The function f:[2,∞)→Y defined by f(x)=x2−4x+5 is both one one and onto if Y =[a,∞), then the value of a is

f(x)=x2−4x+5=(x−2)2+1≥1Range of fx is [1,∞)f is a bijective function. Therefore codomain y= rangeSo Y= [1,∞)

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The function $f : [2, \infty) \to Y$ defined by $f (x) = x^{2} - 4 x + 5$ is both one one and onto if $Y = [a, \infty),$ then the value of $a$ is

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$f (x) = x^{2} - 4 x + 5 = {(x - 2)}^{2} + 1 \geq 1$

Range of $f (x)$ is $[1, \infty)$

f is a bijective function. Therefore codomain y= range

So Y= $[1, \infty)$

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