Question

Let $f_1\left(x\right)=x{\tan }^{-1}x,f_2\left(x\right)=\left\{\begin{array}{cc}\frac{x}{\ln 2};& x\ne 1\\ 1;& x=1\end{array}\right\},f_3\left(x\right)=\left\{\begin{array}{cc}(x+1)e^{(\frac{1}{|x|}+\frac{1}{x})},& x\ne 0\\ 0,& x=0\end{array}\right\}{f}_4\left(x\right)=\frac{2\tan (\frac{\pi }{4}-x)}{\cot 2x}\text{if}x\ne \frac{\pi }{4}$

$\begin{array}{cc}\text{List - I}& \text{List - II}\\ \text{(I) Number of critical points for}{f}_1\left(x\right)& \text{(P) -1}\\ \text{over its domain is}& \\ \text{(II) Derivative of}{f}_2\left(x\right)\text{at}x=1\text{is}& \text{(Q) 0}\\ \text{(III) Number of points of discountinuity}& \text{(R) 1}\\ \text{for}{f}_3\left(x\right)\text{in the interval [-2, 2]}\text{is}& \\ \text{(IV) Value of}{f}_4\left(\frac{\pi }{4}\right)\text{such that the}& \text{(S) 2}\\ \text{function is continuos everywhere}& \\ \text{in the interval}[\frac{\pi }{6},\frac{\pi }{3}]\text{is}& \\ & \text{(T) 3}\\ & \text{(U) None of these}\end{array}$
Which of the following is only CORRECT combination?

• A. $\text{I}\to \text{(R)}$
• B. $\text{II}\to \text{(T)}$
• C. $\text{III}\to \text{(U)}$
• D. $\text{IV}\to \text{(Q)}$

Answer 1

The correct option is A $\text{I}\to \text{(R)}$

$\to f_1^{\prime }\left(x\right)={\tan }^{-1}x+\frac{x}{1+x^2}\begin{array}{c}{f}_1^{\prime }\left(x\right)>0,x>0\\ =0,x>0\\ <0,x<0\end{array}\}$
So, one critical point

$\to f_2\left(x\right)$ is discontinuos at $x=1$

$\to f_3\left(x\right)=\left\{\begin{array}{cc}x+1,& x\in [-2,0)\\ 0,& x=0\\ (x+1)e^{-\frac{2}{x}},& x\in (0,2]\end{array}\right\}$
So, point of discontinuity at $x=0$

$\to \underset{x\to \frac{\pi }{4}}{lim}{f}_4\left(x\right)=\underset{x\to \frac{\pi }{4}}{lim}\frac{2\tan (\frac{\pi }{4}-x)}{\cot 2x}=1$