Question

f(x)=x-1/x+1 prove that f(2x)=3f(x)+1/f(x)+3

Answer 1

GIVEN :–

$\\ \: \: \blacktriangleright \: \: { \bold{F(x) = \dfrac{x - 1}{x + 1} }}$

TO PROVE :–

$\\ \: \: \blacktriangleright \: \: { \bold{F(2x) = \dfrac{3F(x) + 1}{F(x) + 3} }}$

SOLUTION :–

• Let's take R.H.S. –

$\\ \: \: \: \: { \bold{ = \dfrac{3F(x) + 1}{F(x) + 3} }}$

$\\ \: \: \: \: { \bold{ = \dfrac{3 \left \{ \dfrac{x - 1}{x +1 } \right \} + 1}{ \left \{ \dfrac{x - 1}{x +1 } \right \} + 3} }}$

$\\ \: \: \: \: { \bold{ = \dfrac{ \left \{ \dfrac{3x - 3}{x +1 } \right \} + 1}{ \left \{ \dfrac{x - 1}{x +1 } \right \} + 3} }}$

$\\ \: \: \: \: { \bold{ = \dfrac{ \dfrac{3x - 3 + x + 1}{ \cancel{x +1 }}}{ \dfrac{x - 1 + 3(x + 1)}{ \cancel{x +1} }}}}$

$\\ \: \: \: \: { \bold{ = \dfrac{3x - 3 + x + 1}{x - 1 + 3(x + 1)}}}$

$\\ \: \: \: \: { \bold{ = \dfrac{4x -2}{x - 1 +3x+3}}}$

$\\ \: \: \: \: { \bold{ = \dfrac{4x -2}{4x + 2}}}$

$\\ \: \: \: \: { \bold{ = \dfrac{ \cancel2(2x -1)}{ \cancel2(2x + 1)}}}$

$\\ \: \: \: \: { \bold{ = \dfrac{ 2x -1}{2x + 1}}}$

$\\ \: \: \: \: { \bold{ = F(2x)}}$

$\\ \:\:\:\:{ \bold{=F(2x)\:\:\:\:\:\: \left \{ \: \because F(2x)= \dfrac{2x-1}{2x+1} \right \}}}$

$\\ \: \: \: \: { \bold{ = \: L.H.S.}}$

$\\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: { \bold{ \underbrace{ \overbrace{ Hence \: \: prove}}}}$

Answer 2

Step-by-step explanation:

${\red{\underline{\underline{\bold{Given:-}}}}}$

• $f(x) = \frac{x - 1}{x + 1}$

${\blue{\underline{\underline{\bold{To\:Prove:-}}}}}$

• $f(2x) = \frac{3f(x) + 1}{f(x) + 3}$

${\green{\underline{\underline{\bold{Solution:-}}}}}$

• Let us prove by simplifying L.H.S and R.H.S separately

R.H.S =

$\frac{3f(x) + 1}{f(x) + 3}$

$= \frac{3[ \frac{x - 1}{x + 1}] + 1}{ [\frac{x - 1}{x + 1}]+ 3 } \\ \\ = \frac{ [\frac{3(x - 1)}{x + 1} ] + 1}{ [\frac{x - 1}{x + 1} ]+ 3} \\ \\ = \frac{ \frac{3x - 3}{x + 1} + 1}{[\frac{x - 1}{x + 1} ] + 3} \\ \\ = \frac{ \frac{3x - 3 + x + 1}{x + 1} }{ \frac{ x - 1 + 3x + 3}{x + 1} }$

$= \frac{3x - 3 + x + 1}{x - 1 + 3x + 3} \\ \\ = \frac{3x + x - 3 + 1}{x + 3x + 3 - 1} \\ \\ = \frac{4x - 2}{4x + 2} \\ \\ = \frac{2(2x - 1)}{2(2x + 1)} \\ \\ = \frac{2x - 1}{2x + 1} = f(2x)$

L.H.S = R.H.S

Hence,Proved