log + compound angles + qe 11th (pqrs)

QUESTION BANK ON LOGARITHM

COMPOUND ANGLES AND

QUADRATIC EQUATION

Time Limit: 5 Sitting Each of 60 Minutes duration approx.

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MATHEMATICS TARGET IIT JEE 2011

XI (PQRS)

Q.B on Log, Compound angle, Quadratic equation [2]

[STRAIGHT OBJECTIVE TYPE]

Q.1 2/log If 60 a = 3 and 60 b = 5 then the value of ) b 1 ( 2 b a 1

12 − − −

equals (A*) 2 (B) 3 (C) 3 (D) 12

[Sol. 60 a = 3 ⇒ a = log 60 3 [12 th (2082006)] 60 b = 5 ⇒ b = log 60 5

let x = ) b 1 ( 2 b a 1

12 − − −

log 12 x = ) b 1 ( 2 b a 1

− − −

= ) b 1 ( 2 ) b a ( 1

− + −

= ) 5 log 60 (log 2 ) 5 log 3 (log 1

60 60

60 60

− + −

= ( )

( ) 5 log 1 2 15 log 1

60

60

− −

= 12 log 2 4 log

60

60

= 2 1 log 12 4 = log 12 2 (a + b = log 16 15)

∴ log 12 x = log 144 4 = log 12 2 ⇒ x = 2 Ans. ]

Q.2 2/ph1 If x + y = 3 – cos4θ and x – y = 4 sin2θ then

(A) x 4 + y 4 = 9 (B) 16 y x = +

(C) x 3 + y 3 = 2(x 2 + y 2 ) (D*) 2 y x = + [Sol. On adding and subtracting

x = 2 2 sin 4 4 cos 3 θ + θ −

; y = 2 2 sin 4 4 cos 3 θ − θ −

x = 2 ) 4 cos 1 ( ) 2 sin 1 ( 4 θ + − θ +

; y = 2 ) 4 cos 1 ( ) 2 sin 1 ( 4 θ + − θ −

x = 2 (1 + sin2θ ) – cos 2 2θ ; y = 2 (1 – sin2θ) – cos 2 2θ x = 1 + 2 sin2θ + sin 2 2θ ; y = 1 – 2 sin2θ + sin 2 2θ

x = (1 + sin2θ) 2 ; y = (1 – sin2θ) 2 ⇒ 2 y x = + ]

[Alternate : Or put θ = 4 π and verify ]

Q.3 2/qe Let A = x | x 2 + (m – 1)x – 2(m + 1) = 0, x∈ R B = x | (m – 1)x 2 + mx + 1 = 0, x∈ R

Number of values ofm such that A∪B has exactly 3 distinct elements, is (A) 4 (B) 5 (C) 6 (D*) 7

[Sol. CaseI: when B is a quadratic equation [13 th (2782006)] D 1 = (m + 3) 2 and D 2 = (m – 2) 2 roots of 1 st equation are 2, – (m + 1) set A

roots of 2 nd equation are – 1, m 1 1

− set B

For exactly there elements in A∪ B two of the roots must be same note that 2 ≠ – 1 possibilities are

2 = – (m + 1) ⇒ m = – 3

2 = m 1 1

− ⇒ 2 – 2m = 1 ⇒ m = 1/2


– m – 1 = – 1 ⇒ m = 0

– (m + 1) = m 1 1

− ⇒ 1 – m 2 = – 1 ⇒ m = ± 2

m 1 1

− = – 1 ⇒ m – 1 = 1 ⇒ m = 2.

CaseII: Now if m = 1, then B becomes linear roots of B as x = – 1 roots of A are 2 and – 2

⇒ 3 elements in common

∴ all permissiblem are – 3, 2 1 , 2 , – 2 , 2, 0, 1 ]

Q.4 4/qe If a 2 + b 2 + c 2 = 1 then ab + bc + ca lies in the interval (a, b, c ∈ R)

(A)

2 ,2 1

(B) [–1, 2] (C*)

− 1 ,

2 1

(D)

−

2 1 , 1

[Hint: ∑ (a–b) 2 > 0 ⇒ 2∑ a 2 – 2∑ ab > 0 ⇒ ∑ ab < ∑ a 2 ⇒ ab + bc + ca < 1 Also note that (a + b + c) 2 > 0 ]

Q.5 3/ph1 If tanB = A cos n 1 A cos A sin n

2 − then tan(A + B) equals

(A*) A cos ) n 1 ( A sin

− (B) A sin A cos ) 1 n ( −

(C) A cos ) 1 n ( A sin

− (D) A cos ) 1 n ( A sin

+

[Sol. tan(A + B) = B tan A tan 1 B tan A tan

− +

=

A cos n 1 A cos A sin n∙ A tan 1

A cos n 1 A cos A sin n A tan

2

2

− −

− +

= A cos A sin n ) A cos n 1 ( A cos A cos A sin n ) A cos n 1 ( A sin

2 2

2 2

− − + −

= ) A sin n A cos n 1 ( A cos 0 A sin

2 2 − − −

= A cos ) n 1 ( A sin

− ]

Q.6 6/qe Ifα and β are the roots of x 2 + px + q = 0 and α 4 , β 4 are the roots of x 2 − rx + s = 0, then the equation x 2 − 4qx + 2q 2 − r = 0 has always (p, q, r, s ∈ R) : (A*) two real roots (B) two positive roots (C) two negative roots (D) one positive and one negative root.

[Hint : α + β = − p ; α β = q ; α 4 + β 4 = r ; α 4 β 4 = s Now D = 16 q 2 − 4 (2 q 2 − r) = 8 q 2 + 4 r = 4 (4 q 2 + r)

= 2 [ 4 α 2 β 2 + α 4 + β 4 ] = 2 [ (α 2 + β 2 ) 2 + 2 α 2 β 2 ] = 2 [ α + β) 2 − 2 α β 2 + 2 α 2 β 2 ]= 2 (p 2 − 2 q) 2 + 2 q 2 ] > 0

Again consider the product of the roots = 2 q 2 − r which can be either positive or negative. Hence A ]


Q.7 8/qe The natural number n for which the expression y = 5(log 3 n) 2 – log 3 n 12 + 9, has the minimum value is

(A) 2 (B) 3 (C) 3 6/5 (D*) 4 [Sol. Let log 3 n = x [11th, 23122007]

y = 5x 2 – 12x + 9

y is minimum at x = – a 2 b

= 10 12

= 5 6

Here log 3 n = 5 6

⇒ n = 3 6/5 ≅ 3.70s

which is not natural hence minimum occurs at the closest integer now 4 > 3 6/5

4 5 > 3 6 1024 > 729 which is true ]

Q.8 4/ph1 Given a 2 + 2a + cosec 2 π 2 ( ) a x +

F H G

I K J = 0 then, which of the following holds good?

(A) a = 1 ; x I 2

∈ (B*) a = –1 ; x I 2

∈

(C) a ∈ R ; x ∈φ (D) a , x are finite but not possible to find

[Sol. (a+1) 2 + cosec 2 π π a x 2 2

+ F H G

I K J – 1 = 0

or (a+1) 2 + cot 2 π π a x 2 2

+ F H G

I K J = 0

from option [B] If a = –1 ⇒ tan 2 πx/2 = 0 ⇒ x/2∈ I ]

Q.9 9/qe Ifα andβ be the roots of the equation x 2 + 3x + 1 = 0 then the value of 2 2

1 1

+ α β

+

β +

α is equal to

(A) 15 (B*) 18 (C) 21 (D) none [Sol. α + β = – 3; αβ = 1, also α 2 + 3α + 1 = 0 and β 2 + 3β + 1 = 0 [13 th 3072006]

where α 2 = – (3α + 1) and β 2 = – (3β + 1)

E = 2

2

2

2

) 1 ( ) 1 ( + α β

+ β +

α

E = 2

2

2 1 β + β + α

+ 2

2

2 1 α + α + β

=

β − + α − ) 1 3 (

+

α − β + − ) 3 1 (

y = β α + 3 1 +

α β + 3 1 = αβ

β + β + α + α ) 3 1 ( ) 3 1 ( (as αβ = 1)

= 3(α 2 + β 2 ) + (α + β) = 3[9 – 2] + (–3) = 21 – 3 = 18 Ans. ]


Q.10 10/qe If the equation cot 4 x – 2 cosec 2 x + a 2 = 0 has atleast one solution then, sum of all possible integral values of 'a' is equal to (A) 4 (B) 3 (C) 2 (D*) 0

[Sol. cot 4 x – 2(1 + cot 2 x) + a 2 = 0 [11th, 03082008, P1] ⇒ cot 4 x – 2 cot 2 x + a 2 – 2 = 0 ⇒ (cot 2 x – 1) 2 = 3 – a 2 to have atleast one solution

3 – a 2 ≥ 0 ⇒ a 2 – 3 ≤ 0

a ∈ [– 3 , 3 ] integral values – 1, 0, 1 ∴ sum = 0 Ans. ]

Q.11 11/ph1 If A = sin 2 7 π + sin

4 7 π + sin

8 7 π and B = cos

2 7 π + cos

4 7 π + cos

8 7 π then A B 2 2 + is equal

to (A) 1 (B*) 2 (C) 2 (D) 3 [Quiz]

[Hint: A 2 + B 2 = 3 + 2 cos cos cos 2 7

4 7

6 7

π π π + +

= 3 + 2 −

1 2 = 2 ⇒ A B 2 2 + = 2 ]

Q.12 12/qe If the equation 4x 2 – 4(5x + 1) + p 2 = 0 has one root equals to two more than the other, then the value of p is equal to

(A) ± 3 236

(B*) ± 5 (C) 5 or – 1 (D) 4 or – 3

[Hint: 4x 2 – 4(5x + 1) + p 2 = 0 4x 2 – 20x + (p 2 – 4) = 0 two roots are α, α + 2

∴ 2α + 2 = 4 20

= 5 ⇒ α + 1 = 2 5

⇒ α = 1 2 5

− ⇒ α = 2 3

∴ α (α + 2) = 4 4 p 2 −

2 3

+ 2 2 3

= 4 4 p 2 −

⇒ 2 3 ∙ 2 7 =

4 4 p 2 −

⇒ 21 = p 2 – 4

⇒ p 2 = 25 ⇒ p = ± 5 ]

Q.13 20/qe The minimum value of the expression | x – p | + | x – 15 | + | x – p – 15 | for 'x' in the range p ≤ x ≤ 15 where 0 < p < 15, is (A) 10 (B*) 15 (C) 30 (D) 0

[Sol. | x – p | = x – p (since x ≥ p) [11 th (782005)] | x – 15 | = 15 – x (since x ≤ 15) | x – (p + 15) | = (p + 15) – x (as 15 + p > x)


∴ expression reduces to E = x – p + 15 – x + p + 15 – x E = 30 – x

∴ E min occurs when x = 15 ∴ E min = 15 Ans.]

Q.14 22/qe If a, b, c are real numbers satisfying the condition a + b + c = 0 then the roots of the quadratic equation 3ax 2 + 5bx + 7c = 0 are : (A) positive (B) negative (C*) real and distinct (D) imaginary

[Hint: D = 25b 2 – 84 ac = 25(a + c) 2 – 84ac using b = –(a + c) = 21[(a+c) 2 – 4ac] + 4(a+c) 2 > 0 ]

Q.15 25/ph1 The set of angles btween 0 and 2π satisfying the equation 4 cos 2 θ − 2 2 cosθ − 1 = 0 is

(A) π π π π 12

5 12

19 12

23 12

, , , R S T U V W (B*) π π π π

12 7 12

17 12

23 12

, , ,

(C) 5 12

13 12

19 12

π π π , , R S T U V W (D)

π π π π 12

7 12

19 12

23 12

, , , R S T U V W

[Sol. 4 cos 2 θ – 2 2 cosθ – 1 = 0 [11th 15102006 (P, J)]

cosθ = 8

16 8 2 2 + ± =

4 6 2 ±

cosθ = 4

2 6 + 12 23

12 2 ;

12 π

= π

− π π

= θ ⇒

cosθ = 4

2 6 − −

cosθ = cos(π–5π/12) ; cos(π+5π/12) θ = 7π/12 ; 17π/12 ⇒ (B) ]

Q.16 7/log Let ABC be a triangle right angled at C. The value of a log ∙ a log a log a log

b c c b

b c c b

− +

− + + (b + c≠ 1, c – b≠ 1) equals

(A) 1 (B*) 2 (C) 3 (D) 1/2

[Sol. ( )

) b c log( ∙ ) b c log( a log ∙ a log ∙ ) b c log( ∙ ) b c log(

) b c log( ) b c [log( a log

− + − +

+ + − = a log

) b c log( 2 2 − [Transit DPP]

given c 2 = a 2 + b 2 ⇒ c 2 – b 2 = a 2

∴ a log a log 2

= 2 Ans.

N r = ) b c ( log

1 ) c b ( log

1

a a − +

+ = ) b c ( log ∙ ) c b ( log

) b c ( log

a a

2 2 a

− + −


D r = ) b c ( log 1

) c b ( log 1

a a − ×

+

r

r

D N

= log a (c 2 – b 2 ) = log a (a 2 ) = 2 Ans. ]

Q.17 23/qe The roots of the equation a (x − b) (x − c) + b (x − c) (x − a) + c (x − a) (x − b) = 0 (a, b, c are distinct and real ) are always : (A) positive (B) negative (C*) real (D) unreal

[Sol. a(x 2 –(b+c)x + bc) + b(x 2 –(c+a)x + ac ) + c(x 2 – (a+b)x + ab) = 0 (a + b + c)x 2 – 2x(ab + bc + ca) + 3abc = 0

D = 4(ab + bc + ca ) 2 – 12abc (a + b + c) = 4[a 2 b 2 + b 2 c 2 + c 2 a 2 + 2abc(a + b + c) – 3abc(a + b + c) ] = 4[a 2 b 2 + b 2 c 2 + c 2 a 2 – abc(a + b + c) ] = 2 [ (ab – bc) 2 + (bc – ca) 2 + (ca – ab) 2 ] > 0 ]

Q.18 34/ph1 In a triangle ABC, angle B < angle C and the values of B and C satisfy the equation 2 tanx k (1 + tan 2 x) = 0 where (0 < k < 1) . Then the measure of angle A is : (A) π/3 (B) 2π/3 (C*) π/2 (D) 3π/4

[Sol. k = x tan 1 x tan 2 2 + = sin2x ⇒ sin2C = sin2B

But ∠ C > ∠ B 2C = π – 2B ⇒ B + C = π/2 ∴ ∠A = π/2 Ans ]

Q.19 26/qe If one solution of the equation x 3 – 2x 2 + ax + 10 = 0 is the additive inverse of another, then which one of the following inequalities is true? (A) – 40 < a < – 30 (B) – 30 < a < – 20 (C) – 20 < a < – 10 (D*) – 10 < a < 0

[Sol. If α, β, γ are the roots then α + β + γ = 2; also α + β = 0 (where α, β are additive inverse) ∴ γ = 2 which must satisfy the given equation [13th, 05082007] ∴ a = – 5 ⇒ (D) ]

Q.20 29/qe Suppose a, b, and c are positive numbers such that a + b + c = 1. Then the maximum value of ab + bc + ca is

(A*) 3 1

(B) 4 1

(C) 2 1

(D) 3 2

[Sol. a 2 + b 2 + c 2 = 1 – 2 ∑ ab ....(1) also (a – b) 2 ≥ 0 etc. hence a 2 + b 2 + c 2 ≥ ab + bc + ca

1 – 2∑ ab ≥ ∑ ab 1 ≥ 3∑ ab

∴ ∑ ab ≤ 3 1

Ans. ] [18122005, 12 th , 13 th ]


Q.21 30/qe The roots of (x − 1) (x − 3) + K (x − 2) (x − 4) = 0, K > 0 are : (A*) real (B) real and equal (C) imaginary (D) one real and one imaginary

[Hint : check f(1) , f(2) , f(3) and f(4) and interpret note that one root lie between 1 and 2 and the other between 3 and 4 ]

Q.22 36/ph1 If cos α = β − − β

cos 2 1 cos 2 then tan 2

α ∙ cot 2

β has the value equal to where α, β ∈ (0, π)

(A) 2 (B) 2 (C) 3 (D*) 3

[Hint : 1 cosα

= 2 2 1

− −

cos cos

β β

Applying C/D ⇒ 1 1

− + cos cos

α α = ( ) 3 1

1

−

+ cos cos

β

β

⇒ tan 2 α 2 = 3 tan 2 β

2 ⇒ tan 2 α

2 cot 2 β

2 = 3 ]

Q.23 31/qe Assume that p is a real number. In order for 3 1 p 3 x + + – 3 x = 1 to have real solutions, it is necessary that (A) p ≥ 1/4 (B*) p ≥ – 1/4 (C) p ≥ 1/3 (D) p ≥ – 1/3

[Sol. 3 1 p 3 x + + = 3 x + 1 let 3 x = h [12th , 13th 1922006]

3 1 p 3 x + + = h + 1 x + 3p + 1 = h 3 + 3h 2 + 3h + 1 h 3 + 3p + 1 = h 3 + 3h 2 + 3h + 1 3h 2 + 3h – 3p = 0 h 2 + h – p = 0 (for real solution D ≥ 0 i.e. b 2 – 4ac ≥ 0) b 2 – 4ac = 1 + 4p ≥ 0 or p ≥ – 1/4 Ans.

Alternatively : 3 1 p 3 x + + + (– 3 x ) + (–1) = 0 if a + b + c = 0 ⇒ a 3 + b 3 + c 3 = 3abc x + 3p + 1 – x – 1 = 3 [(x + 3p + 1)(x)] 1/3 3p = 3[(x + 3p + 1)(x)] 1/3

⇒ p 3 = x(x + 3p + 1) ∴ x 2 + (3p + 1)x – p 3 = 0 for real roots D ≥ 0 ⇒ 4p 3 + 9p 2 + 6p + 1 ≥ 0

(p + 1) 2 (4p + 1) ≥ 0 p ≥ – 1/4 Ans. ]

Q.24 32/qe PQRS is a common diameter of three circles. The area of the middle circle is the average of the area of the other two. If PQ = 2 and RS = 1 then the length QR is (A) 1 6 + (B*) 1 6 − (C) 5 (D) 4

[Sol. Let QR = x

then the diameters are 2, x + 2, x + 3 ⇒ 2

) 1 x ( 2 2 2 + + = (x + 2) 2 [12 , 13 08012006]


∴ 2(x + 2) 2 = 2 2 + (x + 3) 2 2(x 2 + 4 + 4x) = 4 + (x 2 + 6x + 9) x 2 + 2x – 5 = 0 ⇒ x = 1 6 − Ans.]

Q.25 37/ph1 If a sin x + b cos x = 1 and a 2 + b 2 = 1 (a, b > 0), then consider the following statements: I sin x = a II tan x = a/b III tan x = b (A*) only III is false (B) only I is true (C) All of I, II, III must be true (D) None of I, II or III is correct.

[Sol. b 2 cos 2 x = (1 – a sin x) 2 [12th, 19072007] b 2 (1 – sin 2 x) = 1 + a 2 sin 2 x – 2a sin x (a 2 + b 2 ) sin 2 x – 2a sin x + 1 – b 2 = 0 sin 2 x – 2a sin x + a 2 = 0 (a – sin x) 2 = 0; sin x = a ⇒ tan x = b a Ans. ]

Q.26 33/qe If every solution of the equation 3 cos 2 x – cos x – 1 = 0 is a solution of the equation a cos 2 2x + bcos2x – 1 = 0. Then the value of (a + b) is equal to (A) 5 (B) 9 (C*) 13 (D) 14

[Sol. 1 st equation gives, [12 , 13 0532006]

2 3 (1 + cos 2x) – 1 = cos x ⇒ 3(1 + cos 2x) – 2 = 2 cos x

3 cos 2x + 1 = 2 cos x (3 cos 2x + 1) 2 = 4 cos 2 x = 2(1 + cos 2x) 9 cos 2 2x + 4 cos 2x – 1 = 0 ....(1)

comparing with a cos 2 2x + b cos 2x – 1 = 0 .....(2)

9 a = 4 b = 1 ⇒ b = 4 and a = 9 ⇒ a + b = 13 Ans. ]

Alternatively: For 1 st equation, cos x = 6 13 1± [12 , 13 0532006]

now cos 2x = 2 cos 2 x – 1 = 36 2 (1 + 13 ± 13 2 ) – 1 =

18 18 13 2 14 − ± =

18 4 13 2 − ± =

9 2 13 − ±

cos 2x 1 = 9 2 13 − and cos 2x 2 = – 9

) 2 13 ( +

Now from 2 nd equation a cos 2 2x + b cos 2x – 1 = 0

cos 2x 1 ∙ cos 2x 2 = – a 1

∴ – a 1 = –

−

81 4 13

= – 9 1

⇒ a = 9

and cos 2x 1 + cos 2x 2 = – a b = – 9

b

∴ – 9 b = – 9

4 ⇒ b = 4

∴ a + b = 13 Ans. ]


Q.27 36/qe Let P (x) = kx 3 + 2k 2 x 2 + k 3 . Find the sum of all real numbers k for which x – 2 is a factor of P(x). (A) 4 (B) 8 (C) – 4 (D*) – 8

[Hint: put x = 2, P (2) = 0, k 3 + 8k 2 + 8k = 0 ⇒ k 1 + k 2 + k 3 = – 8 ] [11 th (782005)]

Q.28 43/ph1 If x cos x 3 cos = 3 1 for some angle x, 0 ≤ x ≤ 2

π , then the value of x sin

x 3 sin for some x, is

(A*) 3 7

(B) 3 5

(C) 1 (D) 3 2

[Sol. Consider, x sin x 3 sin – x cos

x 3 cos = x cos x sin

x sin x 3 cos x cos x 3 sin − = x cos x sin

x 2 sin = 2 ∙ x 2 sin

x 2 sin = 2

so x sin x 3 sin – 3 1 = 2

or x sin x 3 sin = 2 + 3

1 = 3 7

Ans. ] [12 th 25062006]

Q.29 39/qe The sum of all the value of m for which the roots x 1 and x 2 of the quadratic equation

x 2 – 2mx + m = 0 satisfy the condition 2 2

2 1

3 2

3 1 x x x x + = + , is

(A) 4 3

(B) 1 (C) 4 9

(D*) 4 5

[Hint: x 1 + x 2 = 2m ; x 1 x 2 = m [12 , 13 th test (29102005)] (x 1 + x 2 ) 3 – 3x 1 x 2 (x 1 + x 2 ) = (x 1 + x 2 ) 2 – 2x 1 x 2 8m 3 – 3m(2m) = 4m 2 – 2m

8m 3 – 10m 2 + 2m = 0 ⇒ m 1 + m 2 + m 2 = 8 10

= 4 5

]

Q.30 41/qe Let r 1 , r 2 and r 3 be the solutions of the equation x 3 – 2x 2 + 4x + 5074 =0 then the value of (r 1 + 2)(r 2 + 2)(r 3 + 2) is

(A) 5050 (B) 5066 (C*) – 5050 (D) – 5066 [Sol. x 3 – 2x 2 + 4x + 5074 = (x – r 1 )(x – r 2 )(x – r 3 ) [12 , 13 th test (29102005)]

put x = – 2 – 8 – 8 – 8 + 5074 = – (2 + r 1 )(2 + r 2 )(2 + r 3 ) ∴ 5050 = – (2 + r 1 )(2 + r 2 )(2 + r 3 )

(2 + r 1 )(2 + r 2 )(2 + r 3 ) = – 5050 Ans.]

Q.31 35/log The set of values of x satisfying simultaneously the inequalities ( ) ( )

( ) ( ) x x − −

−

8 2

5 1 0.3 2 10 7 log log

≥ 0 and

2 x − 3 − 31 > 0 is : (A*) a unit set (B) an empty set (C) an infinite set (D) a set consisting of exactly two elements .

[Hint : ( ) ( )

( ) ( ) x x − −

−

8 2

5 1 0.3 2 10 7 log log

≥ 0

⇒ (x – 8) (2 – x) > 0 and

− ) 1 5 (log 7 10 log 2 3 . 0 > 0 which is true


(x – 8) (x – 2) < 0 so 2 < x < 8 Also 2 x–3 – 31 > 0 or 2 x–3 > 32 ⇒ 2 x–3 > 2 5 ⇒ x – 3 > 5 ⇒ x > 8

so x = 8 is only common solution ]

Q.32 45/ph1 The graphs of y = sin x, y = cos x, y = tan x and y = cosec x are drawn on the same axes from 0 to π/2. A vertical line is drawn through the point where the graphs of y = cos x and y = tan x cross, intersecting the other two graphs at points A and B. The length of the line segment AB is:

(A*) 1 (B) 2 1 5 −

(C) 2 (D) 2 1 5 +

[Sol. Given tan x = cos x [Quiz] or sin x = cos 2 x = 1 – sin 2 x ....(1)

now, cosec x – sin x = x sin x sin 1 2 − = 1 (from (1) ) ]

Q.33 48/qe Ifα and β are the roots of the equation (log 2 x) 2 + 4(log 2 x) – 1 = 0 then the value of log β α+ log α β equals (A) 18 (B) – 16 (C) 14 (D*) – 18

[Sol. log 2 α + log 2 β = – 4; log 2 α ∙ log 2 β = – 1 [Nucleus 2007]

now log β α + log α β = β α

2

2 log log

+ α β

2

2 log log

= β α

β + α

2 2

2 2

2 2

log ∙ log ) (log ) (log

= – [(log 2 α + log 2 β) 2 – 2 log 2 α ∙ log 2 β] = – [16 + 2] = – 18 Ans. ]

Q.34 49/qe If a + b + c = 0 and a 2 + b 2 + c 2 = 1 then the value of a 4 + b 4 + c 4 is (A) 3/ 2 (B) 3/4 (C*) 1/2 (D) 1/4

Q.35 51/qe The graph of a quadratic polynomial y = ax 2 + bx + c (a, b, c∈R) with vertex on yaxis is as shown in the figure. Then which one of the following statement is INCORRECT? (A) Product of the roots of the corresponding quadratic equation is positive. (B) Discriminant of the quadratic equation is negative. (C*) Nothing definite can be said about the sum of the roots, whether positive, negative or zero. (D) Both roots of the quadratic equation are purely imaginary.

[Sol. Roots are purely imaginary i.e. i β and – i β ∴ sum of roots = 0 incorrect (C)

product of roots = – i 2 β 2 = β 2 ⇒ product > 0 ; a c > 0 ⇒ c = + ve

note that – a 2 b

= 0 ⇒ b = 0

hence y = ax 2 + c when x > 0, y = c > 0 ⇒ y = ax 2 + c when c > 0]


Q.36 49/ph1 If 5 2

3 π

π < < x , then the value of the expression 1 1 1 1

− + +

− − + sin sin sin sin

x x x x

is

(A) –cot x 2

(B) cot x 2

(C) tan x 2

(D*) –tan x 2

[Hint: On rationalizing ; we get

x sin 1 x sin 1 | x cos | 2 x sin 1 x sin 1

+ − − + + + −

= ( )

) x (sin 2 | x cos | 1 2

− +

= ) x (sin x cos 1

− −

⇒ (D) ]

Q.37 54/qe If α and β are the roots of the equation ax 2 + bx + c = 0 then the sum of the roots of the equation a 2 x 2 + (b 2 – 2ac)x + b 2 – 4ac = 0 in terms ofα and β is given by (A) – (α 2 – β 2 ) (B) (α + β) 2 – 2αβ (C) α 2 β + βα 2 – 4αβ (D*) – (α 2 + β 2 )

[Hint: α + β = – a b ; α β = a

c ; x 1 + x 2 = 2

2

a b ac 2 −

= a c 2 –

2

a b

= 2 αβ – (α + β) 2 = – (α 2 + β 2 ) ]

Q.38 55/qe The number of solution of the equation e 2x + e x + e –2x + e –x = 3(e –2x + e x ) is (A) 0 (B) 2 (C*) 1 (D) more than 2

[Hint: x = ln 2]

Q.39 60/qe The quadratic equation x 2 – 1088x + 295680 = 0 has two positive integral roots whose greatest common divisor is 16. The least common multiple of the two roots is (A) 18240 (B*) 18480 (C) 18960 (D) 19240

[Sol. x 2 – 1088x + 295680 = 0 β

α [12th, 02122007]

let α = 16k 1 and β = 16k 2 (as HCF of roots is 16) αβ = (HCF) (LCM) 295680 = 16(LCM)

LCM = 16 295680

= 18480 Ans. ]

Q.40 65/qe Given a, b, c are non negative real numbers and if a 2 + b 2 + c 2 = 1, then the value of a + b + c is : (A) ≥ 3 (B) ≥ 2 (C) ≤ 2 (D*) ≤ 3

[Hint : (a + b + c) 2 = a 2 + b 2 + c 2 + 2 ∑ ab = 1 + 2∑ ab put a 2 + b 2 + c 2 ≥ ab + bc + ca

or ∑ ab ≤ 1 ∴ (a + b + c) 2 ≤ 1 + 2

∴ (a + b + c) 2 ≤ 3 Alternatively : using RMS ≥AM in a, b, c

3 c b a 2 2 2 + +

≥ 3 c b a + +

⇒ 3 ≥ (a + b + c) 2 ⇒ a + b + c ≤ 3 ]


Q.41 56/ph1 As shown in the figure AD is the altitude on BC and AD produced meets the circumcircle of ∆ABC at P where DP = x. Similarly EQ = y and FR = z. If a, b, c respectively

denotes the sides BC, CA and AB then z 2 c

y 2 b

x 2 a

+ +

has the value equal to (A*) tanA + tanB + tanC (B) cotA + cotB + cotC (C) cosA + cosB + cosC (D) cosecA + cosecB + cosecC

[Hint: BD = x tanC in ∆PDB and DC = x tanB for ∆PDC

∴ BD + DC = a = x ( tanB + tanC)

x a

= tanB + tanC

⇒ result ]

Q.42 72/qe Number of values of the parameter α ∈ [0, 2π] for which the quadratic function,

(sinα) x 2 + 2 cosα x + 1 2 (cosα + sinα) is the square of a linear function is

(A*) 2 (B) 3 (C) 4 (D) 1

[Hint : Let f (x) = ( ) 2 b x sin + α now compare the coefficient and eliminate b. divide by cos 2 α to get

(tanα – 1) (tanα + 2) = 0 ⇒ α = 4 π

or π – tan –1 ( 2) ]

Q.43 65/ph1 The exact value of 96 80 65 35 20 50 110 sin sin sin

sin sin sin ° ° °

° + ° + ° is equal to

(A) 12 (B*) 24 (C) –12 (D) 48

[Hint : ∑ A sin = 4∏ 2 A cos in Dr. as A + B + C = π ]

Q.44 75/qe The set of values of 'a' for which the inequality, (x− 3a) (x − a − 3) < 0 is satisfied for all x∈ [1, 3] is: (A) (1/3, 3) (B*) (0, 1/3) (C) (− 2, 0) (D) (− 2, 3)

[Hint : Equation is x 2 – (4a + 3)x + 3a(a + 3)

f(1) < 0 and f(3) < 0 (1–3a) (1 – a – 3) < 0 ⇒ 1 – a – 3 – 3a + 3a 2 + 9a < 0

⇒ 3a 2 + 5a – 2 < 0 3a 2 + 6a – a – 2 < 0 3a (a + 2) – (a + 2) < 0

(a + 2 ) (3a – 1) < 0 ⇒


again (3–3a) (–a) < 0 ⇒ (a – 1)a < 0

⇒ 0 < a < 1

Hence a ∈ F H G

I K J 0 1 3 , ]

Q.45 37/log If log 0.3 (x – 1) < log 0.09 (x – 1) , then x lies in the interval (A*) (2 , ∞) (B) (1 , 2) (C) (1, ∞) (D) none of these

[ IIT '85 , 2 ] [Hint: Given log 0.3 (x – 1) < log 0.09 (x – 1)

log 0.3 (x−1) < (1/2) log 0.3 (x – 1) or log 0.3 (x−1) < log 0.3 (x – 1) 1/2

x – 1 > 1 x − and (x–1) 2 > x – 1 ⇒ (x – 2) (x – 1) > 0 ⇒ x > 2 ]

Q.46 67/ph1 The value of cot x + cot (60º + x) + cot (120º + x) is equal to :

(A) cot 3x (B) tan3x (C) 3 tan3x (D*) 3 9 3

2

3

− − tan

tan tan x

x x

[Sol. cotx + ) 60 x sin( ) 60 x cos(

) x 60 sin( ) x 60 cos(

− −

+ + +

= ) 60 x sin( ) 60 x sin( ) x 2 sin(

x sin x cos

− + +

= 3 x sin 4 x cos x sin 8

x sin x cos

2 − + = x sin 3 x sin 4

x cos x sin 8 x cos 3 x cos x sin 4 3

2 2

− + −

= x sin

] x cos 4 x cos 3 [ 3 3

3 − = 3 cot3x ⇒ x tan x tan 3

] x tan 3 1 [ 3 3

2

− −

Ans ]

Q.47 84/qe If the roots of the cubic, x 3 + ax 2 + bx + c = 0 are three consecutive positive integers. Then the value

of 1 b

a 2

+ is equal to

(A*) 3 (B) 2 (C) 1 (D) 1/3 [Sol. n, n + 1, n + 2 [11 th test (2102005)]

sum = 3(n + 1) = – a ∴ a 2 = 9(n + 1) 2 sum of the roots taken 2 at atime = + b ∴ n(n + 1) + (n + 1)(n + 2) + (n + 2)n + 1 = b + 1 (adding 1 both sides)

n 2 + n + n 2 + 3n + 2 + n 2 + 2n + 1 = b + 1 ∴ b + 1 = 3n 2 + 6n + 3 = 3(n + 1) 2

b + 1 = 3(n + 1) 2 = 3 a 2

; ∴ 1 b

a 2

+ = 3 ⇒ (A) ]


Q.48 74/ph1 For every x∈R the value of the expression y = 8 x 2

+ x cos x + cos 2x is never less than

(A*) – 1 (B) 0 (C) 1 (D) 2

[Sol. y = 8 x 2

+ x cos x + 2cos 2 x – 1 [13th, 05082007]

= 8 1 [x 2 + 8x cos x + 16 cos 2 x] – 1

= 8

) x cos 4 x ( 2 + – 1 ⇒ (A) ]

Q.49 90/qe The value of p for which both the roots of the quadratic equation, 4x 2 − 20px + (25p 2 + 15p − 66) are less than 2 lies in : (A) (4/5, 2) (B) (2, ∞) (C) (− 1, 4/5) (D*) (− ∞, −1)

[Hint : D ≥ 0 ; f (2) > 0 ; – a 2 b

< 2 ]

Q.50 97/qe Ifα and β are the roots of a(x 2 – 1) + 2bx = 0 then, which one of the following are the roots of the same equation?

(A) α + β, α – β (B*) β + α 1 2 ,

α + β 1 2 (C) β

+ α 1 ,

α − β 1

(D) β + α 2 1 ,

α − β 2 1

[Hint: Verify and eliminate each option by checking with product of roots. ] [Dpp54]

Q.51 75/ph1 If θ be an acute angle satisfying the equation 8 cos 2θ + 8 sec 2θ = 65, then the value of cos θ is equal to

(A) 8 1

(B) 3 2

(C) 3 3 2

(D*) 4 3

[Sol. Let cos 2θ = t [11th, 03082008, P1]

∴ 8t + t 8 = 65 ⇒ 8t 2 – 65t + 8 = 0 ⇒ 8t 2 – 64t – t + 8 ⇒ 8t(t – 8)– (t – 8) = 0

t = 8 or t = 8 1

(t = 8 is rejected, think ! )

∴ cos 2θ = 8 1 ; 2 cos 2 θ – 1 = 8

1 ⇒ cos 2 θ = 16

9 ⇒ cos θ = 4

3 Ans. ]

Q.52 98/qe If x be the real number such that x 3 + 4x = 8, then the value of the expression x 7 + 64x 2 is (A) 124 (B) 125 (C*) 128 (D) 132

[Sol. Given x 3 + 4x – 8 = 0 [13th, 16122007] now y = x 7 + 64x 2

= 4 4 3 4 4 2 1 zero

3 4 ) 8 x 4 x ( x − + – 4x 5 + 8x 4 + 64x 2 = – 4x 5 + 8x 4 + 64x 2


= – 4x 2 4 43 4 42 1 zero

3 ) 8 x 4 x ( − + + 8x 4 + 16x 3 + 32x 2 = 8x 4 + 16x 3 + 32x 2

= 8x 4 43 4 42 1 zero

3 ) 8 x 4 x ( − + + 16x 3 + 64x = 4 4 3 4 4 2 1 zero

3 ) 8 x 4 x ( 16 − + + 128 Ans. ]

Q.53 124/ph1 One side of a rectangular piece of paper is 6 cm, the adjacent sides being longer than 6 cms. One corner of the paper is folded so that it sets on the opposite longer side. If the length of the crease is l cms and it makes an angleθ with the long side as shown, then l is

(A*) θ θ 2 cos sin

3 (B)

θ θcos sin 6

2

(C) θ θcos sin

3 (D)

θ 3 sin 3

[Sol. sin θ = l x

....(1); also cos 2θ = x x 6 −

1 + cos 2θ = x 6 ;

2 cos 2 θ = θ sin

6 l

substituting x = l sin θ from (1)

l = θ θ 2 cos sin

3 Ans. ] [11 th pqrs, J, 2112007]

Q.54 81/ph1 α, β, γ and δ are the smallest positive angles in ascending order of magnitude which have their sines equal to the positive quantity k . The value of

4 sin α 2 + 3 sin

β 2 + 2 sin

γ 2 + sin

δ 2 is equal to :

(A) 2 1 − k (B*) 2 1 + k (C) 2 k (D) 2 k

[Hint : β = π − α ; γ = 2 π + α ; δ = 3 π − α where 0 < α < π 2

Now E = 4 sin α 2 + 3 cos

α 2

− 2 sin α 2

− cos α 2

= 2 sin cos α α 2 2

+

⇒ E 2 = 4 (1 + sin α) = 4 (1 + k) E = 2 1 + k ]

Q.55 100/qe If the roots of the equation x 3 – px 2 – r = 0 are tan α, tan β and tan γ then the value of sec 2 α ∙ sec 2 β ∙ sec 2 γ is (A) p 2 + r 2 + 2rp + 1 (B*) p 2 + r 2 – 2rp + 1 (C) p 2 – r 2 – 2rp + 1 (D) None

[Sol. ∑ α tan = p ; β α ∑ tan ∙ tan = 0 ; ∏ α tan = r [ 13 th Test (5122004)] now sec 2 α ∙ sec 2 β ∙ sec 2 γ = (1 + tan 2 α) (1 + tan 2 β) (1 + tan 2 γ)

= 1 + ∑ α) (tan 2 + ∑ β α ) tan ∙ (tan 2 2 + tan 2 α ∙ tan 2 β∙ tan 2 γ


now ∑ α 2 tan = ( ) 2 tan ∑ α – 2 ∑ β α tan ∙ tan = p 2

∑ β α 2 2 tan ∙ tan = ( ) 2 tan ∙ tan ∑ β α – 2 tan α ∙ tan β ∙ tan γ ( ) ∑ α tan = 0 – 2rp

∏ α 2 tan = r 2

∴ ∏ α 2 sec = 1 + p 2 – 2rp + r 2 = 1 + (p – r) 2 ]

Q.56 121/ph1 In which one of the following intervals the inequality, sin x < cos x < tan x < cot x can hold good?

(A*)

π

4 , 0 (B)

π

π , 4 3

(C)

π π

2 3 ,

4 5

(D)

π

π 2 , 4 7

[Hint: In 2 nd quadrant sin x < cos x is False (think !) [11th, 03082008, P1] In 4 th quadrant cos x < tan x is False (think !)

in 3 rd quadrant, i.e.

π π

2 3 ,

4 5

if tan x < cot x ⇒ tan 2 x < 1 which is not correct

Hence A can be correct

now sin x < cos x is true in

π

4 , 0 and tan x < cot x is also true

∴ only the value of x for which cos x < tan x is to be determined ∴ now cos x = tan x

i.e. cos 2 x = sin x or 1 – sin 2 x = sin x ⇒ sin 2 x + sin x – 1 > 0

sin x = 2

5 1± − ; sin x =

2 1 5 −

⇒ x = sin –1

−2 1 5

∴ cos x < tan x in

π − −

4 ,

2 1 5 sin 1 and cos x > tan x in

− −

4 1 5 sin , 0 1 ]

Q.57 103/qe The absolute term in the quadratic expression

∑=

− −

+ −

n

1 k 2 k 3 1 x

1 k 3 1 x as n → ∞ is

(A) 1 (B*) 3 1

(C) 3 2

(D) zero

[Hint: T n = ) 2 n 3 )( 1 n 3 ( 1

− + ; T n =

+

− − ) 1 n 3 (

1 ) 2 n 3 (

1 3 1

] [12 th test (05062005)]

Q.58 83/ph1 If A = 340 0 then 2 2 sin A is identical to

(A) 1 1 + + − sin sin A A (B) − + − − 1 1 sin sin A A

(C) 1 1 + − − sin sin A A (D*) − + + − 1 1 sin sin A A


[Hint: A/2 = 170 0 hence 2sinA/2 > 0 now 340 0 lies in IV quadrant. Hence sinA <0. So 1+ sinA < 1 – sinA. Hence B and C are rejected because they give – values. Now we will check A and D. A : | sinA/2 + cosA/2 | + | sinA/2 – cosA/2 |

–ve +ve –sinA/2 – cosA/2 + sinA/2 – cosA/2 = – 2 cosA/2 Hence D is the answer ]

Q.59 107/qe Number of quadratic equations with real roots which remain unchanged even after squaring their roots, is (A) 1 (B) 2 (C*) 3 (D) 4

[ Hint : αβ = α 2 β 2 ....(1) and α 2 + β 2 = α + β ....(2) Hence αβ(1–αβ) = 0 ⇒ α = 0 or β = 0 or αβ = 1 ifα = 0 then from (2) β = 0 or β =1 ⇒ roots are (0,0) or (0,1) if β = 0 then α = 0 or β = 1

if β = 1 α then α

α 2

2 1

+ = α α

+ 1 ⇒ α

α α

α +

L N M

O Q P − = + 1 2 1 2

hence t 2 – t – 2 = 0 ⇒ (t–2) (t+1) = 0 ⇒ t = 2 or t = –1 if t = 2 ⇒ α = 1 and β = 1 , if t = –1 roots are imaginary (ωor ω 2 ) ]

Q.60 108/qe For a, b, c nonzero, real distinct, the equation, (a 2 + b 2 ) x 2 − 2b (a + c)x + b 2 + c 2 = 0 has non zero real roots . One of these roots is also the root of the equation : (A) a 2 x 2 − a (b − c) x + b c = 0 (B*) a 2 x 2 + a (c − b) x − b c = 0 (C) (b 2 + c 2 ) x 2 − 2 a (b + c) x + a 2 = 0 (D) (b 2 − c 2 )x 2 + 2 a (b − c) x − a 2 = 0

[Hint : α, β = 2 4 4 2

2 2 2 2 2 2

2 2

b a c b a c a b b c a b

( ) ( ) ( ) ( ) ( )

+ ± + − + +

+

= b a c b a a c c a b a c b b c

a b ( ) ) ( ( ) + ± + + − + + +

+

2 2 2 2 2 2 2 4 2 2

2 2

2

= b a c b b a c a c

a b ( ) ( ) + ± − − +

+

4 2 2 2

2 2

2 =

b a c b a c a b

( ) ( ) + ± − −

+

2 2

2 2

In order that roots may be real D ≥ 0 ⇒ D = 0 ⇒ b 2 = a c

Hence roots are coincident and is equal to b a c a a c ( ) ++ 2 =

b a

This root satisfies B. ]

Q.61 116/ph1 Let f (x) = a sin x + c, where a and c are real numbers and a > 0. Then f (x) < 0 ∀ x∈R if (A*) c < – a (B) c > – a (C) – a < c < a (D) c < a

[Hint: a sin x + c < 0 [11th, 03082008, P1]

sin x < – a c ; – a

c > sin x; – a

c > 1; – c > a ⇒ a + c < 0 ⇒ (A) ]


Q.62 84/ph1 The value of cosec π 18

– 3 sec π 18

is a

(A) surd (B) rational which is not integral (C) negative natural number (D*) natural number

[Sol. 18 / cos

3 18 / sin

1 π

− π

= ( ) 2 9 sin

18 sin

2 3

18 cos

2 1 2

π

π −

π

[Quiz] [11th 15102006 (P,J)]

=

9 sin

18 sin

6 cos

18 cos

6 sin 4

π

π π

− π π

= 4 Ans ]

Q.63 113/qe If the equation sin 4 x − (k + 2) sin 2 x− (k + 3) = 0 has a solution then k must lie in the interval : (A) (− 4, − 2) (B) [− 3, 2) (C) (− 4, − 3) (D*) [− 3, − 2]

[Sol. sin ( ) ( ) ( ) 2

2 2 2 4 3 2

x k k k

= + ± + + +

= ( ) k k k + ± + + 2 8 16

2

2

= ( ) ( ) k k + ± + 2 4

2 sin 2 x= k+3 or –1 (rejected)

∴ 0 < k+3 < 1 ⇒ –3 < k < –2 ]

Q.64 117/qe Number of solutions of the equation 2 x – 2) 1 x ( − + 2) 2 x ( − = 5 , is (A) 0 (B) 1 (C*) 2 (D) More than 2

[Sol. |x| – |x – 1| + |x – 2| = 5 [11th, 25012009, P2] for x ≥ 2 x – (x – 1) + x – 2 = 5

5 1 x + =

for 1 ≤ x < 2 x – (x – 1) + (2 – x) = 5 3 – x = 5 x = 3 – 5 No solution

0 ≤ x < 1 x – (1 – x) + 2 – x = 5 x = 5 – 1 No solution

for x < 0 –x – (1 – x) + 2 – x = 5

5 1 x − =

Hence x = 5 + 1 or x = 1 – 5 ]


Q.65 119/qe The inequalities y(−1) ≥ −4, y(1) ≤0 and y(3) ≥5 are known to hold for y = ax 2 +bx + c then the least value of 'a' is : (A) − 1/4 (B) − 1/3 (C) 1/4 (D*) 1/8

[Hint : a b + c ≥ − 4 ..... (i) and a + b + c ≤ 0 ⇒ − a − b − c ≥ 0 .... (ii) and 9a + 3b + c ≥ 5 .... (iii) (i) + (ii) ⇒ − 2b ≥ −4 ..... (iv) ; (ii) + (iii) + (iv) ⇒ 8a ≥ 1 ⇒ a ≥1/8 ]

Q.66 87/ph1 If tan x + tan y = 25 and cot x + cot y = 30, then the value of tan(x + y) is (A*) 150 (B) 200 (C) 250 (D) 100

[Sol. Let tan x = a ; tan y = b [11 th (782005)]

⇒ a + b = 25 ....(1) and a 1 + b 1 = 30 ⇒ ab

b a + = 30 ⇒ ab = 30

25 = 6 5

∴ tan (x + y) = ab 1 b a

− +

= ( ) 6 5 1 25

− = 25 × 6 = 150 ] [Quiz]

Q.67 54/log Number of integral values of x the inequality

+ −

1 x 2007 x 2 log 10 ≤ 0 holds true, is

(A) 1004 (B*) 1005 (C) 2007 (D) 2008

[Sol. 0 < 1 x 2007 x 2 +

− ≤ 1 [12 th & 13 th 1132007]

0 < 2x – 2007 ≤ x + 1 0 < x ≤ 2008 ....(1) also 2x – 2007 > 0

x > 1003.5 or x ≥ 1004 ....(2) from (1) and (2)

]

Q.68 115/ph1 For each natural number k , let C k denotes the circle with radius k centimeters and centre at the origin. On the circle C k , a particle moves k centimeters in the counter clockwise direction. After completing its motion on C k , the particle moves to C k+1 in the radial direction. The motion of the particle continues in this manner .The particle starts at (1, 0).If the particle crosses the positive direction of the x axis for the first time on the circle C n then n equal to (A) 6 (B*) 7 (C) 8 (D) 9

[Hint: Total distance travelled = 35 cm; displacement at the instant it crosses the +ve xaxis first time is 6cm ] Angular displacement on each circle is 1 radian.]

Q.69 123/qe Let a, b, c be real numbers , a ≠ 0 , if α is a root of a 2 x 2 + bx + c = 0 , β is a root of a 2 x 2 − b x − c = 0 and 0 < α < β , then the equation a 2 x 2 + 2 b x + 2 c = 0 has a root γ that always satisfies :

(A) γ = α β + 2

(B) γ = α + β 2

(C) γ = α (D*) α < γ < β

[Hint : Given a 2 α 2 + bα + c = 0 and a 2 β 2 – bβ – c = 0 Let f(x) = a 2 x 2 + 2bx + 2c

f(α) = a 2 α 2 + 2bα + 2c = a 2 α 2 – 2a 2 α 2 = – a 2 α 2 < 0 f(β) = a 2 β 2 + 2bβ + 2c = a 2 β 2 + 2a 2 β 2 = 3a 2 β 2 > 0

Hence f(α) and f(β) have opposite signs ⇒ α < γ < β]


Q.70 88/ph1 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

(A) π 3 and

π 6

(B*) π 8 and

3 8 π

(C) π 4 and

π 4

(D) π 5 and

3 10 π

[Sol. p 2 sec 2 θ + p 2 cosec 2 θ = ( ) 2 2 2 p 2

⇒ 8 cos sin 1

2 2 = θ θ

sin 2 2θ = 1/2 = 2

2 1

2θ = nπ + π/4 θ = nπ/2 + π/8 for n = 0 ⇒ θ = π/8 for n=1 ⇒ θ = 3π/8 ]

Q.71 124/qe If a, b∈R, a ≠ 0 and the quadratic equation ax 2 − bx + 1 = 0 has imaginary roots then a + b + 1 is: (A*) positive (B) negative (C) zero (D) depends on the sign of b.

[Hint : D = b 2 − 4a < 0 ⇒ a > 0 ⇒ mouth opens upwards ⇒ f(− 1) > 0 ] [11th, 21092008, P1]

Q.72 125/qe Let a, b, c be three real numbers such that a + b + c = 0 and a 2 + b 2 + c 2 = 2. Then the value of (a 4 + b 4 + c 4 ) is equal to (A*) 2 (B) 5 (C) 6 (D) 8

[Sol. Given a + b + c = 0 [11th, 02092007] and a 2 + b 2 + c 2 = 2

(a 2 + b 2 + c 2 ) 2 = 4 a 4 + b 4 + c 4 + 2[a 2 b 2 + b 2 c 2 + c 2 a 2 ] = 4

let a 4 + b 4 + c 4 = E hence E + 2[(ab + bc + ca) 2 – 2abc(a + b + c)] = 4 ∴ E + 2(ab + bc + ca) 2 = 4 (as a + b + c = 0)

again, (a + b + c) 2 = 0 ⇒ ∑ 2 a + ∑ ab 2 = 0 2 + 2(ab + bc + ca) = 0

∴ ab + bc + ca = – 1 ∴ E + 2 = 4; ∴ E = 2 Ans. ]

Q.73 89/ph1 In ∆ABC, the minimum value of ∏

∑

2 A cot

2 B cot .

2 A cot

2

2 2

is

(A*) 1 (B) 2 (C) 3 (D) non existent

[Hint : E = tan 2 2 A

+ tan 2 2 B

+ tan 2 2 C

now consider 2

2 B tan

2 A tan

− ≥ 0 etc and add to get the result. ]


Q.74 132/qe Consider the two functions f (x) = x 2 + 2bx + 1 and g(x) = 2a(x + b), where the variable x and the constants a and b are real numbers. Each such pair of the constants a and b may be considered as a point (a, b) in an ab – plane. Let S be the set of such points (a, b) for which the graphs of y = f (x) and y = g (x) do not intersect (in the xy – plane.). The area of S is (A) 1 (B*) π (C) 4 (D) 4π

[Sol. We need x 2 + 2bx + 1 = 2ax + 2ab not to have any real solutions, implying that the discriminant is less than or equal to zero. Actually calculating the discriminant and simplifying, we get a 2 + b 2 < 1, which describes a circle of area π in the ab plane] [12 th , 13 th 1132007]

Q.75 90/ph1 The value of cot 7 1 2

0

+ tan 67 1 2

0

– cot 67 1 2

0

– tan7 1 2

0

is :

(A) a rational number (B*) irrational number (C) 2(3 + 2 3 ) (D) 2 (3 – 3 )

Q.76 133/qe The polynomial P(x) = x 3 + ax 2 + bx + c has the property that the mean of its zeroes, the product of its zeroes, and the sum of its coefficients are all equal. If the yintercept of the graph of y = P(x) is 2, then the value of b is (A*) – 11 (B) – 9 (C) – 7 (D) 5

[12 th , 13 th 1132007] [Sol. The yintercept is at x = 0, so we have c = 2, meaning that the product of the roots is – 2. We know that

a is the sum of the roots. The average of the roots is equal to the product, so the sum of the roots is – 6, and a = 6. Finally, 1 + a + b + c = 2 as well, so we have 1 + 6 + b + 2 = – 2 ⇒ b = – 11 Ans.]

Q.77 96/ph1 If m and n are positive integers satisfying

1 + cos 2θ + cos 4θ + cos 6θ + cos 8θ + cos 10θ = θ

θ θsin

n sin ∙ m cos

then (m + n) is equal to (A) 9 (B) 10 (C*) 11 (D) 12

[Sol. Let S = cos 0° + cos 2θ + cos 4θ + .......... + cos 10θ [13 th 3072006] 2 sinθ ∙ S = 2 sinθ [cos0 + cos 2θ + .......... + cos 10θ]

= sin θ + sin θ = sin3θ – sin θ = sin5θ – sin3θ = sin7θ – sin5θ = sin9θ – sin7θ = sin11θ – sin9θ

————————— 2 sinθ ∙ S = sin11θ+ sinθ 2 sinθ ∙ S = 2 sin6θ ∙ sin5θ

S = θ

θ θ sin 2

5 cos 6 sin 2 =

θ θ θ

sin m cos n sin

⇒ n = 6 and m = 5 Ans. ]


Q.78 144/qe Number of values of x satisfying the pair of quadratic equations x 2 – px + 20 = 0 and x 2 – 20x + p = 0 for some p ∈ R, is

(A) 1 (B) 2 (C*) 3 (D) 4 [Sol. x 2 – px + 20 = 0 [11 th , 25022009, P1]

x 2 – 20x + p = 0 if p = 20 both the quadratic equations are identical hence x = 10 + 5 4 or x = 10 – 5 4 satisfy both if p ≠ 20 then x 2 – px + 20 = x 2 – 20x + p ⇒ (20 – p)x + (20 – p) = 0 ⇒ x = – 1 and p = – 21 Hence there are 3 values of x i.e. 1 , 5 4 10 , 5 4 10 − − + ] [11 th , 25022009, P1]

Q.79 98/ph1 The minimum value of the expression x sin x 4 x sin x 9 2 2 + for x∈ (0, π) is

(A) 3 16

(B) 6 (C*) 12 (D) 3 8

[Sol. E = 9x sin x + x sin x 4

[note that x sin x > 0 in (0, π) ] [11 th (782005)] [Quiz]

E = 2

x sin x 2 x sin x 3

− + 12

∴ E min = 12 which occurs when 3 x sin x = 2 ⇒ x sin x = 2/3] note that x sin x is continuous at x = 0 and attains the value π/2 which is greater than 2/3 at x = π/2, hence it must take the 2/3 in ( ) 2 , 0 π ]

[REASONING TYPE] Q.80 301/qe Statement1: If a > b > c and a 3 + b 3 + c 3 = 3abc then the quadratic equation ax 2 + bx + c

= 0 has roots of opposite sign. because Statement2: If roots of a quadratic equation ax 2 + bx + c = 0 are of opposite sign then

product of roots < 0 and | sum of roots | ≥ 0 (A*) Statement1 is true, statement2 is true and statement2 is correct explanation for statement1. (B) Statement1 is true, statement2 is true and statement2 is NOT the correct explanation for statement1. (C) Statement1 is true, statement2 is false. (D) Statement1 is false, statement2 is true.

[Sol. a > b > c ⇒ a, b, c are distinct real [12th, 20072008] also a 3 + b 3 + c 3 – 3abc = 0

+ +

2 c b a

[(a – b) 2 + (b – c) 2 + (c – a) 2 ] = 0 as a, b, c are distinct

∴ a + b + c = 0

hence x = 1 is a root of ax 2 + bx + c = 0 0 a b c

a+b+c=0 and a>b>c ⇒ a and c are of oppsite sign otherwise a+b+c ≠ 0 therefore a c negative]


Q.81 303/qe Consider the following statements Statement1: The equation x 2 + (2m + 1)x + (2n + 1) = 0 wherem and n are integers can not have

any rational roots. because Statement2: The quantity (2m + 1) 2 – 4(2n + 1) wherem, n∈ I can never be a perfect square. (A*) Statement1 is true, statement2 is true and statement2 is correct explanation for statement1. (B) Statement1 is true, statement2 is true and statement2 is NOT the correct explanation for statement1. (C) Statement1 is true, statement2 is false. (D) Statement1 is false, statement2 is true.

[Sol. D =

4 4 4 3 4 4 4 2 1 4 3 42 1 4 3 42 1

odd even odd

2 ) 1 n 2 ( 4 ) 1 m 2 ( + − + [11th, 29062008, P2]

for rational roots D must be a perfect square. As D is odd let D is a perfect square of (2l+ 1) whre l ∈ I ∴ (2m +1) 2 – 4(2n + 1) = (2l + 1) 2 or (2m +1) 2 – (2l + 1) 2 = 4(2n + 1)

[(2m + 1) + (2l + 1)] [2(m – l)] = 4(2n + 1) (m + l + 1)(m – l) = (2n + 1) ....(1)

RHS of (1) is always odd but LHS is always even (think !) Hence D can not be a perfect square ⇒ roots can not be rational hence Statement1 is true and Statement2 is true and is also the correct explanation for Statement1.]

Q.82 305/qe Let f (x) = ax 2 + bx + c, g (x) = ax 2 + qx + r, where a, b, c, q, r ∈R and a < 0. Ifα, β are the roots of f(x) = 0 and α+ δ, β + δ are the roots of g(x) = 0, then Statement1 :Maximum value of f (x) and g(x) are equal. because Statement2 :Discriminants of f(x) = 0 and g(x) = 0 are equal (A*) Statement1 is true, statement2 is true and statement2 is correct explanation for statement1. (B) Statement1 is true, statement2 is true and statement2 is NOT the correct explanation for statement1. (C) Statement1 is true, statement2 is false. (D) Statement1 is false, statement2 is true.

[Sol. | α – β | 2 = | (α + δ) – (β + δ) | 2 [11th, 21122008, P2] ⇒ (α + β) 2 – 4αβ = ( (α + δ) + (β + δ) ) 2 – 4(α + δ) (β + δ)

⇒ 2

2

a b

– a c 4 = 2

2

a q

– a r 4

⇒ b 2 – 4ac = q 2 – 4ar

f max = – a 4 D

= g max , D = b 2 – 4ac ]

Q.83 308/qe Let ax 2 + bx + c = 0, a ≠ 0 (a, b, c ∈ R) has no real root and a + b + 2c = 2. Statement1: ax 2 + bx + c > 0 ∀ x ∈ R. because Statement2: a + b is be positive. (A) Statement1 is true, statement2 is true and statement2 is correct explanation for statement1. (B) Statement1 is true, statement2 is true and statement2 is NOT the correct explanation for statement1. (C*) Statement1 is true, statement2 is false. (D) Statement1 is false, statement2 is true.


[Sol. f (x) = ax 2 + bx + c [11th, 2192008, P1] given f (0) + f (1) = 2 ⇒ f (x) > 0 ∀ x ∈ R ⇒ S1 is true Let f (x) = x 2 – x + 1

a + b = 0 ⇒ S2 is false]

Q.84 313/qe Statement1: If sin 2 x 3 cos 3

y 5 = k 8 – 4k 4 + 5, where x, y∈R then exactly four distinct real

values of k are possible. because

Statement2: sin 2 x 3 and cos 3

y 5 both are less than or equal to one and greater than or equal to – 1.

(A) Statement1 is true, statement2 is true and statement2 is correct explanation for statement1. (B) Statement1 is true, statement2 is true and statement2 is NOT the correct explanation for statement1. (C) Statement1 is true, statement2 is false. (D*) Statement1 is false, statement2 is true.

[Sol. State2 is obviously true. [11th, 30112008, P2] Also k 8 – 4k 4 + 5 = (k 4 – 2) 2 + 1 ∴ k 4 – 2 = 0 ⇒ (k 2 – 2 ) (k 2 + 2 ) = 0

∴ k = ± 2 ⇒ Two real values of k are possible. ∴ state1 is false but state2 is correct. ]

Q.85 Statement1: The quadratic polynomial y = ax 2 + bx + c (a ≠ 0 and b, c∈R) is symmetric about the line 2ax + b = 0.

because Statement2: Parabola is symmetric about its axis of symmetry. (A*) Statement1 is true, statement2 is true and statement2 is correct explanation for statement1. (B) Statement1 is true, statement2 is true and statement2 is NOT the correct explanation for statement1. (C) Statement1 is true, statement2 is false. (D) Statement1 is false, statement2 is true. [11th, 23122007]

Q.86 func Consider a cubic function f (x) = ax 3 + bx + c where a, b, c∈R. Statement1: f (x) can not have 3 non negative real roots. because Statement2: Sum of roots is equal to zero. (A) Statement1 is true, statement2 is true and statement2 is correct explanation for statement1. (B) Statement1 is true, statement2 is true and statement2 is NOT the correct explanation for statement1. (C) Statement1 is true, statement2 is false. (D*) Statement1 is false, statement2 is true.

[Sol. ax 3 + bx + c = 0 α ≥ 0, β ≥ 0, γ ≥ 0 [11th, 27012008]

α + β + γ = 0 ⇒ α = 0, β = 0 γ = 0 ⇒ f (x) = ax 3 ]


[COMPREHENSION TYPE]

Paragraph for Question Nos. 87 to 89 Consider the polynomial P(x) = (x – cos 36°)(x – cos 84°)(x – cos156°)

Q.87 401/qe The coefficient of x 2 is

(A*) 0 (B) 1 (C) – 2 1

(D) 2 1 5 −

Q.88 402/qe The coefficient of x is

(A) 2 3

(B) – 2 3

(C*) – 4 3

(D) zero

Q.89 403/qe The absolute term in P(x) has the value equal to

(A) 4 1 5 −

(B*) 16

1 5 − (C)

16 1 5 +

(D) 16 1

[Sol. (i) cos 36° + cos 84° + cos 156° [12 th (1452006)]

cos(60° – 24°) + cos (60° +24°) – cos 24° = 2 cos 60° cos24° – cos24° = cos24° – cos 24° = 0 Ans. ]

(ii) cos 36° ∙ cos 84° + cos 84° cos 156° + cos36° cos 156° θ = 24° cos(60° – θ)cos(60° + θ) – cos(60° + θ)cosθ – cos(60 – θ)cosθ cos 2 60° – sin 2 θ – cos θ(cos(60° + θ) + cos(60° – θ) ) cos 2 60° – sin 2 θ – 2 cos θ cos60° cosθ

cos 2 60° – 1 = 1 4 1

− = – 4 3

Ans. ]

(iii) cos(60° – θ)cos(60° + θ)cos θ

= 16 1 5

4 ∙ 4 1 5

4 72 cos

4 ) 24 ∙ 3 cos( −

= −

= °

=

as cos 72° = 4 1 5 − Ans. ]

Paragraph for Question Nos. 90 to 92 Let P(x) be quadratic polynomial with real coefficients such that for all real x the relation 2 ( ) ) x ( P 1+ = P(x – 1) + P(x + 1) holds. If P(0) = 8 and P(2) = 32 then

Q.90 404/qe Sum of all the coefficients of P(x) is (A) 20 (B*) 19 (C) 17 (D) 15

Q.91 405/qe If the range of P(x) is [m, ∞) then the value of 'm' is (A) – 12 (B) – 15 (C*) – 17 (D) – 5

Q.92 406/qe The value of P(40) is (A) 2007 (B*) 2008 (C) 2009 (D) 2010


[Sol.(1)Put x = 1 in 2 ( ) ) x ( P 1+ = P(x – 1) + P(x + 1) [11th, 29062008, P2]

( ) ) 1 ( P 1 2 + = P(0) + P(2) ⇒ 2 + 2P(1) = 8 + 32 ⇒ 2P(1) = 38 ⇒ P(1) = 19 hence sum of all the coefficient is 19 Ans.

(2) Let P(x) = ax 2 + bx + c P(0) = c ⇒ c = 8

also P(2) = 32 ⇒ 4a + 2b + 8 = 32 ⇒ 2a + b = 12 and P(1) = 19 ⇒ a + b + c = 19 ⇒ a + b + 8 = 19 ⇒ a + b = 11

a = 1 and b = 10 P(x) = x 2 + 10x + 8

= (x + 5) 2 – 17 ∴ P(x)| min = – 17 ⇒ m = – 17 Ans.

(3) P(40) = 1600 + 400 + 8 = 2008 Ans. ]

[MULTIPLE OBJECTIVE TYPE] Q.93 502/qe The graph of the quadratic polynomial ;

y = ax 2 +bx + c is as shown in the figure . Then : (A*) b 2 − 4ac > 0 (B*) b < 0 (C*) a > 0 (D*) c < 0

Q.94 501/ph1 Let y = x 7 sin x 6 sin x 5 sin x 4 sin x 3 sin x 2 sin x sin x 7 cos x 6 cos x 5 cos x 4 cos x 3 cos x 2 cos x cos

+ + + + + + + + + + + +

then which of the following

hold good? (A) The value of y when x = π/8 is not defined. (B*) The value of y when x = π/16 is 1.

(C) The value of y when x = π/32 is 1 2 − .

(D*) The value of y when x = π/48 is 3 2+ .

[Sol. Numerator =

2 x sin

x 4 cos 2 x 7 sin

; Denominator =

2 x sin

x 4 sin 2 x 7 sin

[11th, 24062007]

∴ f (x) = y = x 4 sin x 4 cos = cot 4x

∴ ( ) 8 π f = cot 2 π = 0 ; ( ) 16 π f = cot 4

π = 1

( ) 32 π f = cot 8 π = 1 2 + ; ( ) 48 π f = cot 12

π = 3 2+ ]

Q.95 507/qe If S is the set of all real x such that (2x−1)/(2x 3 +3x 2 +x) is positive, then S contains (A*) (− ∞, −3/2) (B) (−3/2, −1/4) (C) (−1/4, 1/2) (D*) (+1/2 , 3)

Q.96 530/log If y = log 7–a (2x 2 + 2x + a + 3) is defined ∀ x∈R, then possible integral value(s) of a is/are (A) – 3 (B*) – 2 (C*) 4 (D*) 5


[Sol. 2x 2 + 2x + a + 3 must be positive [11th, 03082008, P2] hence D < 0

i.e. 4 – 8(a + 3) < 0 ⇒ 1 – 2a – 6 < 0 ⇒ – 2a < 5 ⇒ a > – 2 5

....(1)

Also base of the logarithm 7 – a > 0 and 7 – a ≠ 1 a < 7 & a ≠ 6 ....(2)

from (1) and (2) a ∈

− 6 ,

2 5

∪ (6, 7) ⇒ (B), (C) and (D) are correct ]

Q.97 508/ph1 If sin 2 β = sinα cosα then cos2β has the value equal to :

(A) 1 + sin2α (B*) 2 sin 2 π

α 4

−

(C*) 1 − sin 2α (D*) 2 cos 2

π α

4 +

[Sol. sin 2 β = sinα cosα

2 2 sin

2 2 cos 1 α

= β −

cos2β = 1– sin2α ⇒ (A)

= 1 – cos(π/2–2α) = 2sin 2

α −

π 4 ⇒ (B)

= 2cos 2

α +

π 4 ⇒ (D) ]

Q.98 519/qe If the quadratic equation ax 2 + bx + c = 0 (a > 0) has sec 2 θ and cosec 2 θ as its roots then which of the following must hold good? (A*) b + c = 0 (B*) b 2 – 4ac ≥ 0 (C*) c ≥ 4a (D) 4a + b ≥ 0

[Hint: sum = product and roots are real [11th, 21092008, P1]

a c

a b

= − ⇒ b + c = 0 ⇒ b 2 – 4ac ≥ 0 ⇒ A, B, C]

Q.99 522/qe Which of the following statement(s) is/areTrue? (A*) The equation 1 x 1 x − + + = 1 has no real solution. (B*) If 0 < p < π then the quadratic equation, (cosp − 1) x 2 + cos px + sinp = 0 has real roots. (C) If 2a + b + c = 0 (c ≠ 0) then the quadratic equation, ax 2 + bx + c = 0 has no root in (0, 2). (D) If x and y are positive real numbers and m , n are any positive integers then ;

( ) ( ) x y x y

n m

n m

. 1 1

1 4 2 2 + +

> .

[Hint: (A) x = 5/4 is rejected (C) note that f(0) and f(2) have opposing signs under the given condition ]


Q.100 511/ph1 Two parallel chords are drawn on the same side of the centre of a circle of radius R . It is found that they subtend an angle of θ and 2θ at the centre of the circle . The perpendicular distance between the chords is

(A) 2 R sin 3 2 θ sin

θ 2

(B*) 1 2

−

cos θ 1 2

2 +

cos θ R

(C) 1 2

+

cos θ 1 2

2 −

cos θ R (D*) 2 R sin

3 4 θ sin

θ 4

[Sol. OM = p 1 = R cos θ 2

ON = p 2 = R cos θ

MN = p 1 − p 2 = R cos cos θ θ

2 −

= R 2 sin 3 4 θ sin

θ 4

⇒ D

Again convert cos θ = 2 cos 2 θ 2

− 1 and factorise ]

log + compound angles + qe 11th (pqrs)

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