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Question

Question: <img src="https://cdn.pureessence.tech/canvas_581.png?top_left_x=0&top_left_y=600&width=300&height=2...

is equal to -

A

2eπ+422e^{\frac{\pi + 4}{2}}

B

eπ+42e^{\frac{\pi + 4}{2}}

C

eπ42e^{\frac{\pi - 4}{2}}

D

2eπ42e^{\frac{\pi - 4}{2}}

Answer

2eπ42e^{\frac{\pi - 4}{2}}

Explanation

Solution

Let y =

Ž log y =

1n\frac{1}{n} [log(1+12n2)+log(1+22n2)+log(1+n2n2)]\left[ \log \left( 1 + \frac { 1 ^ { 2 } } { \mathrm { n } ^ { 2 } } \right) + \log \left( 1 + \frac { 2 ^ { 2 } } { \mathrm { n } ^ { 2 } } \right) + \ldots \log \left( 1 + \frac { \mathrm { n } ^ { 2 } } { \mathrm { n } ^ { 2 } } \right) \right]

= 1n\frac { 1 } { \mathrm { n } } = 01log\int _ { 0 } ^ { 1 } \log (1 + x2) dx

= [x log (1 + x2) ]01_ { 0 } ^ { 1 }01x\int _ { 0 } ^ { 1 } \mathrm { x } . 2x1+x2\frac { 2 x } { 1 + x ^ { 2 } } dx

= log 2 – 2 01(111+x2)\int _ { 0 } ^ { 1 } \left( 1 - \frac { 1 } { 1 + \mathrm { x } ^ { 2 } } \right) dx

= log 2 – 2 [x – tan–1 x ]01_ { 0 } ^ { 1 } = log 2 + π2\frac { \pi } { 2 } – 2.

\ y = = elog 2 . =.