Calculus – Differential Equations – JEE Main 29 Jan 2025 Shift 1

Solution:

Given the integral equation:

$\int_{0}^{1}f(\lambda x)d\lambda = af(x)$

Let $t = \lambda x$. Then $d\lambda = \frac{1}{x} dt$.

Limits: When $\lambda = 0, t = 0$. When $\lambda = 1, t = x$.

The equation becomes:

$\frac{1}{x} \int_{0}^{x} f(t) dt = af(x)$

$\int_{0}^{x} f(t) dt = ax f(x)$

Differentiate both sides with respect to $x$ using Leibniz rule:

$f(x) = a [x f'(x) + f(x)]$

$f(x) = ax f'(x) + a f(x)$

$(1-a)f(x) = ax f'(x)$

Separating variables:

$\frac{f'(x)}{f(x)} = \frac{1-a}{ax}$

Integrate both sides:

$\ln f(x) = \frac{1-a}{a} \ln x + C$

Using $f(1) = 1$:

$\ln(1) = \frac{1-a}{a} \ln(1) + C \Rightarrow 0 = 0 + C \Rightarrow C = 0$.

So, $f(x) = x^{\frac{1-a}{a}}$.

Using the given condition $f(16) = \frac{1}{8}$:

$\frac{1}{8} = (16)^{\frac{1-a}{a}}$

$2^{-3} = (2^4)^{\frac{1-a}{a}}$

Equating powers of 2:

$-3 = 4\left(\frac{1-a}{a}\right)$

$-3a = 4 – 4a$

$a = 4$.

Now substitute $a=4$ back into the function:

$f(x) = x^{\frac{1-4}{4}} = x^{-\frac{3}{4}}$.

We need to find $16 – f'(\frac{1}{16})$.

Find derivative $f'(x)$:

$f'(x) = -\frac{3}{4} x^{-\frac{3}{4} – 1} = -\frac{3}{4} x^{-\frac{7}{4}}$.

Evaluate at $x = \frac{1}{16} = 2^{-4}$:

$f'(\frac{1}{16}) = -\frac{3}{4} (2^{-4})^{-\frac{7}{4}} = -\frac{3}{4} (2^7) = -\frac{3}{4} (128) = -3(32) = -96$.

Value $= 16 – f'(\frac{1}{16}) = 16 – (-96) = 16 + 96 = 112$.

Ans. (112)