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$Let f (x) = (x + 1)^{2} - 1, x \geq - 1 . Then the set \{x : f (x) = f^{- 1} (x)\} is$

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0,−1,−3+i32,−3−i32

{0,1,−1}

{0,−1}

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Correct option is C

since fx=x+12-1 is continuous function, solution of fx=f-1x lies on the line y=x. Therefore fx=f-1x=x x+12-1=x ⇒x2+x=0⇒ x=0 or -1 Therefore , the required set is 0,-1

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Let f(x)=(x+1)2−1,x≥−1 . Then the set x:f(x)=f−1(x) is

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$Let f (x) = (x + 1)^{2} - 1, x \geq - 1 . Then the set \{x : f (x) = f^{- 1} (x)\} is$