IIT-JEE 2011 · Paper 2 · Mathematics Q60

MathematicsMatrix match+8Question 60 of 60

Match the statements given in Column I with the intervals/union of intervals given in Column II

Column I

A
The set $\left\{\operatorname{Re}\left(\dfrac{2iz}{1-z^2}\right):z\text{ is a complex number},\ |z|=1,\ z\ne\pm1\right\}$ is
B
The domain of the function $f(x)=\sin^{-1}\left(\dfrac{8(3)^{x-2}}{1-3^{2(x-1)}}\right)$ is
C
If $f(\theta)=\begin{vmatrix}1&\tan\theta&1\\-\tan\theta&1&\tan\theta\\-1&-\tan\theta&1\end{vmatrix}$ , then the set $\left\{f(\theta):0\le\theta<\dfrac\pi2\right\}$ is
D
If $f(x)=x^{3/2}(3x-10)$ , $x\ge0$ , then $f(x)$ is increasing in

Column II

P
$(-\infty,-1)\cup(1,\infty)$
Q
$(-\infty,0)\cup(0,\infty)$
R
$[2,\infty)$
S
$(-\infty,-1]\cup[1,\infty)$
T
$(-\infty,0]\cup[2,\infty)$

Answer and worked solution

Official answer: A → S; B → T; C → R; D → R

A For $z=e^{i\theta}$ : $\dfrac{2iz}{1-z^2}=\dfrac{2i}{\bar z-z}=-\dfrac1{\sin\theta}$ , which takes every value with $|{\cdot}|\ge1$ : $(-\infty,-1]\cup[1,\infty)$ . → S.
B With $t=3^x$ the argument is $\dfrac{8t}{9-t^2}$ . Requiring $|{\cdot}|\le1$ : for $t<3$ , $t^2+8t-9\le0$ , so $t\le1$ ; for $t>3$ , $t^2-8t-9\ge0$ , so $t\ge9$ . Hence $x\le0$ or $x\ge2$ : $(-\infty,0]\cup[2,\infty)$ . → T.
C Expanding, $f(\theta)=(1+\tan^2\theta)+(1+\tan^2\theta)=2\sec^2\theta$ . On $[0,\frac\pi2)$ this gives $[2,\infty)$ . → R.
D $f'(x)=\frac32x^{1/2}(3x-10)+3x^{3/2}=\frac{15}2x^{1/2}(x-2)\ge0$ for $x\ge2$ , so $f$ is increasing on $[2,\infty)$ . → R.

Answer: A → S; B → T; C → R; D → R

IIT-JEE 2011 · Paper 2, Mathematics question 60