If the function f ( x ) = { 1 | x | , | x | ? 2 a x 2 + 2 b , | x | < 2 is differentiable on R , then 48 ( a + b ) is equal to ______ . [2024]

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Solution

Q.
If the function $f (x)$ $= {\begin{matrix} \frac{1}{| x |}, & | x | \geq 2 \\ a x^{2} + 2 b, & | x | < 2 \end{matrix}$ is differentiable on $R,$ then $48 (a + b)$ is equal to ______ . [2024]

Ans.
(15)

$f (x)$ $= {\begin{matrix} \frac{1}{| x |}, & | x | \geq 2 \\ a x^{2} + 2 b, & | x | < 2 \end{matrix}$

Since, $f$ is differentiable so $f$ must be continuity

$⇒R.H.L. at 2 = L.H.L. at 2$

$\Rightarrow \frac{1}{2} = 4 a + 2 b$ ...(i)

Also, $f$ is differentiable at $x = 2,$

Now, around 2, $f (x)$ $= {\begin{matrix} \frac{1}{x}, & x \geq 2 \\ a x^{2} + 2 b, & x < 2 \end{matrix}$

$\Rightarrow$ $f' (x)$ $= {\begin{matrix} \frac{- 1}{x^{2}}, & x \geq 2 \\ 2 a x, & x < 2 \end{matrix}$

Now, R.H.D. at $x = 2 =$ L.H.D. at $x = 2$

$\Rightarrow - \frac{1}{4} = 4 a \Rightarrow a = - \frac{1}{16} \Rightarrow 2 b = \frac{1}{2} + \frac{1}{4} = \frac{3}{4} [Using (i)]$

$\Rightarrow b = \frac{3}{8}$

So, $48 (a + b) = 48 (\frac{3}{8} - \frac{1}{16}) = \frac{48 \times 5}{16} = 15$

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