MHT CET · Maths · Functions
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
- A \(\{0,1,-1\}\)
- B \(\{0,-1\}\)
- C \(\left\{0,-1, \frac{-3+\mathrm{i} \sqrt{3}}{2}, \frac{-3-\mathrm{i} \sqrt{3}}{2}\right.\), where \(\left.\mathrm{i}=\sqrt{-1}\right\}\)
- D \(\phi\)
Answer & Solution
Correct Answer
(B) \(\{0,-1\}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \mathrm{f}(x)=\mathrm{f}^{-1}(x) \\ & \Rightarrow \mathrm{f}(\mathrm{f}(x))=x \\ & \Rightarrow(\mathrm{f}(x)+1)^2-1=x \\ & \Rightarrow\left[(x+1)^2-1+1\right]^2-1=x \\ & \Rightarrow(x+1)^4-1=x \\ & \Rightarrow(x+1)^4-(x+1)=0 \\ & \Rightarrow(x+1)\left[(x+1)^3-1\right]=0 \\ & \Rightarrow x+1=0 \text { or }(x+1)^3=1 \\ & \Rightarrow x=-1 \text { or } x=0\end{aligned}\)
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