Question

If $f\left(x\right)=\{\begin{array}{cc}{x}^2\left(\text{sgn}\right[x\left]\right)+\left\{x\right\},& 0\le x<2\\ \sin x+|x-3|,& 2\le x<4\end{array},$
where $\left\{x\right\}$ and $\left[x\right]$ represent the fractional part function and greatest integer function respectively. Then

• A. $f\left(x\right)$ is differentiable at $x=1$.
• B. $f\left(x\right)$ is continuous but non-differentiable at $x=1$.
• C. $f\left(x\right)$ is non-differentiable at $x=2$.
• D. $f\left(x\right)$ is continuous at $x=2$.

Answer 1

Answer: (B), (C)

$f\left(x\right)$ is non-differentiable at $x=2$.

$f\left(x\right)=\{\begin{array}{cc}{x}^2\left(0\right)+\left\{x\right\},& 0\le x<1\\ x^2\left(1\right)+\left\{x\right\},& 1\le x<2\\ \sin x-x+3,& 2\le x<3\\ \sin x+x-3,& 3\le x<4\end{array}$

$f^{\prime }\left(x\right)=\{\begin{array}{cc}1,& 0<x<1\\ 2x+1,& 1<x<2\\ \cos x-1,& 2<x<3\\ \cos x+1,& 3<x<4\end{array}$

Clearly, from the above defining of $f\left(x\right)$ and $f^{\prime }\left(x\right)$, we can conclude that $f\left(x\right)$ is continuous at $x=1$ since $f\left(1\right)=f\left(1^{+}\right)=f\left(1^{-}\right)=1$. But $f\left(x\right)$ is non differentiable at $x=1$ since $f^{\prime }\left(1^{+}\right)=3$ but $f^{\prime }\left(1^{-}\right)=1$.
Also, at $x=2$, $f\left(2^{+}\right)=\sin 2+1$ but $f\left(2^{-}\right)=5$
So, $f\left(x\right)$ is discontinuous at $x=2$ and hence $f\left(x\right)$ is non-differentiable at $x=2.$