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Let f(x)=(x+1)^2-1, xgeq-1. Then the set...

Let `f(x)=(x+1)^2-1, xgeq-1.` Then the set `{x :f(x)=f^(-1)(x)}` is `{0,1,(-3+isqrt(3))/2,(-3-isqrt(3))/2}` (b) `{0,1,-1` `{0,1,1}` (d) `e m p t y`

A

`{0, -1, (-3+isqrt(3))/(2),(-3-isqrt(3))/(2)},i=sqrt(-1)`

B

{0, 1, -1}

C

{0, -1}

D

empty

Text Solution

Verified by Experts

The correct Answer is:
C
`f(x) = (x+1)^(2)-1" "[because xge1]`
`=x^(2)+1+2x-1=x^(2)+2x`
`S={x : f(x)-=f^(-1)(x)}`
S is the set of point of intersection of (y = x) and tf.
Now, solve y = x and `f(x) = x^(2) + 2x`
`x^(2) + 2x = x`
`x^(2) + x = 0`
x(x+1)=0
x = 0 or x = - 1
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