# If for non-zero x, $af(x)+bf(\dfrac{1}{x})=\dfrac{1}{x}-5$ where a not equal to b, then find f(x).

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Hint: Replace x with $\dfrac{1}{x}$ in the given equation and then equate to find the answer.

Given,
$af(x)+bf(\dfrac{1}{x})=\dfrac{1}{x}-5$………(1)
On replacing x by 1/x in the equation we get,
Therefore, after replacing we get,
$\Rightarrow af(\dfrac{1}{x})+bf(x)=x-5$………. (2)
Now for the next step we multiply eq (1) with a and eq (2) with b we get,
$\Rightarrow a^2f(x)+abf(\dfrac{1}{x})=\dfrac{a}{x}-5a$………… (3)
$\Rightarrow abf(\dfrac{1}{x})+b^2f(x)= \dfrac{b}{x}-5b$……. (4)
Now for the next part of the solution we.
Subtract (4) from (3), on doing that we get,
$\Rightarrow (a^2-b^2)f(x)= \dfrac{a}{x}-\dfrac{b}{x}-5a+5b$
Now on simplifying we get,
$\Rightarrow (a-b)(a+b)f(x)=(a-b) \dfrac{1}{x}-5(a-b)$
On taking f(x) as the subject of the formula we get,
$\Rightarrow f(x)=(\dfrac{1-5x}{x(a+b)})$
Before this is the answer to the question,
Therefore, $f(x)=(\dfrac{1-5x}{x(a+b)})$ is the correct answer.

Note: Students often make mistakes in the every step of the question. They are not able to substitute x with the correct variable in the equation. Put correct value while moving to the next step.