Question

The domain of the derivative of the function $f(x) = \begin{cases} \tan^{-1} x & \quad \text{if } | x | \le 1 \\ \frac{1}{2} (|x | - 1) & \quad \text{if } |x | > 1 \end{cases}$ is

• A. R - {0}
• B. R - {1}
• C. R - {-1}
• D. R - {-1, 1}

Answer 1

Answer: (D)

The given function is $f(x) = \begin{cases} \tan^{-1} x & \quad \text{if } | x | \le 1 \\ \frac{1}{2} ( |x | - 1) & \quad \text{if } | x | > 1 \end{cases}$
$\Rightarrow \ f(x) = \begin{cases} \frac{1}{2} (-x -1) & \quad \text{if } x < - 1 \\ \tan^{-1} x & \quad \text{if } - 1 \le x \le 1 \\ \frac{1}{2} (x -1) & \quad \text{if } x > 1 \end{cases} $
Clearly L.H.L. at (x = - 1) = $\displaystyle \lim_{h \to 0} f( - 1 - h) = 0$
R.H.L. at (x = - 1) = $\displaystyle \lim_{h \to 0} \tan^{-1} ( - 1 +h) = 3 \pi /4$
$\therefore $ L.H.L. $\neq$ R.H.L. at x = - 1
$\therefore $ f (x) is discontinuous at x = - 1
Also we can prove in the same way, that f (x) is discontinuous at x = 1
$\therefore $ f ' (x) can not be found for x = $\pm$1 or domain of f ' (x) = R - {- 1, 1}