Question

If $f$ is an even function. Find the real value of $x$ satisfying the equation $f\left(x\right)=f\left(\frac{x+1}{x+2}\right)$

• A. $\{\frac{-1+\sqrt{5}}{2},\frac{-1-\sqrt{5}}{2},\frac{-3+\sqrt{5}}{2},\frac{-3-\sqrt{5}}{2}\}$
• B. $\{\frac{-2+\sqrt{5}}{2},\frac{-1-\sqrt{5}}{2},\frac{2+\sqrt{5}}{2},\frac{-3-\sqrt{5}}{2}\}$
• C. $\{\frac{1+\sqrt{5}}{2},\frac{1-\sqrt{5}}{2},\frac{-3+\sqrt{5}}{2},\frac{-2-\sqrt{5}}{2}\}$
• D. $\{\frac{-1+\sqrt{5}}{2},\frac{-1-\sqrt{5}}{2},\frac{-3+\sqrt{7}}{2},\frac{-2-\sqrt{7}}{2}\}$

Answer 1

The correct option is A $\{\frac{-1+\sqrt{5}}{2},\frac{-1-\sqrt{5}}{2},\frac{-3+\sqrt{5}}{2},\frac{-3-\sqrt{5}}{2}\}$

Since,$f\left(x\right)$ is even, so $f(-x)=f\left(x\right)$

Thus, $x=\frac{x+1}{x+2}$ or $-x=\frac{x+1}{x+2}$

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

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

$\Rightarrow x=\frac{-1\pm \sqrt{5}}{2}$ or $x=\frac{-3\pm \sqrt{5}}{2}$

Thus, $x\{\frac{-1+\sqrt{5}}{2},\frac{-1-\sqrt{5}}{2},\frac{-3+\sqrt{5}}{2},\frac{-3-\sqrt{5}}{2}\}$

Hence, option 'A' is correct.