Question

If $f\left(x\right)=\{\begin{array}{c}a{x}^2-b,if\left|x\right|<-1\\ -\frac{1}{\left|x\right|}if\left|x\right|\ge -1\end{array}$ is differential at $x=1$. Find the values of $a$ and $b$

• A. $a=1/2$; $b=3/2$
• B. $a=1/2$; $b=-3/2$
• C. $a=-1/2$; $b=3/2$
• D. $a=-1/2$; $b=-3/2$

Answer 1

The correct option is A $a=1/2$; $b=3/2$

$f\left(x\right)=\{\begin{array}{c}a{x}^2-b;-1<x<1\\ \frac{-1}{x};x\ge 1\\ \frac{1}{x};x\le -1\end{array}$
Now $f\left(x\right)$ is differentiable at $x=1$
$\Rightarrow 2ax=\frac{1}{x^2}$ at $x=1$
$\Rightarrow a=\frac{1}{2}$
Also $f\left(x\right)$ is continuous at $x=1$
$\Rightarrow a{\left(1\right)}^2-b=-1$
$\Rightarrow b=\frac{3}{2}$