If for non-zero x, c f(x) + d f≤ft(1/x) = ≤ft| |x|| + 3, where c ≠ d, then ₁^{e} f(x) dx = :

Correct Answer is : (c−d)(3e−2)c2−d2\;(c-d)(3e-2)/c²-d²c2−d2(c−d)(3e−2)

Correct Answer is : (c−d)(3e−2)c2−d2\;(c-d)(3e-2)/c²-d²c2−d2(c−d)(3e−2)

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NIMCET 2024 Paper

Section: Mathematics

Question : 17 of 120

Marks: +1, -0

If for non-zero

x, c f(x) + d f\left(\frac{1}{x}\right) = \left|\log |x|\right| + 3

c \neq d

\int\limits_{1}^{e} f(x) dx =

Solution:

Put

x = \frac{1}{x}

in the given equation,

\;c f(x) + d f\left(\frac{1}{x}\right) = \left|\log |x|\right| + 3 \dots

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

We know that

\log\left(\frac{1}{x}\right) = -\log x

\;\text{ Hence }\; c f\left(\frac{1}{x}\right) + d f(x) = \left|\log |x|\right| + 3 \dots (2)

Eliminating

f\left(\frac{1}{x}\right)

from the two equations, we get

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

\Rightarrow \int\limits_{1}^{e} f(x) dx = \frac{c-d}{c^2-d^2} \int\limits_{1}^{e} (\log x + 3) dx

Integrating and putting the values of the limits,

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

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

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