If f(x)= ax^2-b, & if |x|<1 1 |x| , & if |x| 1 is differentiable at x = 1, find a, b.

Here, f(x)= ax^2-b, & if |x|<1 1 |x| , & if |x| 1 = -1 x , & if |x| -1 ^2-b, & if-1<x<1 1 x ,&if x 1 LHL = _ x 1^ - f(x) = _ h 0 f(1-h) = _ h 0 a(1-h)^2-b = a- b RHL = _ x 1^ + f(x) = _ h 0 f(1+h) = _ h 0 1 1+h Since, f(x) is continuous, so LHL = RHL a - b = 1 .......(1) (LHL at x = 1) = _ x 1^ - f(x)-f(1) x-1 _ h 0 f(1-h)-1 1-h-1 = _ h 0 a(1-h)^2-b-1 -h = _ h 0 a(1-h)^2-(a-1)-1 -h Using equation (1), = _ h 0 a+ah^2-2ah-a+1-1 -h = _ h 0 ah^2-2ah -h = _ h 0 (2a-ah) =2a RHL at x = 1 = _ x 1^ + f(x)-f(1) x-1 _ h 0 f(1+h)- f(1) 1+h-1 _ h 0 1 1+h -1 h _ h 0 1-1-h (1+h)h = -1 Since f(x) is differentiable at x = 1, (LHL at x = 1) = (RHL at x = 1) 2a = -1 a=-1 2 Put a=-1 2 in equation (1), a - b = 1 (-1 2 )-b=1 b=-1 2 -1 b=-3 2 a=-1 2

If f(x)= ax^2-b, & if |x|<1 1 |x| , & if |x| 1 is differentiable …

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Question

If $\text{f(x)}=\begin{cases}\text{ax}^2-\text{b}, & \text{if |x|}<1\\\frac{1}{|\text{x}|}, & \text{if |x|}\geq1\end{cases}$ is differentiable at x = 1, find a, b.

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Answer

Here,
$\text{f(x)}=\begin{cases}\text{ax}^2-\text{b}, & \text{if |x|}<1\\\frac{1}{|\text{x}|}, & \text{if |x|}\geq1\end{cases}$
$=\begin{cases}-\frac{1}{\text{x}}, & \text{if |x|}\leq-1\\\text{ax}^2-\text{b}, & \text{if}-1<\text{x}<1\\\frac{1}{\text{x}},&\text{if x}\geq1\end{cases}$
$\text{LHL }=\lim\limits_{\text{x}\rightarrow1^{-}}\text{f(x)}$
$=\lim\limits_{\text{h}\rightarrow0}\text{f}(1-\text{h})$
$=\lim\limits_{\text{h}\rightarrow0}\text{a}(1-\text{h})^2-\text{b}$
$= \text{a}- \text{b}$
$\text{RHL }=\lim\limits_{\text{x}\rightarrow1^{+}}\text{f(x)}$
$=\lim\limits_{\text{h}\rightarrow0}\text{f}(1+\text{h})$
$=\lim\limits_{\text{h}\rightarrow0}\frac{1}{1+\text{h}}$
Since, f(x) is continuous, so
LHL = RHL
a - b = 1 .......(1)
(LHL at x = 1) $=\lim\limits_{\text{x}\rightarrow1^{-}}\frac{\text{f(x)}-\text{f}(1)}{\text{x}-1}$
$\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(1-\text{h})-1}{1-\text{h}-1}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{a}(1-\text{h})^2-\text{b}-1}{-\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{a}(1-\text{h})^2-(\text{a}-1)-1}{-\text{h}}$
Using equation (1),
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{a}+\text{ah}^2-2\text{ah}-\text{a}+1-1}{-\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{ah}^2-2\text{ah}}{-\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}(2\text{a}-\text{ah})$
$=2\text{a}$
RHL at x = 1 $=\lim_\limits{\text{x}\rightarrow1^{+}}\frac{\text{f(x)}-\text{f}(1)}{\text{x}-1}$
$\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(1+\text{h})-{\text{f}(1)}}{1+\text{h}-1}$
$\lim_\limits{\text{h}\rightarrow0}\frac{\frac{1}{1+\text{h}}-1}{\text{h}}$
$\lim_\limits{\text{h}\rightarrow0}\frac{1-1-\text{h}}{(1+\text{h})\text{h}}$
$= -1$
Since f(x) is differentiable at x = 1,
(LHL at x = 1) = (RHL at x = 1)
2a = -1
$\text{a}=\frac{-1}{2}$
Put $\text{a}=\frac{-1}{2}$ in equation (1),
a - b = 1
$\Big(\frac{-1}{2}\Big)-\text{b}=1$
$\text{b}=\frac{-1}{2}-1$
$\text{b}=\frac{-3}{2}$
$\text{a}=\frac{-1}{2}$

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