JEE Main 2025 — 4 April, Evening Shift — Question 25

$f(x)+2 f\left(\frac{1}{x}\right)=x^{2}+5$

$2 f\left(\frac{1}{x}\right)+4 f(x)=2\left(\frac{1}{x^{2}}+5\right)$

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

$f(x)=\frac{1}{3}\left(\frac{2}{x^{2}}-x^{2}+5\right)$

$2 g(x)-3 g\left(\frac{1}{x}\right)=x$

$2 g\left(\frac{1}{x}\right)-3 g(x)=\frac{1}{x}$

Or $4 g(x)-6 g\left(\frac{1}{x}\right)=2 x$

$6 g\left(\frac{1}{x}\right)-9 g(x)=\frac{3}{x}$

$-5 g(x)=2 x+\frac{3}{x}$

Or $g(x)=-\frac{1}{5}\left(2 x+\frac{3}{x}\right)$

$\int_{1}^{2} f(x) d x=\int_{1}^{2} \frac{1}{3}\left(\frac{2}{x^{2}}-x^{2}+5\right) d x$

$$=\frac{1}{3}\left[-\frac{2}{x}-\frac{x^{3}}{3}+5 x\right]_{1}^{2}$$

$=\frac{1}{3}\left[\left(-\frac{2}{2}-\frac{8}{3}+10\right)-\left(-2-\frac{1}{3}+5\right)\right]$

$=\frac{1}{3}\left[-1-\frac{8}{3}+10+2+\frac{1}{3}-5\right]$

$\alpha=\frac{11}{9}$

Now, $2 g(x)=x+3 g\left(\frac{1}{2}\right)$

$2 g\left(\frac{1}{2}\right)=\frac{1}{2}+3 g\left(\frac{1}{2}\right)$

$g\left(\frac{1}{2}\right)=-\frac{1}{2}$

$\therefore \beta=\int_{1}^{2} g(x) d x$

$=\frac{1}{2} \int_{1}^{2}\left(x+3 g\left(\frac{1}{2}\right)\right) d x$

$=\frac{1}{2}\left[\frac{x^{2}}{2}+3 g\left(\frac{1}{2}\right) x\right]_{1}^{2}$ $=0$

$\therefore 9 \alpha+\beta=11$