Question:
If $f(x)=\left\{\begin{array}{cl}\frac{1}{|x|} & ;|x| \geq 1 \\ a x^{2}+b & ;|x|<1\end{array}\right.$ 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 $x=1$ function must be continuous
So, $1=a+b \ldots$ (i)
differentiability at $x=1$
$\left(-\frac{1}{x^{2}}\right)_{x=1}=(2 a x)_{x=1}$
$\Rightarrow-1=2 \mathrm{a} \Rightarrow \mathrm{a}=-\frac{1}{2}$
$(1) \Rightarrow \mathrm{b}=1+\frac{1}{2}=\frac{3}{2}$
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