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

If for x0x\ne 0, af(x)+bf(1x)=1x5af\left( x \right)+bf\left( \dfrac{1}{x} \right)=\dfrac{1}{x}-5, where aba\ne b, then 12xf(x)dx\int\limits_{1}^{2}{xf\left( x \right)dx} is equal to
A. b9a9(a2b2)\dfrac{b-9a}{9\left( {{a}^{2}}-{{b}^{2}} \right)}
B. b9ab(a2b2)\dfrac{b-9a}{b\left( {{a}^{2}}-{{b}^{2}} \right)}
C. b9a6(a2b2)\dfrac{b-9a}{6\left( {{a}^{2}}-{{b}^{2}} \right)}
D. None of these

Explanation

Solution

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

Complete step-by-step answer :
We first use the given function of af(x)+bf(1x)=1x5af\left( x \right)+bf\left( \dfrac{1}{x} \right)=\dfrac{1}{x}-5 to find the main function of f(x)f\left( x \right).
We replace the value of xx with 1x\dfrac{1}{x} and place in the equation of af(x)+bf(1x)=1x5af\left( x \right)+bf\left( \dfrac{1}{x} \right)=\dfrac{1}{x}-5.
So, af(1x)+bf(11/x)=11/x5af(1x)+bf(x)=x5af\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(1x)f\left( \dfrac{1}{x} \right) from the equation of af(1x)+bf(x)=x5af\left( \dfrac{1}{x} \right)+bf\left( x \right)=x-5.
af(1x)+bf(x)=x5 f(1x)=x5bf(x)a \begin{aligned} & 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{aligned}
We replace the value in the equation of af(x)+bf(1x)=1x5af\left( x \right)+bf\left( \dfrac{1}{x} \right)=\dfrac{1}{x}-5.

af(x)+b×x5bf(x)a=1x5 a2f(x)+b[x5bf(x)]=a(1x5) a2f(x)+b(x5)b2f(x)=a(1x5)  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) \\\

a2f(x)b2f(x)=a(1x5)b(x5) \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)f(x) common we have,
f(x)[a2b2]=a(1x5)b(x5)\Rightarrow f(x)\left[ {{a^2} - {b^2}} \right] = a\left( {\dfrac{1}{x} - 5} \right) - b\left( {x - 5} \right)
f(x)[a2b2]=a(1x5)b(x5)\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 a2b2{a^2} - {b^2} on both sides
f(x)=aa2b2(1x5)ba2b2(x5)\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(x)f\left( x \right).
We multiply with xx to find the integral form for 12xf(x)dx\int\limits_{1}^{2}{xf\left( x \right)dx}.

& 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{aligned}$$. We now complete the integration. We have the formula of $$\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{aligned} & \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{aligned}$$ We put the limit values to get $$\begin{aligned} & \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{aligned}$$ 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.