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Question: If we have a function \(af\left( x \right)+bf\left( \dfrac{1}{x} \right)=\dfrac{1}{x}-5,x\ne 0\) and...

If we have a function af(x)+bf(1x)=1x−5,x≠0af\left( x \right)+bf\left( \dfrac{1}{x} \right)=\dfrac{1}{x}-5,x\ne 0 and a≠ba\ne b, then f(2)f\left( 2 \right)is equal to
A. aa2−b2\dfrac{a}{{{a}^{2}}-{{b}^{2}}}
B. (−9a+6b)2(a2−b2)\dfrac{\left( -9a+6b \right)}{2\left( {{a}^{2}}-{{b}^{2}} \right)}
C. 2a+b2(a2−b2)\dfrac{2a+b}{2\left( {{a}^{2}}-{{b}^{2}} \right)}
D. 2ab2(a2−b2)\dfrac{2ab}{2\left( {{a}^{2}}-{{b}^{2}} \right)}

Explanation

Solution

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

Complete step-by-step solution -
In this question, we have been given the equation that af(x)+bf(1x)=1x−5af\left( x \right)+bf\left( \dfrac{1}{x} \right)=\dfrac{1}{x}-5 where x≠0x\ne 0 and a≠ba\ne b. And we have to find the value of f(2)f\left( 2 \right).
First, let us consider the equation, af(x)+bf(1x)=1x−5.........(i)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=1xx=\dfrac{1}{x}. So, we will get equation (i) as follows,
af(1x)+bf(11x)=11x−5af\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(1x)+bf(x)=x−5.........(ii)af\left( \dfrac{1}{x} \right)+bf\left( x \right)=x-5.........(ii)
Now, we will multiply equation (i) by aa so we will get,
a2f(x)+abf(1x)=ax−5a.........(iii){{a}^{2}}f\left( x \right)+abf\left( \dfrac{1}{x} \right)=\dfrac{a}{x}-5a.........(iii)
Now we will multiply equation (ii) by bb. So, we will get,
abf(1x)+b2f(x)=bx−5b.........(iv)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,
a2f(x)+abf(1x)−abf(1x)−b2f(x)=ax−5a−bx+5b{{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,
a2f(x)−b2f(x)=ax−bx−5a+5b{{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(x)[a2−b2]=a−bx2x−5(a−b)f\left( x \right)\left[ {{a}^{2}}-{{b}^{2}} \right]=\dfrac{a-b{{x}^{2}}}{x}-5\left( a-b \right)
⇒f(x)=a−bx2x(a2−b2)−5(a−b)a2−b2\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=2x=2 in the above equation, so we will get,
f(2)=a−b(2)22(a2−b2)−5(a−b)a2−b2f\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,
f(2)=a−b(2)2−10(a−b)2(a2−b2) ⇒f(2)=a−4b−10a+10b2(a2−b2) ⇒f(2)=−9a+6b2(a2−b2) \begin{aligned} & 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{aligned}
Therefore we get the value of f(2)f\left( 2 \right) as −9a+6b2(a2−b2)\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=2x=2 in the equation given in the question, without simplifying and finding the value of f(x)f\left( x \right). Also, in a hurry the students may make mistakes in the calculation.