Question

If for $x\ne 0$, $af\left( x \right)+bf\left( \dfrac{1}{x} \right)=\dfrac{1}{x}-5$, where $a\ne b$, then \[\int\limits_{1}^{2}{xf\left( x \right)dx}\] is equal to
A. $\dfrac{b-9a}{9\left( {{a}^{2}}-{{b}^{2}} \right)}$
B. $\dfrac{b-9a}{b\left( {{a}^{2}}-{{b}^{2}} \right)}$
C. $\dfrac{b-9a}{6\left( {{a}^{2}}-{{b}^{2}} \right)}$
D. None of these

Answer 1

Hint: We first express the main function $f\left( x \right)$. We replace the value of $x$ with $\dfrac{1}{x}$ and place in the equation. From two equations we find the general form of the $f\left( x \right)$. We then complete the integration part to find the solution after multiplying with $x$.

Complete step-by-step answer:
We first use the given function of $af\left( x \right)+bf\left( \dfrac{1}{x} \right)=\dfrac{1}{x}-5$ to find the main function of $f\left( x \right)$.
We replace the value of $x$ with $\dfrac{1}{x}$ and place in the equation of $af\left( x \right)+bf\left( \dfrac{1}{x} \right)=\dfrac{1}{x}-5$.
So, $af\left( \dfrac{1}{x} \right)+bf\left( \dfrac{1}{{}^{1}/{}_{x}} \right)=\dfrac{1}{{}^{1}/{}_{x}}-5\Rightarrow af\left( \dfrac{1}{x} \right)+bf\left( x \right)=x-5$.
We find the value of $f\left( \dfrac{1}{x} \right)$ from the equation of $af\left( \dfrac{1}{x} \right)+bf\left( x \right)=x-5$.
$\begin{align}
  & af\left( \dfrac{1}{x} \right)+bf\left( x \right)=x-5 \\
 & \Rightarrow f\left( \dfrac{1}{x} \right)=\dfrac{x-5-bf\left( x \right)}{a} \\
\end{align}$
We replace the value in the equation of $af\left( x \right)+bf\left( \dfrac{1}{x} \right)=\dfrac{1}{x}-5$.
\[
  af\left( x \right) + b \times \dfrac{{x - 5 - bf\left( x \right)}}{a} = \dfrac{1}{x} - 5 \\
   \Rightarrow {a^2}f\left( x \right) + b\left[ {x - 5 - bf\left( x \right)} \right] = a\left( {\dfrac{1}{x} - 5} \right) \\
   \Rightarrow {a^2}f\left( x \right) + b\left( {x - 5} \right) - {b^2}f\left( x \right) = a\left( {\dfrac{1}{x} - 5} \right) \\
 \]
\[ \Rightarrow {a^2}f\left( x \right) - {b^2}f\left( x \right) = a\left( {\dfrac{1}{x} - 5} \right) - b\left( {x - 5} \right)\]
Now taking \[f(x)\] common we have,
\[ \Rightarrow f(x)\left[ {{a^2} - {b^2}} \right] = a\left( {\dfrac{1}{x} - 5} \right) - b\left( {x - 5} \right)\]
\[ \Rightarrow f\left( x \right)\left[ {{a^2} - {b^2}} \right] = a\left( {\dfrac{1}{x} - 5} \right) - b\left( {x - 5} \right)\]
Now dividing by \[{a^2} - {b^2}\] on both sides
\[ \Rightarrow f\left( x \right) = \dfrac{a}{{{a^2} - {b^2}}}\left( {\dfrac{1}{x} - 5} \right) - \dfrac{b}{{{a^2} - {b^2}}}\left( {x - 5} \right)\]
We found the main function of $f\left( x \right)$.
We multiply with $x$ to find the integral form for \[\int\limits_{1}^{2}{xf\left( x \right)dx}\].
\[\begin{align}
  & xf\left( x \right) \\
 & =x\left[ \dfrac{a}{{{a}^{2}}-{{b}^{2}}}\left( \dfrac{1}{x}-5 \right)-\dfrac{b}{{{a}^{2}}-{{b}^{2}}}\left( x-5 \right) \right] \\
 & =\dfrac{a}{{{a}^{2}}-{{b}^{2}}}\left( 1-5x \right)-\dfrac{b}{{{a}^{2}}-{{b}^{2}}}\left( {{x}^{2}}-5x \right) \\
\end{align}\].
We now complete the integration. We have the formula of \[\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}\].
\[\int\limits_{1}^{2}{xf\left( x \right)dx}=\int\limits_{1}^{2}{\left[ \dfrac{a}{{{a}^{2}}-{{b}^{2}}}\left( 1-5x \right)-\dfrac{b}{{{a}^{2}}-{{b}^{2}}}\left( {{x}^{2}}-5x \right) \right]dx}\].
Breaking it into two parts we get
\[\begin{align}
  & \int\limits_{1}^{2}{xf\left( x \right)dx} \\
 & =\dfrac{a}{{{a}^{2}}-{{b}^{2}}}\int\limits_{1}^{2}{\left( 1-5x \right)dx}-\dfrac{b}{{{a}^{2}}-{{b}^{2}}}\int\limits_{1}^{2}{\left( {{x}^{2}}-5x \right)dx} \\
 & =\left[ \dfrac{a}{{{a}^{2}}-{{b}^{2}}}\left( x-\dfrac{5{{x}^{2}}}{2} \right)-\dfrac{b}{{{a}^{2}}-{{b}^{2}}}\left( \dfrac{{{x}^{3}}}{3}-\dfrac{5{{x}^{2}}}{2} \right) \right]_{1}^{2} \\
\end{align}\]
We put the limit values to get
\[\begin{align}
  & \left[ \dfrac{a}{{{a}^{2}}-{{b}^{2}}}\left( x-\dfrac{5{{x}^{2}}}{2} \right)-\dfrac{b}{{{a}^{2}}-{{b}^{2}}}\left( \dfrac{{{x}^{3}}}{3}-\dfrac{5{{x}^{2}}}{2} \right) \right]_{1}^{2} \\
 & =\dfrac{a}{{{a}^{2}}-{{b}^{2}}}\left( 2-10+\dfrac{5}{2}-1 \right)-\dfrac{b}{{{a}^{2}}-{{b}^{2}}}\left( \dfrac{8}{3}-10+\dfrac{5}{2}-\dfrac{1}{3} \right) \\
 & =\dfrac{-13a}{2\left( {{a}^{2}}-{{b}^{2}} \right)}-\dfrac{-31b}{6\left( {{a}^{2}}-{{b}^{2}} \right)} \\
 & =\dfrac{31b-39a}{6\left( {{a}^{2}}-{{b}^{2}} \right)} \\
\end{align}\]
Therefore, the correct option is D.
So, the correct answer is “Option D”.

Note: The change of function can also be done for $f\left( \dfrac{1}{x} \right)$. But in that case, we need to find the change of the limit. if we get the function form of $f\left( \dfrac{1}{x} \right)$, then all the function would have to be written in $\dfrac{1}{x}$ form to get the general form.