137880897 Maths Bansal Classes

PRACTICE TEST MATHS

MATHEMATICS

PART-A [STRAIGHT OBJECTIVE TYPE] Q.1 to Q.13 has four choices (A), (B), (C), (D) out of which ONLY ONE is correct. Q.1

If r, s are the roots of Ax2 + Bx + C = 0 (A ≠ 0) and r2, s2 are the roots of x2 + px + q = 0, then p is equal to

B2 − 2AC (A) A2 [Sol.

We have r + s = − r2

Q.2

Q.3 [Sol.

+

s2

= – p;

r2s2

B2 − 4AC (B) A2

AC − 2B2 (C) A2

2AC − B2 (D*) A2

B C ; rs = A A

=q;

Now, – p = (r +

Which of the following is a graph of f (x) =

s)2

B2 2C – 2rs = 2 − ⇒ A A

2AC − B2 p= Ans.] A2

1 sin(2x) 2

(A)

(B)

(C*)

(D)

The value of 'k' for which the equation x3 + kx2 + 3 = 0 and x2 + kx + 3 = 0 have a common root, is (A) 4 (B) 1 (C*) – 4 (D) – 1 Let α be a common root. then α3 + Kα2 + 3 = 0 ..............(1) and α2 + Kα + 3 = 0 ...............(2) Now, (1) – α × (2) ⇒ 3– 3α = 0 ; ∴ α = 1 ; So, from (1), we get 1 + k + 3 = 0 ∴k=–4 ] A A + 1 − sin 2 2 is equal to A A 1 + sin − 1 − sin 2 2 1 + sin

Q.4

If A ∈ (π,2π), then the value of

(A) – tan

[Sol.

We have

A 4

(B) – cot

A 4

A A 1 + sin + 1 − sin 2 2 = A A 1 + sin − 1 − sin 2 2

(C) cot

A 4 2

2

2

2

A A⎞ A A⎞ ⎛ ⎛ ⎜ cos + sin ⎟ + ⎜ cos − sin ⎟ 4 4⎠ 4 4⎠ ⎝ ⎝ A A⎞ A A⎞ ⎛ ⎛ ⎜ cos + sin ⎟ − ⎜ cos − sin ⎟ 4 4⎠ 4 4⎠ ⎝ ⎝

XI (P1-P7) PRACTICE PAPER [MATHS]

A 4

(D*) tan

Page # 2

MATHEMATICS A A⎞ ⎛ A A⎞ A A A A ⎛ A + sin + cos − sin ⎜ cos + sin ⎟ − ⎜ cos − sin ⎟ 2 sin 4 4⎠ ⎝ 4 4⎠ 4 4 4 4 A ⎝ 4 = = = = tan ] A A⎞ ⎛ A A⎞ A A A A A ⎛ 4 cos + sin − cos − sin ⎜ cos + sin ⎟ + ⎜ cos − sin ⎟ 2 cos 4 4⎠ ⎝ 4 4⎠ 4 4 4 4 4 ⎝ cos

If AD, BE, CF are medians of a triangle ABC and [(AD)2 + (BE)2 + (CF)2] : [(BC)2 + (CA)2 + (AB)2] is equal to p q , where p and q are in lowest form then p + q equals (A*) 7 (B) 10 (C) 6 (D) 15 [Hint: Sum of the square of the median is 3/4 times sum of the square of the sides ⇒ p = 3; q = 4 ⇒ p + q = 7 Ans. ] Q.5

Q.6 [Sol. Q.7

[Sol.

The value of x ∈ (0, 90°) and satisfying cos x° = sin 61° + sin 47° – sin 25° – sin 11°, is (A*) 7° (B) 11° (C) 13° (D) 17° RHS = 2 sin 54° cos 7° – 2 sin 18° cos 7° = 2 cos 7°[sin 54° – sin 18°] = cos 7° ⇒ x = 7° Ans. ] If in a triangle PQR, sin P, sin Q, sin R are in A.P., then (A) the altitudes are in A.P. (B*) the altitudes are in H.P. (C) the medians are in G.P. (D) the medians are in A.P. sinP, sinQ, sinR → AP (given) ⇒ sides are in AP (using sin law) Let altitude be h1, h2, h3 to sides p, q, r respectively. Now, Δ = ½ × p × h1 = ½ × q × h2 = ½ × r × h3 ⇒

2Δ 2Δ 2Δ p= h ; q= h ; r= h 1 2 3

p, q, r → AP ⇒

2Δ 2Δ 2Δ h1 , h 2 , h 3 → AP⇒

1 1 1 , , h1 h 2 h 3 → AP ⇒ h1, h2, h3 are in HP]

Q.8

If α, β are the roots of the quadratic equation x2 + 2(1 – cos 3θ) x – 2 sin23θ = 0 (θ ∈ R), then the maximum value of α2 + β2 is equal to (A) 0 (B) 4 (C) 8 (D*) 16

[Sol.

We have

Q.9

The number of integral value(s) of 'p' for which the equation 99 cos 2θ – 20 sin 2θ = 20p + 35, will have a solution is (A) 8 (B) 9 (C*) 10 (D) 11 We have 99 cos 2θ – 20 sin 2θ = 20p + 35 ....(1) As – 101 ≤ 99 cos 2θ – 20 sin 2θ ≤ 101 ∴ Equation (1) will have a solution, If – 101 ≤ 20p + 35 ≤ 101 ⇒ – 136 ≤ 20p ≤ 66 ⇒ – 6·8 ≤ p ≤ 3·3 ∴ Possible integral value(s) of 'p' are – 6, – 5, – 4, – 3, – 2, – 1, 0, 1, 2, 3 Hence number of integral value(s) of 'p' = 10 ]

[Sol.

Q.10

[Sol.

x2 + 2(1 – cos 3θ)x – 2sin23θ = 0 ; α + β = – 2(1 – cos 3θ) ; αβ = – 2sin23θ α β 2 2 Now, α + β = (α + β)2 – 2αβ = 4 (1 – cos 3θ)2 + 4 sin23θ = 4(1 – 2 cos 3θ + cos23θ + sin23θ) ⇒ α2 + β2 = 4 (2 – 2 cos 3θ) ;Clearly maximum value of α2 + β2 is 16. ]

A point 'P' is an arbitrary interior point of an equilateral triangle of side 4. If x, y, z are the distances of 'P' from sides of the triangle then the value of (x + y + z)2 is equal to (A) 3 (B*) 12 (C) 18 (D) 48 We have ar. (Δ ABC) = ar. (ΔPBC) + ar. (ΔPAC) + ar. (ΔPAB)

1 1 1 3 2 ( 4) = (4) (x) + (4) (y) + (4) (z) 2 2 2 4 XI (P1-P7) PRACTICE PAPER [MATHS]

⇒

Page # 3

MATHEMATICS ⇒

⇒

4 3 = 2 (x + y + z)

A

x+y+z= 2 3

Hence (x + y + z)2 = 12

F

] B

Q.11

[Sol.

z y P• x

E

C

D

If α, β, γ are the roots of the cubic 2009x3 + 2x2 + 1 = 0, then the value of α–2 + β–2 + γ–2 is equal to (A) 4 (B) – 2 (C) 2 (D*) – 4 We have 2009

x3

+

2x2

α β ...... (1) γ

+1=0

Put x =

1 in equation (1), we get ⇒ t

∴

t3

and

1 1 1 + + =2 αβ βγ γα

2009 2 + 2 +1=0 3 t t

1/α 1/β ; So, 1/γ

+ 2t + 2009 = 0

1 1 1 + + =0 γ α β

2 ⎛ 1 1 1 ⎞ ⎛ 1 1 1 ⎞ ⎛ 1 1 1⎞ + + ⎟⎟ Now, we know that ⎜⎜ + + ⎟⎟ = ⎜⎜ 2 + 2 + 2 ⎟⎟ + 2 ⎜⎜ ⎝ αβ βγ γα ⎠ ⎝α β γ ⎠ ⎝α β γ⎠

⇒

⎛ 1 1 1 ⎞ 0 = ⎜⎜ 2 + 2 + 2 ⎟⎟ + (2 × 2) ; ∴ γ ⎠ ⎝α β

[Alternative We have 2009

Also,

and

x3

+

2x2

α–2 + β–2 + γ–2 = – 4

]

α β γ

+1=0

−2 2009 αβ + βγ + γα = 0

α+β+γ=

αβγ =

..... (1)

−1 2009

α 2β 2 + β 2 γ 2 + γ 2α 2 Now, 2 + 2 + 2 = ....... (2) β γ α 2β 2 γ 2 α But (αβ + βγ + γα)2 = α2β2 + β2γ2 + γ2α2 + 2αβγ (α + β + γ) Using (1), we get α2β2 + β2γ2 + γ2α2 = – 2 αβγ (α + β + γ) ..... (3)

1

1

1

Putting (3) in (2), we have

1 α2

1

+

β2

+

XI (P1-P7) PRACTICE PAPER [MATHS]

γ2

=

(αβγ ) 2

−2 2009 =–4 −1 2009

− 2×

− 2αβγ (α + β + γ )

1

=

]

Page # 4

MATHEMATICS

Q.12

1 1+ 2 1+ 2 + 3 + ...... + n If Sn = 3 + 3 3 +...... + 3 3 3 , n = 1, 2, 3,...... Then Sn is not greater than 1 1 +2 1 + 2 + 3 + ...... + n 3

(A) 1/2 [Sol.

Sn =

(B) 1

(C*) 2

1 1+ 2 1+ 2 + 3 + + + ............. ; 13 13 + 23 13 + 23 + 33

Tn =

(D) 4

1 + 2 + 3 + 4 + 5.............n 13 + 23 + 33 + ...........n 3

n (n + 1) 1 ⎞ ⎛1 2 2 2 − = ⎜ ⎟ ] = S = ∑ ⇒ 2 n n (n + 1) ⎝ n n +1⎠ ⎡ n (n + 1) ⎤ ⎢⎣ 2 ⎥⎦

Q.13

Let f(θ) =

sin 4 θ + 4 cos 2 θ –

2− 2 2 We have

(A) [Sol.

f(θ) =

∴

(B) –

⎛ 1° ⎞ cos 4 θ + 4 sin 2 θ , then the value of f ⎜11 ⎟ is equal to ⎝ 4 ⎠

2+ 2 2

(C) –

2− 2 2

(D*)

2+ 2 2

sin 4 θ + 4(1 − sin 2 θ) – cos 4 θ + 4(1 − cos 2 θ) = (2 − sin 2 θ) 2 – = (2 – sin2θ) – (2 – cos2θ) = cos2θ – sin2θ = cos 2θ

(2 − cos 2 θ) 2

1° 1 + cos 2θ ⎛ 1° ⎞ f ⎜11 ⎟ = cos 22 ; Now, cos2θ = ...........(1) 2 2 ⎝ 4⎠

1 1+ 2 +1 2 +1 1° 1° 2 = = × Put θ = 22 in equation (1), we get cos2 22 = 2 2 2 2 2 2 2 ∴

⎛ 1° ⎞ 1° f ⎜11 ⎟ = cos 22 = 2 ⎝ 4⎠

2 2+ 2 = 2 4

π 2+ 2 (If 0 < θ < then cos θ is positive) 2 2

]

[COMPREHENSION TYPE] Q.14 to Q.16 are based upon a paragraph. Each question has four choices (A), (B), (C), (D) out of which ONLY ONE is correct. Paragraph for question nos. 14 to 16

x 2 + 3x + 1 Consider a rational function f(x) = 2 and a quadratic function x + x +1 g(x) = x2 – (m + 1) x + m – 1, where m is a parameter. Q.14 Q.15 Q.16

Number of integral value(s) of 'm' so that g(x) is always positive, is (A*) 0 (B) 1 (C) 2

(D) more than 2

Number of integral value(s) in the range of f (x), is (A) 1 (B) 2 (C*) 3

(D) more than 3

If both roots of g(x) = 0 are greater than the smallest value of the function f (x), then 'm' lies in the interval (A) ( – ∞, – 2)

1⎞ ⎛ (B) ⎜ − ∞, − ⎟ 4⎠ ⎝

XI (P1-P7) PRACTICE PAPER [MATHS]

(C) (– 2, ∞)

⎛ 1 ⎞ (D*) ⎜ − , ∞ ⎟ ⎝ 2 ⎠

Page # 5

MATHEMATICS [Sol.(i)

We have g(x) = x2

– (m + 1) x + m – 1 Now, discriminant = (m + 1)2 – 4(m – 1) = m2 – 2m + 5, which is always positive. ⇒ g(x) always positive is not possible for any integral value of m.

x 2 + 3x + 1 We have y = 2 x + x +1 Since x ∈ R, so D ≥ 0

(ii)

⇒

(y – 1) x2 + (y – 3) x + (y – 1) = 0

⇒ (y – 3)3 – 4(y – 1)2 ≥ 0 ⇒ (– y – 1) (3y – 5) ≥ 0 ⇒ (y + 1) (3y – 5) ≤ 0 ⇒ – 1 ≤ y ≤

5 3

5⎤ ⎡ Range of f(x) = ⎢ − 1, ⎥ ;Clearly range of f(x) contains three integral values viz. – 1, 0, 1. 3⎦ ⎣ Different possibilities are as follows : –

∴ (iii)

g(x)

g(x) x

or

x

–1

–1

If both roots of g(x) = 0 are greater than – 1 then 3 conditions should be satisfied simultaneously. (1) D ≥ 0

(B)

−B >–1 2A

Now, (1) ⇒ (m + 1)2 – 4 (m – 1) ≥ 0 (2) ⇒

m +1 >–1 2

⇒

(C) g(– 1) > 0 ⇒

m2 – 2m + 5 ≥ 0, ∀m∈ R .

m>–3

⎛ 1 ⎞ 1 ; Hence from (1), (2) & (3), we get m ∈ ⎜ − , ∞ ⎟ ] 2 ⎝ 2 ⎠ [REASONING TYPE]

and (3) ⇒ 1 + (m + 1) + m – 1 > 0 ⇒ m > –

Q.17 & Q.18 are Reasoning type questions, 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. Statement-1: In Δ ABC, sin 2A + sin 2B + sin 2C is always positive. because Statement-2: In Δ ABC, sin 2A + sin 2B + sin2C = 8 sinA sinB sinC. (A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1. (B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1. (C*) Statement-1 is true, statement-2 is false. (D) Statement-1 is false, statement-2 is true. [Sol. We know that in Δ ABC, ; sin 2A + sin 2B + sin 2C = 4 sinA sinB sinC As A,B,C∈(0, π) ; ∴ sin 2A + sin 2B + sin 2C is always positive ⇒ Statement-1 is true, statement-2 is false] Q.18 Consider f(x) = x2 – (a + b)x + 2, where a, b ∈ R. Statement-1: If f(x) = 0 does not have two distinct real roots then the minimum value of a + b is 3. because Statement-2: If ax2 + bx + c = 0 (a, b , c ∈ R and a ≠ 0) does not have two distinct real roots then either f(x) ≥ 0 ∀ x ∈ R or f(x) ≤ 0 ∀ x ∈ R. (A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1. (B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1. (C) Statement-1 is true, statement-2 is false. (D*) Statement-1 is false, statement-2 is true. [Sol. We have f(x) = x2 – (a + b)x + 2 Q.17

XI (P1-P7) PRACTICE PAPER [MATHS]

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MATHEMATICS As f(x) = 0 does not have 2 distinct real roots and f(0) = 2 ∴ f(x) ≥ 0 ∀ x ∈ R In particular f(–1) ≥ 0 ⇒ 1 + (a + b) + 2 ≥ 0 ⇒ a + b ≥ – 3 Hence the least value of a + b is –3 Statement-1 is false and Statement -2 is true. ] [MULTIPLE OBJECTIVE TYPE] Q.19 to Q.25 has four choices (A), (B), (C), (D) out of which ONE OR MORE THAN ONE is/are correct. ⎛

6

Q.19

If f(θ) =

∑ cosec ⎜⎝ θ + n =1

π (n − 1)π ⎞ nπ ⎞ ⎛ ⎟ cosec ⎜ θ + ⎟ , where 0 < θ < , 2 4 ⎠ 4 ⎠ ⎝

then minimum value of f (A) lies between 3 and 4 (C*) occurs when θ = [Sol.

f(θ) =

=

Q.21

π 6

2

∑

n =1

=

6

2

2 (cot θ + tan θ) =

π⎞ ⎛ f min . ⎜ θ = ⎟ = 2 2 4⎠ ⎝

∑

n =1

⎡ ⎛ π⎞ nπ ⎞⎤ ⎛ ⎢cot ⎜ θ + (n − 1) 4 ⎟ − cot ⎜ θ + 4 ⎟⎥ ⎠ ⎝ ⎠⎦ ⎣ ⎝

(

)

2 ⎡ ⎤ 2 ⎢⎣ tan θ − cot θ + 2⎥⎦

]

x 2 − 2x + 2 For all real x, the value of the rational function f (x) = can lie in the interval 2x − 2 (A*) (– ∞, – 1) (B) (–1, 1) (C*) (1, 2) (D*) (2, ∞) Y x 2 − 2x + 2 y= 2x − 2 (0,1) ∴ x2 – 2x + 2 = 2xy – 2y 2 ⇒ x – 2x(y + 1) + 2(y + 1) = 0 O As x ∈ R, so D ≥ 0 (–1,0) 2 ⇒ (y + 1) – 2(y + 1) ≥ 0 ⇒ (y + 1)(y – 1) ≥ 0 hence y can not lie in (–1, 1) ]

Let E = cos2 (A*)

[Sol.

(D) occurs when θ =

⎡⎛ π ⎞⎤ nπ ⎞ ⎛ sin ⎢⎜ θ + ⎟ − ⎜ θ + (n − 1) ⎟⎥ 4 ⎠ ⎝ 4 ⎠⎦ ⎣⎝ π⎞ nπ ⎞ = ⎛ ⎛ sin ⎜ θ + (n − 1) ⎟ sin ⎜ θ + ⎟ 4⎠ 4 ⎠ ⎝ ⎝

⎛ ⎛ 3π ⎞ ⎞ 2 ⎜⎜ cot θ − cot ⎜ θ + 2 ⎟ ⎟⎟ ⎝ ⎠⎠ ⎝

∴

[Sol.

π 4

We have

6

Q.20

(B*) lies between 2 and 3

(2,1) 1

2

⇒

X

y ≥ 1 or y ≤ – 1

π 2π 3π + cos2 + cos2 . Then which of the following alternative(s) is/are incorrect? 7 7 7

1 3 cos x · sin3x, 0 ≤ x ≤ 2π, is

(P)

3π ⎤ ⎡ π π ⎤ ⎡ 3π ⎤ ⎡ ⎢ − π, − 4 ⎥ ∪ ⎢− 4 , 4 ⎥ ∪ ⎢ 4 , π⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦

(B)

4 sin2x – 8 sin x + 3 ≤ 0, 0 ≤ x ≤ 2π, is

(Q)

⎡ 3π ⎤ ⎢ 2 , 2π⎥ ∪ {0} ⎣ ⎦

(C)

| tan x | ≤ 1 and x ∈ [– π, π] is

(R)

⎛ π⎞ ⎜ 0, ⎟ ⎝ 4⎠

(D)

cos x – sin x ≥ 1 and 0 ≤ x ≤ 2π is

(S)

⎡ π 5π ⎤ ⎢6 , 6 ⎥ ⎣ ⎦ [Ans. (A) R; (B) S; (C) P; (D) Q]

(A)

sin x · cos x (cos2x – sin2x) > 0

⇒

1 sin 2x · cos 2x > 0 ⇒ sin 4x > 0; ∴ 0 < 4x < π ⇒ 2 Here, (2 sin x – 1)(2 sin x – 3) ≤ 0 ; but 2 sin x – 3 is always negative.

(B)

∴

(C)

2 sin x – 1 ≥ 0 ⇒

sin x ≥ 1/2

∴ from the figure, π/6 ≤ x ≤ 5π/6 ⇒ (S) – 1 ≤ tan x ≤ 1. The value scheme for this is as shown below:

from the figure, – ∴ (D)

(R)

π π ≤x≤ 4 4

or

3π ⎤ ⎡ π π ⎤ ⎡ 3π ⎤ ⎡ x ∈ ⎢ − π, − ⎥ ∪ ⎢ − , ⎥ ∪ ⎢ , π⎥ 4 ⎦ ⎣ 4 4⎦ ⎣ 4 ⎦ ⎣

–π≤x≤– ⇒

3π 4

or

3π ≤ x ≤ π] 4

(P)

1 π⎞ ⎛ cos ⎜ x + ⎟ > . The value for this is as shown below: 4⎠ 2 ⎝

XI (P1-P7) PRACTICE PAPER [MATHS]

Page # 10

∴

MATHEMATICS π π π from the figure − π ≤ x + π ≤ π and, in general 2nπ – ≤ x + ≤ 2nπ + 4 4 4 4 4 4 π π 3π 2nπ – ≤ x ≤ 2nπ ; for n = 0 – ≤ x ≤ 0; for n = 1, ≤ x ≤ 2π⇒ (Q) ] 2 2 2

Q.2 is "Match the Column" type. Column-I contains Four entries and column-II contains Five entries. Entry of column-I are to be matched with one or more than one entries of column-II or vice versa. Q.2 (A)

(B)

Column-I If exactly one root of the quadratic equation (a2 – 4a + 3)x2 + (a2 – 5a + 6)x + (a + 5) = 0 lies at infinity then the possible integral value(s) of 'a' is equal to The smallest natural number 'n' so that (2 – n) x2 – 8x – 4 – n < 0, ∀ x ∈ R is equal to

Column-II (P) 1

(Q)

2

π then the maximum value of 4 sin B sin C is equal to (R) 3 2 (D) Number of values of 'k' so that the equation (S) 4 2 2 3 2 2 (k – 3) (k – 4) (k + 1) x – (k – 5k + 6k) x + (k – 9) = 0 has more than two unequal roots is equal to (T) 5 [Ans. (A) P ; (B) T ; (C) Q ; (D) P ] 2 [Sol.(A) If one root of the quadratic equation Ax + Bx + C = 0, lies at infinity then A = 0 and B ≠ 0. ∴ Coefficient of x2 = 0 ⇒ a2 – 4a + 3 = 0 ⇒ (a – 1) (a – 3) = 0 ∴ a=1;a=3 But for a = 3, coefficient of x = 0; ∴ a = 3 (rejected think ?) ⇒ a = 1 (B) We have (n – 2)x2 + 8x + (n + 4) > 0. ∀x∈ R ⇒ n – 2 > 0 and D < 0 ⇒ 64 – 4(n – 2) (n + 4) < 0 ⇒ n2 + 2n – 24 > 0 ⇒ (n + 6) (n – 4) > 0 ⇒ n > 4 as n ∈ N ∴ nsmallest = 5

(C)

If in ΔABC, ∠A =

π π , so B + C = 2 2 Now, 4sin B sin C = 2(2sin B sin C) = 2[cos (B – C) – cos (B + C)] = 2 cos (B – C) ≤ 2

(C)

A=

π (each). 4 We must have (k – 3) (k2 – 4) (k + 1) = 0, k3 – 5k2 + 6k = 0 and k2 – 9 = 0 Hence clearly k = 3 only. ]

∴

(D)

Maximum value = 2, when B = C =

PART-C [SUBJECTIVE] Q.1 & Q.6 are "Subjective" type questions. (The answer to each of the questions are upto 4 digit) Q.1

If in a Δ ABC , a = 6, b = 3 and cos(A − B) = 4/5 then find its area.

[Sol:

⎛A−B⎞ ⎛a −b ⎞ ⎟ = ⎜ ⎟ cot C / 2 tan ⎜ ⎝ 2 ⎠ ⎝a+b⎠

XI (P1-P7) PRACTICE PAPER [MATHS]

[Ans. 0009 sq. unit]

(using Napier's Analogy)

Page # 11

MATHEMATICS 1 − cos( A − B) ⎛3⎞ = ⎜ ⎟ cot C/2 1 + cos(A − B) ⎝9⎠

(using tanθ =

1 1− 4 / 5 = cot C/2 3 1+ 4 / 5

Now, Δ = ½ ab sinC

C C 1 1 = cot ⇒ cot = 1 3 3 2 2 ⇒ C = 90º

Q.2

1 − cos 2θ ) 1 + cos 2θ

⇒ ½ × 3 × 6 sin 90º =9

]

Let A denotes the value of expression x4 + 4x3 + 2x2 – 4x + 7, when x = cot and B denotes the value of the expression

11π 8

1 − cos 8θ 1 + cos 8θ + , when θ = 9° tan 2 4θ cot 2 4θ

Find the value of (AB). [Ans 12] [Sol. (A) We have 3π ⎞ ⎛ 11π 3π = cot ⎜ π + ⎟ = cot = 8 ⎠ 8 8 ⎝ ⇒ (x + 1)2 = 2 ∴ x2 + 2x – 1 = 0 Now, consider x4 + 4x3 + 2x2 – 4x + 7

x = cot

2 −1

x 2 + 2 x − 1) + 2x3 + 3x2 – 4x + 7 = x2 (1 4243 ( = 0)

x 2 + 2 x − 1) + 6 = 2x3 + 3x2 – 4x + 7 = 2x ( x 2 + 2 x − 1) – x2 – 2x + 7 = – x2 – 2x + 7 = – (1 4243 14243 ( = 0)

( = 0)

∴ (B)

A=6

1 − cos 8θ 1 + cos 8θ 2 sin 2 4θ 2 cos 2 4θ + = + = 2 (cos2 4θ + sin2 4θ) = 2 2 2 tan 2 4θ cot 2 4θ tan 4θ cot 4θ AB = 12 Ans.]

We have, ∴

Q.3

Find the sum of all integers satisfying the inequalities log5(x – 3) +

[Sol.

1 1 log53 < log5(2x2 – 6x + 7) and log3x + log 2 2

3

x + log 1 x < 6.[Ans.0039 ] 3

⎛ 2 x 2 − 6x + 7 ⎞ 1 ⎟ We have log5(x – 3) < log5 ⎜⎜ ⎟ 3 2 ⎝ ⎠ 2x 2 − 6 x + 7 ⇒ (x – < ⇒ x2 – 12x + 20 < 0 ⇒ (x – 2) (x – 10) < 0 3 ⇒ 2 < x < 10 But for the domiain , x > 3, so x ∈ (3, 10) Also log3x + 2log3x – log3x < 6 ⇒ 2log3x < 6 ∴ log3x < 3

3)2

XI (P1-P7) PRACTICE PAPER [MATHS]

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MATHEMATICS ⇒ 0 < x < 27 so x ∈ (0, 27) ; ∴ Possible integers are 4, 5, 6, 7, 8, 9. ;Hence sum of integers = 39.] 2π π⎞ ⎛ 4π 7π A denotes the value of expression 4 ⎜ cos + cos – cos – cos ⎟ 15 15 ⎠ 15 15 ⎝ and B denotes the value of 8 cot (α + β + γ), where tan α, tan β, tan γ are the real roots of the cubic x3 – 8(a – b) x2 + (2a – 3b) x – 4(b + 1) = 0. Find absolute value of (AB). [Ans. 0004 ] [Sol.A) We have 4[(cos 24° + cos 48°) – (cos 84° + cos 12°)] = 4[2 cos 36° cos 12° – 2 cos 48° cos 36°] = 8 cos 36° [cos 12° – cos 48°] = 8 cos 36° [2 sin 30° sin 18°]

Q.4

Let

1 5 +1 × × 2 4

= 16 × (B)

5 − 1 5 −1 4 = = =2 2 2 4

x3 – 8(a – b) x2 + (2a – 3b) x – 4(b + 1) = 0

We have

tan α tan β tan γ Σ tan α = 8(a – b) ;

Σ tan α tan β = (2a – 3b) and ∏ tan α = 4(b + 1)

Σ tan α − ∏ tan α 8(a − b) − 4(b + 1) 4(2a − 2b − b − 1) Now, tan (α + β + γ) = 1 − Σ tan α tan β = = 1 − (2a − 3b) (1 − 2a + 3b)

4(2a − 3b − 1) 1 = – 4 ⇒ cot (α + β + γ) = – ; ∴ 8 cot (α + β + γ) = – 2Hence | AB | = 4 ] (1 − 2a + 3b) 4 If k1 and k2 are the two values of 'k' where k1 < k2 for which the expression f(x, y) = x2 + 2xy + 4y2 + 2kx – 6y + 3 can be resolved as a product of two linear factors then find the value of 5k2 – 4k1. [Ans. 0006 ] We have A = 1 ; B = 4 ; C = 3 ; F = –3 ; G = k ; H = 1

=

Q.5

[Sol.

Now, ABC + 2FGH – AF2 – BG2 – CH2 = 0 ⇒ ∴ Q.6 [Sol.

3 k1 = – , k2 = 0 ⇒ 2

5k2 – 4k1 = 6 Ans.

k = 0, –

3 2

]

The set of values of ‘c’ for which the equation x 2 − 4x − c − 8x 2 − 32 x − 8c = 0 has exactly two distinct real solutions, is (a, b) then find the value of (b – a). [Ans. 8] 2 We have x – 4x – c ≥ 0 ⇒ D≤0 ∴ 16 + 4c ≤ 0 ⇒ 4c ≤ – 16 ⇒ c ≤ – 4 Let

f(x) = x2 – 4x – c ; ∴ x2 – 4x – c –

8x 2 − 32x − 8c = 0

(

)

8f (x ) = 0 ; ∴ f (x) f (x) − 8 = 0 ∴ f(x) = 0 or f(x) = 8 Case-I f(x) = 0 has two distinct real solutions and f(x) = 8 has no real solution. ∴ For f(x) = 0, D > 0 and for f(x) = 8, D < 0 ⇒16 + 4c > 0 and 16 + 4(c + 8) < 0 ⇒ c > – 4 and c < – 12 ; Q No common solution for 'c' exist. ; ∴ c ∈ φ Case-II f(x) = 0 has no real roots but f(x) = 8 has two distinct real roots ∴ For f(x) = 0, D < 0 and for f(x) = 8, D > 0 ⇒ 16 + 4c < 0 and 16 + 4(c + 8) > 0 ⇒ c < – 4 and c > – 12 ∴ c ∈ (– 12, – 4) Hence b – a = – 4 – (– 12) = – 4 + 12 = 8 ] ⇒

f(x) –

XI (P1-P7) PRACTICE PAPER [MATHS]

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