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GATE CSE 2015 | Set 3 | Question: 45

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If for non-zero $x, \: af(x) + bf(\frac{1}{x}) = \frac{1}{x} - 25$ where $a \neq b \text{ then } \int\limits_1^2 f(x)dx$ 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. $\frac{1}{a^2 - b^2} \begin{bmatrix} a(\ln 2 - 25) - \frac{47b}{2} \end{bmatrix}$

3 Answers

Best answer
137 137 votes

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

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

• selected by
8 8 votes
Just do option elimination...first put a=1,b=0 you will get f(x)=1/x-25 and do integration...option b,c elininated....

now put a=0, b=1...you will get answer after integration
0 0 votes

The problem gives you this equation:
$$ a \cdot f(x) \mathbf{+} b \cdot f(1/x) = \text{Right Hand Side} $$

This Gives you the following result. Better to remember this !!!

( 1 / (a² - b²) ) times [ ( a times Integral of Original RHS ) MINUS ( b times Integral of FLIPPED RHS ) ]

FLIPPED RHS = x - 25 

Integral limit is 0 to 2. That's why, Small number - 25(Big number) = Negetive Number 

Then the formula becomes : 

( 1 / (a² - b²) ) times [ ( a times Integral of Original RHS ) ADD ( b times Integral of FLIPPED RHS ) ]

Now, Options A and D have this. 

Integral of 1/x = ln (x)

Integral of -25 =  - 25x

Limit is 1 to 2 : 

Then :

[ ln(2) - 25*2 ] - [ ln(1) - 25 ] =  ln(2) - 25

 Option A has ln(2) - 25. Hence we Eliminate D. 

Option A = Correct 

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