Let f(x)= 1 |x| & for |x| 1 ^2+b & for |x|<1 if f(x) is continuou… - Vidyadip

MCQ

Let $\text{f(x)}=\begin{cases}\frac{1}{|\text{x}|} & \text{for |x|}\geq1\\\text{ax}^2+\text{b} & \text{for |x|}<1\end{cases}$ if f(x) is continuous and differentiable at any point, then:

A
$\text{a}=\frac{1}{2},\text{b}=-\frac{3}{2}$
✓
$\text{a}=-\frac{1}{2},\text{b}=\frac{3}{2}$
C
$\text{a}=1,\text{b}=-1$
D
None of these.

✓

Answer

Correct option: B.

$\text{a}=-\frac{1}{2},\text{b}=\frac{3}{2}$

Given function is continuous at x = 1.
$\Rightarrow\lim\limits_{\text{x}\rightarrow1^{+}}\text{f(x)}=\lim\limits_{\text{x}\rightarrow1^{-}}\text{f(x)}$
$\Rightarrow\lim\limits_{\text{x}\rightarrow1}\frac{1}{\text{x}}=\lim\limits_{\text{x}\rightarrow1^{-}}\text{ax}^2+\text{b}$
$\Rightarrow1=\text{a}+\text{b}\ \dots(1)$
Function is derivable at x = 1.
$\Rightarrow\lim\limits_{\text{h}\rightarrow0^{+}}\frac{\text{f}(1+\text{h}-\text{f}(1))}{\text{h}}=\lim\limits_{\text{h}\rightarrow0^{-}}\frac{\text{f}(0+\text{h}-\text{f}(1))}{\text{h}}$
$\Rightarrow\lim\limits_{\text{h}\rightarrow0^{+}}\frac{\frac{1}{1+\text{h}}+1}{\text{h}}=\lim\limits_{\text{h}\rightarrow0^{-}}\frac{\text{a}(1+\text{h})^2-\text{a}}{\text{h}}$
$\Rightarrow-1=\lim\limits_{\text{h}\rightarrow0^{-}}\frac{\text{h}(2\text{a}+\text{h})}{\text{h}}$
$\Rightarrow2\text{a}=-1$
$\Rightarrow\text{a}=\frac{-1}{2}$
$\text{a}+\text{b}=1\ (\text{From(1)})$
$\frac{-1}{2}+\text{a}=1$
$\Rightarrow\text{b}=\frac{3}{2}$

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