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If \mathrm{f(x)=\int_{-1}^t|1| d t, x \geq-1}, then

Option: 1

f and f ' are continuous for x + 1 > 0


Option: 2

f is continuous but f ' is not so for x + 1 > 0


Option: 3

f and f ' are continuous at x=0.


Option: 4

f is continuous at x=0 but f ' is not so.


Answers (1)

best_answer

If \mathrm{-1 \leq x<0}, then
\mathrm{ f(x)=\int_{-1}^x|t| d t=\int_{-1}^x-t d t=-\frac{1}{2}\left(x^2-1\right) . }
If \mathrm{ x \geq 0 }, then
\mathrm{ f(x) =\int_{-1}^0-t d t+\int_0^x t d t=\frac{1}{2}\left(x^2+1\right) . \\ }
\mathrm{ \therefore f(x)= \begin{cases}-\frac{1}{2}\left(x^2-1\right), & -1 \leq x<0 \\ \frac{1}{2}\left(x^2+1\right), & 0 \leq x .\end{cases} }

It can be easily seen that f(x) is continuous at x=0. So, it is continuous for all x > -1.

Also \mathrm{ R f^{\prime}(0)=0=L f^{\prime}(0) }. So, f(x) is differentiable at x=0.

\mathrm{ \therefore f^{\prime}(x)=\left\{\begin{array}{rrr} -x, & -1<x<0 \\ 0, & x=0 \\ x, & x>0 \end{array}\right. }

Clearly f ' (x) is continuous at x=0

Consequently, it is continuous for all x > -1 i.e. for x + 1 > 0.

Hence, f and f ' are continuous for x + 1 > 0.

Posted by

Divya Prakash Singh

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