Continuity And Differentiability Question 6
Question 6 - 2024 (30 Jan Shift 1)
If the function $f(x)=\begin{cases}\frac{1}{|x|} & ,|x| \geq 2 \\ a x^{2}+2 b, & |x|<2\end{cases}$ is differentiable on $R$, then $48(a+b)$ is equal to
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Answer (15)
Solution
$f(x)\begin{cases}{c} \frac{1}{x} ; x \geq 2 \\ ax^{2}+2 b ;-2<x<2 \\ -\frac{1}{x} ; x \leq-2 \end{cases}$
Continuous at $x=2 \Rightarrow \frac{1}{2}=\frac{a}{4}+2 b$
Continuous at $x=-2 \Rightarrow \frac{1}{2}=\frac{a}{4}+2 b$
Since, it is differentiable at $x=2$
$-\frac{1}{x^{2}}=2 a x$
Differentiable at $x=2 \Rightarrow \frac{-1}{4}=4 a \Rightarrow a=\frac{-1}{16}, b=\frac{3}{8}$
