Let P(x)= {dx}{e^{x}+8e^{-x}+4e^{-3x}}Q(x)= {dx}{e³x}+8e^{x}+4e^{-x}} and R(x)=P(x)-2Q(x)If R(x)=1/2A≤ft({B+2e^{-x}}{C})+K, then the value of (A,B,C)is

Correct Answer is : (tan⁡−1,ex,2)(^{-1},e^{x},2)(tan−1,ex,2)

Correct Answer is : (tan⁡−1,ex,2)(^{-1},e^{x},2)(tan−1,ex,2)

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BITSAT 2018 Slot 2 Solved Paper

Section: Mathematics

Question : 143 of 150

Marks: +1, -0

Let

P(x)=\int \frac{dx}{e^{x}+8e^{-x}+4e^{-3x}}

Q(x)=\int \frac{dx}{e^{3x}+8e^{x}+4e^{-x}}

R(x)=P(x)-2Q(x)

R(x)=\frac{1}{2}A\left(\frac{B+2e^{-x}}{C}\right)+K

(A,B,C)

Solution:

We have,

R(x)=\int (e^{x}+8e^{-x}+4e^{-3x})\,dx

-2\int (e^{3x}+8e^{x}+4e^{-x})\,dx

\Rightarrow R(x)=\int \frac{e^{x}(e^{2x}-2)}{e^{4x}+8e^{2x}+4}\,dx

On substituting

e^{x}=t \Rightarrow e^{x}dx=dt

we get

R(t)=\int \frac{(t^{2}-2)\,dt}{t^{4}+8t^{2}+4} = \int \frac{(1-2t^{-2})\,dt}{(t+2t^{-1})^{2}+4}

=\frac{1}{2}\tan^{-1}\left(\frac{t+2t^{-1}}{2}\right)+K

⇒

R(x)=\frac{1}{2}\tan^{-1}\left(\frac{e^{x}+2e^{-x}}{2}\right)+K

Hence,

(A,B,C)=(\tan^{-1},e^{x},2)

∴ Option (b) is correct.

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