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.
Text Solution
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f(x) = (x – 1) 2 + 1, x ≥ 1
f : [1, ∞ ) → [1, ∞ ) is a bijective function ,
⇒ y = (x – 1) 2 + 1 ⇒ (x – 1) 2 = y – 1 ⇒ x = 1 ± 
⇒ f –1 (y) = 1 ± 
⇒ f –1 (x) = 1 + 
{ ∴ x ≥ 1}
so statement-2 is correct
Now f(x) = f –1 (x) ⇒ f(x) = x ⇒ (x – 1) 2 + 1 = x ⇒ x 2 – 3x + 2 = 0 ⇒ x = 1, 2
so statement-1 is correct
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