edited by
8,035 views
42 votes
42 votes

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. $\frac{1}{a^2 - b^2} \begin{bmatrix} a(\ln 2 - 25) - \frac{47b}{2} \end{bmatrix}$
edited by

1 Answer

Best answer
101 votes
101 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
Answer:

Related questions

32 votes
32 votes
6 answers
1
go_editor asked Feb 14, 2015
13,113 views
The value of $\displaystyle \lim_{x \rightarrow \infty} (1+x^2)^{e^{-x}}$ is$0$$\frac{1}{2}$$1$$\infty$
21 votes
21 votes
2 answers
2
makhdoom ghaya asked Feb 13, 2015
7,563 views
Compute the value of:$$ \large \int \limits_{\frac{1}{\pi}}^{\frac{2}{\pi}}\frac{\cos(1/x)}{x^{2}}dx$$
31 votes
31 votes
4 answers
3
go_editor asked Sep 28, 2014
8,043 views
The value of the integral given below is$$\int \limits_0^{\pi} \: x^2 \: \cos x\:dx$$$-2\pi$$\pi$$-\pi$$2\pi$
35 votes
35 votes
3 answers
4
Kathleen asked Sep 14, 2014
6,343 views
Let $S = \sum_{i=3}^{100} i \log_{2} i$, and $T = \int_{2}^{100} x \log_{2}x dx$.Which of the following statements is true?$S T$$S = T$$S < T$ and $2S T$$2S ≤ T$