15-05-2025
7501CJA101021250028 JA
MATHEMATICS
SECTION-I(i)
1) Let and I1 = sin(cosθ) + cos(cosθ), I2 = sin(sinθ) + cos(sinθ), I3 = sin(cosθ) +
cos(sinθ) and I4 = cos(cosθ) + sin(sinθ), then
(A) I3 > I2 > I1 > I4
(B) I3 > I1 > I2 > I4
(C) I1 > I3 > I4 > I2
(D) I1 > I2 > I3 > I4
2) where αβ < 0, then number of possible integral values of
x is -
(A) 1
(B) 2
(C) 4
(D) 5
3) Sum of values of x for which is :
(A) zero
(B)
(C)
(D)
4) A function f satisfies f(3x) = 3f(x) ∀ x ∈ [1, ∞) and f(x) = 1 – for x ∈ [1, 3]. The smallest x for
which f(x) = f(2001) is
(A) 429
(B) 430
(C) 432S
(D) None of these
5) Let ƒ(x) = x – [x], x ≠ 0, x ∈ R and [.] is greatest integer function, then the number of solutions of
is :
(A) 1
(B) 0
(C) 2
(D) Infinite
6) Let ƒ(x) = & g(x) = 1 – x then number of solutions for ƒ(x) = g(x) when x > 0
is/are -
(A) 0
(B) 1
(C) 2
(D) 3
SECTION-I(ii)
1) If f(x) = sin–1(x + 2) + cos–1 and g is the inverse of f, then
(A)
(B)
(C)
(D)
2) Let , then
(A) minimum value of ƒ(x) is π2
(B)
maximum value of ƒ(x) is
(C) maximum value of ƒ(x) does not exist
(D) minimum value of ƒ(x) does not exist.
3) If tan–1x2 + tan–1y2 = [xy] , then (where x,y > 0 and [.] denotes greatest integer function)-
(A) minimum integral value of x2 + y2 is 2
(B) minimum integral value of x2 + y2 is 8
(C) maximum integral value of x2 + y2 is 14
(D) maximum integral value of x2 + y2 is 15
4) Consider the equation |(cot–1x)2 – (tan–1x)2| = . This equation has -
(A)
No solution for
(B)
Exactly one solution for
(C)
Exactly two solutions for
(D)
Atleast one solution for
5) If x and y are positive integers which satisfies the equation x2 – x(10 – 2y) + y2 – 10y = 2000, then
-
(A) Sum of all possible values of x is 1225.
(B) Sum of all possible values of x is 1250.
(C) Maximum value of (1 + x)(y – 1) is 624.
(D) Maximum value of (1 + x)(y – 1) is 625.
6) Let A1, A2, A3, ........, A9 are different numbers in arithmetic progression and A3, A5, A8, B1,B2,B3,.......
are in geometric progression. If A9 = 40, then
(A)
(B)
(C)
(D)
SECTION-I(iii)
Common Content for Question No. 1 to 2
Let , , then answer the following
1) Range of ƒ is -
(A)
(B)
(C)
(D) (–π, π)
2) Number of solutions of the equation ƒ(x) = 1 is -
(A) 1
(B) 2
(C) 3
(D) 4
Common Content for Question No. 3 to 4
Let roots of equation are , tan–1α & tan–1β where
α & β are positive integers.
On the basis of above information, answer the following questions :
3) is equal to
(A)
(B) –1
(C)
(D)
4) tan–1(α) + tan–1(β) + tan–1(α + β) is equal to-
(A)
(B)
(C) π
(D)
5) If α > β, then number of solutions of equation sin–1(x + β) + sin–1(αx) = , is
(A) 0
(B) 1
(C) 2
(D) more than two
SECTION-II
1) If , then value of is
2) Let A, G2 and H be the roots of cubic equation x3 + px2 + qx + r = 0 which are in G.P., where p, q
are integers and A, G, H are respectively AM, GM, HM of two positive numbers. If p, q ∈ (–100, 100),
then number of all possible ordered triplets (p, q, r) is
3) If tanA, tanB, tanC are roots of the equation ,
then sum of all possible values of A + B + C [where A + B + C ∈ (0, 2π) & A, B, C > 0] is (where
a & b are coprime numbers), then a + b is equal to
ANSWER KEYS
MATHEMATICS
SECTION-I(i)
Q. 1 2 3 4 5 6
A. B B A A D B
SECTION-I(ii)
Q. 7 8 9 10 11 12
A. A,B A,C B,C B,C,D A,D A,D
SECTION-I(iii)
Q. 13 14 15 16 17
A. A D D C B
SECTION-II
Q. 18 19 20
A. 7.00 97.00 9.00
SOLUTIONS
MATHEMATICS
1) cos (cosθ) < cos(sinθ)
⇒ I4 < I2 and I1 < I3
sin(cosθ) > sin(sinθ) ⇒ I1 > I4 and I2 < I3
I2 = sin(sinθ) + cos(sinθ)
I1 = sin(cosθ) + cos(cosθ)
= sin(cosθ) +
2)
If a > 0 ⇒
&b<0
⇒ , then number of integer in x is 1.
If ⇒
&β>0⇒
⇒ , then number of integer in x is 1.
3)
⇒ tan–16x =
taking tan of both sides :
⇒
2 4 2
⇒ x + 1 – 9x = x ( x ≠ 0)
⇒ 1 – 9x4 = 0
⇒
Both values satisfy given equation
∴ Sum of values = 0
4)
Since 2 × 36 < 2001 < 37
⇒ f(2001) = 37 – 2001 = 186
Now, f(x) ≤ 81, when x ≤ 35
And 0 < f(x) < 35 = 243
when 35 < x < 2 × 35
So, 186 = x – 35
⇒ x = 243 + 186 = 429
5)
= integer ......(1)
But n ≠ 2, –2 as it does not satisfy (i)
⇒ n can be any integer in (–∞, –2) ∪ (2, ∞)
So infinite solutions.
6)
= 2θ (because x > 0
⇒ 0 < cosθ ≤ 1)
Number of solutions are 1
7) Domain of f(x) = {–1}
Range of f(x) =
Now, verify alternatives.
8) Domain of ƒ(x) = x ∈ (–1, 1)
Put sin–1x = t ;
But fmax. = does not exist
9) If xy < 1, then equation has no solution
If xy = 1, then equation has no solution
If xy > 1 ⇒ x2y2 > 1
possible when 3 ≤ x y < 4
or x2 + y2 = x2y2 – 1
⇒ 9 ≤ x2 + y2 + 1 < 16
10) |(cot–1x + tan–1x) (cot–1x – tan–1x)|
⇒ |cot–1 x – tan–1x| = k
graph of
11) Given equation is (x + y)2 – 10(x + y) – 2000 = 0
Let x + y = t
⇒ t2 – 10t – 2000 = 0
⇒ (t – 50)(t + 40) = 0
⇒ x + y = 50 or x + y = – 40 (reject x,y ∈ R+)
⇒ x + y = 50
∴ Possible values of x are 1, 2, 3, .... 49
∴ Sum of all values =
Use A.M. ≥ G.M.
⇒ (25)2 ≥ (x + 1)(y – 1)
⇒ (x + 1)(y – 1) ≤ 625
12) Let A1 = A (first term) and common difference = D
Now, A9 = 40 ⇒ A + 8D = 40 ...(i)
Also, A + 2D, A + 4D , A + 7D, B1, B2, B3, .............. in G.P.
⇒ (A + 4D)2 = ( A + 2D) (A + 7D)
∴ A = 2D
⇒ A = 8, D = 4
and
15)
⇒α=2&β=1
= tan(2tan–12)
16)
⇒α=2&β=1
= tan(2tan–12)
tan (α) + tan (β) + tan–1(α + β) = tan–1(1) + tan–1(2) + tan–1(3) = π
–1 –1
17)
⇒α=2&β=1
α =2&β=1
possible when
⇒ x = 0 or
but is not solution of equation.
18)
rationalise
Now
19) x3 + px2 + qx + r = 0
A, G2, H are in G.P.
⇒ (G2)2 = AH
⇒ G4 = G2 ( AH = G2 for two positive reals)
⇒ G = 0, 1, –1
⇒ G = 1 ( G is GM of two positive numbers)
⇒ Product of roots = – r
⇒ A . 1. ⇒ r = –1
Sum of the roots = –p
⇒A+1+ ........(1)
⇒ –p = ( A > 0)
⇒p≤–3
and sum of roots taken two at a time = q
⇒A.1+ ........(2)
⇒q≥3
from (1) and (2) p + q = 0
Now p, q ∈ (–100, 100)
⇒ possible values of q are 3, 4, 5, ...... 99
for which p must be fixed as –3, –4, –5, ..... –99 respectively.
Number of ways of choosing q is 97
for every q the value of p is fixed
⇒ number of ways of choosing p is 1
and r is fixed as –1
⇒ Number of ordered triplets (p, q, r) is 97.
20)
S3 < 0 ⇒ one root is –ve
⇒a+b=9
