Question

Examine for continuity and differentiability at the points $x=1,x=2$, the function $f$ defined by $f\left(x\right)=\{\begin{array}{c}x\left[x\right],0\le x<2\\ (x-1)\left[x\right],2\le x\le 3\end{array}$ where $\left[x\right]=$ greatest integer less than or equal to $x$

• A. discontinuous and not derivable at $x=1,2$
• B. discontinuous and not derivable at $x=1$, continuous but not derivable at $x=2$
• C. continuous and not derivable at $x=1,2$.
• D. continuous and not derivable at $x=1$, discontinuous but not derivable at $x=2$.

Answer 1

Answer: (B)

discontinuous and not derivable at $x=1$, continuous but not derivable at $x=2$

$f\left(1^{-}\right)=\underset{h\to 0}{lim}(1-h)0=0$
$f\left(1^{+}\right)=\underset{h\to 0}{lim}1+h=1$
Since f is not continuous at $x=1$, its also not differentiable at that point.
$f\left(2^{-}\right)=\underset{h\to 0}{lim}(2-h)\left(1\right)=2$
$f\left(2^{+}\right)=\underset{h\to 0}{lim}(2+h-1)2=2$
$f^{\prime }\left(2^{-}\right)=\underset{h\to 0}{lim}\frac{2-h-2}{-h}=1$
$f\left(2^{+}\right)=\underset{h\to 0}{lim}\frac{2+2h-2}{h}=2$
Hence, the function is continuous but not differentiable at $x=2.$