Mathematics - Continuity - Differentiability Question with Solution | TestHub

$f\left(x\right)=\left\{\begin{array}{cl}\frac{1}{\left|x\right|},& \left|x\right|\ge 1\\ a{x}^2+b,& \left|x\right|<1\end{array}\right.$

at $x=1$ function must be continuous

So, $1=a+b \dots \left(1\right)$

differentiability at $x=1$

${\left(-\frac{1}{x^2}\right)}_{x=1}=(2·ax{)}_{x=1}$

$\Rightarrow -1=2a\Rightarrow a=-\frac{1}{2}$

Put in $\left(1\right)\Rightarrow b=1+\frac{1}{2}=\frac{3}{2}$