Let f(x)=(x+1)^2-1 for x geq -1.Statement-1: | vedclass

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(A) Given $f(x) = (x+1)^2 - 1$ for $x \geq -1$.To find $f^{-1}(x)$,set $y = (x+1)^2 - 1$. Since $x \geq -1$,$y \geq -1$.Solving for $x$: $(x+1)^2 = y+

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Statement-$1$ is true,Statement-$2$ is true; Statement-$2$ is a correct explanation for Statement-$1$

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(A) Given $f(x) = (x+1)^2 - 1$ for $x \geq -1$.To find $f^{-1}(x)$,set $y = (x+1)^2 - 1$. Since $x \geq -1$,$y \geq -1$.Solving for $x$: $(x+1)^2 = y+1 \Rightarrow x+1 = \sqrt{y+1} \Rightarrow x = \sqrt{y+1} - 1$.Thus,$f^{-1}(x) = \sqrt{x+1} - 1$. Since $f$ is strictly increasing on $[-1, \infty)$,it is a bijection (one-one and onto). So,Statement-$2$ is true.To find $S = \{x : f(x) = f^{-1}(x)\}$,we solve $f(x) = x$ because $f$ is increasing.$(x+1)^2 - 1 = x \Rightarrow x^2 + 2x + 1 - 1 = x \Rightarrow x^2 + x = 0 \Rightarrow x(x+1) = 0$.Thus,$x = 0$ or $x = -1$. So,$S = \{0, -1\}$. Statement-$1$ is true.Since $f(x) = f^{-1}(x)$ is equivalent to $f(x) = x$ for increasing functions,Statement-$2$ explains Statement-$1$.

Let $f(x)=(x+1)^2-1$ for $x \geq -1$.
Statement-$1$: $S=\{x:f(x)=f^{-1}(x)\}=\{0, -1\}$
Statement-$2$: $f$ is a bijection.

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Let $f(x)=(x+1)^2-1$ for $x \geq -1$.Statement-$1$: $S=\{x:f(x)=f^{-1}(x)\}=\{0, -1\}$Statement-$2$: $f$ is a bijection.