Question

Each of the following function is defined to be zero at x = 0
$f_1\left(x\right)=x^{2}sgn\left(x\right)$
$f_2\left(x\right)={\int }_0^x{t}^{2}sin\left(\frac{1}{t}\right)dt$
$f_3\left(x\right)=x^{-\frac{1}{3}}\left[sinx\right]$
$f_4\left(x\right)=x^3[-x]$
[where sgn(x) denotes signum function and [.] denotes greatest integer function]
$$\begin{array}{cccc}& \text{List - I}& & \text{List - II}\\ \text{I}& \text{The function}{f}_1& \text{(P)}& \text{continuous but not}\\ & & & \text{differentiable at x = 0}\\ \text{II}& \text{The function}{f}_2\text{is}& \text{(Q)}& \text{first derivative exists at x = 0 but}\\ & & & \text{second derivative does not exist}\\ \text{III}& \text{The function}{f}_3\text{is}& \text{(R)}& \text{second derivative exists at x = 0,}\\ & & & \text{but it is not continuous there}\\ \text{IV}& \text{The function}{f}_4\text{is}& \text{(S)}& \text{second derivative exists at x = 0}\\ & & & \text{and is continuous at x = 0.}\end{array}$$
Which of the following is the only CORRECT combination?

• A. (I), (P)
• B. (II), (Q)
• C. (III), (R)
• D. (IV), (S)

Answer 1

The correct option is D (IV), (S)

$f_1\left(x\right)=\{\begin{array}{cc}{x}^2& x⩾0\\ -x^2,& x<0\end{array}$

$f_1^{\prime }\left(x\right)=\{\begin{array}{cc}2x,& x>0\\ -2x,& x<0\end{array}$

$f_1^{\prime \prime }\left(x\right)=\{\begin{array}{cc}2,& x>0\\ -2,& x<0\end{array}$

So first derivative exists at x = 0 but second derivative does not exist.
$f_2\left(x\right)={\int }_0^x{t}^{2}sin\left(\frac{1}{t}\right)dt,f_2\left(0\right)=0$
$f_2^{\prime }\left(x\right)=x^{2}sin\left(\frac{1}{x}\right)\text{and is not defined at x=0}$

$f_3\left(x\right)=x^{\frac{-1}{3}}|sinx|,f_3\left(0\right)=0$
$f_3^{\prime }\left(0^{+}\right)=\underset{x\to 0}{lim}\frac{x^{-\frac{1}{3}}\sin x}{x}$= does not exist.
Therefore, $f_3$ is continuous but not differentiable at x =0.

$f_4\left(x\right)=\{\begin{array}{cc}0,& -1<x\le 0\\ -x^3,& 0<x<1\end{array}$

$f_4^{\prime }\left(x\right)=\{\begin{array}{cc}0,& -1<x<0\\ -3x^2,& 0<x<1\end{array}$

$f_4^{\prime \prime }\left(x\right)=\{\begin{array}{cc}0,& -1<x<0\\ -6x,& 0<x<1\end{array}$

$f_4^{\prime \prime }\left(0^{+}\right)=0,f_4^{\prime \prime }\left(0^{-}\right)=0$
second derivative exists and continuous at x = 0.