Question

Let $f(x)(x\ge 1)$ be a differentiable function satisfying $f(x)={\left({\log }_{x}x\right)}^2-\underset{1}{\overset{e}{\int }}\frac{f(t)}{t}dt$.
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 $C=\underset{1}{\overset{e}{\int }}\frac{f(t)}{t}dt$
$f(x)={\ln }^{2}x-C$
$\therefore C=\underset{1}{\overset{e}{\int }}\left(\frac{{\ln }^2(t)-C}{t}\right)dt=\underset{1}{\overset{e}{\int }}\frac{{\ln }^2(t)}{t}dt-C\underset{1}{\overset{e}{\int }}\frac{1}{t}dt$
${C=\underset{0}{\overset{1}{\int }}{y}^{2}dy-C\ln t]}_1^e$
$C=\frac{1}{3}-C\Rightarrow C=\frac{1}{6}$

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