# GATE2015-3-45

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If for non-zero $x, \: af(x) + bf(\frac{1}{x}) = \frac{1}{x} - 25$ where a $a \neq b \text{ then } \int_1^2 f(x)dx$ is

1. $\frac{1}{a^2 - b^2} \begin{bmatrix} a(\ln 2 - 25) + \frac{47b}{2} \end{bmatrix}$
2. $\frac{1}{a^2 - b^2} \begin{bmatrix} a(2\ln 2 - 25) - \frac{47b}{2} \end{bmatrix}$
3. $\frac{1}{a^2 - b^2} \begin{bmatrix} a(2\ln 2 - 25) + \frac{47b}{2} \end{bmatrix}$
4. $\frac{1}{a^2 - b^2} \begin{bmatrix} a(\ln 2 - 25) - \frac{47b}{2} \end{bmatrix}$
in Calculus
edited
5

In the equation, substitute (1/x) for x, then you get another equation. Now solve the given equation and the obtained equation. You will get f(x). And then integrate. Simple calculations.

$af\left ( x \right )+bf\left ( \frac{1}{x} \right )=\frac{1}{x} -25$ --- $\left ( 1 \right )$

Integrating both sides,

$a\int_{1}^{2}f\left ( x \right )dx+b\int_{1}^{2}f\left ( \frac{1}{x} \right )dx=\left [ \log\left ( x \right )-25x \right ]_{1}^{2}=\log2-25$ --- $\left ( 2 \right )$

Replacing  $x$ by $\frac{1}{x}$ in $\left ( 1 \right )$, we get

$af\left ( \frac{1}{x} \right )+bf\left ( x \right )=x-25$

Integrating both sides, we get

$a\int_{1}^{2}f\left ( \frac{1}{x} \right )dx+b\int_{1}^{2}f\left ( x \right )dx=\left [ \frac{x^{2}}{2}-25x \right ]_{1}^{2}=-\frac{47}{2}$ --- $\left ( 3 \right )$

Eliminate $\int_{1}^{2}f\left ( \frac{1}{x} \right )$ between $\left ( 2 \right )$ and $\left ( 3 \right )$ by multiplying $\left ( 2 \right )$ by $a$ and $\left ( 3 \right )$ by $b$ and subtracting

$\therefore \left ( a^{2}-b^{2} \right )\int_{1}^{2}f\left ( x \right )dx=a\left ( \log2-25 \right )+b\times\frac{47}{2}$

$\therefore \int_{1}^{2}f\left ( x \right )dx=\frac{1}{\left ( a^{2}-b^{2} \right )}\left [ a\left ( \log2-25 \right )+\frac{47b}{2} \right ]$

Answer: A. $\frac{1}{\left ( a^{2}-b^{2} \right )}\left [ a\left ( \log2-25 \right )+\frac{47b}{2} \right ]$

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