Question

Let $f\left(x\right)=\{\begin{array}{cc}ta{n}^{-1}x:& |x|\ge 1\\ \frac{x^2-1}{4}:& |x|<1.\end{array}$ Then domain of f'(x) is:

• A. $R-\left\{1\right\}$
• B. $R-\{-1,0,1\}$
• C. $R-\{-1,1\}$
• D. $-\{-1\}$

Answer 1

The correct option is C $R-\{-1,1\}$

$f\left(x\right)=\{\begin{array}{cc}ta{n}^{-1}x:& |x|\ge 1\\ \frac{x^2-1}{4}:& |x|<1\end{array}=\{\begin{array}{cc}ta{n}^{-1}x:& x\le -1\\ \frac{x^2-1}{4}:& -1<x<1\\ ta{n}^{-1}x:& x\ge 1\end{array}$

Clearly f(x) is discontinuous at change points –1 and 1 and therefore not differentiable at x = –1, 1.
$\therefore$ Domain $f^{\prime }\left(x\right)=R-\{-1.1\}$