal at point (a(1+ cos ), a sin ) is y – a sin =

(x – a (1 + cos ))

y cos = (x – a)sin It is clear that normal passes through fixed point (a, 0) 6.

y = x2 – 5x + 6 = 2x – 5

= – 1,

product of slopes = –1

=1

angle between

tangents is 7.

Let f1(x) = x3 + 6x2 + 6 Increasing in (–, – 4] Let

f2 (x) = 3x2 – 2x + 1

f1(x) = 3x (x + 4)

f2(x) = 6x – 2

Not increasing in Let

f3(x) = 2x3 – 3x2 – 12x + 6

f3(x) = 6x2 – 6x – 12 = 6(x +1) (x – 1)

Increasing in [2, ) Let f4(x) = x3 – 3x2 + 3x + 3 Increasing in (–, ) 8.

V=

4 (10 + r)2

, 0 r 15 = – 50. = – 50

=

9.

Any point on ellipse is P(x, y) = (a cos , b sin ) Area of rectangle = 4ab cos sin = 2ab sin 2 Maximum area = 2ab.

10.

By LMVT and f(x) 2

f(x) = 3(x –1)2 0

= f(x),

x (1, 6)

11*.

(where r = 5)

2

2

f(6) 8.

Equation of normal is y – a(sin – cos ) =

(x – a (cos + sin ))

x cos + y sin = a. Distance from origin = |a| (constant) Slope of normal = –

RESONANCE

= tan

angle made with x-axis is

+ . S OLUTIONS (XII) # 123

12.

f (x) = f (x) changes sign as x crosses 2. f(x) has minima at x = 2.

13.

= f(c) 1 < c < 3

c =

14.

f(x) =

15.

Graph of y = x3 – px + q cuts x-axis at three distinct pints

(cos x – sin x)

=0 16.

= 2 log3e

x =

Maxima at x = –

, minima at x =

Condition for shortest distance is slope of tangent to x = y2 must be same as slope of line y = x +1.

=1

y=

,

x=

,x–y+1=0

Shortest distance =

17.

P (x) = 4x 3 + 3ax 2 + 2bx + c and P(0) = 0 c=0 P (x) = x(4x 2 + 3ax + 2b) D = 9a2 – 32b < 0

b>

>0

(P(x) = 0 has only one root x = 0)

P (– 1) < P (1) a>0 P(x) has only one change of sign. x = 0 is a point of minima. P(–1) = 1 – a + b + d, P(0) = d P(1) = 1 + a + b + d P(–1) < P(1), P (0) < P(1), P (–1) > P(0) P(–1) is not minimum but P(1) is maximum.

18.

y=x+ y = 1 –

=0

y=2+

=3

x3 = 8

x=2

(2, 3) is point of contact Thus y = 3 is tangent Hence correct option is (3)

RESONANCE

S OLUTIONS (XII) # 124

19.

f(x) = 1 f(–1) = k + 2 f(x) = k + 2 f has a local minimum at x = – 1 1k+2k+2 possible value of k is – 1 Hence correct option is (3)

20.

ex + 2e–x 2

f(–1+) f(–1) f(–1–) k–1

(AM GM)

f(x) > 0

so statement- 2 is correct

As f(x) is continuous and

f(c) =

21.

belongs to range

for some C.

of f(x),

Hence correction option is (4).

y–x=1 y2 = x 2y

=1

=

y =

= 1 , x=

tangent at

y=

y=x+

y–x=

distance =

22.

=

=

f(x) =

In right neighbourhood of ‘0’ tan x > x

RESONANCE

S OLUTIONS (XII) # 125

In left neighbourhood of ‘0’ tan x < x as(x < 0) at x = 0 , f(x) = 1 x = 0 is point of minima so statement 1 is true. statement 2 obvious

23.

r3

V=

4500 =

= 4r2

45 × 25 × 3 = r3

r = 15 m after 49 min = (4500 – 49.72) =972 m3 972 =

r3

r3 = 3×243 = 3× 35 r=9 72 = 4 × 9 × 9

=

24.

f '(x) = at x = – 1

+ 2bx + a –1 – 2b + a = 0 a – 2b = 1

at x = 2

+ 4b + a = 0

a + 4b =

On solving (i) and (ii)

f '(x) =

...(i)

=

a=

...(ii)

, b=

=

So maxima at x = – 1, 2

RESONANCE

S OLUTIONS (XII) # 126

SOLUTION OF TRIANGLE EXERCISE # 1 PART - I Section (A) : A-1.

(i)

L.H.S. = a sin (B – C) + b sin (C – A) + c sin (A – B) = k sin A sin (B – C) + k sin B sin (C – A) + k sin C sin (A – B) = k (sin2 B – sin2 C) + k (sin2C – sin2 A) + k (sin2 A – sin2 B) = 0 = R.H.S.

(ii)

L.H.S. =

first term =

= = k 2 sin (B + C) sin (B – C) = k 2 (sin2 B – sin2 C) = k 2 (sin2 C – sin2 A)

Similarly

= k 2 (sin2 A – sin2 B)

and

L.H.S. = k 2 (sin2 B – sin2C + sin2C – sin2A + sin2 A – sin2 B) = 0 = R.H.S. (iii) L.H.S. = 2bc cos A + 2ca cos B + 2ab cos C = b2 + c 2 – a2 + a2 + c 2 – b2 + a2 + b2 – c 2 = a2 + b2 + c 2 = R.H.S

(iv) L.H.S. = a2

– 2ab

= a2 + b2 – 2ab cos C = a2 + b2 – (a2 + b2 – c 2) = c 2 = R.H.S. (v) L.H.S. = b2 sin 2C + c2 sin 2B = 2b2 sin C cos C + 2c 2 sin B cos B = 2k 2 sin2 B cos C sin C + 2k 2 sin2 C sin B cos B = 2k 2 sin B sin C [sin B cos C + cos B sin C] = 2(k sin B) (k sin C) sin (B + C) = 2bc sin A (vi) R.H.S =

(b = ksin B, c = ksin C)

c = a cos B + b cos A, b = c cos A + a cos C

=

=

=

A–4.

= L.H.S.

= sin(B + C) sin(B – C) = sin(A + B) sin(A – B) sin2 B – sin2 C = sin2 A – sin2 B 2 sin2 B = sin2 A + sin2 C 2b2 = a2 + c 2 a2, b2, c 2 are in A.P.

RESONANCE

S OLUTIONS (XII) # 127

A–7.

x 3 – Px 2 + Qx – R = 0

a2 + b2 + c 2 = P a 2b 2 + b 2c 2 + c 2a 2 = Q a2b2c 2 = R abc =

+

+

=

[a2 + b2 + c 2] =

Section (B) : B–1.

(i)

L.H.S. = 2a sin2 = = = =

(ii)

+ 2 c sin2

a(1 – cos c) + c(1 – cos A) a + c – (a cos C + c cos A) a+c–b R.H.S.

L.H.S. =

+

=

.

+ +

= (iii)

.

+

=

.

.

L.H.S. = 2bc(1 + cos A) + 2ca(1 + cos B) + 2ab(1 + cos C) = 2bc + 2ca + 2ab + 2bc cos A + 2ca cos B + 2 ab cos C + a2 + b2 + c 2 = (a + b + c)2

=2 = R.H.S. (iv)

L.H.S. = (b – c)

(b – c) cot

+ (c – a)

+ (a – b)

= k(sin B – sin C)

= 2k cos

sin

= 2k sin

sin

= k [cos C – cos B] similarly (c – a) cot and (a – b) cot

= k[cos A – cos C]

= k[cos B – cos A]

L.H.S. = k[cos C – cos B + cos A – cos C + cos B – cos A] =0 = R.H.S.

RESONANCE

S OLUTIONS (XII) # 128

(v) L.H.S. = 4 (cot A + cot B + cot C) = 4 = 2bc cos A + 2 ca cos B + 2ab cos C = a2 + b2 + c 2 = R.H.S. (vi) L.H.S. =

cos

. cos

. cos

= =

= = R.H.S.

B–3.

Let ADB = we have to prove that tan = if we aply m – n rule, then (1 + 1) cot= 1.cot C – 1.cotA. =

–

=

–

= =

[2(a2 – c 2)]

2cot =

tan =

Section (C) : C–2.

(i)

r. r1 .r2 .r3 =

(ii)

r1 + r2 – r3 + r = 4R cosC

= 2

L.H.S. = =

=

=

RESONANCE

S OLUTIONS (XII) # 129

=

=

=c

=

L.H.S. =

(iii)

=

= R.H.S.

=

(s – a + s – b + s – c) =

·

=

=

=

=

=

=

=

= 24 sq. cm 2s = 24 r1, r2, r3 are in H.P.

=

=

=r

=r

.... (i) s = 12 .... (ii)

are in A.P..

[4s 2 – 2s(a + b + c) +a2] =

R.H.S. =

similarly we can show that

= 4RcosC

L.H.S. =

=

C–4.

=

(s + s – a + s – b + s – c)2 = 4

(v)

=

L.H.S. =

=

cos C =

[s 2 + (s – a)2 + (s – b)2 + (s – c)2] =

(iv)

=

=

a, b, c are in A.P. 2s = 24 a + b + c = 24 3b = 24 b=8

are in A.P..

2b = a + c

a + c = 16

But = = 24 × 24 = 12 × (12 – a) × 4 × (12 – c) 2 × 6 = 144 – 12 (a + c) + ac 12 = 144 – 192 + ac ac = 60 and a + c = 16 a= 10, c = 6 or a = 6, c = 10 and b = 8

RESONANCE

S OLUTIONS (XII) # 130

Section (D) : D–1.

(i)

,=

=

, =

=

R.H.S. =

=

=

L.H.S. = R.H.S.

(ii)

=

=

=

R.H.S. =

=

= L.H.S.= R.H.S.

PART - II Section (A) : A–4.

(b + c)2 – a2 = kbc

(a + b + c) (b + c – a) = kbc

b2 + c2 – a2 = (k – 2) bc

In a ABC –1 < cos A < 1

–1 <

b

=

= cos A

0, c = 5x + 5y is the largest side C is the largest angle. Now cos C =

=

r2 > r3

>

>

s – a < s – b < s – c –a < –b < –c;

3.

tan

=

; sin

r+R=

RESONANCE

a>b>c

=

r+R=

.cot

S OLUTIONS (XII) # 143

4.

a

=

=

a + c = 2b 5.

a + b + c = 3b.

a, b, c are in A.P.

AD = 4

AG =

Area of ABG =

=

Area of ABC = 3(Area of ABG)

6.

cos =

7.

C = /2

×4=

×

×

× AB × AG sin 30º

×

=

Sin 60º =

AB =

=

= 120º

=

=–

r = (s – c) tan

C = 90º

r = s – 2R 2r + 2R = 2 (s – 2R) + 2R. = 2s – 2R = (a + b + c) –

C = 90º

=a+b+c–c =a+b

8.

are in H.P.

are in A.P.

9.

a,b,c are in A.P.

= cos

Let cos

=

As

for some n 3, n N

cos

cos

cos

3 n < 4, which is not possible so option (2) is the false statement so it will be the right choice Hence correct option is (2)

RESONANCE

S OLUTIONS (XII) # 144

ADVANCE LEVEL PROBLEMS PART - I 1.

From figure, AD = c sin B Hence number of triangle is 0 if b < c sin B one triangle for b = c sin B two triangles for b > c sin B

2.

C = 60° Hence c2 = a2 + b2 – ab

3.

=

= 2 cos

Using properties of pedal triangle, we have

MLN = 180° – 2A LMN = 180° – 2B MNL = 180° – 2C

Hence the required sum =

sin2A + sin2B + sin2C

=

4.

5.

4sinA sinB sinC

From figure, we can observe that

OGD is directly similar to PGA

BD = s – b, CE = s – c and AF = s – a Hence BD + CE + AF = s

6.

, as cos

= cos

A = B, in either case

7.

, Using cosine rule in

ABO, we get

h=

8.

In

ABD,

RESONANCE

S OLUTIONS (XII) # 145

Comprehension # 1

9.

10.

+

+

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

k = 2R

cot A + cot B + cot C =

=

(b2 + c2 + a2) =

(b2 + c2 – a2 + c2 + a2 – b2 + a2 + b2 – c2)

=

=

.

=

11.

=

k=

=6

Comprehension # 2 (12 to 14) 12.

PG =

AD

= =

= PG = =

.ab sin C

or

b sin C

( =

ac sin B)

ac sin B c sin B

13.

Area of GPL =

and

Area of ALD =

(PL) (PG) (DL) (AD)

=

PL =

=

DL and PG =

AD

=

14.

Area of PQR = Area of PGQ + Area of QGR + Area of RGP ...(1)

Area of PGQ = = =

RESONANCE

×

×

PG.GQ.sin(PGQ) AD ×

×

BE sin ( – C) sin C S OLUTIONS (XII) # 146

=

×

=

bc sin A ×

ac sin B × sin C

sin A.sin B.sin C

Similarly Area of QGR =

sin A.sin B .sin C and Area of RGP =

sin A.sin B.sin C

From equation (1), we get (a2 + b2 + c 2) sin A.sin B.sin C

Area of PQR = 15.

In

CDB ,

=

Also from same triangle 16.

17.

BD =

cosAcosB + sinAsinBsinC = 1

(cosA – cosB)2 + (sinA – sinB)2 + 2sinAsinB(1 – sinC) = 0

a:b:c=1:1:

A = B & C = 90°

We have

18.

=

a:b:c=5:4:3

from figure, OO = ON – ON = R –

ZO = ZM + = from

RcosA +

OZO, using Pythagorous theorem,

we get (R –

)2 = (RcosA +

)2 +

=

PART - II 1.

from

from

ABC ,

=

AB = 2Rsin(A + )

ACB,

=

AC’ = 2Rsin( – A)

=

4RcossinA = 2acos

similarly CA = 2bcos =

RESONANCE

area

BC = 2R(sin (A + ) – sin( – A))

ABC = =

4cos2. S OLUTIONS (XII) # 147

2.

c2 – 2bc cosA + (b2 – a2) = 0 c1 & c2 are roots of this quadratic equation Hence (c1 – c2)2 + (c1 + c2)2tan2A = 4a2

3.

Area =

=

= =

2Rs

= 4.

We know that OA = R, HA = 2RcosA and applying Appoloneous theorem to

AOH, we get

2.(AQ)2 + 2(OQ)2 = OA2 + (HA)2

2.(AQ)2 = R2 + 4R2cos2A –

5.

=

+

using sine rule, diameter of required circle =

6.

=

= 20

radius = 10

L.H.S. =

(a2 (b + c – a) + b2 (c + a – b) + c2 (a + b – c))

=

=

=

abc

=

4R

RESONANCE

S OLUTIONS (XII) # 148

7.

from the parellelogram ABAC, AA = 21 , from

AAC, AA < b + c 21 < b + c

similarly 22 < c + a

...(1) ...(2)

and 23 < a + b ...(3) (1) + (2) + (3) gives 1 + 2 + 3 < 2s 8.

ZXY =

and

Area of

=

Area of

Area of XYZ = 2R2 cos

9.

cos

cos

=

Drop a perpendicular from the apex P to the base

ABC.

The foot of perpendicular is at circum centre O of

ABC

Using given data, we get from, right angle

POB, we get

= = 8.83 m 10.

from cyclic quadrileteral CQFP, we get

from cyclic quadriletral AQMF, we get FQM = FAM = 90º – B

AQM = 90º + 90º – B = 180º – B

P, Q, M are collinear

similarly P, Q, N are collinear hence, P, Q, M, N are collinear

RESONANCE

S OLUTIONS (XII) # 149

View more...
MATHEMATICS CLASS : XII Preface

1.

Functions Exercise

2.

79 - 100

Application of derivatives Exercise

6.

54 - 78

Method of differentiation Exercise

5.

29 - 53

Continuity and derivability Exercise

4.

01 - 28

Limits Exercise

3.

Page No.

101 - 126

Solution of triangle Exercise

127 - 149

© Copyright reserved 2012-13. All rights reserved. Any photocopying, publishing or reproduction of full or any part of this study material is strictly prohibited. This material belongs to only the enrolled student of RESONANCE. Any sale/resale of this material is punishable under law. Subject to Kota Jurisdiction only.

FUNCTIONS EXERCISE # 1 PART - I Section (A) : A–3.

(i)

f(x) =

f(x) =

x 3 5x 3 x2 1 x 3 5x 3 ( x 1)( x 1)

Division by zero is undefined Domain x R – {1, –1}

x±1 x (–, –1) (–1, 1) (1, )

(ii)

f(x) =

sin1 x x For sin–1x, x [–1, 1] and division by zero is undefined x 0

Domain x [–1, 0) (0, 1]

1 (iii)

(iv)

(v)

f(x) =

x | x |

for function to be defined x + |x| > 0 for x > 0, x + |x| = 2x > 0 for x 0, x + |x| = 0 Domain is x (0, ) f(x) = ex + sin x Domain x R as there is no restriction for exponent of e.

1 f(x) = log (1 x ) + 10

x2

1 – x > 0 and x + 2 0 x (– , 1) – {0} and x – 2

(vi)

f(x) =

and

1–x1 x [–2, 0) (0, 1)

x

3x 1 –1 1 2x + 3 sin 2

1 – 2x 0 and – 1

3x 1 1 2

1 2

and –

1 x1 3

Taking intersection

1 1 Domain x , 3 2

(vii)

f(x) = 2 sin

1

x

x

1

1

x

x (0, 1) (1, ) and 0 < x –

1 f(x) = logx log 2 x 1/ 2 In case of composite function in log. We start with outer log. x > 0, x 1 and

1/2

–

x2 – 1 x 1 and x > 2 (viii)

–1/3

1

1 >1 x 2 1 3 x (0, ) – {1} and 0 Range 0 < sin–1 x

(iv)

1

>1+

x 1

Range y (0, 1]

x (0, 1]

2

– < n (sin–1x) n 2 Inequality doesn't change as n is increasing function f(x) = 2 – 3x – 5x2 Domain x R Method 1 y = – 5x2 – 3x + 2 opening downward parabola 0

D Range y , 4a

49 y , 20 Method 2 5x2 + 3x + (y – 2) = 0

D0 (v)

– D/4a 2

9 – 20 (y – 2) 0

20y – 49 0

y

49 20

f(x) = 3 |sin x| – 4|cos x| f(x) is a periodic function with period . So analysis is limited in [0, ]

, |sin x| = 1, |cos x| = 0 2 fmin = 3.0 – 4.1 = – 1 at x = 0, |sin x| = 0, |cos x| = 1 fmax = 3.1 – 4.0 = + 3 at x =

(vi)

f(x) =

sin x

Range y [–4, 3]

cos x

1 tan x 1 cot 2 x f(x) = sin x |cos x| + cos x |sin x| periodic period = 2 sin 2x 0 f(x) = sin 2x 0

+

2

x 0, 2 , x , 2 , x , 2 3 , x , 2 2 ,

Range y [–1, 1]

RESONANCE

S OLUTIONS (XII) # 2

Section (B) : B–1.

(ii)

2 x 2 and g(x) = ( x ) Domain x R, Domain x [0, ) non-identical functions f(x) = sec(sec –1x) and g(x) = cosec (cosec –1x) Domain x (–, –1] [1, ), Domain x (–, –1] [1, ) f(x) = x g(x) = x Identical functions

(iii)

f(x) =

(i)

(iv)

B–5.

(i)

(ii)

(iii)

(iv)

f(x) =

1 cos x and g(x) = cos x 2 f(x) = |cos x| non-identical function f(x) = x and g(x) = enx, Domain x R+ Domain x R non-identical function f(x) = ex and g(x) = n x fog(x) = en x = x, x > 0 gof(x) = n ex = x, x R f(x) = |x| and g(x) = sin x fog(x) = f(sin x) = |sin x| gof (x) = g(|x|) = sin |x| f(x) = sin–1x and g(x) = x2 fog(x) = sin–1(g(x)) = sin–1x2 gof(x) = (f2(x)) = (sin–1x)2 f(x) = x2 + 2, g(x) =

fog(x) = g2(x) + 2 =

gof(x) =

B–6.

x x 1

x2 ( x 1)2

+2=

3x 2 4x 2 ( x 1)2

f (x) x2 2 = 2 f (x) 1 x 1

1 x 2 , x 1 f(x) = 1 x , 1 x 2 g(x) = 1 – x, – 2 x 1

1 g2 , g( x ) 1 x [0,1] fog(x) = 1 g( x ) , 1 g( x ) 2 x [–1,0)

1 (1 – x )2 fog (x) = 1 (1 – x )

,

x [0,1]

, x [–1, 0)

2 – 2x x 2 , x [0,1] fog (x) = 2 – x , x [–1, 0)

Section (C) : C–1.

(i)

y = |(x + 2) (x + 3)| many - one function

(ii)

y = |nx| many - one function

RESONANCE

S OLUTIONS (XII) # 3

(iii)

f(x) =sin 4x, x – , 8 8

2 one-one function period =

1 1 , x (0, ) x x many one function

(iv)

f(x) = x +

(v)

f(x) =

1 –1

1 – e x 1 –1 .

e x f =

2 1– e

1 x2 0

1 –1 x

increasing function

Hence one - one

C–5.

3x 2 – cos( x ) 4 Hence many - one

(vi)

f(x) =

(vii)

f(x) = sin–1 x – cos–1 x = 2 sin–1 x –

(i) (ii) (iii)

f(x) = sin (x2 +1) f(x) = x + x2 f(x) = x – x3

f(– x) = f (x) = even function f(– x) = x2 – x f (x) or – f(x) Neither even nor odd function f(–x) = – x + x3 = – f(x) odd function

(iv)

a x – 1 f(x) = x x a 1

ax – 1 f(–x) = – x – x a 1

even function

monotonically increasing. 2

a x – 1 = f(x) even function f(–x) = x x a 1

(v)

f(x) = log (x +

(vi)

2 2 f(x) + f(– x) = log ( x x 1)(– x x 1) = log [(x2 + 1) – x2] = 0 hence odd function f(x) = sin x + cos x f(– x) = – sin x + cos x f(x) or – f(x) Neither even nor odd. f(x) = (x2 – 1) |x| f(–x) = f(x) even function.

(vii)

(viii)

x2 1 )

f(–x) = log (–x +

x2 1 )

| tan(tan 1 x ) | x 2 x [ 2 x ] 1 x 1 f(x) = sec(sec 1 x ) x x | x | x x0 f(x) = 3 0 x 1 x x

RESONANCE

x x x x0 f(x) = 3 0 x 1 x x

S OLUTIONS (XII) # 4

C–6.

(i)

x 2 – sin x , – 1 x 0 even extension of f(x) = f(–x) = – x e x , x –1

(ii)

– x 2 sin x , – 1 x 0 odd extension of f(x) = – f(– x) = x – e x , x –1

Section (D) : D–2.

(i)

f(x) = 2 + 3 cos (x – 2)

(ii)

f(x) = sin 3x + cos2 x + |tan x|

fundamental period = 2

2 period of f(x) = L.C.M. , , = 2 3 f(x + ) = – sin x + cos2 x + |tan x| f(x)

f(x) = sin

(iv)

f(x) = cos

for fundamental period

fundament period = 2

3x sin 2x – 5 7

period

10 ,7 3

10 , 7 = 70 period of f(x) = L.C.M. 3

Fundament period = 70

f(x) = [sin 3x] – |cos 6x|

period

2 2 period of f(x) = L.C.M. , = 3 3 3

Fundamental period =

(vi)

f(x)=

(vii)

f(x) =

2 3

3 2 3

1 fundamental period = 2 1 cos x sin12x

period of f(x) = L.C.M. , = 6 3 3

2

1 cos 6 x for fundamental period

fx = 6

(viii)

2 , , 3

x x + sin 4 3 period 8, 6 period of f(x) = L.C.M. (8, 6) = 24 fundamental period = 24

(iii)

(v)

period

sin12 x 6 1 cos 2 6 x 6

= f(x)

f(x) = sec3 x + cosec3 x period 2 Fundamental period = L.C.M. (2, 2) = 2

Fundament period =

6

2

Section (E) : E–1.

(i)

f:DR f(x) = 1 – 2–x

f (x) = 2– x n2 > 0 increasing function one one function

D : [x R), Range : (–, 1) codomain

function is not bijective

RESONANCE

f –1 does not exist S OLUTIONS (XII) # 5

(ii)

f(x) = (4 – (x – 7)3)1/5 f (x) =

1 (4 – (x – 7)3) 5

– 4/5

. (– 3 (x – 7)2) 0 decreasing function one one function

Lim f ( x ) – x

Lim f ( x )

x –

D:R

Range : R = codomain

onto function

function is bijective (invertible)

y = (4 – (x – 7)3)1/5 4 – y5 = (x – 7)3 x = 7 + (4 – y5)1/3 (iii)

f –1 (x) =

E–6.

f –1(x) = 7 + (4 – x 5)1/3

or

x=

f(x) = 1 ± xn

or

f(x) = 1 – x3 f(1) = – 3

f(x) = n x 1 x 2 D : x R, Range : R y = n x 1 x 2

E–5.

or

e y ey 2

e x e x 2

1 1 f(x) . f = f(x) + f x x f(3) = – 26 f(x) = – 3x2

f(x + y) = f(x) . f(y) and f(1) = 2 10

f (n) = f(1) + f(2) + ........... + f(10)

n 1

210 1 = 21 + 22 + 23 + ....... + 210 = 2 2 1 = 2046

PART - II Section (A) : A–2.

For domain – log0.3(x – 1) 0 log0.3(x – 1) 0 (x – 1) 1 x2 Taking intersection x [2, )

A–3.

f(x) = cot–1

x( x 3 ) + cos –1

for domain x(x + 3) 0 x (–, –3] [0, ) Taking intersection x {–3, 0}

RESONANCE

and and and

x2 + 2x + 8 > 0 (x + 1)2 + 7 > 0 xR

x 2 3x 1

and and

0 x 2 + 3x + 1 1 x 2 + 3x + 1 0 and

x 2 + 3x 0 x [–3, 0]

S OLUTIONS (XII) # 6

A–5.

f(x) = 4x + 2x + 1 Let 2x = t > 0, x R

f(x) = g(t) = t2 + t + 1,

t>0

2

1 3 g(t) = t + 2 4 2

1 1 t > 2 2

1 1 t > 2 4

2

1 3 t + >1 2 4

Range is (1, )

Section (B) : B–2.*

–1

f(x) = en(sec x ) = sec–1x, x (–, – 1] (1, ) –1 g(x) = sec x, x (–, – 1] [1, ) non-identical functions f(x) = tan (tan–1 x) = x, x R g(x) = cot (cot–1 x) = x, x R identical functions

(A)

(B)

1 x 0 f(x) = sgn (x) = 0 x 0 – 1 x 0

(C)

1 x 0 g(x) = sgn(sgn x) = 0 x 0 – 1 x 0

Identical functions f(x) = cot2 x . cos2 x, x R – {n }, g(x) = cot2 x – cos2 x = cot2 x (1 – sin2 x) = cot2 x. cos2 x Identical functions

(D)

B–4.

B–6.*

Domain of f(g(x)) Range of g(x) Domain of f(x) – 5 |2x + 5| 7 – 12 2x 2 f(x) =

1– x , 1 x fog(x) =

0x1

n I x R – {n },

n I

0 |2x + 5| 7 –6x1

g(x) = 4x (1 – x),0 x 1

–7 2x + 5 7

1 – g( x ) 1 – 4 x(1 – x ) 1 – 4 x 4x 2 = = 1 g( x ) 1 4 x(1 – x ) 1 4x – 4x 2

1– x 1– x gof(x) = 4f(x) . (1 – f(x)) = 4 1 x 1 – 1 x

=

8 x(1 – x ) (1 x )2

Section (C) : C–2.

One One / Many One f(x) =

f(x) =

f(x) =

2x 2 x 5 7 x 2 2x 10

, Domain x R

( 4x 1)(7 x 2 2x 10) (14 x 2)(2x 2 x 5) (7 x 2 2x 10)2 11x 2 30x 20 2

(7 x 2x 10)

2

30 > 0 x (– , 0) , 11 30 f (x) < 0 x 0, 11

f(x) = 0

RESONANCE

x = 0,

30 11 S OLUTIONS (XII) # 7

Function is increasing and decreasing in different intervals, so non monotonic Many one function. Onto / Into f(x) =

2x 2 x 5

7 x 2 2x 10 2x 2 – x + 5 > 0, x R and 7x 2 + 2x + 10 > 0 x R a = 2 > 0 and a = 7 and D = 4 – 280 < 0 D = 1 – 40 = – 39 < 0 f(x) > 0 x R Also f(x) never tends to ± as 7x 2 + 2x + 10 has no real roots, Range Codomain so into function.

C–3.

f(x) = x 3 + x 2 + 3x + sin x, f(x) = 3x 2 + 2x + 3 + cos x

3x 2 + 2x + 3

–1 cos x 1

lim f(x) = +

xR

32 as a = 3 > 0 and D < 0 12 so f(x) > 0 x R

lim f(x) = –

x

x

Hence f(x) is one-one and onto function (as f(x) is continuous function) C–6.

f(g(x1)) = f(g(x2)) as f is one - one function hence f(g(x1)) = f(g(x2)) x1 = x2

g(x1) = g(x2) x1 = x2

f(g(x)) is one - one function

as

g is one - one function

Section (D) : 2 D–2.

f(x) = sin

[a] x .

[a] = 4 D–3.

Period =

[ a]

=

a [4, 5)

f(x) = x + a – [x + b] + sin x + cos 2x + sin (3x) + cos (4x) + ........ + sin (2n – 1) + cos (2px) f(x) = {x + b} + a – b + sin (x) + cos (2x) + sin (3x) + cos (4x) + .... + sin (2n – 1) + cos (2nx) Period of f(x) = L.C.M (1, 2,

2 2 2 2 , , ........., , )=2 3 4 2n 1 2n

period of f(x) = 2 since f(1 + x) f(x) , hence fundamental period is 2 D–7.*

(A) (B)

f(x) = cos (cos–1 x) = x, x [–1, 1] odd function f (x + ) = cos (sin (x +)) + cos (cos (x + )) f (x + ) = cos (sin x) + cos (cos x) = f(x)

f x = cos sin x + cos cos x 2 2 2 = cos (cos x) + cos (sin x) = f(x) fundamental period = (C)

2

f(x) = cos (3 sin x), x [–1, 1] – 3 sin1 3 sin x 3 sin 1 cos (3 sin 1) cos (3 sin x) 1

Range is [cos (3 sin1), 1]

1 x f–1(x) = n 1 x

Section (E) : E–1.

ex ex y = x 1 e e x By compnendo and dividendo

1 y 2e x = 1 y 2

RESONANCE

1 y x = n 1 y

S OLUTIONS (XII) # 8

E–7.

E–8.

f(1) = 1 = 2 – 1 f(n + 1) = 2f(n) + 1 f(3) = 7 = 23 – 1 f(4) = 15 = 24 – 1 Similarly f(n) = 2n – 1

f(2) = 2f(1) + 1 = 2. 1 + 1 = 3 = 22 – 1

Method 1 : (usual but lengthy) x2 f(x) + f(1 – x) = 2x – x4 .....(1) replace x by (1 – x) in equation (1) (1 – x)2 f(1 – x)+ f(x) = 2 (1– x) – (1 – x)4 .....(2) eliminate f(1 – x) by equation (1) and (2) we get f(x) = 1 – x2 Method 2 : Since R.H.S. is polynomial of 4th degree and also by options consider f(x) = ax2 + bx + c x2 f(x) + f(1 – x) = 2x – x4 x 2 (ax2 + bx + c) + a (1 – x)2 + b (1 – x) + c = 2x – x4 by comparing coefficients a=–1 b=0 c=1 f(x) = – x2 + 1

EXERCISE # 2 PART - I 1.

(i)

f(x) =

or

3 2 x 2 .2 x 3 – 2x – 2 .2–x 0 (2x – 1) (2x – 2) 0

(ii)

f(x) =

(iii)

1 – 1 x 2 0 f(x) = (x2 + x + 1)–3/2 D:xR

(iv)

f(x) =

or

(2x)2 – 3.2x + 2 0 2x [1, 2] x [0, 1]

1 1 x2

x2 + x2

1 x 2 1

0 1 – x2 1

or

x

or

x 2n

or

x (0, 1] [4, 5)

x [– 1, 1]

1 x 1 x

x2 1 x 0 and 0 x2 1 x x (– , –2) [2, ) and x (–1, 1] D: (v)

f(x) =

tan x tan 2 x

tan x – tan2x 0

or

0 tan x 1

n

n , n 4

1

(vi)

(vii)

f(x) = 2 sin x 2

f(x) =

sin

x 0 2

5x x 2 log1 / 4 4

5x x 2 1 4

RESONANCE

and

5x – x2 > 0

S OLUTIONS (XII) # 9

(viii) or 2.

(i)

f(x) = log10 (1 – log10(x2 – 5x + 16)) 1 – log10 (x2 – 5x + 16) > 0 x (2, 3) f(x) = 1 – |x – 2| |x – 2| [0, )

or

x2 – 5x + 6 < 0

f(x) (– , 1]

1 f(x) , 1 3

1 (ii)

(iii)

f(x) =

x5 D : x (5, ) R : f(x) (0, ) 1 f(x) = 2 cos 3 x range of cos 3x is [–1, 1] cos 3x [–1, 1]

(iv)

x2

f(x) =

=y x 8x 4 x + 2 = yx2 – 8yx – 4y for x to be real D 0 (8y + 1)2 + 4y (4y + 2) 0 64y2 + 16y + 1 + 16y2 + 8y 0 2

80y2 + 24y + 1 0 (v)

f(x) =

(viii) (ix)

f(x) = 3 sin

(i)

1 1 , y , 4 20

or

1 y , 3 3

2 x2 16

D : x , 4 4

f(x) = x3 – 2x2 + 5 = (x2 – 1)2 + 4 R : [4, ) f(x) = x3 – 12x , x [–3, 1] = x (x2 – 12) f(x) = 3x2 – 12 = 0 or f(x) = sin2x + cos4x = sin2x + 1 + sin4x – 2 sin2x = sin 4x – sin2x + 1 1 2 3 = sin x + 2 4

7.

or

=y x 2 2x 4 x2 – 2x + 4 = yx2 + 2xy + 4y x2 (1 – y) – 2x(1 + y) + 4(1 – y) = 0 D0

2 x 2 0 , 4 16

(vii)

yx2 – x (8y + 1) – (4y + 2) = 0

x 2 2x 4

4(1 + y)2 – 16(1 – y)2 0

(vi)

or

f(–x) =

3 f(x) 0 , 2

x=±2

R : [–11, 16]

3 R : , 1 . 4

(2 x 1)7 (2 x )6

neither even non add (ii) (iii)

sec x x 2 9 = f(x) x sin x f(–x) = – f(x) odd f(–x) =

RESONANCE

even

S OLUTIONS (XII) # 10

(iv)

x 1 x2 2 [ x ] [ x ] 1 x 1 , even by graph of function f(x) = 2 x x 1

(v)

f(x) =

2x(sin x tan x ) x 2 1

if x = n, f(n) = 0 if x n

8.

x x = – 1

f(–x) = – f(x) odd function

(i)

f(x) = 1 –

cos2 x sin2 x – fx = 1 – f(x) 2 1 – tan x 1 – cot x fundamental period =

(ii)

(iii)

(iv)

sin2 x cos2 x – 1 cot x 1 tan x period of f(x) = L.C.M. (, ) = For fundamental period

f(x) = tan [ x ] : [x] 2n + 1 2 f (x) = 0 By graph fundamental period = 2 f(x) = log (2 + cos 3x) fundamental period of f(x) = fundamental period of (2 + cos 3x) (as log is a monotonic function) f(x) = en sin x + tan3x – cos (3x – 5) f(x) = sin x + tan3x – cos (3x – 5), sin x > 0

period 2 ; ,

2 3

2 = 2 Period of f(x) = L.C.M. 2 , , 3

(v)

x x x x x x x f(x) = sin x sin 2 sin 4 .... sin n –1 tan tan 3 tan 5 ... tan n 2 2 2 2 2 2 2

Period of f(x) = L.C.M. of 2, 23 , 25 ,......2n , 2, 23 .....2n = 2 n

(vi)

f(x) =

sin x sin 3 x cos x cos 3 x

2 2 period of f(x) = L.C.M. 2. , 2, = 2 3 3 For fundamental period

f(x + ) = 11.

f(x) =

sin ( x ) sin(3 x 3 ) – sin x – sin 3 x = cos( x ) cos(3 x 3) – cos x – cos 3 x

1 x

1 x2

f(x) = 0 at x = 1 ±

Fundamental period =

2

for x 2 1, 1 2 f is bijective function hence f is invertible.

RESONANCE

S OLUTIONS (XII) # 11

1 x 1 x2 or

or

=y x2y + x + (y – 1) = 0 x=

1 1 4 y( y 1) 2y

=

1 4y 4 y 2 1

1 4x 4x 2 1 , f–1(x) = 2x , 1

2y

x0 x0

as f (1) 0

n

f (a k ) = 16 (2

n

13.

k 1

or

or Now or 15.

(i)

(ii)

– 1)

f(a + 1) + f(a + 2) + ......... + f(a + n) = 16 (2n – 1) f(x + y) = f(x) . f(y) f(0) = 1, f(1) = 2 f(x) = 2x f(a + 1) + f(a + 2) + ........ + f(a + n) = 2a [2 + 4 + ......... + 2n] = 2a . 2(2n – 1) 16 = 2a + 1 or a=3 f(x) = Ax2 + Bx + C x and f(x) at x = 0, f(0) = C at x = 1, f(1) = A + B + C C is integer at x = –1, f(–1) = A – B + C f(1) + f(–1) = 2A + 2C 2A is also integer f(x) = A x(x – 1) + (A + B) x + C f(x) = 2A

C is integer A + B is also integer C is integer

x( x 1) + (A + B)x + C 2

x( x 1) is also an integer and 2A, (A+ B), C 2 f(x) is also an integer.

If x is an integer then

PART - II 2.

f(x) = here

1

x

1 cos 1 (2 x 1) tan 3 x

– 1 2x + 1 < 1

– 2 2x < 0

–1x 3x > –

2

or

x , 0 6

Domain : , 0 (–1, 0) , 0 6 6

RESONANCE

S OLUTIONS (XII) # 12

3.

1 x3 f(x) = sin 3 / 2 + sin(sin x ) + log(3{x} + 1) (x 2 + 1) 2x Domain : 3{x} + 1 1 or 0 x –1

and

6.

1 x 3

1 2x 3 / 2 – 2x 3/2 1 + x 3 2x 3/2 1 + x 3 + 2x 3/2 0 (1 + x 3/2)2 0 xR 1 + x 3 – 2x 3/2 0 or (1 – x 3/2)2 0 3/2 or 1–x =0 or x=1 Hence domain x f(x) = (sin–1x + cos –1x)3 – 3 sin–1x cos –1x (sin–1x + cos –1x) –1

=

3 3 1 3 2 – 3 sin–1x cos x = – sin–1 x + 3 (sin–1x)2 2 2 8 8 2 4

3 3 = + 8 2

2 2 2 1 1 3 3 3 3 1 sin x (sin x ) sin x – = + 2 16 4 32 32 2

maximum value of f(x) at x = – 1 8.

f maximum =

9 3 7 3 3 3 + × = 32 2 16 8

Here (2 – log2 (16 sin2x + 1) > 0

0 < 16 sin2x + 1 < 4

1 16 sin2x + 1 4

3 16 0 log2 (16 sin2x + 1) < 2

log

2

2 2 – log2 (16 sin x + 1) > 0

0 sin2x <

2

2 log

2

(2 – log2 (16 sin2x + 1)) > –

2y>– Hence range is y (– 2] 11.

(A)

f(x) = e1/2 n x = g(x) =

12.

D:x>0

x,D:x0

D : x ± (2n +1)

2

(B)

tan–1 (tan x) = x

(C)

cot–1 (cot x) = x D : x ± n f(x) = cos 2x + sin4x = cos 2x + (1 – cos 2x)2 = 1 – cos 2x + cos 4x = sin2x + cos 4x g(x) = sin2x + cos 4x

(D)

f(x) =

|x| , D:x0 x g(x) = sgn (x), D : x R

f(6{x}2 – 5{x} + 1) (3{x} – 1) (2{x} – 1) 0

13.

x,

f((3{x} – 1) (2 {x} – 1)) or

1 1 {x} , 3 2

x

n

1 1 n 3 , n 2

R+ 0 , 2 g(x) = 2x – x2 R R f(g(x)) = cot–1 (2x – x2), where x (0, 1]

f(x) = cot–1x

hence f(g(x)) , 4 2

RESONANCE

S OLUTIONS (XII) # 13

20.

21.

25.

1 1 2 1 f x = x + x + [x + 1] – 3 x + 15 3 3 3 3 1 2 = x + x + [x] – 3x + 15 = f(x) 3 3 f(x) = |x – 1| f : R+ R x g(x) = e , g : [–1, ) R fog(x) = f[g(x)] = |ex – 1| D : [–1, ) R : [0, )

f(x) = (A) (B) (C)

(D)

fundamental period is 1/3

sin( [ x ]) =0,x { x} By graph fundamental period is one f(–x) = 0 = f(x) even function Range y {0} { x } y = sgn sgn – 1, x { x } y = sgn (1) – 1 y=1–1 y = 0, x Identical to f(x)

28.

f(x) = sin x + tan x + sgn (x2 – 6x + 10) f(x) = sinx + tan x + sgn ((x – 3)2 +1) f(x) = sin x + tan x + 1 period = L.C.M. (2, ) = 2 fundamental period = 2

29.

f:NI

n – 1 , n odd f(n) = 2 n – , n even 2 For For

n odd numbers f(n) 0, 1, 2, 3, ...... n even numbers f(n) –1, –2, –3, ...... range I

f(n) is one -one onto function.

EXERCISE # 3 2.

x = 2

(A)

sin–1 x + cos–1

(B)

2 sin–1 x + cos–1 1 x = 0

(C)

x [0, 1]

x [–1, 0]

x [0, )

1 x tan–1 x + tan–1= tan–1 1– x

1 – x2 g 2 = 2h(x) 1 x

(D)

cos–1 1 x 2 = – sin–1(x)

cos

–1

1 – x2 –1 1 x 2 = 2 tan x

1 x h(x) + h (1) = h 1– x x (–, 1)

RESONANCE

S OLUTIONS (XII) # 14

Comprehension # 2 (6, 7, 8) Period of e

x tan 4

is 4

(1 2 [ x]) =0 2

cos

xR

[x]

Period of sin 2 is 4 then y =

8 2[ x ] [ x ] 2

p=4

[x]2 – 2[x] – 8 0 – 2 [x] 4 q=–2 , r=5 r–q–1=5+2–1=6 x 2 , f2 (x) = 2 x ,

Period of f(x) is 4

– [x]2 + 2 [x] + 8 0

i.e.,

([x] – 4) ([x] + 2) 0 –2 xx–1 f(x + 1) < f(x – 1) as f(x) is M.D. g(f(x + 1) < g(f(x – 1)) as g(x) is M..

Let

f(x) =

, x

f(x) = Let

f(x) 0

.

g(x) = x sec2x – tan x g(x) = 2x sec2x tanx > 0 x>0 g(x) > g(0), g(x) > 0 x1 < x2 f(x1) < f(x2) <

RESONANCE

f(x) > 0

f(x) is M.I.

< S OLUTIONS (XII) # 102

Section (D) : D-2.

f(x) = 3x 3 f(x) = 0 x=0 x = – 2, f(–2) = – 8 x = 0, f(0) = 0 x = 2, f(2) = 8 Minimum = – 8, maximum = 8 (ii) f(x) = cos x – sin x (i)

f(x) = 0

x=

x = 0, x=

f(0) = 1 ,

x = ,

f

=

f() = – 1

Minimum = –1, Maximum = (iii) f(x) = 4 – x f(x) = 0 x=4 x = – 2,f(–2) = – 10 x = 4, f(4) = 8 x=

,f

=

Minimum = –10, (iv)

Maximum = 8

f(x) = cos x – sin 2x f(x) = 0

x = 0, f(0) = Minimum = D-6_.

cos x = 0, sin x =

x=

, x=

x=

f

=

x=

, f

=

, Maximum =

Let No. of children of john & anglina = y x + (x + 1) + y = 24 y = 23 – 2x Number of fights F = x(x + 1) + x(23 – 2x) + (x + 1) (23 – 2x) F = – 3x2 + 45x + 23 But 'x' wil be integral.

D-8.

,

2

2

= 0 – 6x + 45 = 0

x = 7.5

check x = 6 or x = 7, F = 191

2

R +r =h R2 = h2 – r2 volume of cylinder , V = R2 (2h) = (2h) (

)2

= 2 (r2 – h2) + 2h(–2h) = 0

r2 = 3h2

h=

< 0 at h =

Vmax = 2

RESONANCE

= S OLUTIONS (XII) # 103

D-10.

2 + 2r = 440 A = 2r = – 2r2 + 440r = – 4r + 440 = 0

D-12.

at

r=

Let Let

x, y be dimensions of rectangle. 2x + 2y = 36. V be volume sweeped V = x 2y V = x 2(18 – x) = x.3.(12 – x)

At

x = 12, V has maximum value

y=6

Section (E) : E-2.

Let (h, k) be point of inflection h sin h = k y = sin x + x cos x y = cos x + cos x – x sin x y = 0 2 cos h – h sin h = 0 2 cos h = k sin2 h + cos 2 h = 1 = 1 4k 2 + h2k 2 = 4h2

+

...(1)

...(2)

locus y2 (4 + x 2) = 4x 2

Section (F) : F-2.

F-4.

Let

f(x) = 3x2 + px –1

f(x) = x3 +

f(x) satisfies conditions in Rolle's theorem 3x2 + px –1 = 0 has atleast one root in (–1,1).

–x+c

f(–1) =

+ c = f(1)

f(c) = 0 for atleast one c (–1,1)

Let h(x) = f(x) g(x) h(a) = 0 = h(b) By Rolle’s theorem on [a, b] h(x) = 0, for at least one c (a, b). f (c) g(c) + f(c) g(c) = 0

PART - II Section (A) : A-1.

V=

V=

=

77 × 103 =

× 70 × 70 ×

( 1 litre = 103 c.c.)

= 20 cm/min.

RESONANCE

S OLUTIONS (XII) # 104

A-3_.

= Let x = 25 and x = 0.2 such that f(x) = f (x) =

f(x + x) = f(x) + f(x). x

=

+

.x

= A-4_.

+

× 0.2 =

=5+

= 5 + 0.02 = 5.02

V = x3 x = (3x 2) x = (3x 2) (0.04x) = 0.12x 3m 3

V =

Section (B) : B-5*.

2y3 = ax 2 + x 3 6y2

= 2ax + 3x 2

=

=

Tangent at (a, a) is 5x – 6y = – a =

,

2 + 2 = 61

= = 61

a2 = 25.36 a = ± 30 B-8.

P1 : y2 = 8x C1 : x 2 + (y + 6)2 = 1

Equation of normal of parabola y = mx – 2am – am 3 if passes through (0,–6) –6 = – 2am – am 3 a=2 3 = 2m + m 3 3 m + 2m – 3 = 0 m = 1. Point on parabola (am 2 , – 2am) (2, –4).

Section (C) : C-2.

5x 4 – 3

f(x) =

It is sufficient to solve for p, the condition f(x) 0

xR

5x 4 – 3 0 x R

RESONANCE

S OLUTIONS (XII) # 105

Case -

1–p1 Inequality holds true. 1–p>0 p 0 and for x cos x – x < 0,

RESONANCE

f(x) has maxima at x 0

S OLUTIONS (XII) # 106

D-8*.

f(x) =

,x>0

=

x > 0.

f(x) is decreasing On

x > 0.

, greatest value is

f(

)=

–

n

f

=

n

and least value is

.

D-10. S = 2rh = 2H

= 2H

Maximum at r =

D-13.

Let d be distance between (k, 0) and any point (x, y) on curve. d= ( y2 = 2x – 2x 2).

d=

Maximum d =

Maximum d =

Section (E) : E-3.

x=1

a=–

E-4.

3=a+b

=0

,

b=

3a + b = 0

f(x) = n(x – 2) –

f(x) =

=

=

=

.

As n(x – 2) is defined when x > 2 f(x) is M.. for x (2, )

f –1(x) is M.. wherever defined

Also

f(x) is always concave downward

f(x) =

RESONANCE

0

If x > 0, (x) > 0 Hence (x) is increasing As we know ex x + 1 (ex) (x + 1) ex + 13.

x+1+

Let x > –1 Consider f(x) = (1 + x)n(1 + x) – tan–1 x f(x) = n(1 + x) + 1 – f(x) =

>0

f(x) is increasing f(x) < 0

For x < 0, f(x) < f(0) f(x) decreasing f(x) > f(0)

(1 + x)n(1 + x) – tan–1 x > 0 For x > 0,

f(x) > f(0) f(x) > 0

f(x) > 0

n(1 + x) > f(x) is increasing

f(x) > f(0)

f(x) > 0

n(1 + x) > Hence larger of these is n(1 + x). 15.

f(x) = a0 + a1x + a2x 2 + a3x 3 + a4x 4 + a5x 5 + a6x 6 a0 = a1 = a2 = a3 = 0 = e2

a4 = 2

f(x) = 2x 4 + a5x 5 + a6x 6 f(x) = x 3 ( 8 + 5a5x + 6a6x 2) f(1) = 0, f(2) = 0 , 17.

f(x) = 2x 4

xy = 18 Area of printed space

=

=

Maximum when x=

RESONANCE

y= S OLUTIONS (XII) # 109

21.

f(x) = 0 sin x=

=0

= n

,nN

x = .......,

,

, .......

,

Consider interval

, 1.

=0=

By Rolle’s theorem f(x) vanishes at least once in Infinite number of such intervals are there. Hence f(x) vanishes at infinite number of points in (0, 1) 24.

Let g(x) =

, x [a, b]

By Rolle’s theorem, g’(x0) = 0 =0

25.

f(x0) =

Let f(x) = (x + a) – (x) f(x) = (x + a) – (x) + k f(0) = (a) – (0) + k f(2a) = (3a) – (2a) + k f(0) = f(2a) By Rolle's theorem on [0, 2a], f(c) = 0 for at least one c (0, 2a) (x + a) = (x) has at least one root in (0, 2a)

PART - II 3.

f(x) =

>0

f(x) increasing hence g(x) is also increasing function 7.

Let

be quantity y=

y2 = y

=

=

=a 9.

= = a

f(x) = ; g(x) = for a > 1, a 1 and x R n a h(x) = n f(x) + ng(x)

(n a) h(x) =

h(x) =

na +

n a

+ |x|

h(x) = a sgn x Now h(–x) = a|–x| sgn (–x) = –h(x) h(x) is an odd function Also graph of h(x) is It is clear from the graph that h(x) is an increasing function

RESONANCE

S OLUTIONS (XII) # 110

12.

f(x) =

f(x) = For 13.

16.

0 < x < 1,

tan

f(1–) f(1) and f(1+) f(1) – 2 + log2 (b2 – 2) 5

f (x) > 0

0 < b2 – 2 128

f(x) is increasing.

2 < b2 130

2 = h2 + x 2 Area of base (triangle) is

3x =

a

Volume V = h

. 4 . 3.x 2 = 3

=h

=3 18.

a2

(2 – 3h2)

h (2 – h2)

V is maximum when h =

Maximum of f(x) is Given expression is f(x 1) + f(x 2)

20.

.

f(x 1)

f(x 2)

f(x 1) + f(x 2)

f (x) = 1 +1– 2– At x = 0, f(x) is least. Least value = f(0) = 1

23.

f(x) = m – n is odd. f (x) < 0

25.

>0

x (–, 0)

f (x) > 0

x (0, )

(x) = (3(f(x))2 – 6(f(x)) + 4)f(x) + 5 + 3 cos x – 4 sin x 5–

5 + 3 cosx – 4 sin x 5 +

adding (3(f(x))2 – 6(f(x)) + 4)f(x) (3(f(x))2 – 6(f(x)) + 4)f(x) (x) (3(f(x))2 – 6(f(x)) + 4)f(x) + 10 3(f(x))2 – 6f(x) + 4 = 3 (f(x) – 1)2 + 1 > 0 2 (3(f(x)) – 6(f(x)) + 4)f(x) 0 when ever f(x) is increasing. (x) 0 (x) is increasing, when ever f(x) is increasing. If f(x) = – 11 then (3(f(x))2 – 6f(x) + 4) f(x) + 10 = – 33 (f(x) – 1)2 – 1 < 0 (x) < 0 (x) is decreasing.

RESONANCE

S OLUTIONS (XII) # 111

28.

f(x) = f(x) = 0 at x = 0, x = – 3, x = 1 so at x = 0, f(x) has local minima. and at x = –3, x = 1 ; f(x) has local maxima f(1) =

,

f(– 3) =

. f(–3) < 0, f(1) > 0 and f(x) 0

f(x) is undefined at point(s) in (–3, 1). Hence f(x) has no absolute maxima.

EXERCISE # 3 PART - I 2.

(A) (B)

f(x) is continuous and differentiable f(0) = f() Hence condition in Rolle’s theorem and LMVT are satisfied. f(1–) = –1, f(1) = 0, f(1+) = 1 f(x) is not continuous at x = 1, belonging to Hence, atleast one condition in LMVT and Rolle’s theorem is not satisfied

(C)

f’(x) =

(x – 1)–3/5, x 1

(D)

At x = 1, f(x) is not differentiable. Hence at least one condition in LMVT and Rolle’s theorem is not satisfied. At x = 0

L.H.D. =

3.

=

= –1

R.H.D. = 1 At x = 0, f(x) is not differentiable Hence at least one condition in LMVT and Rolle’s theorem is not satisfied. (A) Let PQ = x Then BP =

PS =

tan60º =

area A of rectangle =

=

=–

(4 – x) x

(4 – 2x) = 0

x=2

0

g(x) = –

× f(x).g(x)

g(x) > 0 g(x) = f –1 (x) concave upward

15.

f(x) =

17.

(1 – nx) f(x) 0, when x e f(x) is decreasing function, when x e >e f() < f(e) 1/ < e1/e e > e Statement-1 is True, Statement-2 is False

f ’ (x) = 50x49 – 20x19 = 10x19(5x30 – 2) x = 0 is stationary point. Statement-2 is ture. f(0) = 0 =

–

0 increasing

gmax = g(1) = e +

h(x) = x2

+

+

hmax = h(1) = e +

RESONANCE

h(x) = 2x ,

+ 2x3 so

– 2x

= 2x

>0

a=b=c S OLUTIONS (XII) # 120

26.

(A)

Re

= Re

= –1 y 1 =

1 or

= Re

= Re (–1/y) =

– 1

Alternate

Re

= Re

= Re

= Re

= Re

= Re as –1 sin 1 (– , 0 ) (0, ) (B)

(C) (D)

–1

1

–1

0

1

–1

0

0

0

1

– 1 0

t (– , –9] [–1 , 1] [9, ) x (– , 0) [2 , ) f() = 2 sec2 f() 2 f(x) = x3/2 (3x – 10)

f ’(x) = x3/2 3 +

0

f() [2, )

x1/2 (3x –10)

asf ’(x) 0 0

27.

– 15 0

x 2

3x +

– 15 0

x [2, )

f(x) = x4 – 4x3 + 12x2 + x – 1 f(x) = 4x3 – 12x2 + 24x + 1 f(x) = 12x2 – 24x + 24 = 12 (x2 – 2x + 2) > 0 xR f(x) is S.I. function Let is a real root of the eqution f(x) = 0 f(x) is MD for x (– , ) and M.I. for x (, ) where < 0 f(0) = – 1 and < 0 f() is also negative f(x) = 0 has two real & distinct roots.

RESONANCE

S OLUTIONS (XII) # 121

28.

p = (x – 1) (x – 3) = (x2 – 4x + 3) p(x) = (x3/3 –2x2 + 3x) + p(1) = 6 6 = (1/3 – 2 + 3) + 6 = (1/3 + 1) + 18 = 4 + 3 ...(i) p(3) = 2 2 = (27/3 – 2 × 9 + 9) + 2= =2=3 p(x) = 3(x – 1) (x – 3) p(0) = 3(–1)(–3) =9

29.

f(x) = |x| + |x2 – 1|

f(x) =

f(x) =

PART - II 1.

Let

f(x) = x +

f(x) = Minimum occures at x = 1. 2.

3.

f(x) = 6x2 – 18ax + 12a2 f(x) = 0 x = a, 2a f(x) = 12 x – 18a f(a) = – 6a < 0 and f (2a) = 6a > 0 p = a, q = 2a p2 = q a = 2 (a > 0) Let

f (x) = ax2 + bx + c

f(x) =

a2 – 2a = 0

a = 0, 2

f(0) = d = f(1) (2a + 3b + 6c = 0) By Rolle's theorem, at least one root of f(x) = 0 lies in (0, 1) 4.

y2 = 18x Differentiating w.r.t. t

...... (1)

From equation (1), x =

RESONANCE

2y.2 = 18

y=

Required point is S OLUTIONS (XII) # 122

5.

x = a (1 + cos ), y = a sin

Equation of normal at point (a(1+ cos ), a sin ) is y – a sin =

(x – a (1 + cos ))

y cos = (x – a)sin It is clear that normal passes through fixed point (a, 0) 6.

y = x2 – 5x + 6 = 2x – 5

= – 1,

product of slopes = –1

=1

angle between

tangents is 7.

Let f1(x) = x3 + 6x2 + 6 Increasing in (–, – 4] Let

f2 (x) = 3x2 – 2x + 1

f1(x) = 3x (x + 4)

f2(x) = 6x – 2

Not increasing in Let

f3(x) = 2x3 – 3x2 – 12x + 6

f3(x) = 6x2 – 6x – 12 = 6(x +1) (x – 1)

Increasing in [2, ) Let f4(x) = x3 – 3x2 + 3x + 3 Increasing in (–, ) 8.

V=

4 (10 + r)2

, 0 r 15 = – 50. = – 50

=

9.

Any point on ellipse is P(x, y) = (a cos , b sin ) Area of rectangle = 4ab cos sin = 2ab sin 2 Maximum area = 2ab.

10.

By LMVT and f(x) 2

f(x) = 3(x –1)2 0

= f(x),

x (1, 6)

11*.

(where r = 5)

2

2

f(6) 8.

Equation of normal is y – a(sin – cos ) =

(x – a (cos + sin ))

x cos + y sin = a. Distance from origin = |a| (constant) Slope of normal = –

RESONANCE

= tan

angle made with x-axis is

+ . S OLUTIONS (XII) # 123

12.

f (x) = f (x) changes sign as x crosses 2. f(x) has minima at x = 2.

13.

= f(c) 1 < c < 3

c =

14.

f(x) =

15.

Graph of y = x3 – px + q cuts x-axis at three distinct pints

(cos x – sin x)

=0 16.

= 2 log3e

x =

Maxima at x = –

, minima at x =

Condition for shortest distance is slope of tangent to x = y2 must be same as slope of line y = x +1.

=1

y=

,

x=

,x–y+1=0

Shortest distance =

17.

P (x) = 4x 3 + 3ax 2 + 2bx + c and P(0) = 0 c=0 P (x) = x(4x 2 + 3ax + 2b) D = 9a2 – 32b < 0

b>

>0

(P(x) = 0 has only one root x = 0)

P (– 1) < P (1) a>0 P(x) has only one change of sign. x = 0 is a point of minima. P(–1) = 1 – a + b + d, P(0) = d P(1) = 1 + a + b + d P(–1) < P(1), P (0) < P(1), P (–1) > P(0) P(–1) is not minimum but P(1) is maximum.

18.

y=x+ y = 1 –

=0

y=2+

=3

x3 = 8

x=2

(2, 3) is point of contact Thus y = 3 is tangent Hence correct option is (3)

RESONANCE

S OLUTIONS (XII) # 124

19.

f(x) = 1 f(–1) = k + 2 f(x) = k + 2 f has a local minimum at x = – 1 1k+2k+2 possible value of k is – 1 Hence correct option is (3)

20.

ex + 2e–x 2

f(–1+) f(–1) f(–1–) k–1

(AM GM)

f(x) > 0

so statement- 2 is correct

As f(x) is continuous and

f(c) =

21.

belongs to range

for some C.

of f(x),

Hence correction option is (4).

y–x=1 y2 = x 2y

=1

=

y =

= 1 , x=

tangent at

y=

y=x+

y–x=

distance =

22.

=

=

f(x) =

In right neighbourhood of ‘0’ tan x > x

RESONANCE

S OLUTIONS (XII) # 125

In left neighbourhood of ‘0’ tan x < x as(x < 0) at x = 0 , f(x) = 1 x = 0 is point of minima so statement 1 is true. statement 2 obvious

23.

r3

V=

4500 =

= 4r2

45 × 25 × 3 = r3

r = 15 m after 49 min = (4500 – 49.72) =972 m3 972 =

r3

r3 = 3×243 = 3× 35 r=9 72 = 4 × 9 × 9

=

24.

f '(x) = at x = – 1

+ 2bx + a –1 – 2b + a = 0 a – 2b = 1

at x = 2

+ 4b + a = 0

a + 4b =

On solving (i) and (ii)

f '(x) =

...(i)

=

a=

...(ii)

, b=

=

So maxima at x = – 1, 2

RESONANCE

S OLUTIONS (XII) # 126

SOLUTION OF TRIANGLE EXERCISE # 1 PART - I Section (A) : A-1.

(i)

L.H.S. = a sin (B – C) + b sin (C – A) + c sin (A – B) = k sin A sin (B – C) + k sin B sin (C – A) + k sin C sin (A – B) = k (sin2 B – sin2 C) + k (sin2C – sin2 A) + k (sin2 A – sin2 B) = 0 = R.H.S.

(ii)

L.H.S. =

first term =

= = k 2 sin (B + C) sin (B – C) = k 2 (sin2 B – sin2 C) = k 2 (sin2 C – sin2 A)

Similarly

= k 2 (sin2 A – sin2 B)

and

L.H.S. = k 2 (sin2 B – sin2C + sin2C – sin2A + sin2 A – sin2 B) = 0 = R.H.S. (iii) L.H.S. = 2bc cos A + 2ca cos B + 2ab cos C = b2 + c 2 – a2 + a2 + c 2 – b2 + a2 + b2 – c 2 = a2 + b2 + c 2 = R.H.S

(iv) L.H.S. = a2

– 2ab

= a2 + b2 – 2ab cos C = a2 + b2 – (a2 + b2 – c 2) = c 2 = R.H.S. (v) L.H.S. = b2 sin 2C + c2 sin 2B = 2b2 sin C cos C + 2c 2 sin B cos B = 2k 2 sin2 B cos C sin C + 2k 2 sin2 C sin B cos B = 2k 2 sin B sin C [sin B cos C + cos B sin C] = 2(k sin B) (k sin C) sin (B + C) = 2bc sin A (vi) R.H.S =

(b = ksin B, c = ksin C)

c = a cos B + b cos A, b = c cos A + a cos C

=

=

=

A–4.

= L.H.S.

= sin(B + C) sin(B – C) = sin(A + B) sin(A – B) sin2 B – sin2 C = sin2 A – sin2 B 2 sin2 B = sin2 A + sin2 C 2b2 = a2 + c 2 a2, b2, c 2 are in A.P.

RESONANCE

S OLUTIONS (XII) # 127

A–7.

x 3 – Px 2 + Qx – R = 0

a2 + b2 + c 2 = P a 2b 2 + b 2c 2 + c 2a 2 = Q a2b2c 2 = R abc =

+

+

=

[a2 + b2 + c 2] =

Section (B) : B–1.

(i)

L.H.S. = 2a sin2 = = = =

(ii)

+ 2 c sin2

a(1 – cos c) + c(1 – cos A) a + c – (a cos C + c cos A) a+c–b R.H.S.

L.H.S. =

+

=

.

+ +

= (iii)

.

+

=

.

.

L.H.S. = 2bc(1 + cos A) + 2ca(1 + cos B) + 2ab(1 + cos C) = 2bc + 2ca + 2ab + 2bc cos A + 2ca cos B + 2 ab cos C + a2 + b2 + c 2 = (a + b + c)2

=2 = R.H.S. (iv)

L.H.S. = (b – c)

(b – c) cot

+ (c – a)

+ (a – b)

= k(sin B – sin C)

= 2k cos

sin

= 2k sin

sin

= k [cos C – cos B] similarly (c – a) cot and (a – b) cot

= k[cos A – cos C]

= k[cos B – cos A]

L.H.S. = k[cos C – cos B + cos A – cos C + cos B – cos A] =0 = R.H.S.

RESONANCE

S OLUTIONS (XII) # 128

(v) L.H.S. = 4 (cot A + cot B + cot C) = 4 = 2bc cos A + 2 ca cos B + 2ab cos C = a2 + b2 + c 2 = R.H.S. (vi) L.H.S. =

cos

. cos

. cos

= =

= = R.H.S.

B–3.

Let ADB = we have to prove that tan = if we aply m – n rule, then (1 + 1) cot= 1.cot C – 1.cotA. =

–

=

–

= =

[2(a2 – c 2)]

2cot =

tan =

Section (C) : C–2.

(i)

r. r1 .r2 .r3 =

(ii)

r1 + r2 – r3 + r = 4R cosC

= 2

L.H.S. = =

=

=

RESONANCE

S OLUTIONS (XII) # 129

=

=

=c

=

L.H.S. =

(iii)

=

= R.H.S.

=

(s – a + s – b + s – c) =

·

=

=

=

=

=

=

=

= 24 sq. cm 2s = 24 r1, r2, r3 are in H.P.

=

=

=r

=r

.... (i) s = 12 .... (ii)

are in A.P..

[4s 2 – 2s(a + b + c) +a2] =

R.H.S. =

similarly we can show that

= 4RcosC

L.H.S. =

=

C–4.

=

(s + s – a + s – b + s – c)2 = 4

(v)

=

L.H.S. =

=

cos C =

[s 2 + (s – a)2 + (s – b)2 + (s – c)2] =

(iv)

=

=

a, b, c are in A.P. 2s = 24 a + b + c = 24 3b = 24 b=8

are in A.P..

2b = a + c

a + c = 16

But = = 24 × 24 = 12 × (12 – a) × 4 × (12 – c) 2 × 6 = 144 – 12 (a + c) + ac 12 = 144 – 192 + ac ac = 60 and a + c = 16 a= 10, c = 6 or a = 6, c = 10 and b = 8

RESONANCE

S OLUTIONS (XII) # 130

Section (D) : D–1.

(i)

,=

=

, =

=

R.H.S. =

=

=

L.H.S. = R.H.S.

(ii)

=

=

=

R.H.S. =

=

= L.H.S.= R.H.S.

PART - II Section (A) : A–4.

(b + c)2 – a2 = kbc

(a + b + c) (b + c – a) = kbc

b2 + c2 – a2 = (k – 2) bc

In a ABC –1 < cos A < 1

–1 <

b

=

= cos A

0, c = 5x + 5y is the largest side C is the largest angle. Now cos C =

=

r2 > r3

>

>

s – a < s – b < s – c –a < –b < –c;

3.

tan

=

; sin

r+R=

RESONANCE

a>b>c

=

r+R=

.cot

S OLUTIONS (XII) # 143

4.

a

=

=

a + c = 2b 5.

a + b + c = 3b.

a, b, c are in A.P.

AD = 4

AG =

Area of ABG =

=

Area of ABC = 3(Area of ABG)

6.

cos =

7.

C = /2

×4=

×

×

× AB × AG sin 30º

×

=

Sin 60º =

AB =

=

= 120º

=

=–

r = (s – c) tan

C = 90º

r = s – 2R 2r + 2R = 2 (s – 2R) + 2R. = 2s – 2R = (a + b + c) –

C = 90º

=a+b+c–c =a+b

8.

are in H.P.

are in A.P.

9.

a,b,c are in A.P.

= cos

Let cos

=

As

for some n 3, n N

cos

cos

cos

3 n < 4, which is not possible so option (2) is the false statement so it will be the right choice Hence correct option is (2)

RESONANCE

S OLUTIONS (XII) # 144

ADVANCE LEVEL PROBLEMS PART - I 1.

From figure, AD = c sin B Hence number of triangle is 0 if b < c sin B one triangle for b = c sin B two triangles for b > c sin B

2.

C = 60° Hence c2 = a2 + b2 – ab

3.

=

= 2 cos

Using properties of pedal triangle, we have

MLN = 180° – 2A LMN = 180° – 2B MNL = 180° – 2C

Hence the required sum =

sin2A + sin2B + sin2C

=

4.

5.

4sinA sinB sinC

From figure, we can observe that

OGD is directly similar to PGA

BD = s – b, CE = s – c and AF = s – a Hence BD + CE + AF = s

6.

, as cos

= cos

A = B, in either case

7.

, Using cosine rule in

ABO, we get

h=

8.

In

ABD,

RESONANCE

S OLUTIONS (XII) # 145

Comprehension # 1

9.

10.

+

+

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

k = 2R

cot A + cot B + cot C =

=

(b2 + c2 + a2) =

(b2 + c2 – a2 + c2 + a2 – b2 + a2 + b2 – c2)

=

=

.

=

11.

=

k=

=6

Comprehension # 2 (12 to 14) 12.

PG =

AD

= =

= PG = =

.ab sin C

or

b sin C

( =

ac sin B)

ac sin B c sin B

13.

Area of GPL =

and

Area of ALD =

(PL) (PG) (DL) (AD)

=

PL =

=

DL and PG =

AD

=

14.

Area of PQR = Area of PGQ + Area of QGR + Area of RGP ...(1)

Area of PGQ = = =

RESONANCE

×

×

PG.GQ.sin(PGQ) AD ×

×

BE sin ( – C) sin C S OLUTIONS (XII) # 146

=

×

=

bc sin A ×

ac sin B × sin C

sin A.sin B.sin C

Similarly Area of QGR =

sin A.sin B .sin C and Area of RGP =

sin A.sin B.sin C

From equation (1), we get (a2 + b2 + c 2) sin A.sin B.sin C

Area of PQR = 15.

In

CDB ,

=

Also from same triangle 16.

17.

BD =

cosAcosB + sinAsinBsinC = 1

(cosA – cosB)2 + (sinA – sinB)2 + 2sinAsinB(1 – sinC) = 0

a:b:c=1:1:

A = B & C = 90°

We have

18.

=

a:b:c=5:4:3

from figure, OO = ON – ON = R –

ZO = ZM + = from

RcosA +

OZO, using Pythagorous theorem,

we get (R –

)2 = (RcosA +

)2 +

=

PART - II 1.

from

from

ABC ,

=

AB = 2Rsin(A + )

ACB,

=

AC’ = 2Rsin( – A)

=

4RcossinA = 2acos

similarly CA = 2bcos =

RESONANCE

area

BC = 2R(sin (A + ) – sin( – A))

ABC = =

4cos2. S OLUTIONS (XII) # 147

2.

c2 – 2bc cosA + (b2 – a2) = 0 c1 & c2 are roots of this quadratic equation Hence (c1 – c2)2 + (c1 + c2)2tan2A = 4a2

3.

Area =

=

= =

2Rs

= 4.

We know that OA = R, HA = 2RcosA and applying Appoloneous theorem to

AOH, we get

2.(AQ)2 + 2(OQ)2 = OA2 + (HA)2

2.(AQ)2 = R2 + 4R2cos2A –

5.

=

+

using sine rule, diameter of required circle =

6.

=

= 20

radius = 10

L.H.S. =

(a2 (b + c – a) + b2 (c + a – b) + c2 (a + b – c))

=

=

=

abc

=

4R

RESONANCE

S OLUTIONS (XII) # 148

7.

from the parellelogram ABAC, AA = 21 , from

AAC, AA < b + c 21 < b + c

similarly 22 < c + a

...(1) ...(2)

and 23 < a + b ...(3) (1) + (2) + (3) gives 1 + 2 + 3 < 2s 8.

ZXY =

and

Area of

=

Area of

Area of XYZ = 2R2 cos

9.

cos

cos

=

Drop a perpendicular from the apex P to the base

ABC.

The foot of perpendicular is at circum centre O of

ABC

Using given data, we get from, right angle

POB, we get

= = 8.83 m 10.

from cyclic quadrileteral CQFP, we get

from cyclic quadriletral AQMF, we get FQM = FAM = 90º – B

AQM = 90º + 90º – B = 180º – B

P, Q, M are collinear

similarly P, Q, N are collinear hence, P, Q, M, N are collinear

RESONANCE

S OLUTIONS (XII) # 149

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