Question:
If $f(x)=\left\{\begin{array}{ll}\frac{1}{|x|} & ;|x| \geq 1 \\ a x^{2}+b & ;|x|<1\end{array}\right.$ is differentiable at
every point of the domain, then the values of a and b are respectively :
Correct Option: , 4
Solution:
$f(x)=\left\{\begin{array}{cc}\frac{1}{|x|}, & |x| \geq 1 \\ a x^{2}+b, & |x|<1\end{array}\right.$
at $\mathrm{x}=1$ function must be continuous
So, $1=a+b$ $\ldots(1)$
differentiability at $x=1$
$\left(-\frac{1}{x^{2}}\right)_{x=1}=(2 a x)_{x=1}$
$\Rightarrow-1=2 \mathrm{a} \Rightarrow \mathrm{a}=-\frac{1}{2}$
(1) $\Rightarrow \mathrm{b}=1+\frac{1}{2}=\frac{3}{2}$