Question

Let $f(x)(x \geq 1)$ be a differentiable function satisfying $f(x)=\left(\log _x x\right)^2-\int\limits_1^e \frac{f(t)}{t} d t$.
Area bounded by tangent line of $y=f(x)$ at the point $(e, f(e))$, the curve $y=f(x)$ and the line $x=1$, is

• A. $e +\frac{1}{ e }$
• B. $e +\frac{1}{ e }-1$
• C. $e +\frac{1}{ e }-2$
• D. $e +\frac{1}{ e }-3$

Answer 1

Answer: (D)

Let $C =\int\limits_1^{ e } \frac{ f ( t )}{ t } dt$
$f ( x )=\ln ^2 x - C $
$\therefore C =\int\limits_1^{ e }\left(\frac{\ln ^2( t )- C }{ t }\right) dt =\int\limits_1^{ e } \frac{\ln ^2( t )}{ t } dt - C \int\limits_1^{ e } \frac{1}{ t } dt $
$\left. C =\int\limits_0^1 y ^2 dy - C \ln t \right]_1^{ e } $
$C =\frac{1}{3}- C \Rightarrow C =\frac{1}{6}$

$\Rightarrow f ( x )=\ln ^2 x -\frac{1}{6} $
$f ^{\prime}( x )=\frac{2 \ln x }{ x } \uparrow \forall x \geq 1$
$f ^{\prime \prime}( x )=\frac{2(1-\ln x )}{ x ^2}=0 \Rightarrow x = e$
$A =\int\limits_1^e\left(\left(\ln ^2 x-\frac{1}{6}\right)-\left(\frac{2 x}{e}-\frac{7}{6}\right)\right) d x=\int\limits_1^e\left(\ln ^2 x-\frac{2 x}{e}+1\right) d x $
$ \left.\left.=x \ln ^2 x-2 x \ln x+2 x\right]_1^e-\frac{x^2}{e}+x\right]_1^e=(e-2 e+2 e)-(0-0+2)+\left(e-\frac{e^2}{e}\right)-\left(1-\frac{1}{e}\right)$
$=( e -2)+ e - e +\frac{1}{ e }-1= e +\frac{1}{ e }-3$.