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

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(D) Given $f(x) = (x - 1)^2 + 1$ for $x \ge 1$.To find $f^{-1}(x)$,set $y = (x - 1)^2 + 1$.Then $y - 1 = (x - 1)^2$. Since $x \ge 1$,we have $x - 1 =

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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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(D) Given $f(x) = (x - 1)^2 + 1$ for $x \ge 1$.To find $f^{-1}(x)$,set $y = (x - 1)^2 + 1$.Then $y - 1 = (x - 1)^2$. Since $x \ge 1$,we have $x - 1 = \sqrt{y - 1}$,so $x = 1 + \sqrt{y - 1}$.Thus,$f^{-1}(x) = 1 + \sqrt{x - 1}$ for $x \ge 1$. So,Statement-$2$ is true.To find $S = \{x : f(x) = f^{-1}(x)\}$,we solve $f(x) = x$.$(x - 1)^2 + 1 = x \implies x^2 - 2x + 1 + 1 = x \implies x^2 - 3x + 2 = 0$.$(x - 1)(x - 2) = 0$,so $x = 1$ or $x = 2$.Since both values satisfy $x \ge 1$,$S = \{1, 2\}$. So,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 $x \ge 1$.
Statement-$1$: $S = \{x : f(x) = f^{-1}(x)\} = \{1, 2\}$.
Statement-$2$: $f$ is a bijection and $f^{-1}(x) = 1 + \sqrt{x - 1}$ for $x \ge 1$.

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