Question

If $f(x)=\begin{cases}\int\limits_{0}^{x}(5+|1-t|) d t, & x > 2 \\ 5 x+1, & x \leq 2\end{cases}$, then

• A. f(x) is not continuous at x=2
• B. f(x) is everywhere differentiable
• C. f(x) is continuous but not differentiable at x=2
• D. f(x) is not differentiable at x=1

Answer 1

Answer: (C)

$f(x)=\int\limits_{0}^{1}(5+(1-t)) d t+\int\limits_{0}^{x}(5+(t-1)) d t$
$=6-\frac{1}{2}+\left(4 t+\frac{t^{2}}{2}\right)_{1}^{x}$
$=\frac{11}{2}+4 x+\frac{x^{2}}{2}-4-\frac{1}{2}$
$=\frac{x^{2}}{2}+4 x+1$
$f\left(2^{+}\right)=2+8+1=11$
$f(2)=f(2)=5 \times 2+1=11$
$\Rightarrow$ continuous at $x=2$
Clearly differentiable at $x=1$
$L f'(2)=5$
$R f'(2)=6$
$\Rightarrow$ not differentiable at $x=2$