Question

If $f\left(x\right)=$$\{\begin{array}{c}\frac{1}{|x|};|x|\ge 1\\ a{x}^2+b;|x|<1\end{array}$ is a differentiable at every point of the domain, then the values of $a$ and $b$ are respectively:

• A. $\frac{5}{2},-\frac{3}{2}$
• B. $-\frac{1}{2},\frac{3}{2}$
• C. $\frac{1}{2},\frac{1}{2}$
• D. $\frac{1}{2},-\frac{3}{2}$

Answer 1

The correct option is B $-\frac{1}{2},\frac{3}{2}$

$f\left(x\right)$ is continuous at $x=1\Rightarrow 1=a+b$
$f\left(x\right)$ is differentiable at $x=1\Rightarrow -1=2a$
$\Rightarrow a=-\frac{1}{2}$
$\therefore b=\frac{3}{2}$