Let be a twice differentiable function. If for some , f(1) = 1 and , then is equal to __________. [2025]

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Solution

Q.
Let $f : (0, \infty) \to R$ be a twice differentiable function. If for some $a \neq 0, \int_{0}^{1} f (λ x) d λ = a f (x)$ , f(1) = 1 and $f (16) = \frac{1}{8}$ , then $16 - f' (\frac{1}{16})$ is equal to __________. [2025]

Ans.
(112)

We have $\int_{0}^{1} f (λ x) d λ = a f (x)$

$\Rightarrow \frac{1}{x} \int_{0}^{x} f (t) d t = a f (x) [Put λ x = t \Rightarrow d λ = \frac{1}{x} d t]$

$\Rightarrow \int_{0}^{x} f (t) d t = a x f (x)$

$\Rightarrow f (x) = a (x f' (x) + f (x)) \Rightarrow (1 - a) f (x) = a x f' (x)$

$\Rightarrow \frac{f' (x)}{f (x)} = \frac{(1 - a)}{a} \frac{1}{x}$ $\Rightarrow \ln f (x) = \frac{1 - a}{a} \ln x + c$ (On integrating)

Now, x = 1, f(1) = 1 $\Rightarrow$ c = 0

Again x = 16, $f (16) = \frac{1}{8} \Rightarrow \frac{1}{8} = {(16)}^{\frac{1 - a}{a}} \Rightarrow - 3 = \frac{4 - 4 a}{a} \Rightarrow a = 4$

Therefore f(x) = $x^{- 3 / 4}$

$\Rightarrow f' (x) = - \frac{3}{4} x^{- \frac{7}{4}}$

Hence, $16 - f' (\frac{1}{16}) = 16 - (- \frac{3}{4} {(2^{- 4})}^{- 7 / 4})$

= 16 + 96 = 112.

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