JEE Main 2025 Apr 4 Shift 2, Mathematics Q18: Definite Integration And Fundamental Theorem

Find $f$. Replace $x$ by $1/x$ in the given relation:

$f(1/x)+2f(x)=\frac{1}{x^2}+5$.

Combine with the original $f(x)+2f(1/x)=x^2+5$ and eliminate $f(1/x)$ to get

$f(x)=\frac{2}{3x^2}-\frac{x^2}{3}+\frac{5}{3}$.

Integrate from 1 to 2:

$\alpha ={\left[-\frac{2}{3x}-\frac{x^3}{9}+\frac{5x}{3}\right]}_1^2=\left(-\frac{1}{3}-\frac{8}{9}+\frac{10}{3}\right)-\left(-\frac{2}{3}-\frac{1}{9}+\frac{5}{3}\right)=\frac{11}{9}$.

Find $g$. Setting $x=1/2$ in $2g(x)-3g(1/2)=x$ gives $g(1/2)=-\frac{1}{2}$. So $g(x)=\frac{x}{2}-\frac{3}{4}$.

$\beta ={\int }_1^2\left(\frac{x}{2}-\frac{3}{4}\right)dx={\left[\frac{x^2}{4}-\frac{3x}{4}\right]}_1^2=(1-3/2)-(1/4-3/4)=-\frac{1}{2}+\frac{1}{2}=0$.

$\therefore 9\alpha +\beta =9\cdot \frac{11}{9}+0=11$.