GATE | GATE-CS-2015 (Set 3) | Question 55

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Discuss

(A) A
(B) B
(C) C
(D) D

Answer: (A)

Explanation: Given-
$af(x) + bf(\frac{1}{x}) = \frac{1}{x} - 25$
Putting $x=\frac{1}{x}$ we get-
$af(\frac{1}{x}) + bf(x) = x - 25$
Subtracting the second equation from the first one we get-
$\Rightarrow\:a(f(x)-f(\frac{1}{x})) - b(f(x)-f(\frac{1}{x})) = \frac{1}{x} - x\\\\ \Rightarrow\:(f(x)-f(\frac{1}{x}))(a-b) = \frac{1}{x} - x\\\\ \Rightarrow\:f(\frac{1}{x}) = f(x) - (\frac{1}{x}-x)\frac{1}{(a-b)}$
Putting this value back in the first equation-
$af(x) + b\Big( f(x) - (\frac{1}{x}-x)\frac{1}{(a-b)} \Big) = \frac{1}{x} - 25\\ af(x) + bf(x) = (\frac{1}{x}-x)\frac{b}{(a-b)} + \frac{1}{x} - 25\\ f(x) = \frac{1}{(a^2-b^2)} \Big( b(\frac{1}{x}-x) + \frac{a-b}{x} - 25(a-b)\Big)\\$
Now that we have $f(x)$ we can find $\int \limits_{1}^{2} f(x)\:dx$

$\begin{flalign*} \int f(x)\:dx &= \int \frac{1}{(a^2-b^2)} \Big( b(\frac{1}{x}-x) + \frac{a-b}{x} - 25(a-b)\Big)\:dx \\ &= \frac{1}{(a^2-b^2)} \Big( b(\ln{x}-\frac{x^2}{2}) + (a-b)\ln{x} - 25(a-b)x\Big) \\ &= \frac{1}{(a^2-b^2)} \Big( \frac{-bx^2}{2} + a\ln{x} - 25(a-b)x\Big) \\ &\text{Putting the limits}\\ &= \frac{1}{(a^2-b^2)} \Big[ \frac{-bx^2}{2} + a\ln{x} - 25(a-b)x\Big] \limits_{1}^{2} \\ &= \frac{1}{(a^2-b^2)} \Big[ \big(-2b + a\ln{2} - 50(a-b)\big) - \big(-\frac{b}{2} + 0 - 25(a-b)\big) \Big] \\ &= \frac{1}{(a^2-b^2)} \Big( \frac{-3b}{2} + a\ln{2} - 25(a-b)\Big) \\ &= \frac{1}{(a^2-b^2)} \Big( \frac{47b}{2} + a(\ln{2} - 25)\Big) \\ \end{flalign*}$

Therefore the correct option is A.

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Last Updated : 13 Nov, 2017

GATE | GATE-CS-2015 (Set 3) | Question 55

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