Q116 / 160 · AP EAMCET 2024 (19 May Shift 2) · Mathematics

If $\mathrm{I}_{\mathrm{n}}=\int_0^{\frac{\pi}{4}} \tan ^n x d x$, then $\mathrm{I}_{13}+\mathrm{I}_{11}=$

  • A
    $\frac{1}{13}$
  • Correct answer
    B
    $\frac{1}{12}$
  • C
    $\frac{1}{10}$
  • D
    $\frac{1}{11}$

Solution

Given, $I_n=\int_0^{\pi / 4} \tan ^n x d x$ and $I_{n+1}=\int_0^{\pi / 4} \tan ^{n+1} x d x \quad(n \rightarrow n+1)$ $I_{n-1}=\int_0^{\pi / 4} \tan ^{n-1} x d x \quad(n \rightarrow n-1)$ Since, we know $I_{n-1}+I_{n+1}=\frac{1}{n} \Rightarrow I_{11}+I_{13}=\frac{1}{12} \quad(\because n=12)$.

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

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