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 (A) 2 2 A AC 2 B − (B) 2 2 A AC 4 B − (C) 2 2 A B 2 AC− (D*) 2 2 A B AC 2 − [Sol. We have r + s = A B − ; rs = A C r2 + s2 = – p; r2s2 = q ; Now, – p = (r + s)2 – 2rs = A C 2 A B 2 2 − ⇒ p = 2 2 A B AC 2 − Ans.]
Q.2 Which of the following is a graph of f (x) = 2 1
sin(2x)
(A) (B)
(C*) (D)
Q.3 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
[Sol. 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 ]
Q.4 If A ∈ (π,2π), then the value of
2 A sin 1 2 A sin 1 2 A sin 1 2 A sin 1 − − + − + + is equal to (A) – tan 4 A (B) – cot 4 A (C) cot 4 A (D*) tan 4 A [Sol. We have 2 A sin 1 2 A sin 1 2 A sin 1 2 A sin 1 − − + − + + = 2 2 2 2 4 A sin 4 A cos 4 A sin 4 A cos 4 A sin 4 A cos 4 A sin 4 A cos ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − − ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ +
MATHEMATICS = 4 A sin 4 A cos 4 A sin 4 A cos 4 A sin 4 A cos 4 A sin 4 A cos − − + − + + = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − − ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + 4 A sin 4 A cos 4 A sin 4 A cos 4 A sin 4 A cos 4 A sin 4 A cos = 4 A cos 2 4 A sin 2 = tan 4 A ]
Q.5 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.6 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°
[Sol. RHS = 2 sin 54° cos 7° – 2 sin 18° cos 7° = 2 cos 7°[sin 54° – sin 18°] = cos 7° ⇒ x = 7° Ans. ]
Q.7 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.
[Sol. 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 ⇒ p = 1 h 2Δ ; q = 2 h 2Δ ; r = 3 h 2Δ p, q, r → AP ⇒ 1 h 2Δ , 2 h 2Δ , 3 h 2Δ → AP⇒ 3 2 1 h 1 , h 1 , h 1 → 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 x + 2(1 – cos 3 )x – 2sin 3 = 02 θ 2 θ
α β
; α + β = – 2(1 – cos 3θ) ; αβ = – 2sin23θ
Now, α2 + β2 = (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. ]
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
[Sol. 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 ]
Q.10 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
[Sol. We have ar. (Δ ABC) = ar. (ΔPBC) + ar. (ΔPAC) + ar. (ΔPAB)
⇒ 2 ) 4 ( 4 3 = 2 1 (4) (x) + 2 1 (4) (y) + 2 1 (4) (z)
MATHEMATICS ⇒ 4 3 = 2 (x + y + z) ⇒ x + y + z = 2 3 • A D B C y x P z E F Hence (x + y + z)2 = 12 ]
Q.11 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 [Sol. We have 2009 x3 + 2x2 + 1 = 0 α β γ ... (1) Put x = t 1 in equation (1), we get ⇒ 3 t 2009 + 2 t 2 + 1 = 0 ∴ t3 + 2t + 2009 = 0 1/α 1/β 1/γ ; So, α 1 + β1 + 1γ = 0 and αβ1 + βγ1 + γα1 = 2
Now, we know that
2 1 1 1 ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ γ + β + α = ⎟⎟⎠ ⎞ ⎜⎜ ⎝ ⎛ γ + β + α2 2 2 1 1 1 + 2 ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ γα + βγ + αβ 1 1 1 ⇒ 0 = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ γ + β + α2 2 2 1 1 1 + (2 × 2) ; ∴ α–2 + β–2 + γ–2 = – 4 ] [Alternative We have 2009 x3 + 2x2 + 1 = 0 α β γ Also, α + β + γ = 2009 2 − αβ + βγ + γα = 0 ... (1) and αβγ = 2009 1 − Now, 12 α + 2 1 β + 2 1 γ = 2 2 2 2 2 2 2 2 2 γ β α α γ + γ β + β α ... (2) 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) Putting (3) in (2), we have 12 α + 2 1 β + 2 1 γ = 2 ) ( ) ( 2 αβγ γ + β + α αβγ − = 2009 1 2009 2 2 − − × − = – 4 ]
MATHEMATICS Q.12 If Sn = 1 1 1 2 1 2 3 + 3 3 + + +... + 1 2 3 13 23 33 3 + + + + + + + + ... ... n
n , n = 1, 2, 3,... Then Sn is not greater than
(A) 1/2 (B) 1 (C*) 2 (D) 4 [Sol. Sn = ... 3 2 1 3 2 1 2 1 2 1 1 1 3 3 3 3 3 3 + + + + + + + + + ; T n = 3 3 3 3 n . ... 3 2 1 n ... ... 5 4 3 2 1 + + + + + + + = ) 1 n ( n 2 2 ) 1 n ( n 2 ) 1 n ( n 2 = + ⎥⎦ ⎤ ⎢⎣ ⎡ + + ⇒Sn =
∑
⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + − 1 n 1 n 1 2 ]Q.13 Let f(θ) = sin4θ+4cos2θ – cos4θ+4sin2θ, then the value of ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ° 4 1 11 f is equal to (A) 2 2 2− (B) – 2 2 2+ (C) – 2 2 2− (D*) 2 2 2+ [Sol. We have
f(θ) = sin4θ+4(1−sin2θ) – cos4θ+4(1−cos2θ) = 2 2 ) sin 2 ( − θ – 2 2 ) cos 2 ( − θ
= (2 – sin2θ) – (2 – cos2θ) = cos2θ – sin2θ = cos 2θ
∴ f ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ° 4 1 11 = cos 22 2 1° ; Now, cos2θ = 2 2 cos 1+ θ ...(1) Put θ = 22 2 1°
in equation (1), we get cos2 22 2 1° = 2 2 1 1+ = 2 2 1 2+ = 2 2 1 2+ × 2 2 = 4 2 2+ ∴ f ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ° 4 1 11 = cos 22 2 1° = 2 2 2+ (If 0 < θ < 2 π
then cos θ is positive) ] [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 Consider a rational function f(x) =
1 x x 1 x 3 x 2 2 + + + +
and a quadratic function g(x) = x2 – (m + 1) x + m – 1, where m is a parameter.
Q.14 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
Q.15 Number of integral value(s) in the range of f (x), is
(A) 1 (B) 2 (C*) 3 (D) more than 3
Q.16 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) (B) ⎟ ⎠ ⎞ ⎜ ⎝ ⎛−∞ − 4 1 , (C) (– 2, ∞) (D*) ⎟ ⎠ ⎞ ⎜ ⎝ ⎛− ,∞ 2 1
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.
(ii) We have y = 1 x x 1 x 3 x 2 2 + + + + ⇒ (y – 1) x2 + (y – 3) x + (y – 1) = 0 Since x ∈ R, so D ≥ 0
⇒ (y – 3)3 – 4(y – 1)2≥ 0 ⇒ (– y – 1) (3y – 5) ≥ 0 ⇒ (y + 1) (3y – 5) ≤ 0 ⇒ – 1 ≤ y ≤ 3 5 ∴ Range of f(x) = ⎢⎣⎡− ⎥⎦⎤ 3 5 ,
1 ;Clearly range of f(x) contains three integral values viz. – 1, 0, 1. (iii) Different possibilities are as follows : –
x g(x) –1 or x g(x) –1
If both roots of g(x) = 0 are greater than – 1 then 3 conditions should be satisfied simultaneously.
(1) D ≥ 0 (B) A 2 B − > – 1 (C) g(– 1) > 0 Now, (1) ⇒ (m + 1)2 – 4 (m – 1) ≥ 0 ⇒ m2 – 2m + 5 ≥ 0, ∀m∈R. (2) ⇒ 2 1 m+ > – 1 ⇒ m > – 3 and (3) ⇒ 1 + (m + 1) + m – 1 > 0 ⇒ m > – 2 1
; Hence from (1), (2) & (3), we get m ∈ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛− ,∞ 2 1 ] [REASONING TYPE]
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.
Q.17 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.
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.
Q.19 If f(θ) = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛θ+ π ⎟ ⎠ ⎞ ⎜ ⎝ ⎛θ+ − π
∑
= 4 n ec cos 4 ) 1 n ( ec cos 6 1 n , where 0 < θ < 2 π , then minimum value of f(A) lies between 3 and 4 (B*) lies between 2 and 3 (C*) occurs when θ = 4 π (D) occurs when θ = 6 π [Sol. We have f(θ) =
∑
= 6 1 n 2 ⎟ ⎠ ⎞ ⎜ ⎝ ⎛θ+ π ⎟ ⎠ ⎞ ⎜ ⎝ ⎛θ+ − π ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛θ+ − π − ⎟ ⎠ ⎞ ⎜ ⎝ ⎛θ+ π 4 n sin 4 ) 1 n ( sin 4 ) 1 n ( 4 n sin =∑
= 6 1 n 2 ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛θ+ π − ⎟ ⎠ ⎞ ⎜ ⎝ ⎛θ+ − π 4 n cot 4 ) 1 n ( cot = 2 ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛θ+ π − θ 2 3 cotcot = 2(cot θ + tan θ) = 2
(
)
⎥⎦ ⎤ ⎢⎣ ⎡ tanθ− cotθ 2+2 ∴ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛θ=π 4 f . min = 2 2 ]
Q.20 For all real x, the value of the rational function f (x) =
2 x 2 2 x 2 x2 − + −
can lie in the interval
(A*) (– ∞, – 1) (B) (–1, 1) (C*) (1, 2) (D*) (2, ∞) [Sol. y = 2 x 2 2 x 2 x2 − + − ∴ x2 – 2x + 2 = 2xy – 2y ⇒ x2 – 2x(y + 1) + 2(y + 1) = 0 1 2 O (–1,0) (0,1) Y X (2,1) As x ∈ R, so D ≥ 0 ⇒ (y + 1)2 – 2(y + 1) ≥ 0 ⇒ (y + 1)(y – 1) ≥ 0 ⇒ y ≥ 1 or y ≤ – 1 hence y can not lie in (–1, 1) ]
Q.21 Let E = cos2 7 π + cos2 7 2π + cos2 7 3π
. Then which of the following alternative(s) is/are incorrect?
(A*) 2 1 < E < 4 3 (B*) 4 3 < E < 1 (C) 1 < E < 2 3 (D*) 2 3 < E < 4 7
[Sol. We have E = cos2 7 π + cos2 7 2π + cos2 7 3π
MATHEMATICS = 2 7 6 cos 1 2 7 4 cos 1 2 7 2 cos 1 + π + π + + π + = 2 1 2 3 + 4 4 4 4 3 4 4 4 4 2 1 S 7 6 cos 7 4 cos 7 2 cos ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ π + π + π
Now, S = cos 2θ + cos 4θ + cos 6θ , where θ = 7 π
⇒ 2S sin θ = sin 3θ – sin θ + sin 5θ – sin 3θ + sin 7θ – sin 5θ = sin7θ sin− θ zero23 1 ∴ S = – 2 1 ⇒ E = 4 5 4 1 2 3− = ; Clearly 1 < E < 2 3 ]
Q.22 The following figure shows the graph of f(x) = ax2 + bx – c. Then which of the following alternative(s) is/are correct ?
(A*) c b
< 0
(B*) a and b are of same sign.
X Y
•α β•
f(x)
(C*) a and c are of opposite sign. (D) f(1) > 0
[Sol. We have ax2 + bx – c = 0
α β
As the parabola is opening downward so a < 0. Also both roots of f(x) = 0 are negative. Sum of roots = α + β = –
a b
< 0 ⇒ 'b' must be negative. ... (1) ∴ a and b are of same sign.
Product of roots = αβ = – a c
> 0 ⇒ 'c' must be positive. ... (2) ∴ a and c are of opposite sign.
Hence from (1) & (2), c b
< 0 and from above figure, f(1) < 0 ] Q.23 If tan2
8 3π
is a root of the equation 2x2 – 3ax + 4b = 0, where a, b ∈ Q, then the value of a + 4b can not be equal to
(A*) 5 (B*) 8 (C) 6 (D*) 4
[Sol. tan2 8 3π
= ( 2 + 1)2 = 3 + 2 2
∴ x = 3 + 2 2 satisfy the equation 2x2 – 3ax + 4b = 0, we get
⇒ 2(3 + 2 2)2 – 3a (3 + 2 2) + 4b = 0 ⇒ 2(17 + 12 2) – 9a – 6 2a + 4b = 0 ⇒ (34 – 9a + 4b) + 2 (24 – 6a) = 0
Since a and b are rational ⇒ 24 – 6a = 0 and 34 – 9a + 4b = 0 ∴ a = 4 and b = 2 1 ⇒ a + 4b = 6. ] Q.24 If 2 tan e 1 e 1 2 tan φ − + = θ
, where e ∈ (0, 1) then cos θ is equal to
(A*) − φ−φ cos e 1 e cos (B) + φ−φ cos e 1 e cos (C) − φ+φ cos e 1 e cos (D) + φ+φ cos e 1 e cos
MATHEMATICS [Sol. Given 2 tan e 1 e 1 2 tan2 2 φ − + = θ ⇒ tan21( 2) (1+e)1tane2(φ 2) − = θ ) 2 ( tan ) e 1 ( ) e 1 ( ) 2 ( tan ) e 1 ( ) e 1 ( ) 2 ( tan 1 ) 2 ( tan 1 2 2 2 2 φ + + − φ + − − = θ + θ − ; cos θ =
(
(
) (
) (
)
)
) 2 ( tan 1 e ) 2 ( tan 1 ) 2 ( tan 1 e ) 2 ( tan 1 2 2 2 2 φ − − φ + φ + − φ −On dividing numerator and denominator of RHS by ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + φ 2 tan 1 2 , we get cos θ = φ − − φ cos e 1 e cos Ans.] OR
Q.24 If a, b, c are +ve numbers such that a + b + c = 1 then minimum value of
ca 1 bc 1 ab 1 + + is (A*) 27 (B) 9 (C) 3 (D) None [Sol: ca 1 bc 1 ab 1 + + = abc 1 (a .+ b + c = 1) ; 3 c b a+ + > (abc)1/3⇒ abc 1 > 27 ]
Q.25 If S is the set of all real 'x' such that
) 2 x ( ) 1 x 5 ( ) x 2 1 ( ) x 5 ( x2 + + − − is negative and x x x 6 1 x 3 2 3+ − +
is positive, then S contains
(A*) (1, 4) (B) (5, 11) (C*) ⎟ ⎠ ⎞ ⎜ ⎝ ⎛− − 2 1 , 2 3 (D) (– 10, – 4) [Sol. We have ) 1 x x 6 ( x 1 x 3 2+ − + > 0 ⇒ ) 1 x 2 ( ) 1 x 3 ( x 1 x 3 + − + > 0 ... (1) + – + – + –1 2 –1 3 0 1 3 Now, ) 2 x ( ) 1 x 5 ( ) 1 x 2 ( ) 5 x ( x2 + + − − < 0 ... (2) + – + + – –2 –1 5 0 1 2 5 +
MATHEMATICS
PART-B
[MATCH THE COLUMN]
Q.1 is "Match the Column" type. Column-I contains Four entries and column-II contains Four entries. Entry of column-I are to be uniquely matched with only one entry of column-II.
Q.1 Column-I Column-II
The solution of following inequalities :
(A) sin x · cos3x > cos x · sin3x, 0 ≤ x ≤ 2π, is (P)
⎥⎦ ⎤ ⎢⎣ ⎡ π π ∪ ⎥⎦ ⎤ ⎢⎣ ⎡ π π − ∪ ⎥⎦ ⎤ ⎢⎣ ⎡ π − π − , 4 3 4 , 4 4 3 , (B) 4 sin2x – 8 sin x + 3 ≤ 0, 0 ≤ x ≤ 2π, is (Q) ⎥⎦ ⎤ ⎢⎣ ⎡ π π 2 , 2 3 ∪ {0} (C) | tan x | ≤ 1 and x ∈ [– π, π] is (R) ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ π 4 , 0
(D) cos x – sin x ≥ 1 and 0 ≤ x ≤ 2π is (S) ⎢⎣⎡π π⎥⎦⎤
6 5 , 6
[Ans. (A) R; (B) S; (C) P; (D) Q] [Sol. (A) sin x · cosx (cos2x – sin2x) > 0
⇒
2 1
sin 2x · cos 2x > 0 ⇒ sin 4x > 0; ∴ 0 < 4x < π ⇒ (R)
(B) Here, (2 sin x – 1)(2 sin x – 3) ≤ 0 ; but 2 sin x – 3 is always negative. ∴ 2 sin x – 1 ≥ 0 ⇒ sin x ≥ 1/2
∴ from the figure, π/6 ≤ x ≤ 5π/6 ⇒ (S)
(C) – 1 ≤ tan x ≤ 1. The value scheme for this is as shown below:
from the figure, –
4 π ≤ x ≤ 4 π or – π ≤ x ≤ – 4 3π or 4 3π ≤ x ≤ π] ∴ x ∈ ⎡⎢⎣−π − π⎤⎥⎦∪⎡⎢⎣−π π⎤⎥⎦∪⎡⎢⎣ π,π⎤⎥⎦ 4 3 4 , 4 4 3 , ⇒ (P) (D) cos ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ +π 4 x > 2 1
MATHEMATICS
from the figure
4 4 x 4 π ≤ π + ≤ π − and, in general 2nπ – 4 π ≤ x + 4 π ≤ 2nπ + 4 π ∴ 2nπ – 2 π ≤ x ≤ 2nπ ; for n = 0 – 2 π ≤ x ≤ 0; for n = 1, 2 3π ≤ x ≤ 2π⇒ (Q) ]
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 Column-I Column-II
(A) If exactly one root of the quadratic equation (P) 1
(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
(B) The smallest natural number 'n' so that (2 – n) x2 – 8x – 4 – n < 0, ∀x ∈ R (Q) 2 is equal to
(C) If in ΔABC, ∠A = 2 π
then the maximum value of 4 sin B sin C is equal to (R) 3
(D) Number of values of 'k' so that the equation (S) 4
(k – 3) (k2 – 4) (k + 1) x2 – (k3 – 5k2 + 6k) x + (k2 – 9) = 0
has more than two unequal roots is equal to (T) 5
[Ans. (A) P ; (B) T ; (C) Q ; (D) P ] [Sol.(A) If one root of the quadratic equation Ax2 + 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) A = 2 π , so B + C = 2 π
Now, 4sin B sin C = 2(2sin B sin C) = 2[cos (B – C) – cos (B + C)] = 2 cos (B – C) ≤ 2 ∴ Maximum value = 2, when B = C =
4 π
(each).
(D) We must have (k – 3) (k2 – 4) (k + 1) = 0, k3 – 5k2 + 6k = 0 and k2 – 9 = 0
Hence clearly k = 3 only. ]
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. [Ans. 0009 sq. unit]
[Sol: tan ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − 2 B A = cot C/2 b a b a ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + −
MATHEMATICS ) B A cos( 1 ) B A cos( 1 − + − − = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ 9 3
cot C/2 (using tanθ =
θ + θ − 2 cos 1 2 cos 1 ) 5 / 4 1 5 / 4 1 + − = 3 1
cot C/2 Now, Δ = ½ ab sinC
2 C cot 3 1 3 1 = ⇒ 2 C cot = 1 ⇒ ½ × 3 × 6 sin 90º ⇒ C = 90º = 9 ]
Q.2 Let A denotes the value of expression x4 + 4x3 + 2x2 – 4x + 7, when x = cot 8 11π and B denotes the value of the expression
θ θ − 4 tan 8 cos 1 2 + θ θ + 4 cot 8 cos 1 2 , when θ = 9° Find the value of (AB).
[Ans 12] [Sol. (A) We have
x = cot 8 11π = cot ⎟ ⎠ ⎞ ⎜ ⎝ ⎛π+ π 8 3 = cot 8 3π = 2−1 ⇒ (x + 1)2 = 2 ∴ x2 + 2x – 1 = 0 Now, consider x4 + 4x3 + 2x2 – 4x + 7 = x2 43 42 1 ) 0 ( 2 ) 1 x 2 x ( = − + + 2x3 + 3x2 – 4x + 7 = 2x3 + 3x2 – 4x + 7 = 2x 43 42 1 ) 0 ( 2 2x 1) x ( = − + – x2 – 2x + 7 = – x2 – 2x + 7 = – 43 42 1 ) 0 ( 2 ) 1 x 2 x ( = − + + 6 ∴ A = 6 (B) We have, θ θ − 4 tan 8 cos 1 2 + θ θ + 4 cot 8 cos 1 2 = θ θ 4 tan 4 sin 2 2 2 + θ θ 4 cot 4 cos 2 2 2 = 2 (cos2 4θ + sin2 4θ) = 2 ∴ AB = 12 Ans.]
Q.3 Find the sum of all integers satisfying the inequalities log5(x – 3) + 2 1 log53 < 2 1
log5(2x2 – 6x + 7) and log3x + log 3x+ log x 3
1 < 6.[Ans.0039 ]
[Sol. We have log5(x – 3) < 2 1 log5 ⎟⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − + 3 7 x 6 x 2 2 ⇒ (x – 3)2 < 3 7 x 6 x 2 2− + ⇒ x2 – 12x + 20 < 0 ⇒ (x – 2) (x – 10) < 0 ⇒ 2 < x < 10
But for the domiain , x > 3, so x ∈ (3, 10) Also log3x + 2log3x – log3x < 6 ⇒ 2log3x < 6 ∴ log3x < 3
MATHEMATICS ⇒ 0 < x < 27 so x ∈ (0, 27) ; ∴ Possible integers are 4, 5, 6, 7, 8, 9. ;Hence sum of integers = 39.] Q.4 Let A denotes the value of expression 4⎜
⎝ ⎛ π 15 2 cos + cos 15 4π – cos 15 7π – cos ⎟ ⎠ ⎞ π 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°]
= 16 × 4 1 5+ × 2 1 × 4 1 5− = 2 1 5− = 2 4 = 2
(B) We have x – 8(a – b) x + (2a – 3b) x – 4(b + 1) = 03 2 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) Now, tan (α + β + γ) = −Σ α β α − α Σ ∏ tan tan 1 tan tan = ) b 3 a 2 ( 1 ) 1 b ( 4 ) b a ( 8 − − + − − = ) b 3 a 2 1 ( ) 1 b b 2 a 2 ( 4 + − − − − = ) b 3 a 2 1 ( ) 1 b 3 a 2 ( 4 + − − − = – 4 ⇒ cot (α + β + γ) = – 4 1 ; ∴ 8 cot (α + β + γ) = – 2Hence | AB | = 4 ] Q.5 If k1 and k2 are the two values of 'k' where k1 < k2 for which the expressionf(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 ]
[Sol. We have A = 1 ; B = 4 ; C = 3 ; F = –3 ; G = k ; H = 1 Now, ABC + 2FGH – AF2 – BG2 – CH2 = 0 ⇒ k = 0, – 2 3 ∴ k1 = – 2 3 , k2 = 0 ⇒ 5k2 – 4k1 = 6 Ans. ]
Q.6 The set of values of ‘c’ for which the equation x2−4x−c− 8x2 −32x−8c =0 has exactly two distinct real solutions, is (a, b) then find the value of (b – a). [Ans. 8] [Sol. We have x2 – 4x – c ≥ 0 ⇒ D ≤ 0 ∴ 16 + 4c≤ 0 ⇒ 4c ≤ – 16 ⇒ c ≤ – 4 Let f(x) = x2 – 4x – c ; ∴ x2 – 4x – c – c 8 x 32 x 8 2− − = 0 ⇒ f(x) – 8f(x) = 0 ; ∴ f(x)
(
f(x)− 8)
= 0 ∴ f(x) = 0 or f(x) = 8Case-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)
