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Let $f(x)=(x+1)^2-1, x \geq-1$
Statement-1: The set $\left\{x: f(x)=f^{-1}(x)\right\}=\{0,-1\}$
Statement-2 : $\mathrm{f}$ is a bijection.
Options:
Statement-1: The set $\left\{x: f(x)=f^{-1}(x)\right\}=\{0,-1\}$
Statement-2 : $\mathrm{f}$ is a bijection.
Solution:
2554 Upvotes
Verified Answer
The correct answer is:
Statement-1 is true, Statement-2 is false
Statement-1 is true, Statement-2 is false
There is no information about co-domain therefore $f(x)$ is not necessarily onto.
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