Let {N} be the set of natural numbers. For {n} \\in {N}, define I_{n}=\\int_{0} ^{\\pi} {x \sin ^{2 n}(x)}/{\sin ^{2 n}(x)+\cos ^{2 n}(x)} d x.Then for {m}, {n} \\in{N}

Correct Answer is : Im=In for all m≠n

{I}_{{m}}>{I}_{{n}} for all {m}

I_{{m}}=I_{{n}} for all {m} \\neq{n}

{I}_{{m}} {I}_{{n}} for some {m}

KVPY 2020-21 SB SX Exam Previous Papers

Question : 86 of 120

Marks: +1, -0

Let

{N}

be the set of natural numbers.
For

{n} \∈ {N}

, define

I_{n}=\∫↙{0} ↖{\π} {x \sin ^{2 n}(x)}/{\sin ^{2 n}(x)+\cos ^{2 n}(x)} d x

.Then for

{m}, {n} \∈{N}

$I_{m} < I_{n}$ for all $m < n$
${I}_{{m}}>{I}_{{n}}$ for all ${m} < {n}$
$I_{{m}}=I_{{n}}$ for all ${m} \≠{n}$
${I}_{{m}} < {I}_{{n}}$ for some ${m} < {n}$ and ${I}_{{m}}>{I}_{{n}}$ for some ${m} < {n}$

Solution:

I_{n}={1}/{2} \∫↙{0} ↖{\π} ({x \sin ^{2 n} x}/{\sin ^{2 n} x+\cos ^{2 n} x}+{(\π-x) \sin ^{2 n} x}/{\sin ^{2 n} x+\cos ^{2 n} x}) d x

={\π}/{2} \∫↙{0} ↖{\π} {\sin ^{2 n} x d x}/{\sin ^{2 n} x+\cos ^{2 n} x}

=2 \×{\π}/{2}\∫↙{0} ↖{\π\/2} {\sin ^{2 n} x d x}/{\sin ^{2 n} x+\cos ^{2 n} x}

={\π}/{2}\∫↙{0} ↖{\π\/2}{\sin ^{2 n} x+\cos ^{2 n} x}/{\sin ^{2 n} x+\cos ^{2 n} x} d x

={\π}/{2} \×{\π}/{2}={\π^{2}}/{4}

\⇒{I}_{{m}}={I}_{{n}} \∀{m}, {n}

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