Question

$\text{List I}$ has four entries and $\text{List II}$ has five entries. Each entry of $\text{List I}$ is to be correctly matched with one or more than one entries of $\text{List II}.$

$\begin{array}{cccc}& \text{List I}& & \text{List II}\\ \text{(A)}& \text{Let}f:A\to B\text{be a function defined by}& \text{(P)}& -1\\ & f\left(x\right)=\log (x^2-7|x|+12)\text{. If}C=Z-A& & \\ & \text{is a set, then an element in}C\text{is}& & \\ \text{(B)}& \text{A solution of the inequation}& \text{(Q)}& 0\\ & \left|\frac{2}{x-4}\right|>1\text{is}& & \\ \text{(C)}& \text{If}f\left(x\right)=\log \left(\frac{1-|x|}{|x-2|}\right),\text{then an}& \text{(R)}& 1\\ & \text{integer which is not in the domain}& & \\ & \text{of}f\text{, is}& & \\ \text{(D)}& \text{An element in the domain of the}& \text{(S)}& 2\\ & \text{function}f\left(x\right)=\frac{e^{x-4x^2}}{\sqrt{4x-x^2}}\text{is}& & \\ & & \text{(T)}& 3\end{array}$

Which of the following is the only CORRECT combination?

Answer 1

$\left(A\right)$
$f\left(x\right)=\log (x^2-7|x|+12)$
For $f$ to be defined,
$x^2-7|x|+12>0$
$\Rightarrow (|x|-3)(|x|-4)>0$
$\Rightarrow |x|\notin [3,4]$
$x\in (-\infty ,4)\cup (-3,3)\cup (4,\infty )$
$\therefore C=Z-A=\{\pm 3,\pm 4\}$
$\text{(A)}\to \text{(T)}$

$\left(B\right)$
$\left|\frac{2}{x-4}\right|>1$
$\Rightarrow 2>|x-4|$
$\Rightarrow -2<x-4<2$
$\Rightarrow 2<x<6$ and $x\ne 4$
$\Rightarrow x\in (2,4)\cup (4,6)$
$\text{(B)}\to \text{(T)}$