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प्रश्न
If \[f\left( x \right) = \begin{cases}a x^2 - b, & \text { if }\left| x \right| < 1 \\ \frac{1}{\left| x \right|} , & \text { if }\left| x \right| \geq 1\end{cases}\] is differentiable at x = 1, find a, b.
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उत्तर
Given:
It is given that the given function is differentiable at x = 1.
We know every differentiable function is continuous. Therefore it is continuous at x =1. Then,
\[\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) \]
\[ \Rightarrow \lim_{x \to 1} a x^2 - b = \lim_{x \to 1} \frac{1}{x}\]
\[ \Rightarrow a - b = 1 . . . (i)\]
It is also differentiable at x=1. Therefore,
(LHD at x = 1) = (RHD at x = 1)
\[\Rightarrow \lim_{x \to 1^-} \frac{f(x) - f(1)}{x - 1} = \lim_{x \to 1^+} \frac{f(x) - f(1)}{x - 1}\]
\[ \Rightarrow \lim_{x \to 1} \frac{a x^2 - b - 1}{x - 1} = \lim_{x \to 1} \frac{\frac{1}{x} - 1}{x - 1} \]
\[ \Rightarrow \lim_{x \to 1} \frac{a x^2 + 1 - a - 1}{x - 1} = \lim_{x \to 1} \frac{- (x - 1)}{x - 1} \left[ \text { Using }(i) \right]\]
\[ \Rightarrow \lim_{x \to 1} a (x + 1) = \lim_{x \to 1} - 1 \]
\[ \Rightarrow 2a = - 1 \]
\[ \Rightarrow a = - \frac{1}{2}\]
From (i), we have:
\[a - b = 1\]
\[ \Rightarrow - \frac{1}{2} - b = 1\]
\[ \Rightarrow b = - \frac{3}{2}\]
Hence, when
the function is differentiable at x = 1.
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