Question

Match $\text{List I}$ with the $\text{List II}$ and select the correct answer using the code given below the lists :

Let $\left[k\right]$ denote the greatest integer less than or equal to $k$ and $\text{sgn}$ denote the signum function.
$\begin{array}{cccc}& \text{List I}& & \text{List II}\\ \text{(A)}& \text{If}f\left(x\right)=\text{sgn}(x^2-ax+1)\text{has exactly one point of discontinuity,}& \text{(P)}& 1\\ & \text{then value(s) of}a\text{can be}& & \\ \text{(B)}& \text{If}f\left(x\right)=[2+3|n|\sin x]\text{has exactly}11\text{points of discontinuity in}& \text{(Q)}& 2\\ & x\in (0,\pi ),\text{then}n\text{cannot be}& & \\ \text{(C)}& \text{If}f\left(x\right)=|||x|-2|-p|\text{has exactly three points of non-differentiability,}& \text{(R)}& -1\\ & \text{then value(s) of}p\text{can be}& & \\ \text{(D)}& \text{If}\underset{x\to -\infty }{lim}\frac{\sqrt{4x^2+3}-x}{3x-2}=L,\text{then}L\text{equals}& \text{(S)}& -2\\ & & \text{(T)}& 3\end{array}$

Which of the following is a CORRECT combination?

• A. $\left(A\right)\to \left(P\right),\left(R\right);\left(B\right)\to \left(Q\right),\left(S\right)$
• B. $\left(A\right)\to \left(Q\right),\left(S\right);\left(B\right)\to \left(Q\right),\left(R\right),\left(T\right)$
• C. $\left(A\right)\to \left(P\right),\left(R\right);\left(B\right)\to \left(Q\right),\left(T\right)$
• D. $\left(A\right)\to \left(Q\right),\left(S\right);\left(B\right)\to \left(P\right),\left(R\right),\left(T\right)$

Answer 1

The correct option is D $\left(A\right)\to \left(Q\right),\left(S\right);\left(B\right)\to \left(P\right),\left(R\right),\left(T\right)$

$\left(A\right)$
$f\left(x\right)=\text{sgn}(x^2-ax+1)$ has exactly one point of discontinuity.
It means $y=x^2-ax+1$ touches $x-$axis.
$\Rightarrow D=0$
$\Rightarrow a^2-4=0$
$\Rightarrow a=-2,2$

$\left(B\right)$
$[2+3|n|\sin x]=2+\left[3|n|\sin x\right]$
We know that $\left[x\right]$ is discontinuous at $\forall x\in Z$
Let $g\left(x\right)=3|n|\sin x$
So, $f\left(x\right)$ is discontinuous where $g\left(x\right)$ becomes an integer.
Given that $x\in (0,\pi )$
Clearly, at $x=\frac{\pi }{2},$ $g\left(x\right)$ is discontinuous.
And, due to symmetry in the graph of $\sin x$ around $x=\frac{\pi }{2},$ we should have $5$ points of discontinuity in $x\in (0,\frac{\pi }{2})$ and $x\in (\frac{\pi }{2},\pi )$
So, now choose $n$ in such a way that there are exactly $5$ natural numbers less than $3|n|$
For $n=1$ or $-1,$ $3|n|=3$ (rejected)
For $n=2$ or $-2,$ $3|n|=6$ (accepted)
For $n=3,$ $3|n|=9$ (rejected)

Hence, $n$ cannot be $1,-1,3$