Let f(x)=(x+1)2-1, x ≥-1 then the set x: f(x)=f-1(x) is equal to

Question

Let f(x)=(x+1)2-1, x ≥-1 then the set x: f(x)=f-1(x) is equal to (A)  0,-1  (B)  0,1  (C)  -1,1  (D)  0

Tardigrade Admin · Accepted Answer

Let y=(x+1)2-1, x ≥-1 ⇒ (x+1)2=y+1 ⇒ x+1=√y+1 text as x ≥-1 ⇒ x=-1+√y+1, y ≥-1 text Thus f-1(x)=-1+√x+1 text So f(x)=f-1(x) ⇒ (x+1)2-1=-1+√x+1 ⇒ √x+1=0 text or (x+1)3 / 2=1 ⇒ x=-1 text or x=0

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

${0, - 1}$

${0, 1}$

${- 1, 1}$

${0}$

Q. Let f(x)=(x+1)2−1,x≥−1 then the set {x:f(x)=f−1(x)} is equal to

{0,−1}

{0,1}

{−1,1}

{0}

Let y=(x+1)2−1,x≥−1 ⇒(x+1)2=y+1 ⇒x+1=y+1​ as x≥−1 ⇒x=−1+y+1​,y≥−1 Thus f−1(x)=−1+x+1​ So f(x)=f−1(x) ⇒(x+1)2−1=−1+x+1​ ⇒x+1​=0 or (x+1)3/2=1 ⇒x=−1 or x=0

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

${0, - 1}$

${0, 1}$

${- 1, 1}$

${0}$