Question

If $f$is an even function defined on the interval $(-5,5)$, then real values of $x$ satisfying the equation $f\left(x\right)=f\left[\frac{(x+1)}{(x+2)}\right]$are

• A. $\frac{\left[-1+\sqrt{5}\right]}{2},\frac{\left[3\pm \sqrt{5}\right]}{2}$
• B. $\frac{\left[-2+\sqrt{3}\right]}{2},\frac{\left[-3\pm \sqrt{5}\right]}{2}$
• C. $\frac{\left[3\pm \sqrt{5}\right]}{2},\frac{\left[-3\pm \sqrt{5}\right]}{2}$
• D. None of these

Answer 1

The correct option is C

$\frac{\left[3\pm \sqrt{5}\right]}{2},\frac{\left[-3\pm \sqrt{5}\right]}{2}$

Explanation for the correct option:

Step 1. Find the real value of $x$:

Given, $f\left(x\right)=f\left[\frac{(x+1)}{(x+2)}\right]$

$f(-x)=f\left[\frac{(-x+1)}{(-x+2)}\right]$ ….(1)

$f\left(x\right)=f\left[\frac{(-x+1)}{(-x+2)}\right]$ $\left[\mathbf{\because }\mathit{f}\mathbf{(}\mathbf{-}\mathit{x}\mathbf{)}\mathbf{}\mathbf{=}\mathbf{}\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\right]$

Step 2. On Taking $f^{-1}$ on the both side, we get

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

$\Rightarrow$ $-x^2+2x=-x+1$

$\Rightarrow$$x^2–3x+1=0$

$\Rightarrow$ $x=\frac{[3\pm \surd 5]}{2}$ $\left[\mathbf{\because }\mathit{x}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{-}\mathbf{b}\mathbf{}\mathbf{\pm }\sqrt{{\mathbf{b}}^{\mathbf{2}}\mathbf{}\mathbf{-}\mathbf{}\mathbf{4}\mathbf{a}\mathbf{c}}}{\mathbf{2}}\right]$

Again, $f\left(x\right)=f\left[\frac{(x+1)}{(x+2)}\right]$

$f(-x)=f\left[\frac{(x+1)}{(x+2)}\right]$ $\left[\mathbf{\because }\mathit{f}\mathbf{(}\mathbf{-}\mathit{x}\mathbf{)}\mathbf{}\mathbf{=}\mathbf{}\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\right]$

Step 3. On taking $f^{-1}$ on the both side, we get

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

$\Rightarrow$ $-x^2-2x=x+1$

$\Rightarrow$$x^2+3x+1=0$

$\Rightarrow$ $x=\frac{[-3\pm \surd 5]}{2}$ $\left[\mathbf{\because }\mathit{x}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{-}\mathbf{b}\mathbf{}\mathbf{\pm }\sqrt{{\mathbf{b}}^{\mathbf{2}}\mathbf{}\mathbf{-}\mathbf{}\mathbf{4}\mathbf{a}\mathbf{c}}}{\mathbf{2}}\right]$

$\therefore$The real values of $x$ are $\frac{\mathbf{[}\mathbf{3}\mathbf{}\mathbf{\pm }\mathbf{}\mathbf{\surd }\mathbf{5}\mathbf{]}}{\mathbf{2}}$, $\frac{\mathbf{[}\mathbf{-}\mathbf{3}\mathbf{}\mathbf{\pm }\mathbf{}\mathbf{\surd }\mathbf{5}\mathbf{]}}{\mathbf{2}}$

Hence, Option ‘C’ is Correct.