Question

Let \[f\left( x \right) = \begin{cases}\frac{1}{\left| x \right|} & for \left| x \right| \geq 1 \\ a x^2 + b & for \left| x \right| < 1\end{cases}\] If f (x) is continuous and differentiable at any point, then

 

 

 

विकल्प

• \[a = \frac{1}{2}, b = - \frac{3}{2}\]

• \[a = - \frac{1}{2}, b = \frac{3}{2}\]

• a = 1, b = − 1

• none of these

Answer 1

(b) \[a = - \frac{1}{2}, b = \frac{3}{2}\]

We have,
`f(x) = {((-1)/x , xle-1),(ax^2 +b, -1 <x<1):}`
\[\text{Given:} f\left( x \right)\text {  is differentiable and continuous at every point} . \]
\[\text{Consider a point x} = 1\]
\[ \lim_{x \to 1^-} f\left( x \right) = \lim_{x \to 1^+} f\left( x \right)\]
\[ \Rightarrow \lim_{x \to 1^-} \left( a x^2 + b \right) = \lim_{x \to 1^+} \frac{1}{x}\]
\[ \Rightarrow a + b = 1 . . . \left( i \right)\]
\[\text{It is also differentiable at x} = 1\]
\[ \lim_{x \to 1^-} \frac{f\left( x \right) - f\left( 1 \right)}{x - 1} = \lim_{x \to 1^+} \frac{f\left( x \right) - f\left( 1 \right)}{x - 1}\]
\[ \Rightarrow \lim_{x \to 1^-} \frac{a x^2 + b - 1}{x - 1} = \lim_{x \to 1^+} \frac{\frac{1}{x} - 1}{x - 1}\]
\[ \Rightarrow \lim_{x \to 1^-} \frac{a x^2 - a}{x - 1} = \lim_{x \to 1^+} \frac{1 - x}{\left( x - 1 \right)x} \left[ \text { Using } \left( i \right) \right]\]
\[ \Rightarrow \lim_{x \to 1^-} a\left( x + 1 \right) = \lim_{x \to 1^+} \left( - x \right)\]
\[ \Rightarrow 2a = - 1\]
\[ \Rightarrow a = \frac{- 1}{2}\]
\[\text{ Plugging }a = \frac{- 1}{2} \text { in } \left( i \right) \text{ we get }, \]
\[b = \frac{3}{2}\]
\[ \therefore a = \frac{- 1}{2}, b = \frac{3}{2}\]