Question

A differentiable function $f(x)$ satisfying integral equation $f(x)={\left({\log }_{c}x\right)}^2-\underset{1}{\overset{e}{\int }}\frac{f(t)}{t}dt$. Suppose $f(x)+\frac{1}{6}=g(x)$.
If $\frac{dz}{dx}-\frac{1}{x\ln x}z=g(x)$, then $z$ is given by
(where $c$ is constant of integration.)

• A. $x(\ln x{)}^2+(c-x)\ln x$
• B. $(x+c)\ln x$
• C. $x(\ln x-1)$
• D. $x(\ln x{)}^2+(c-x)$

Answer 1

Answer: (A)

$f(x)=(\ln x{)}^2-\underset{1}{\overset{e}{\int }}\frac{f(t)dt}{t}$.....(1)
Let $f(x)=(\ln x{)}^2-a$
$\left(\text{ Let }a={\int }_1^c\frac{f(t)}{t}dt\right)$
$\therefore f(x)=(\ln x{)}^2-\frac{1}{6}\Rightarrow g(x)=(\ln x{)}^2$
Now, $\frac{dz}{dx}-\frac{1}{x\ln x}z=(\ln x{)}^2$
$\text{ I.F. }=e^{-\int \frac{1}{x\ln x}dx}=\frac{1}{\ln x}$
Solution is $z-\frac{1}{\ln x}=\int \ln xdx+c$
$\Rightarrow z=(\ln x)(x\ln x-x)+(\ln x)c$
$z=x(\ln x{)}^2+(c-x)\ln x$