ExamBro
ExamBro
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:

  1. A \(1\)
  2. B \(0\)
  3. C \(10\)
  4. D \(11\)
Verified Solution

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…

From JEE Mains
Explore more questions on app