If for non-zero x, c f(x)+d f (\frac{1}{x}) =|\log | x| |+3, where c \neq d, then \int_{1}^{e} f(x) \dx= :

Correct Answer is : (c−d)(3e−2)c2−d2

NIMCET 2024 Paper

Section: Mathematics

Question : 17 of 120

Marks: +1, -0

If for non-zero

x, c f(x)+d f (\;{1}/{x}) =|\log | x| |+3

, where

c ≠ d

, then

∫↙{1}↖{e} f(x) \dx=

Solution:

Put

x=\;{1}/{x}

in the given equation,

\;c f(x)+d f (\;{1}/{x}) =|\log | x| |+3 . . . . .

\;c f (\;{1}/{x}) +d f(x)=|\log | \;{1}/{x}| |+3

We know that

\log (\;{1}/{x}) =-\log x

\;\text " Hence "\; c f (\;{1}/{x}) +d f(x)=|\log | x| |+3 . . . .(2)

Eliminating

f (\;{1}/{x})

from the two equations, we get

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

\;⇒ ∫↙{1}↖{e} f(x) \dx=\;{c-d}/{c^2-d^2} ∫↙{1}↖{e(\log} x+3) \dx

Integrating and putting the values of the limits,

\;\;{c-d}/{c^2-d^2}[x \log x-x+3 x]_1^e

\;=\;{c-d}/{c^2-d^2}[3 e-2]

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