$\begin{aligned}
& f(x)+2 f\left(\frac{1}{x}\right)=x^2+5 \
& f\left(\frac{1}{x}\right)+2 f(x)=\frac{1}{x^2}+5 \
& f(x)=\frac{2}{3 x^2}-\frac{x^2}{3}+\frac{5}{3} \
& \alpha=\int_1^2\left(\frac{2}{3 x^2}-\frac{x^2}{3}+\frac{5}{3}\right) \mathrm{dx} \
& \left(-\frac{2}{3 x}-\frac{x^3}{9}+\frac{5 x}{3}\right)_1^2 \
& -\frac{1}{3}-\frac{8}{9}+\frac{10}{3}+\frac{2}{3}+\frac{1}{9}-\frac{5}{3} \
& \alpha=2-\frac{7}{9}=\frac{11}{9} \
& 2 g(x)-3 g\left(\frac{1}{2}\right)=x \
& \mathrm{g}\left(\frac{1}{2}\right)=-\frac{1}{2} \
& \mathrm{g}(\mathrm{x})=\frac{\mathrm{x}}{2}-\frac{3}{4} \
& \beta=\int_1^2\left(\frac{x}{2}-\frac{3}{4}\right) d x \
& \left(\frac{x^2}{4}-\frac{3 x}{4}\right)_1^2=1-\frac{3}{2}-\frac{1}{4}+\frac{3}{4}=0 \
& 9 \alpha+\beta=11 \
& \text { option (4) }
\end{aligned}$
Let f(x)+2 f ( 1 x )=x^2+5 and 2 g(x)-3 g ( 1 2 )=x, x gt 0. If α=∫_1^2 f(x) d x, and β=∫_1^2 g(x) d x, then the value of 9 α+β is:
JEE Main 2025 — Mathematics Algebra
Let f(x)+2f(x1)=x2+5 and 2g(x)−3g(21)=x,x>0. If α=∫12f(x)dx, and β=∫12g(x)dx, then the value of 9α+β is:
Held on 4 Apr 2025 · Verified 6 Jul 2026.
Options
- A
1
- B
0
- C
10
- D
11
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