If f(x)={cases} ^{-1} x, & { when } |x| ≤ 1 \\ 1/2(|x|-1), & { when } |x| > 1 {cases}, then the domain of d/dx f(x) is

Correct Answer is : R−{−1}R-\{-1\}R−{−1}

Correct Answer is : R−{−1}R-\{-1\}R−{−1}

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AP EAPCET 08-Jul-2022 Shift 1 Paper

Section: Mathematics

Question : 63 of 160

Marks: +1, -0

f(x)=\begin{cases} \tan^{-1} x, & \text{ when } |x| \le 1 \\ \frac{1}{2}(|x|-1), & \text{ when } |x| > 1 \end{cases}

\frac{d}{dx} f(x)

Solution: 👈: Video Solution

f(x) = \begin{cases} \tan^{-1}x, & \text{ when }|x|<1 \\ \frac{1}{2}(|x|-1), & \text{ when }|x|>1 \end{cases}

= \begin{cases} \frac{1}{2}(x-1), & \text{ when } x>1 \\ \tan^{-1}, & \text{ when } -1 < x<1 \\ \frac{1}{2}(-x-1), & \text{ when } x<-1 \end{cases}

f'(x) = \begin{cases} \frac{1}{2}, & \text{ when } x>1 \\ \frac{1}{1+x^2}, & \text{ when } -1 < x<1 \\ \frac{-1}{2}, & \text{ when } x<-1 \end{cases}

\lim\limits_{x\to -1^{-}} f'(x) = \frac{-1}{2}

\lim\limits_{x\to -1^{+}} f'(x) = \lim\limits_{x\to -1^{+}} \frac{1}{1+x^2} = \frac{1}{2}

∴

f(x)

is not differentiable at

-1

\lim\limits_{x\to -1^{+}} f'(x) = \frac{1}{2}

\lim\limits_{x\to -1^{-}} f'(x) = \lim\limits_{x\to -1^{-}} \frac{1}{1+x^2} = \frac{1}{2}

f(x)

is differentiable at

x=1

∴

f'(-1)

does not exist.

∴ The domain of the function is

R-(-1)

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