Question

Suppose f, f' and f'' are continuous on [0,e] and that f'(e)=f(e)=f(1)=1and ${\int }_1^e\frac{f\left(x\right)}{x^2}dx=\frac{1}{2}$ then the value of ${\int }_1^e{f}^{\prime \prime }\left(x\right)lnxdx$ equals

• A. 0
• B. 1
• C. 2
• D. none of these

Answer 1

The correct option is D none of these

Let $I={\int }_1^e{f}^{\prime \prime }\left(x\right)lnxdx={\left[lnx{f}^{{}^{\prime }}\left(x\right)\right]}_1^e-{\int }_1^e\frac{f^{{}^{\prime }}\left(x\right)}{x}dx$
And $I=1-I_1$
Where $I_1={\int }_1^e\frac{f^{\prime }\left(x\right)}{x}dx={\left[\frac{1}{x}f\left(x\right)\right]}_1^e+{\int }_1^e\frac{f\left(x\right)}{x^2}dx$
$=(\frac{1}{e}-1)+\frac{1}{2}-\frac{1}{e}-\frac{1}{2}$
$\therefore I=1-\frac{1}{e}+\frac{1}{2}=\frac{3}{2}-\frac{1}{e.}$