NIMCET 2024 — Mathematics PYQ

Given:

$cf(x)+df\left(\frac{1}{x}\right)=|\log |x||+3$ ---(1)

Replacing $x$ with $\frac{1}{x}$:

$cf\left(\frac{1}{x}\right)+df(x)=\left|\log \left|\frac{1}{x}\right|\right|+3$

$df(x)+cf\left(\frac{1}{x}\right)=|\log |x||+3$ ---(2)

Multiply (1) by $c$ and (2) by $d$:

$c^{2}f(x)+cdf\left(\frac{1}{x}\right)=c(|\log |x||+3)$

$d^{2}f(x)+cdf\left(\frac{1}{x}\right)=d(|\log |x||+3)$

Subtracting the equations:

$(c^2-d^2)f(x)=(c-d)(|\log |x||+3)$

$f(x)=\frac{(c-d)(|\log |x||+3)}{c^2-d^2}$

Integrating from $1$ to $e$:

$I={\int }_1^{e}f(x)dx={\int }_1^e\frac{(c-d)(|\log |x||+3)}{c^2-d^2}dx$

$I=\frac{c-d}{c^2-d^2}{\int }_1^e(\log x+3)dx$

Evaluating the integral:

${\int }_1^e(\log x+3)dx=[x\log x-x+3x{]}_1^e$

$=[x\log x+2x{]}_1^e$

$=(e\log e+2e)-(1\log 1+2(1))$

$=(e+2e)-(0+2)$

$=3e-2$

Substituting back:

$I=\frac{(c-d)(3e-2)}{c^2-d^2}$

Correct Option: (B)