Question

If we have a function $af\left( x \right)+bf\left( \dfrac{1}{x} \right)=\dfrac{1}{x}-5,x\ne 0$ and $a\ne b$, then $f\left( 2 \right)$is equal to
A. $\dfrac{a}{{{a}^{2}}-{{b}^{2}}}$
B. $\dfrac{\left( -9a+6b \right)}{2\left( {{a}^{2}}-{{b}^{2}} \right)}$
C. $\dfrac{2a+b}{2\left( {{a}^{2}}-{{b}^{2}} \right)}$
D. $\dfrac{2ab}{2\left( {{a}^{2}}-{{b}^{2}} \right)}$

Answer 1

Hint: In this question, we have to find the value of $f\left( x \right)$ by putting $x$ as $\dfrac{1}{x}$ in the given equation. After calculating, $f\left( x \right)$ we will put $x=2$ and then after simplifying we will get the value of $f\left( x \right)$.

Complete step-by-step solution -
In this question, we have been given the equation that $af\left( x \right)+bf\left( \dfrac{1}{x} \right)=\dfrac{1}{x}-5$ where $x\ne 0$ and $a\ne b$. And we have to find the value of $f\left( 2 \right)$.
First, let us consider the equation, $af\left( x \right)+bf\left( \dfrac{1}{x} \right)=\dfrac{1}{x}-5.........(i)$
Now in equation (i), we will put the value of $x=\dfrac{1}{x}$. So, we will get equation (i) as follows,
$af\left( \dfrac{1}{x} \right)+bf\left( \dfrac{1}{\dfrac{1}{x}} \right)=\dfrac{1}{\dfrac{1}{x}}-5$
The above equation can be also written as,
$af\left( \dfrac{1}{x} \right)+bf\left( x \right)=x-5.........(ii)$
Now, we will multiply equation (i) by $a$ so we will get,
${{a}^{2}}f\left( x \right)+abf\left( \dfrac{1}{x} \right)=\dfrac{a}{x}-5a.........(iii)$
Now we will multiply equation (ii) by $b$. So, we will get,
$abf\left( \dfrac{1}{x} \right)+{{b}^{2}}f\left( x \right)=bx-5b.........(iv)$
Now, we will subtract equation (iv) from equation (iii). So, we will get,
${{a}^{2}}f\left( x \right)+abf\left( \dfrac{1}{x} \right)-abf\left( \dfrac{1}{x} \right)-{{b}^{2}}f\left( x \right)=\dfrac{a}{x}-5a-bx+5b$
Now, in the above equation, we can cancel the like terms. So, we will get,
${{a}^{2}}f\left( x \right)-{{b}^{2}}f\left( x \right)=\dfrac{a}{x}-bx-5a+5b$
We can further simplify the above equation, so we will get,
$f\left( x \right)\left[ {{a}^{2}}-{{b}^{2}} \right]=\dfrac{a-b{{x}^{2}}}{x}-5\left( a-b \right)$
$\Rightarrow f\left( x \right)=\dfrac{a-b{{x}^{2}}}{x\left( {{a}^{2}}-{{b}^{2}} \right)}-\dfrac{5\left( a-b \right)}{{{a}^{2}}-{{b}^{2}}}$
Now, we will put $x=2$ in the above equation, so we will get,
$f\left( 2 \right)=\dfrac{a-b{{\left( 2 \right)}^{2}}}{2\left( {{a}^{2}}-{{b}^{2}} \right)}-\dfrac{5\left( a-b \right)}{{{a}^{2}}-{{b}^{2}}}$
Now, we will take the LCM, so we will get,
$\begin{align}
  & f\left( 2 \right)=\dfrac{a-b{{\left( 2 \right)}^{2}}-10\left( a-b \right)}{2\left( {{a}^{2}}-{{b}^{2}} \right)} \\
 & \Rightarrow f\left( 2 \right)=\dfrac{a-4b-10a+10b}{2\left( {{a}^{2}}-{{b}^{2}} \right)} \\
 & \Rightarrow f\left( 2 \right)=\dfrac{-9a+6b}{2\left( {{a}^{2}}-{{b}^{2}} \right)} \\
\end{align}$
Therefore we get the value of $f\left( 2 \right)$ as $\dfrac{-9a+6b}{2\left( {{a}^{2}}-{{b}^{2}} \right)}$.
Hence, option (B) is the correct answer.

Note: The possible mistakes that the students can make in this question is, by directly substituting the value of $x=2$ in the equation given in the question, without simplifying and finding the value of $f\left( x \right)$. Also, in a hurry the students may make mistakes in the calculation.