Question

The values of $a$ and $b$ such that the function defined as $f\left(x\right)=\{\begin{array}{cc}a{x}^2-b,& \left|x\right|<1\\ -1/\left|x\right|,& \left|x\right|\ge 1\end{array}$ is differentiable are ?

• A. $a=1,b=-1$
• B. $a=\frac{1}{2},b=\frac{1}{2}$
• C. $a=\frac{1}{2},b=\frac{3}{2}$
• D. $a=\frac{3}{2},b=\frac{3}{2}$

Answer 1

Answer: (C)

$a=\frac{1}{2},b=\frac{3}{2}$

For differentiability $f\left(x\right)$ should be continou and differentiable at $x=1$

$f\left(x\right)=\{\begin{array}{cc}\frac{1}{x}& x\le -1\\ a{x}^2-b& -1<x<1\\ -\frac{1}{x}& x\ge 1\end{array}$
$f\left(1^{-}\right)=f\left(1^{+}\right)$ for continuity
$a-b=-1$
$f^{\prime }\left(1^{-}\right)=f^{\prime }\left(1^{+}\right)$

$2ax=\frac{1}{x^2}$

$2a=1$
$a=\frac{1}{2},b=\frac{3}{2}$