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

$\left\{0,-1, \frac{-3+i \sqrt{3}}{2}, \frac{-3-i \sqrt{3}}{2}\right\}, i=\sqrt{-1}$

JEE MATHEMATICS MCQ: SET THEORY

1 Easy

1 of 1

Analyzing +4 | -1

Let

f(x)=(x+1)^2-1,(x \geq-1)

. Then, the set

S=\left\{x: f(x)=f^{-1}(x)\right\}

View answer and explanation

Correct option: C

\{0,-1\}

Option C (

\{0,-1\}

) is correct because these are the points where

f(x) = x

, which are the intersection points of

f(x)

and

f^{-1}(x)

Practice All Chapter MCQs →

S

2 a_1 a_2 a_3 \ldots . . a_{2006}

a_1, a_2, a_3, \ldots ., a_{2006}

S

a N=\{a n: n \in N: b N=\{n b: n \in N\}

c N=\{c N: n \in N\}

b N \cap c N=d N

a, b, c \in N

b, c

A=\{x: x

\}

B=\{x: x

A-B

F(n)

n

[\mathrm{F}(20) \cap \mathrm{F}(16)] \subseteq \mathrm{F}(\mathrm{y})

40 \%

A, 20 \%

10 \%

C

A

C

4 \%

A

C

2 \%

A