Question

If $f(x)=\{\begin{array}{ll}\underset{0}{\overset{x}{\int }}(5+|1-t|)dt,& x>2\\ 5x+1,& x\le 2\end{array}$, 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)

$f(x)=\underset{0}{\overset{1}{\int }}(5+(1-t))dt+\underset{0}{\overset{x}{\int }}(5+(t-1))dt$
$=6-\frac{1}{2}+{\left(4t+\frac{t^2}{2}\right)}_1^x$
$=\frac{11}{2}+4x+\frac{x^2}{2}-4-\frac{1}{2}$
$=\frac{x^2}{2}+4x+1$
$f\left(2^{+}\right)=2+8+1=11$
$f(2)=f(2)=5\times 2+1=11$
$\Rightarrow$ continuous at $x=2$
Clearly differentiable at $x=1$
$L{f}^{\prime }(2)=5$
$R{f}^{\prime }(2)=6$
$\Rightarrow$ not differentiable at $x=2$