Let $\mathrm{f}(x)=(x+1)^2-1, x \geqslant-1$, then the set $\left\{x /…
Let $\mathrm{f}(x)=(x+1)^2-1, x \geqslant-1$, then the set $\left\{x / \mathrm{f}(x)=\mathrm{f}^{-1}(x)\right\}$ is
- $\{0,1,-1\}$
- $\{0,-1\}$
- $\left\{0,-1, \frac{-3+\mathrm{i} \sqrt{3}}{2}, \frac{-3-\mathrm{i} \sqrt{3}}{2}\right.$, where $\left.\mathrm{i}=\sqrt{-1}\right\}$
- $\phi$
Solution
$\begin{aligned} & \mathrm{f}(x)=\mathrm{f}^{-1}(x) \\ & \Rightarrow \mathrm{f}(\mathrm{f}(x))=x \\ & \Rightarrow(\mathrm{f}(x)+1)^2-1=x \\ & \Rightarrow\left[(x+1)^2-1+1\right]^2-1=x \\ & \Rightarrow(x+1)^4-1=x \\ & \Rightarrow(x+1)^4-(x+1)=0 \\ & \Rightarrow(x+1)\left[(x+1)^3-1\right]=0 \\ & \Rightarrow x+1=0 \text { or }(x+1)^3=1 \\ & \Rightarrow x=-1 \text { or } x=0\end{aligned}$
Asked in: MHT CET 2024 (15 May Shift 1)
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