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Question : 103 of 118
Marks:
+1,
-0
Solution:
Given
f(x)=f(x)+f(‌1/x),
where
f(x)=‌log‌t‌/‌1+t‌dt‌∴F(e)=f(e)+f(‌1/e)‌⇒F(e)‌=‌log‌t‌/‌1+t‌dt+‌log‌t‌/‌1+t‌dt...(A) Now for solving,
I=∈t11/e‌‌dt ∴ Put
=z⇒−‌‌dt=dz‌⇒dt=−‌and limit for
t=1⇒z=1 and for
‌t=1/e⇒z=e‌∴I=‌(−‌)‌=‌(−‌)‌=(−‌)‌[‌ as ‌log‌1=0] =‌dz∴I=‌‌dt [ By property
f(t)‌dt=f(x)‌dx]
Equation
(A) becomes
F(e)=‌‌dt+‌‌dt=‌| t⋅log‌t+log‌t |
| t(1+t) |
‌dt=‌⇒F(e)=‌‌dt Let
log‌t=x∴‌‌dt=dx [ for limit
t=1,x=0 and
t=e,x=log‌e=1]∴F(e)=x‌dx⇒F(e)=[‌]01⇒F(e)=‌
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