KVPY 2020-21 SB SX Exam Previous Papers

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Question : 15 of 120

Marks: +1, -0

Let

f(x)=\{\table {x}/{\sin x}{,}, x \∈(0,1);1{,}, x=0

,Consider the integral

{I}_{{n}}=\√{{n}} \∫↙{0} ↖{1 \/ {n}} {f}({x}) {e}^{-{nx}} {dx}

Then

\lim ↙{n \→\∞} {I}_{{n}}

does not exist
exists and is 0
exists and is 1
exists and is $1 – e^{–1}$

Solution:

ƒ(x) is an increasing function.
so,

f({x}) \∈[1, {1}/{\sin 1})\∀ {x} \∈[0,1)

Now,

\√{n}∫↙{0} ↖{1 \/ {n}} e^{-n x} d x \≤ \√ {n}∫↙{0} ↖{1 \/ {n}} f(x) e^{-n x} d x \

≤ {\√{n}}/{\sin 1}∫↙{0} ↖{1 \/ {n}} e^{-n x} d x

\⇒\lim ↙{n \→\∞} {1-{1}/{{e}}}/{\√{{n}}} ≤ \ \lim ↙{n \→\∞}{I}_{{n}}

\≤ {1-{1}/{{e}}}/{(\sin 1) \√{{n}}}

\⇒0 \≤ \lim ↙{n \→\∞}{I}_{{n}} \≤ 0

\⇒\lim ↙{n \→\∞}{I}_{{n}}=0

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