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Question: The domain of the derivative of the function f(x) = \(\left\{ \begin{matrix} \tan^{- 1}x & if|x| \l...

The domain of the derivative of the function f(x)

= {tan⁡−1xif∣x∣≤112(∣x∣−1)if∣x∣>1 \left\{ \begin{matrix} \tan^{- 1}x & if|x| \leq 1 \\ \frac{1}{2}\left( |x| - 1 \right) & if|x| > 1 \end{matrix} \right.\ is

A

R − {0}

B

R − {1}

C

R − {−1}

D

R − {−1, 1}

Answer

R − {−1, 1}

Explanation

Solution

The given function is

f(x) = {tan⁡−1xif∣x∣≤112(∣x∣−1)if∣x∣>1 \left\{ \begin{matrix} \tan^{- 1}x & if|x| \leq 1 \\ \frac{1}{2}\left( |x| - 1 \right) & if|x| > 1 \end{matrix} \right.\

⇒ f(x) = {12(−x−1)tan⁡−1x12(x−1) ifx<−1if−1≤x≤1ififx>1\left\{ \begin{matrix} \frac{1}{2}( - x - 1) \\ \tan^{- 1}x \\ \frac{1}{2}(x - 1) \end{matrix} \right.\ \begin{matrix} ifx < - 1 \\ if - 1 \leq x \leq 1 \\ ififx > 1 \end{matrix}

Clearly L.H.L. at (x = −1) = lim⁡h→0f(−1−h)\lim_{h \rightarrow 0}f( - 1 - h)

R.H.L. at (x = −1) = lim⁡h→0f(−1+h)=lim⁡h→0tan⁡−1(−1+h)\lim_{h \rightarrow 0}f( - 1 + h) = \lim_{h \rightarrow 0}\tan^{- 1}( - 1 + h)

= −3 π/4

∴ L.H.L. ≠ R.H.L. at x = −1

∴ f(x) is discontinuous at x = − 1

Also we can prove in the same way, that f(x) is discontinuous at x = 1

∴ f’(x) can not be found for x = ±\pm1 or domain of f’(x)

= R−{−1, 1}