Q250 / 270 · NDA 2024 (Phase 1) · Mathematics

Let $3 f(x)+f\left(\frac{1}{x}\right)=\frac{1}{x}+1$ What is $8 \int_1^2 f(x) d x$ equal to?

  • Correct answer
    A
    $\ln (8 \sqrt{e})$
  • B
    $\ln (4 \sqrt{e})$
  • C
    $\ln 2$
  • D
    $\ln 2-1$

Solution

$\begin{aligned} 8 \int_1^2 f(x) d x & =8 \int_1^2\left(\frac{3}{8 x}-\frac{x}{8}+\frac{1}{4}\right) d x \\ & =\int_1^2\left(\frac{3}{x}-x+2\right) d x \\ & =\left[3 \ln x-\frac{x^2}{2}+2 x\right]_1^2 \\ & =\left(3 \ln 2-\frac{4}{2}+4\right)-\left(3 \ln 1-\frac{1}{2}+2\right) \\ & =3 \ln 2+2-\frac{3}{2}=\frac{1}{2}+3 \ln 2 \\ & =\frac{1}{2} \ln e+\ln 2^3=\ln \sqrt{e}+\ln 8 \\ & =\ln (8 \sqrt{e})\end{aligned}$

Hence, the correct answer is option: 'A'.

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