Questions & Answers

Question

Answers

Answer

Verified

92.1k+ views

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.