Question

A differentiable function $f(x)$ satisfying integral equation $f(x)=\left(\log _c x\right)^2-\int\limits_1^e \frac{f(t)}{t} d t$. 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)

$f(x)=(\ln x)^2-\int\limits_1^e \frac{f(t) d t}{t}$.....(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, $\frac{ dz }{ dx }-\frac{1}{ x \ln x } z =(\ln x )^2$
$\text { I.F. }=e^{-\int \frac{1}{x \ln x} d x}=\frac{1}{\ln x}$
Solution is $z -\frac{1}{\ln x }=\int \ln xdx + c$
$\Rightarrow z =(\ln x )( x \ln x - x )+(\ln x ) c $
$z = x (\ln x )^2+( c - x ) \ln x$