Question

If for non-zero x, $af\left( x \right) + bf\left( {\dfrac{1}{x}} \right) = \dfrac{1}{x} - 5$ , where $a \ne b$ , find $f\left( x \right)$ .

Answer 1

We can replace the $x$in the equation with $\dfrac{1}{x}$ . Then we get 2 equations. Then we can multiply the equations with a or b to get the term $abf\left( {\dfrac{1}{x}} \right)$ in both the equations. Then we can take their difference. After simplification, we can write the resulting equation in terms of $f\left( x \right)$ . Thus, we can obtain the required value of the function.

Complete step-by-step answer:
It is given that $af\left( x \right) + bf\left( {\dfrac{1}{x}} \right) = \dfrac{1}{x} - 5$ …. (1)
As x is not equal to zero, we can substitute $\dfrac{1}{x}$ in the place of $x$. So, we get,
$af\left( \dfrac{1}{x} \right) + bf\left( {\dfrac{1}{\dfrac{1}{x}}} \right) = \dfrac{1}{\dfrac{1}{x}} - 5$
On simplification, we get,
$\Rightarrow af\left( {\dfrac{1}{x}} \right) + bf\left( x \right) = x - 5$ … (2)
Now we can multiply (1) with a. Then we obtain,
${a^2}f\left( x \right) + abf\left( {\dfrac{1}{x}} \right) = a\left( {\dfrac{1}{x} - 5} \right)$
On simplification, we get,
$\Rightarrow {a^2}f\left( x \right) + abf\left( {\dfrac{1}{x}} \right) = \dfrac{a}{x} - 5a$ … (3)
Now we can multiply equation (2) with b. So, we get,
$\Rightarrow abf\left( {\dfrac{1}{x}} \right) + {b^2}f\left( x \right) = b\left( {x - 5} \right)$
On simplification, we get,
$\Rightarrow abf\left( {\dfrac{1}{x}} \right) + {b^2}f\left( x \right) = bx - 5b$ …. (4)
Now subtract equation (4) from equation (3)

$$\Rightarrow abf\left( {\dfrac{1}{x}} \right) + {a^2}f\left( x \right) = \dfrac{a}{x} - 5a \\\ \underline {\left( - \right)abf\left( {\dfrac{1}{x}} \right) + {b^2}f\left( x \right) = bx - 5b} \\\ {a^2}f\left( x \right) - {b^2}f\left( x \right) = \dfrac{a}{x} - 5a - \left( {bx - 5b} \right) \\\$$

On simplification, we get,
$\Rightarrow \left( {{a^2} - {b^2}} \right)f\left( x \right) = \dfrac{a}{x} - 5a - bx + 5b$
On dividing both sides of the equation with $\left( {{a^2} - {b^2}} \right)$ and taking the common factors from numerator, we get,
$\Rightarrow f\left( x \right) = \dfrac{{\dfrac{a}{x} - bx - 5\left( {a - b} \right)}}{{\left( {{a^2} - {b^2}} \right)}}$
Now we can multiply the numerator and denominator with x. so we will get,
$\Rightarrow f\left( x \right) = \dfrac{{a - b{x^2} - 5x\left( {a - b} \right)}}{{x\left( {{a^2} - {b^2}} \right)}}$
Now we have the value of $f\left( x \right)$ which is dependent on the variable x. So, the required function is,
$f\left( x \right) = \dfrac{{a - b{x^2} - 5x\left( {a - b} \right)}}{{x\left( {{a^2} - {b^2}} \right)}}$ , $x = 0$

Note: We must substitute the $\dfrac{1}{x}$ in the place of $x$, only inside the function and corresponding the values. Then we must change all the x in the equation to $\dfrac{1}{x}$ . We must never take the value of x equal to $\dfrac{1}{x}$ while simplifying the equation. We must make sure that the function $f\left( x \right)$ must contain only the variable x and constants. As the denominator is a multiple of x, we must note that the function is not defined when x is equal to zero. But as it is given in the question that x is non zero, this will be the required function.