For x>0, if fx=∫1xlog_et/(1+t)dt, then fe+f1/e is equal to

Correct answer: (B) 1/2. fx=∫1xlog_et/(1+t)dt f1/x=∫11/x(ℓnt)/(1+t)dt, let t=1/y =∫1x(ℓn1/y)/(1+1/y).(-1y^2dy=+∫1xℓny)/(1+y).(yy^2d)/(y=∫1xℓnyy1+ydyhence fx+f1x)=∫1x(1+tℓnt)/(t1+t)dt=∫1x(ℓnt)/tdt =1/2ln^2x so fe+f1/e=1/2 ...3

Q40 / 90 · JEE Main 2021 (26 Feb Shift 2) · Mathematics

For $x > 0$ , if $f (x) = \int_{1}^{x} \frac{\log_{e} t}{(1 + t)} d t$ , then $f (e) + f (\frac{1}{e})$ is equal to

A
$0$
Correct answer
B
$\frac{1}{2}$
C
$- 1$
D
$1$

Solution

$f (x) = \int_{1}^{x} \frac{\log_{e} t}{(1 + t)} d t$

$f (\frac{1}{x}) = \int_{1}^{1 / x} \frac{ℓ n t}{1 + t} d t$ , let $t = \frac{1}{y}$

$= \int_{1}^{x} \frac{ℓ n (1 / y)}{1 + (1 / y)} . (\frac{- 1}{y^{2}}) d y$

$= + \int_{1}^{x} \frac{ℓ n y}{1 + y} . \frac{y}{y^{2}} d y$

$= \int_{1}^{x} \frac{ℓ n y}{y (1 + y)} d y$

hence

$f (x) + f (\frac{1}{x}) = \int_{1}^{x} \frac{(1 + t) ℓ n t}{t (1 + t)} d t = \int_{1}^{x} \frac{ℓ n t}{t} d t$

$= \frac{1}{2} \ln^{2} (x)$

so $f (e) + f (\frac{1}{e}) = \frac{1}{2} . . . (3)$

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

For x>0, if fx=∫1xloget1+tdt, then fe+f1e is equal to

Solution

Relevant Questions

For $x > 0$ , if $f (x) = \int_{1}^{x} \frac{\log_{e} t}{(1 + t)} d t$ , then $f (e) + f (\frac{1}{e})$ is equal to