JEE Main 2025 Jan 29 Shift 1, Mathematics Q21: Methods of Solving First Order, First Degree Differential Equation

Substitute $t=\lambda x$, $d\lambda =dt/x$ in the integral:

$\frac{1}{x}{\int }_0^{x}f(t)dt=af(x)⟹{\int }_0^{x}f(t)dt=axf(x).$

Differentiate both sides with respect to $x$: $f(x)=af(x)+ax{f}^{\prime }(x)$, i.e. $(1-a)f(x)=ax{f}^{\prime }(x).$

This separable ODE gives $\frac{f^{\prime }(x)}{f(x)}=\frac{1-a}{a}\cdot \frac{1}{x}$, so $\ln f(x)=\frac{1-a}{a}\ln x+c.$ Equivalently $f(x)=C{x}^{(1-a)/a}.$

$f(1)=1\Rightarrow C=1.$ $f(16)=1/8\Rightarrow 16^{(1-a)/a}=2^{-3}$, i.e. $\frac{4(1-a)}{a}=-3$, giving $a=4.$

Therefore $f(x)=x^{-3/4}$ and $f^{\prime }(x)=-\frac{3}{4}{x}^{-7/4}.$

$f^{\prime }(1/16)=-\frac{3}{4}(2^{-4}{)}^{-7/4}=-\frac{3}{4}\cdot 2^7=-96.$

$16-f^{\prime }(1/16)=16-(-96)=112.$