Let f be a function defined by $$f(x)=(x$$−$$1)2+1$$,(x$$≥$$1)Statement-1$$: The set $${x$$ : $$f(x) $$=$$ f –$$1(x)}$$ $$=$$ $${1$$, $$2}Statement-2$$:f is a bijection and $$f-1(x)$$ $$=$$ $$1+x-1$$ , x ≥ $$1$$

Question

Let f be a function defined by $$f(x)=(x$$−$$1)2+1$$,(x$$≥$$1)Statement-1$$: The set $${x$$ : $$f(x) $$=$$ f –$$1(x)}$$ $$=$$ $${1$$, $$2}Statement-2$$:f is a bijection and $$f-1(x)$$ $$=$$ $$1+x-1$$ , x ≥ $$1$$

Infinity Learn · Accepted Answer

Let y $$=$$ $$f(x)$$ $$=$$ (x$$ – $$1)2$$ $$+$$ $$1y$$ – $$1$$ $$=$$ $$(x$$ – $$1)2$$⇒ x $$=$$ $$1$$ $$+$$ y $$-$$ $$1$$ , y ≥ $$1Thus$$ $$f-1(x) $$=$$ $$1$$ $$+$$ x $$-$$ $$1$$ , x ≥ $$1$$. So $$statement-2$$ is trueNow $$f(x)$$ $$=$$ f –$$1(x)$$ ⇒ $$(x$$ – $$1)2$$ $$=$$ x $$-$$ $$1$$ ⇒x−$$1(x$$−$$1)3/2$$−$$1=0$$ ⇒ $$x=1$$,$$2So$$ $$statement-1$$ is true and $$statement-2$$ is a correct explanation for $$statement-1$$.

Let f be a function defined by
$f (x) = (x - 1)^{2} + 1, (x \geq 1)$
Statement-1: The set {x : f(x) = f ^–1(x)} = {1, 2}
Statement-2: $f$ is a bijection and $f^{- 1} (x) = 1 + \sqrt{x - 1}, x \geq 1$

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Let f be a function defined by f(x)=(x−1)2+1,(x≥1)Statement-1: The set {x : f(x) = f –1(x)} = {1, 2}Statement-2:f is a bijection and f-1(x) = 1+x-1 , x ≥ 1

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Let f be a function defined by
$f (x) = (x - 1)^{2} + 1, (x \geq 1)$
Statement-1: The set {x : f(x) = f ^–1(x)} = {1, 2}
Statement-2: $f$ is a bijection and $f^{- 1} (x) = 1 + \sqrt{x - 1}, x \geq 1$