Let f(x)+2 f((1)/(x))=x^2+5 and 2 g(x)-3 g((1)/(2))=x, x 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:

Correct answer: D. 11 aligned & f(x)+2 f((1)/(x))=x^2+5 & f((1)/(x))+2 f(x)=(1)/(x^2)+5 & f(x)=(2)/(3 x^2)-(x^2)/(3)+(5)/(3) & α=∫_1^2((2)/(3 x^2)-(x^2)/(3)+(5)/(3)) dx & (-(2)/(3 x)-(x^3)/(9)+(5 x)/(3))_1^2 & -(1)/(3)-(8)/(9)+(10)/(3)+(2)/(3)+(1)/(9)-(5)/(3) & α=2-(7)/(9)=(11)/(9) & 2 g(x)-3 g((1)/(2))=x & g((1)/(2))=-(1)/(2) & g(x)= x 2 -(3)/(4) & β=∫_1^2((x)/(2)-(3)/(4)) d x & ((x^2)/(4)-(3 x)/(4))_1^2=1-(3)/(2)-(1)/(4)+(3)/(4)=0 & 9 α+β=11 & option (4) aligned

JEE Main 20254 AprilEvening Shift

Exam: JEE Main
Year: 2025
Date: 4 April
Shift: Evening Shift
Subject: Mathematics
Chapter: Functions

JEE Main 2025 · 4 April · Evening Shift · Mathematics · Functions

Let f(x)+2 f((1)/(x))=x^2+5 and 2 g(x)-3 g((1)/(2))=x, x 0 . If...

Let

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

2 g(x)-3 g\left(\frac{1}{2}\right)=x, x \gt 0

. If

\alpha=\int_1^2 f(x) d x

, and

\beta=\int_1^2 g(x) d x

, then the value of

9 \alpha+\beta$ is:

Options

A $1$
B $0$
C $10$
D $11$

Correct answer

D. $11$

Step-by-step solution

\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}

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