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JEE Advanced · Mathematics · 2014

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1
Let f1:R→Rf_1:\mathbb{R}\to\mathbb{R}, f2:[0,∞)→Rf_2:[0,\infty)\to\mathbb{R}, f3:R→Rf_3:\mathbb{R}\to\mathbb{R} and f4:R→[0,∞)f_4:\mathbb{R}\to[0,\infty) be defined by f1(x)={∣x∣if x<0,exif x≥0;f_1(x)=\begin{cases}|x| & \text{if }x<0,\\ e^x & \text{if }x\ge0;\end{cases} f2(x)=x2f_2(x)=x^2; f3(x)={sin⁡xif x<0,xif x≥0;f_3(x)=\begin{cases}\sin x & \text{if }x<0,\\ x & \text{if }x\ge0;\end{cases} and f4(x)={f2(f1(x))if x<0,f2(f1(x))−1if x≥0.f_4(x)=\begin{cases}f_2(f_1(x)) & \text{if }x<0,\\ f_2(f_1(x))-1 & \text{if }x\ge0.\end{cases} Match each entry of List I\textbf{List I} to its value in List II\textbf{List II}. List I\textbf{List I} P.\textbf{P.} f4f_4 is Q.\textbf{Q.} f3f_3 is R.\textbf{R.} f2∘f1f_2\circ f_1 is S.\textbf{S.} f2f_2 is List II\textbf{List II} 1.\textbf{1.} onto but not one-one 2.\textbf{2.} neither continuous nor one-one 3.\textbf{3.} differentiable but not one-one 4.\textbf{4.} continuous and one-one Which of the following is the correct combination?
Not attempted
A
P→3, Q→1, R→4, S→2P{\to}3,\ Q{\to}1,\ R{\to}4,\ S{\to}2
B
P→1, Q→3, R→4, S→2P{\to}1,\ Q{\to}3,\ R{\to}4,\ S{\to}2
C
P→3, Q→1, R→2, S→4P{\to}3,\ Q{\to}1,\ R{\to}2,\ S{\to}4
D
P→1, Q→3, R→2, S→4P{\to}1,\ Q{\to}3,\ R{\to}2,\ S{\to}4
2
Match each entry of List I\textbf{List I} to its value in List II\textbf{List II}. List I\textbf{List I} P.\textbf{P.} Let y(x)=cos⁡(3cos⁡−1x)y(x)=\cos(3\cos^{-1}x), x∈[−1,1]x\in[-1,1], x≠±32x\ne\pm\dfrac{\sqrt3}{2}. Then 1y(x){(x2−1)d2y(x)dx2+xdy(x)dx}\dfrac{1}{y(x)}\left\{(x^2-1)\dfrac{d^2y(x)}{dx^2}+x\dfrac{dy(x)}{dx}\right\} equals Q.\textbf{Q.} Let A1,A2,…,AnA_1,A_2,\dots,A_n (n>2n>2) be the vertices of a regular polygon of nn sides with centre at the origin. Let a⃗k\vec{a}_k be the position vector of AkA_k. If ∣∑k=1n−1(a⃗k×a⃗k+1)∣=∣∑k=1n−1(a⃗k⋅a⃗k+1)∣\left|\displaystyle\sum_{k=1}^{n-1}(\vec a_k\times\vec a_{k+1})\right|=\left|\displaystyle\sum_{k=1}^{n-1}(\vec a_k\cdot\vec a_{k+1})\right|, the minimum value of nn is R.\textbf{R.} If the normal from the point P(h,1)P(h,1) on the ellipse x26+y23=1\dfrac{x^2}{6}+\dfrac{y^2}{3}=1 is perpendicular to the line x+y=8x+y=8, then the value of hh is S.\textbf{S.} Number of positive solutions satisfying the equation tan⁡−1 ⁣(12x+1)+tan⁡−1 ⁣(14x+1)=tan⁡−1 ⁣(2x2)\tan^{-1}\!\left(\dfrac{1}{2x+1}\right)+\tan^{-1}\!\left(\dfrac{1}{4x+1}\right)=\tan^{-1}\!\left(\dfrac{2}{x^2}\right) is List II\textbf{List II} 1.\textbf{1.} 11 2.\textbf{2.} 22 3.\textbf{3.} 88 4.\textbf{4.} 99 Which of the following is the correct combination?
Not attempted
A
P→4, Q→3, R→2, S→1P{\to}4,\ Q{\to}3,\ R{\to}2,\ S{\to}1
B
P→2, Q→4, R→3, S→1P{\to}2,\ Q{\to}4,\ R{\to}3,\ S{\to}1
C
P→4, Q→3, R→1, S→2P{\to}4,\ Q{\to}3,\ R{\to}1,\ S{\to}2
D
P→2, Q→4, R→1, S→3P{\to}2,\ Q{\to}4,\ R{\to}1,\ S{\to}3
3
Match each entry of List I\textbf{List I} to its value in List II\textbf{List II}. List I\textbf{List I} P.\textbf{P.} The number of polynomials f(x)f(x) with non-negative integer coefficients of degree ≤2\le2, satisfying f(0)=0f(0)=0 and ∫01f(x) dx=1\displaystyle\int_0^1 f(x)\,dx=1, is Q.\textbf{Q.} The number of points in the interval [−13,13][-\sqrt{13},\sqrt{13}] at which f(x)=sin⁡(x2)+cos⁡(x2)f(x)=\sin(x^2)+\cos(x^2) attains its maximum value, is R.\textbf{R.} ∫−223x21+ex dx\displaystyle\int_{-2}^{2}\dfrac{3x^2}{1+e^x}\,dx equals S.\textbf{S.} ∫−1/21/2cos⁡2x log⁡ ⁣(1+x1−x)dx∫01/2cos⁡2x log⁡ ⁣(1+x1−x)dx\dfrac{\displaystyle\int_{-1/2}^{1/2}\cos2x\,\log\!\left(\dfrac{1+x}{1-x}\right)dx}{\displaystyle\int_{0}^{1/2}\cos2x\,\log\!\left(\dfrac{1+x}{1-x}\right)dx} equals List II\textbf{List II} 1.\textbf{1.} 88 2.\textbf{2.} 22 3.\textbf{3.} 44 4.\textbf{4.} 00 Which of the following is the correct combination?
Not attempted
A
P→3, Q→2, R→4, S→1P{\to}3,\ Q{\to}2,\ R{\to}4,\ S{\to}1
B
P→2, Q→3, R→4, S→1P{\to}2,\ Q{\to}3,\ R{\to}4,\ S{\to}1
C
P→3, Q→2, R→1, S→4P{\to}3,\ Q{\to}2,\ R{\to}1,\ S{\to}4
D
P→2, Q→3, R→1, S→4P{\to}2,\ Q{\to}3,\ R{\to}1,\ S{\to}4
4
Let zk=cos⁡(2kπ10)+isin⁡(2kπ10)z_k=\cos\left(\dfrac{2k\pi}{10}\right)+i\sin\left(\dfrac{2k\pi}{10}\right), k=1,2,…,9k=1,2,\dots,9 (the nonzero-index 10th10^{\text{th}} roots of unity). Match each entry of List I\textbf{List I} to its value in List II\textbf{List II}. List I\textbf{List I} P.\textbf{P.} For each zkz_k there exists a zjz_j such that zk⋅zj=1z_k\cdot z_j=1. Q.\textbf{Q.} There exists a k∈{1,2,…,9}k\in\{1,2,\dots,9\} such that z1⋅z=zkz_1\cdot z=z_k has no solution zz in the set of complex numbers. R.\textbf{R.} ∣1−z1∣∣1−z2∣⋯∣1−z9∣10\dfrac{|1-z_1||1-z_2|\cdots|1-z_9|}{10} equals S.\textbf{S.} 1−∑k=19cos⁡(2kπ10)1-\displaystyle\sum_{k=1}^{9}\cos\left(\dfrac{2k\pi}{10}\right) equals List II\textbf{List II} 1.\textbf{1.} True 2.\textbf{2.} False 3.\textbf{3.} 11 4.\textbf{4.} 22 Which of the following is the correct combination?
Not attempted
A
P→1, Q→2, R→4, S→3P{\to}1,\ Q{\to}2,\ R{\to}4,\ S{\to}3
B
P→2, Q→1, R→3, S→4P{\to}2,\ Q{\to}1,\ R{\to}3,\ S{\to}4
C
P→1, Q→2, R→3, S→4P{\to}1,\ Q{\to}2,\ R{\to}3,\ S{\to}4
D
P→2, Q→1, R→4, S→3P{\to}2,\ Q{\to}1,\ R{\to}4,\ S{\to}3
5
Comprehension\textbf{Comprehension}: Box 1 contains three cards bearing numbers 1,2,31,2,3; box 2 contains five cards bearing numbers 1,2,3,4,51,2,3,4,5; and box 3 contains seven cards bearing numbers 1,2,3,4,5,6,71,2,3,4,5,6,7. A card is drawn from each of the boxes. Let xix_i be the number on the card drawn from the ithi^{\text{th}} box, i=1,2,3i=1,2,3. The probability that x1+x2+x3x_1+x_2+x_3 is odd, is
Not attempted
A
29105\dfrac{29}{105}
B
53105\dfrac{53}{105}
C
57105\dfrac{57}{105}
D
12\dfrac12
6
Comprehension\textbf{Comprehension}: Box 1 contains three cards bearing numbers 1,2,31,2,3; box 2 contains five cards bearing numbers 1,2,3,4,51,2,3,4,5; and box 3 contains seven cards bearing numbers 1,2,3,4,5,6,71,2,3,4,5,6,7. A card is drawn from each of the boxes. Let xix_i be the number on the card drawn from the ithi^{\text{th}} box, i=1,2,3i=1,2,3. The probability that x1,x2,x3x_1,x_2,x_3 are in an arithmetic progression, is
Not attempted
A
9105\dfrac{9}{105}
B
10105\dfrac{10}{105}
C
11105\dfrac{11}{105}
D
7105\dfrac{7}{105}
7
Comprehension\textbf{Comprehension}: Given that for each a∈(0,1)a\in(0,1), lim⁡h→0+∫h1−ht−a(1−t)a−1 dt\displaystyle\lim_{h\to0^+}\int_h^{1-h}t^{-a}(1-t)^{a-1}\,dt exists. Let this limit be g(a)g(a). It is given that g(a)g(a) is differentiable on (0,1)(0,1). The value of g ⁣(12)g\!\left(\dfrac12\right) is
Not attempted
A
π\pi
B
2π2\pi
C
π2\dfrac{\pi}{2}
D
π4\dfrac{\pi}{4}
8
Comprehension\textbf{Comprehension}: Given that for each a∈(0,1)a\in(0,1), lim⁡h→0+∫h1−ht−a(1−t)a−1 dt\displaystyle\lim_{h\to0^+}\int_h^{1-h}t^{-a}(1-t)^{a-1}\,dt exists. Let this limit be g(a)g(a). It is given that g(a)g(a) is differentiable on (0,1)(0,1). The value of g′ ⁣(12)g'\!\left(\dfrac12\right) is
Not attempted
A
π2\dfrac{\pi}{2}
B
π\pi
C
−π2-\dfrac{\pi}{2}
D
00
9
Comprehension\textbf{Comprehension}: Let a,r,s,ta,r,s,t be nonzero real numbers. Let P(at2,2at), Q, R(ar2,2ar)P(at^2,2at),\ Q,\ R(ar^2,2ar) and S(as2,2as)S(as^2,2as) be distinct points on the parabola y2=4axy^2=4ax. Suppose that PQPQ is the focal chord and lines QRQR and PKPK are parallel, where KK is the point (2a,0)(2a,0). The value of rr is
Not attempted
A
−1t-\dfrac1t
B
t2+1t\dfrac{t^2+1}{t}
C
1t\dfrac1t
D
t2−1t\dfrac{t^2-1}{t}
10
Comprehension\textbf{Comprehension}: Let a,r,s,ta,r,s,t be nonzero real numbers. Let P(at2,2at), Q, R(ar2,2ar)P(at^2,2at),\ Q,\ R(ar^2,2ar) and S(as2,2as)S(as^2,2as) be distinct points on the parabola y2=4axy^2=4ax. Suppose that PQPQ is the focal chord and lines QRQR and PKPK are parallel, where KK is the point (2a,0)(2a,0). If st=1st=1, then the tangent at PP and the normal at SS to the parabola meet at a point whose ordinate is
Not attempted
A
(t2+1)22t3\dfrac{(t^2+1)^2}{2t^3}
B
a(t2+1)22t3\dfrac{a(t^2+1)^2}{2t^3}
C
a(t2+1)2t3\dfrac{a(t^2+1)^2}{t^3}
D
a(t2+2)2t3\dfrac{a(t^2+2)^2}{t^3}
11
The quadratic equation p(x)=0p(x)=0 with real coefficients has purely imaginary roots. Then the equation p(p(x))=0p(p(x))=0 has
Not attempted
A
Only purely imaginary roots
B
All real roots
C
Two real and two purely imaginary roots
D
Neither real nor purely imaginary roots
12
Three boys and two girls stand in a queue. The probability that the number of boys ahead of every girl is at least one more than the number of girls ahead of her, is
Not attempted
A
12\dfrac12
B
13\dfrac13
C
23\dfrac23
D
34\dfrac34
13
Six cards and six envelopes are numbered 1,2,3,4,5,61,2,3,4,5,6, and cards are to be placed in envelopes so that each envelope contains exactly one card and no card is placed in the envelope bearing the same number, and moreover the card numbered 11 is always placed in envelope numbered 22. Then the number of ways it can be done is
Not attempted
A
264264
B
265265
C
5353
D
6767
14
For x∈(0,π)x\in(0,\pi), the equation sin⁡x+2sin⁡2x−sin⁡3x=3\sin x+2\sin2x-\sin3x=3 has
Not attempted
A
Infinitely many solutions
B
Three solutions
C
One solution
D
No solution
15
In a triangle the sum of two sides is xx and the product of the same two sides is yy. If x2−c2=yx^2-c^2=y, where cc is the third side of the triangle, then the ratio of the in-radius to the circum-radius of the triangle is
Not attempted
A
3y2x(x+c)\dfrac{3y}{2x(x+c)}
B
3y2c(x+c)\dfrac{3y}{2c(x+c)}
C
3y4x(x+c)\dfrac{3y}{4x(x+c)}
D
3y4c(x+c)\dfrac{3y}{4c(x+c)}
16
The common tangents to the circle x2+y2=2x^2+y^2=2 and the parabola y2=8xy^2=8x touch the circle at the points P,QP,Q and the parabola at the points R,SR,S. Then the area of the quadrilateral PQRSPQRS is
Not attempted
A
33
B
66
C
99
D
1515
17
Let f:[0,2]→Rf:[0,2]\to\mathbb{R} be a function which is continuous on [0,2][0,2] and differentiable on (0,2)(0,2) with f(0)=1f(0)=1. Let F(x)=∫0x2f(t) dtF(x)=\displaystyle\int_0^{x^2}f(\sqrt t)\,dt for x∈[0,2]x\in[0,2]. If F′(x)=f′(x)F'(x)=f'(x) for all x∈(0,2)x\in(0,2), then F(2)F(2) equals
Not attempted
A
e2−1e^2-1
B
e4−1e^4-1
C
e−1e-1
D
e4e^4
18
Coefficient of x11x^{11} in the expansion of (1+x2)4(1+x3)7(1+x4)12(1+x^2)^4(1+x^3)^7(1+x^4)^{12} is
Not attempted
A
10511051
B
11061106
C
11131113
D
11201120
19
The following integral ∫π/4π/2(2 cosec x)17 dx\displaystyle\int_{\pi/4}^{\pi/2}(2\,\text{cosec }x)^{17}\,dx is equal to
Not attempted
A
∫0log⁡(1+2)2(eu+e−u)16 du\displaystyle\int_0^{\log(1+\sqrt2)}2(e^u+e^{-u})^{16}\,du
B
∫0log⁡(1+2)(eu+e−u)17 du\displaystyle\int_0^{\log(1+\sqrt2)}(e^u+e^{-u})^{17}\,du
C
∫0log⁡(1+2)(eu−e−u)17 du\displaystyle\int_0^{\log(1+\sqrt2)}(e^u-e^{-u})^{17}\,du
D
∫0log⁡(1+2)2(eu−e−u)16 du\displaystyle\int_0^{\log(1+\sqrt2)}2(e^u-e^{-u})^{16}\,du
20
The function y=f(x)y=f(x) is the solution of the differential equation dydx+xyx2−1=x4+2x1−x2\dfrac{dy}{dx}+\dfrac{xy}{x^2-1}=\dfrac{x^4+2x}{\sqrt{1-x^2}} in (−1,1)(-1,1) satisfying f(0)=0f(0)=0. Then ∫−3232f(x) dx\displaystyle\int_{-\frac{\sqrt3}{2}}^{\frac{\sqrt3}{2}} f(x)\,dx is
Not attempted
A
π3−32\dfrac{\pi}{3}-\dfrac{\sqrt3}{2}
B
π3−34\dfrac{\pi}{3}-\dfrac{\sqrt3}{4}
C
π6−34\dfrac{\pi}{6}-\dfrac{\sqrt3}{4}
D
π6−32\dfrac{\pi}{6}-\dfrac{\sqrt3}{2}
21
Let f:[0,4π]→[0,π]f:[0,4\pi]\to[0,\pi] be defined by f(x)=cos⁡−1(cos⁡x)f(x)=\cos^{-1}(\cos x). The number of points x∈[0,4π]x\in[0,4\pi] satisfying the equation f(x)=10−x10f(x)=\dfrac{10-x}{10} is
Not attempted
Your answer: —
Correct answer: 3
22
Let x⃗,y⃗\vec x,\vec y and z⃗\vec z be three vectors each of magnitude 2\sqrt 2 and the angle between each pair of them is π3\frac{\pi}{3}. If a⃗\vec a is a nonzero vector perpendicular to x⃗\vec x and y⃗×z⃗\vec y\times\vec z, and b⃗\vec b is a nonzero vector perpendicular to y⃗\vec y and z⃗×x⃗\vec z\times\vec x, then:
Not attempted
A
b⃗=(b⃗⋅z⃗)(z⃗−x⃗)\vec b = (\vec b\cdot\vec z)(\vec z-\vec x)
B
a⃗=(a⃗⋅y⃗)(y⃗−z⃗)\vec a = (\vec a\cdot\vec y)(\vec y-\vec z)
C
a⃗⋅b⃗=−(a⃗⋅y⃗)(b⃗⋅z⃗)\vec a\cdot\vec b = -(\vec a\cdot\vec y)(\vec b\cdot\vec z)
D
a⃗=(a⃗⋅y⃗)(z⃗−y⃗)\vec a = (\vec a\cdot\vec y)(\vec z-\vec y)
23
Let a⃗,b⃗\vec a,\vec b, and c⃗\vec c be three non-coplanar unit vectors such that the angle between every pair of them is π3\frac{\pi}{3}. If a⃗×b⃗+b⃗×c⃗=pa⃗+qb⃗+rc⃗\vec a\times\vec b+\vec b\times\vec c=p\vec a+q\vec b+r\vec c, where p,qp,q and rr are scalars, then the value of p2+2q2+r2q2\dfrac{p^2+2q^2+r^2}{q^2} is
Not attempted
Your answer: —
Correct answer: 4
24
From a point P(λ,λ,λ)P(\lambda,\lambda,\lambda), perpendiculars PQPQ and PRPR are drawn respectively on the lines y=x, z=1y=x,\,z=1 and y=−x, z=−1y=-x,\,z=-1. If PP is such that ∠QPR\angle QPR is a right angle, then the possible value(s) of λ\lambda is(are):
Not attempted
A
2\sqrt{2}
B
11
C
−1-1
D
−2-\sqrt{2}
25
A circle SS passes through the point (0,1)(0,1) and is orthogonal to the circles (x−1)2+y2=16(x-1)^2+y^2=16 and x2+y2=1x^2+y^2=1. Then:
Not attempted
A
radius of SS is 88
B
radius of SS is 77
C
centre of SS is (−7,1)(-7, 1)
D
centre of SS is (−8,1)(-8, 1)
26
For a point PP in the plane, let d1(P)d_1(P) and d2(P)d_2(P) be the distances of the point PP from the lines x−y=0x-y=0 and x+y=0x+y=0 respectively. The area of the region RR consisting of all points PP lying in the first quadrant of the plane and satisfying 2≤d1(P)+d2(P)≤42\leq d_1(P)+d_2(P)\leq4, is
Not attempted
Your answer: —
Correct answer: 6
27
Let f:(0,∞)→Rf:(0,\infty)\to\mathbb{R} be given by f(x)=∫1/xxe−(t+1t)dttf(x)=\displaystyle\int_{1/x}^{x} e^{-\left(t+\frac{1}{t}\right)}\frac{dt}{t}. Then:
Not attempted
A
f(x)f(x) is monotonically increasing on [1,∞)[1,\infty).
B
f(x)f(x) is monotonically decreasing on (0,1)(0,1).
C
f(x)+f(1x)=0f(x)+f\left(\frac{1}{x}\right)=0, for all x∈(0,∞)x \in (0,\infty).
D
f(2x)f(2^x) is an odd function of xx on R\mathbb{R}.
28
Let f:[a,b]→[1,∞)f:[a,b]\to[1,\infty) be a continuous function and let g:R→Rg:\mathbb{R}\to\mathbb{R} be defined as g(x)={0if x<a,∫axf(t) dtif a≤x≤b,∫abf(t) dtif x>b.g(x)=\begin{cases}0 & \text{if } x<a,\\ \displaystyle\int_a^x f(t)\,dt & \text{if } a\leq x\leq b,\\ \displaystyle\int_a^b f(t)\,dt & \text{if } x>b.\end{cases} Then:
Not attempted
A
g(x)g(x) is continuous but not differentiable at aa.
B
g(x)g(x) is differentiable on R\mathbb{R}.
C
g(x)g(x) is continuous but not differentiable at bb.
D
g(x)g(x) is continuous and differentiable at either aa or bb but not both.
29
The value of ∫014x3{d2dx2(1−x2)5}dx\displaystyle\int_0^1 4x^3\left\{\frac{d^2}{dx^2}(1-x^2)^5\right\}dx is
Not attempted
Your answer: —
Correct answer: 2
30
For every pair of continuous functions f,g:[0,1]→Rf, g : [0,1] \to \mathbb{R} such that max⁡{f(x):x∈[0,1]}=max⁡{g(x):x∈[0,1]}\max\{f(x) : x \in [0,1]\} = \max\{g(x): x \in [0,1]\}, the correct statement(s) is(are):
Not attempted
A
(f(c))2+3f(c)=(g(c))2+3g(c)(f(c))^2+3f(c) = (g(c))^2+3g(c) for some c∈[0,1]c \in [0,1].
B
(f(c))2+f(c)=(g(c))2+3g(c)(f(c))^2+f(c) = (g(c))^2+3g(c) for some c∈[0,1]c \in [0,1].
C
(f(c))2+3f(c)=(g(c))2+g(c)(f(c))^2+3f(c) = (g(c))^2+g(c) for some c∈[0,1]c \in [0,1].
D
(f(c))2=(g(c))2(f(c))^2 = (g(c))^2 for some c∈[0,1]c \in [0,1].
31
Let a∈Ra \in \mathbb{R} and let f:R→Rf:\mathbb{R}\to\mathbb{R} be given by f(x)=x5−5x+af(x)=x^5-5x+a. Then:
Not attempted
A
f(x)f(x) has three real roots if a>4a > 4.
B
f(x)f(x) has only one real root if a>4a > 4.
C
f(x)f(x) has three real roots if a<−4a < -4.
D
f(x)f(x) has three real roots if −4<a<4-4 < a < 4.
32
Let f:R→Rf:\mathbb{R}\to\mathbb{R} and g:R→Rg:\mathbb{R}\to\mathbb{R} be respectively given by f(x)=∣x∣+1f(x)=|x|+1 and g(x)=x2+1g(x)=x^2+1. Define h:R→Rh:\mathbb{R}\to\mathbb{R} by h(x)={max⁡{f(x),g(x)}if x≤0,min⁡{f(x),g(x)}if x>0.h(x)=\begin{cases}\max\{f(x),g(x)\} & \text{if } x\leq 0,\\ \min\{f(x),g(x)\} & \text{if } x> 0.\end{cases} The number of points at which h(x)h(x) is not differentiable is
Not attempted
Your answer: —
Correct answer: 3
33
The slope of the tangent to the curve (y−x5)2=x(1+x2)2(y-x^5)^2=x(1+x^2)^2 at the point (1,3)(1,3) is
Not attempted
Your answer: —
Correct answer: 8
34
The largest value of the nonnegative integer aa for which lim⁡x→1{−ax+sin⁡(x−1)+ax+sin⁡(x−1)−1}1−x1−x=14\displaystyle\lim_{x\to1}\left\{\frac{-ax+\sin(x-1)+a}{x+\sin(x-1)-1}\right\}^{\frac{1-x}{1-\sqrt x}}=\frac14 is
Not attempted
Your answer: —
Correct answer: 0
35
Let a,b,ca,b,c be positive integers such that ba\frac{b}{a} is an integer. If a,b,ca,b,c are in geometric progression and the arithmetic mean of a,b,ca,b,c is b+2b+2, then the value of a2+a−14a+1\dfrac{a^2+a-14}{a+1} is
Not attempted
Your answer: —
Correct answer: 4
36
Let n≥2n\geq 2 be an integer. Take nn distinct points on a circle and join each pair of points by a line segment. Colour the line segment joining every pair of adjacent points blue and the rest red. If the number of red and blue line segments are equal, then the value of nn is
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Your answer: —
Correct answer: 5
37
Let n1<n2<n3<n4<n5n_1<n_2<n_3<n_4<n_5 be positive integers such that n1+n2+n3+n4+n5=20n_1+n_2+n_3+n_4+n_5=20. Then the number of such distinct arrangements (n1,n2,n3,n4,n5)(n_1,n_2,n_3,n_4,n_5) is
Not attempted
Your answer: —
Correct answer: 7
38
Let MM and NN be two 3×33\times 3 matrices such that MN=NMMN = NM. Further, if M≠N2M \neq N^2 and M2=N4M^2 = N^4, then which of the following statement(s) is(are) correct?
Not attempted
A
Determinant of (M2+MN2)(M^2 + MN^2) is 00.
B
There is a 3×33\times 3 non-zero matrix UU such that (M2+MN2)U(M^2+MN^2)U is the zero matrix.
C
Determinant of (M2+MN2)≥1(M^2+MN^2) \geq 1.
D
For a 3×33\times 3 matrix UU, if (M2+MN2)U(M^2+MN^2)U equals the zero matrix then UU is the zero matrix.
39
Let MM be a 2×22\times 2 symmetric matrix with integer entries. Then MM is invertible if:
Not attempted
A
the first column of MM is the transpose of the second row of MM
B
the second row of MM is the transpose of the first column of MM
C
MM is a diagonal matrix with nonzero entries in the main diagonal
D
the product of entries in the main diagonal of MM is not the square of an integer
40
Let f:(−π2,π2)→Rf:\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\to\mathbb{R} be given by f(x)=(log⁡(sec⁡x+tan⁡x))3f(x)=(\log(\sec x+\tan x))^3. Then:
Not attempted
A
f(x)f(x) is an odd function.
B
f(x)f(x) is a one-one function.
C
f(x)f(x) is an onto function.
D
f(x)f(x) is an even function.
Tip\textbf{Tip}: First show u(x)=log⁡(sec⁡x+tan⁡x)u(x)=\log(\sec x+\tan x) is odd and strictly increasing using u′(x)=sec⁡x>0u'(x)=\sec x>0 and the identity (sec⁡x−tan⁡x)(sec⁡x+tan⁡x)=1(\sec x-\tan x)(\sec x+\tan x)=1; cubing an odd, strictly increasing bijection preserves all three properties.
Deconstruction\textbf{Deconstruction}: Let u(x)=log⁡(sec⁡x+tan⁡x)u(x)=\log(\sec x+\tan x), so f(x)=u(x)3f(x)=u(x)^3. We must check odd/even-ness, injectivity, and surjectivity of ff. Strategy\textbf{Strategy}: Establish the properties of uu first, then transfer them to f=u3f=u^3 using that t↦t3t\mapsto t^3 is an odd, strictly increasing bijection of R\mathbb{R}. Execution\textbf{Execution}: uu is odd: u(−x)=log⁡(sec⁡x−tan⁡x)u(-x)=\log(\sec x-\tan x). Since (sec⁡x−tan⁡x)(sec⁡x+tan⁡x)=sec⁡2x−tan⁡2x=1(\sec x-\tan x)(\sec x+\tan x)=\sec^2x-\tan^2x=1, sec⁡x−tan⁡x=1sec⁡x+tan⁡x\sec x-\tan x = \frac{1}{\sec x+\tan x}, so u(−x)=−u(x)u(-x)=-u(x). uu is strictly increasing: u′(x)=sec⁡x>0u'(x)=\sec x>0 throughout (−π/2,π/2)\left(-\pi/2,\pi/2\right). Since uu is odd, f(−x)=u(−x)3=(−u(x))3=−f(x)f(-x)=u(-x)^3=(-u(x))^3=-f(x): ff is odd — proves (A)\textbf{(A)}, and rules out (D)\textbf{(D)} (an odd function that is also even must be identically 00, which ff is not). Since uu is strictly increasing and t↦t3t\mapsto t^3 is strictly increasing on R\mathbb{R}, the composite f=u3f=u^3 is strictly increasing, hence one-one — proves (B)\textbf{(B)}. Since u:(−π/2,π/2)→Ru:\left(-\pi/2,\pi/2\right)\to\mathbb{R} is continuous, strictly increasing, with u(x)→±∞u(x)\to\pm\infty as x→±π/2x\to\pm\pi/2, uu is onto R\mathbb{R}; and t↦t3t\mapsto t^3 is a bijection of R\mathbb{R}, so ff is also onto R\mathbb{R} — proves (C)\textbf{(C)}. Verification\textbf{Verification}: f(0)=(log⁡1)3=0f(0)=(\log 1)^3=0, consistent with an odd function; and ff being a strictly increasing bijection of an interval onto R\mathbb{R} automatically forces both one-one and onto. Final Answer\textbf{Final Answer}: (A), (B), and (C).