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JEE Main · 2025 · 29 Jan 2025 · Shift 1 · Differential Equations

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 = af (x)

,

f (1) = 1

and

f (16) = \frac{1}{8}

, then

16 - f^{'} (\frac{1}{16})

is equal to ______.

Answer: 112

Solution

Substituting

t = λ x

in

\int_{0}^{1} f (λ x) d λ = a f (x)

gives

\frac{1}{x} \int_{0}^{x} f (t) d t = a f (x) ⟹ \int_{0}^{x} f (t) d t = a x f (x) .

Differentiating,

f (x) = a (x f^{'} (x) + f (x)) ⟹ (1 - a) f (x) = a x f^{'} (x) ⟹ \frac{f ^{'} ( x )}{f ( x )} = \frac{1 - a}{a} \cdot \frac{1}{x} .

Integrating,

lo g_{e} f (x) = \frac{1 - a}{a} lo g_{e} x + c

;

f (1) = 1

gives

c = 0

, so

f (x) = x^{(1 - a) / a}

.

Using

f (16) = 1/8

:

1 6^{(1 - a) / a} = 2^{- 3} ⟹ \frac{4 ( 1 - a )}{a} = - 3 ⟹ a = 4.

So

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

and

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

. Then

f^{'} (\frac{1}{16}) = - \frac{3}{4} (\frac{1}{16})^{- 7/4} = - \frac{3}{4} \cdot 1 6^{7/4} = - \frac{3}{4} \cdot 128 = - 96,

so

16 - f^{'} (\frac{1}{16}) = 16 - (- 96) = 112.

JEE Main · 2025 · 29 Jan 2025 · Shift 1 · Integer answer

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