 # If for non-zero X, $af(x) + bf\left( {\dfrac{1}{x}} \right) = \dfrac{1}{x} - 5,$ where $a \ne b,$ then $f(x).$ Verified
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Hint: Take the given expression, as the given function is with respect to “x” then create the expression in the form of $\dfrac{1}{x}$ and then simplify both the equation finding the correlation between the two for the resultant answer.

Take the given expression –
$af(x) + bf\left( {\dfrac{1}{x}} \right) = \dfrac{1}{x} - 5$ .... (A)
Replace $x$ by $\left( {\dfrac{1}{x}} \right)$ in the above equation –
$\Rightarrow af\left( {\dfrac{1}{x}} \right) + bf\left( x \right) = x - 5$ .... (B)
Multiply Equation (A) with “a” on both the sides of the equation –
$a\left[ {af(x) + bf\left( {\dfrac{1}{x}} \right)} \right] = a\left[ {\dfrac{1}{x} - 5} \right]$
Simplify the above equation –
$\Rightarrow \left[ {{a^2}f(x) + abf\left( {\dfrac{1}{x}} \right)} \right] = \left[ {\dfrac{a}{x} - 5a} \right]$ ..... (C)
Multiply Equation (B) with “b” on both the sides of the equation –
$\Rightarrow b\left[ {af\left( {\dfrac{1}{x}} \right) + bf\left( x \right)} \right] = b\left[ {x - 5} \right]$
Simplify the above equation –
$\Rightarrow \left[ {abf\left( {\dfrac{1}{x}} \right) + {b^2}f\left( x \right)} \right] = \left[ {bx - 5b} \right]$ ..... (D)
Subtract equation (D) from the equation (C)
$\Rightarrow \left[ {{a^2}f(x) + abf\left( {\dfrac{1}{x}} \right)} \right] - \left[ {abf\left( {\dfrac{1}{x}} \right) + {b^2}f\left( x \right)} \right] = \left[ {\dfrac{a}{x} - 5a} \right] - \left[ {bx - 5b} \right]$
When you open the brackets and if there is a negative sign outside it, then the sign of all the terms inside the bracket changes. Positive term becomes negative and negative term becomes positive.
$\Rightarrow {a^2}f(x) + abf\left( {\dfrac{1}{x}} \right) - abf\left( {\dfrac{1}{x}} \right) - {b^2}f\left( x \right) = \dfrac{a}{x} - 5a - bx + 5b$
Make pair of like terms in the above equation –
$\Rightarrow \underline {{a^2}f(x) - {b^2}f\left( x \right)} + \underline {abf\left( {\dfrac{1}{x}} \right) - abf\left( {\dfrac{1}{x}} \right)} = \dfrac{a}{x} - 5a - bx + 5b$
Like terms with equal values and opposite sign cancel each other.
$\Rightarrow \left( {{a^2} - {b^2}} \right)f\left( x \right) = \dfrac{a}{x} - 5a - bx + 5b$
The above equation can be written as –
$\Rightarrow \left( {{a^2} - {b^2}} \right)f\left( x \right) = \left[ {\dfrac{a}{x} - 5a - bx + 5b} \right]$
When the term multiplicative on one side is moved to the opposite side then it goes to the denominator on the opposite side.
$\Rightarrow f\left( x \right) = \dfrac{{\left[ {\dfrac{a}{x} - 5a - bx + 5b} \right]}}{{\left( {{a^2} - {b^2}} \right)}}$
This is the required solution.
So, the correct answer is “ $\Rightarrow f\left( x \right) = \dfrac{{\left[ {\dfrac{a}{x} - 5a - bx + 5b} \right]}}{{\left( {{a^2} - {b^2}} \right)}}$ ”.

Note: Be very careful while simplifying the terms. When you open the brackets and there is a negative sign outside the bracket then the sign of the terms inside the bracket will be changed. Positive terms will become negative and negative terms become positive. Also, remember when there is a positive sign outside the bracket then the sign of the terms inside the bracket do not change.