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

$\Rightarrow \quad \frac{1}{x} \int_{0}^{x} f(\lambda) d\lambda = a f(x)$

On diff. $-\frac{1}{x^2} \int_{0}^{x} f(\lambda) d\lambda + \frac{f(x)}{x} = a f'(x)$

$\Rightarrow \quad -\frac{1}{x^2} (ax f(x)) + \frac{f(x)}{x} = a f'(x)$

$\Rightarrow \quad \frac{y}{x} (1 - a) = a \frac{dy}{dx} \qquad \dots \text{(i)}$

$\Rightarrow \quad (1 - a) \frac{dx}{x} = \frac{a \, dy}{y}$

$\Rightarrow \quad (1 - a) \ln x = a \ln y + c$

$\because \quad f(1) = 1 \implies 0 - 0 = c \implies c = 0$

$\therefore \quad (1 - a) \ln x = a \ln y$

$\quad \;\; (1 - a) \ln x = a \ln(f(x)) \qquad \dots \text{(ii)}$

Put $x = 16 \quad \Rightarrow \quad (1 - a) \ln 16 = a \ln f(16)$

$\Rightarrow \quad 4(1 - a) \ln 2 = a \ln\left(\frac{1}{8}\right) \quad \Rightarrow \quad a = 4$

From (ii), put $x = \frac{1}{16}, a = 4$

$-3 \ln\left(\frac{1}{16}\right) = 4 \ln\left(f\left(\frac{1}{16}\right)\right)$

$12 \ln 2 = 4 \ln f\left(\frac{1}{16}\right)$

$\Rightarrow \quad f\left(\frac{1}{16}\right) = 8$

$\Rightarrow \quad y\left(\frac{1}{16}\right) = 8$

Now from (i),
$\frac{dy}{dx} = \frac{-3}{4} \frac{y}{x} \quad \Rightarrow \quad f'\left(\frac{1}{16}\right) = -\frac{3}{4} \left( \frac{8}{1/16} \right) = -96$

$\therefore \quad 16 - f'\left(\frac{1}{16}\right) = 112$

JEE MainsJEE AdvanceBITSATJEE Mains | PYQ

JEE Mains JEE Advance BITSAT
JEE Mains | PYQ