Prove the following


If $f(x)=\left\{\begin{array}{ll}\int_{0}^{x}(5+|1-t|) d t, & x>2 \\ 5 x+1, & x \leq 2\end{array}\right.$, then

  1. $f(x)$ is not continuous at $x=2$

  2. $f(x)$ is everywhere differentiable

  3. $f(x)$ is continuous but not differentiable at $x=2$

  4. $f(x)$ is not differentiable at $x=1$

Correct Option: , 3


$f(\mathrm{x})=\int_{0}^{1}(5+(1-\mathrm{t})) \mathrm{dt}+\int_{1}^{\mathrm{x}}(5+(\mathrm{t}-1)) \mathrm{dt}$

$=6-\frac{1}{2}+\left.\left(4 \mathrm{t}+\frac{\mathrm{t}^{2}}{2}\right)\right|_{1} ^{\mathrm{x}}$

$=\frac{11}{2}+4 x+\frac{x^{2}}{2}-4-\frac{1}{2}$

$=\frac{x^{2}}{2}+4 x+1$


$f(2)=f\left(2^{-}\right)=5 \times 2+1=11$

$\Rightarrow$ continuous at $x=2$

Clearly differentiable at $x=1$



$\Rightarrow$ not differentiable at $x=2$

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