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  • Let   <img alt="\mathrm{f(x)=\left\{\begin{array}{cc}\int_a^x|5+| 1-t|| d t, & \text { if } x>2 \\ 5 x+1 & \text {, if } x \leq 2\end{array}\right.}" src="https://entrancecorne

Let   \mathrm{f(x)=\left\{\begin{array}{cc}\int_a^x|5+| 1-t|| d t, & \text { if } x>2 \\ 5 x+1 & \text {, if } x \leq 2\end{array}\right.}  , then

Option: 1

f(x) is continuous at x=2


Option: 2

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


Option: 3

f(x) is everywhere differentiable
 


Option: 4

the right derivative of f(x) at x=1 does not exist.


Answers (1)

best_answer

\mathrm{For~ x>2, ~we ~have }

\mathrm{\begin{aligned} f(x) & =\int_0^x(5+|1-t|) d t=\int_0^1\left(5+(1-t) d t+\int_1^x(5-(1-t)) d t\right. \\ & =\int_0^1(6-t) d t+\int_1^x(4+t) d t=\left[6 t-\frac{t^2}{2}\right]_0^1+\left[4 t+\frac{t^2}{2}\right]_1^x \\ & =1+4 x+\frac{x^2}{2} \end{aligned} }

\mathrm{ \text { Thus, we have } f(x)=\left\{\begin{array}{r} 5 x+1, \text { if } x \leq 2 \\ \frac{x^2}{2}+4 x+1, \text { if } x>2 \end{array}\right. \text {. } }

Clearly, f(x) is everywhere continuous and differentiable except possibly at x=2.
Now,   \mathrm{ \lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2} 5 x+1=11 }

and,    \mathrm{ \lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2}\left(\frac{x^2}{2}+4 x+1\right)=11 \\ }
\mathrm{ \therefore \lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2^{+}} f(x) }
So, f(x) is continuous at x=2.
We have : \mathrm{ (LHD ~at~ x=2)=\lim _{x \rightarrow 2^{-}} f^{\prime}(x)=\lim _{x \rightarrow 2} 5=5 }.

Posted by

Ajit Kumar Dubey

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