Question

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.

Answer 1

Given:  

If \[f\left( x \right) = \begin{cases}a x^2 +b, & \left| x \right| < 1 \\ \frac{1}{\left| x \right|} , & \left| x \right| \geq 1\end{cases}\] 

 

If \[f\left( x \right) = \begin{cases}-\frac{1}{x}, & x  < -1 \\ a x^2 +b , & -1  < x <  1 \\ \frac{1}{ x } , &  x  \geq 1\end{cases}\] 

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 

\[a = - \frac{1}{2}\] and 

\[b = - \frac{3}{2}\]

 the function is differentiable at x = 1.