Question

if f(x)=ax+b/bx+a, prove that f(x).f(1/x)=1​

Answer 1

Given:-

• $\sf f(x) = \dfrac{ax + b}{bx + a}$

To show:-

• $\sf f(x).f \bigg( \dfrac{1}{x} \bigg) = 1$

Answer:-

For f(x):

$\sf f(x) = \dfrac{ax + b}{bx + a}$

For f(1/x):

Given that,

$\sf f(x) = \dfrac{ax + b}{bx + a}$

For f(1/x) we just have to substitute x as 1/x in f(x).

$\sf \longrightarrow \: f \bigg( \dfrac{1}{x} \bigg) = \dfrac{ \bigg(a \times \dfrac{1}{x} \bigg) + b}{ \bigg(b \times \dfrac{1}{x} \bigg)+ a}$

$\sf \longrightarrow \: f \bigg( \dfrac{1}{x} \bigg) = \dfrac{ \quad \bigg(\dfrac{a}{x} + b \bigg) \quad}{ \quad \bigg( \dfrac{b}{x} + a \bigg) \quad}$

$\sf \longrightarrow \: f \bigg( \dfrac{1}{x} \bigg) = \dfrac{ \quad \bigg(\dfrac{a + bx}{x} \bigg) \quad}{ \quad \bigg( \dfrac{b + ax}{x} \bigg) \quad}$

$\sf \longrightarrow \: f \bigg( \dfrac{1}{x} \bigg) = \dfrac{ \quad \bigg(\dfrac{a + bx}{ \cancel{x}} \bigg) \quad}{ \quad \bigg( \dfrac{b + ax}{ \cancel{x}} \bigg) \quad}$

$\sf \longrightarrow \: f \bigg( \dfrac{1}{x} \bigg) = \dfrac{a + bx}{b + ax}$

So f(x).f(1/x):

$\sf f(x).f \bigg( \dfrac{1}{x} \bigg) = \dfrac{ ax + b}{ bx + a} \times \dfrac{a + bx}{b + ax}$

$\sf \longrightarrow \: f(x).f \bigg( \dfrac{1}{x} \bigg) = \dfrac{ \cancel{ ax + b}}{ \cancel{ bx + a}} \times \dfrac{ \cancel{a + bx}}{ \cancel{b + ax}}$

$\longrightarrow \underline{\underline{ \sf{\green{f(x).f\bigg(\dfrac{1}{x}\bigg) = 1}}}}$