Let f(x) = (x + 1)2 –1, x –1 Statement – 1 : The set {x : f(x) = f–1(x)} = {0, –1}. Statement – 2 : f is a bijection.

Question

Let f(x) = (x + 1)2 –1, x > –1 Statement – 1 : The set {x : f(x) = f–1(x)} = {0, –1}. Statement – 2 : f is a bijection.

Arun · Answer

Dear Jayant Question is incomplete. Please check and repost the question. You can also attach an image. I will be happy to help you.

Vikas TU · Answer

f(x is not cont.................... Bijection - one - one function f(x) = x (x+1)^2 -1 = x f(x) -x = 0 x^2 -2x + 2-1 -x = 0 x(x+1) =0 roots are 0,1 statement 1 is true …...........

Vikas TU · Answer

Dear student Roots are (0, -1 )...............................................................................................................

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Let f(x) = (x + 1)2 –1, x > –1 Statement – 1 : The set {x : f(x) = f–1(x)} = {0, –1}. Statement – 2 : f is a bijection.

Let f(x) = (x + 1)2 –1, x > –1
Statement – 1 : The set {x : f(x) = f–1(x)} = {0, –1}. Statement – 2 : f is a bijection.

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Let f(x) = (x + 1)2 –1, x > –1 Statement – 1 : The set {x : f(x) = f–1(x)} = {0, –1}. Statement – 2 : f is a bijection.

Let f(x) = (x + 1)2 –1, x > –1 Statement – 1 : The set {x : f(x) = f–1(x)} = {0, –1}. Statement – 2 : f is a bijection.

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Let f(x) = (x + 1)2 –1, x > –1
Statement – 1 : The set {x : f(x) = f–1(x)} = {0, –1}. Statement – 2 : f is a bijection.