JEE Mains · Maths · STD 12 - 7.2 definite integral
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:
- A \(1\)
- B \(0\)
- C \(10\)
- D \(11\)
Answer & Solution
Correct Answer
(D) \(11\)
Step-by-step Solution
Detailed explanation
\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…
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