Question

The function $f(x)=\{\begin{array}{ll}|x-3|,& x\ge 1\\ \frac{x^2}{4}-\frac{3x}{2}+\frac{13}{4},& x<1\end{array}$ is

• A. continuous at x = 1
• B. differentiable at x = 1
• C. discontinuous at x = 1
• D. differentiable at x = 3

Answer 1

Answer: (A), (B)

Here $f(x)=\{\begin{array}{ll}|x-3|,& x\ge 1\\ \frac{x^2}{4}-\frac{3x}{2}+\frac{13}{4},& x<1\end{array}$ is
$\therefore RHL$ at $x=1$,
$\underset{h\to 0}{\lim }|1+h-3|=2$
LHL at $x=1$.
$\underset{h\to 0}{\lim }\frac{(1-{)}^2}{4}-\frac{3(1-h)}{2}+\frac{13}{4}$
$=\frac{1}{4}-\frac{3}{2}+\frac{13}{4}=\frac{14}{4}-\frac{3}{2}=2$
$\therefore f(x)$ is continuous at $x=1$
Again, $f(x)=\{\begin{array}{ll}-(x-3),& 1\le x<3\\ (x-3),& x\ge 3\\ \frac{x^2}{4}-\frac{3x}{2}+\frac{13}{4},& x<1\end{array}$
$\therefore f^{\prime }(x)=\{\begin{array}{ll}-1,& 1\le x<3\\ 1,& x\ge 3\\ \frac{x}{2}-\frac{3}{2},& x<<br/>1\end{array}$