Test Index

Functions Part 1

Section: Mathematics

Question : 85 of 100

Marks: +1, -0

Let

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

\{x: f(x)=f^{-1}(x)\} = \{0,-1\}

-2: f

Statement $-1$ is true, Statement $-2$ is true; Statement $-2$ is a correct explanation for Statement - 1
Statement $-1$ is true, Statement $-2$ is true; Statement $-2$ is not a correct explanation for Statement - 1
Statement $-1$ is true, Statement $-2$ is false
Statement $-1$ is false, Statement - 2 is true

Solution:

Given that

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

Clearly

D_f=[-1, \infty)

but co-domain is not given

\therefore f(x)

need not be necessarily onto.

But if

f(x)

is onto then as

f(x)

is one one also,

(x+1)

being something

+v e, f^{-1}(x)

will exist where

\; (x+1)^2-1=y

\; \Rightarrow x+1=\sqrt{y+1}

(+v e

square root as

x+1 \ge 0)

\; \Rightarrow x=-1+\sqrt{y+1}

\; \Rightarrow f^{-1}(x)=\sqrt{x+1}-1

Then

\; \; f(x)=f^{-1}(x)

\; \Rightarrow (x+1)^2-1=\sqrt{x+1}-1

\; \Rightarrow (x+1)^2=\sqrt{x+1}

\; \Rightarrow (x+1)^4=(x+1)

\; \Rightarrow (x+1)[(x+1)^3-1]=0

\; \Rightarrow x=-1,0

\therefore

The statement

-1

is correct but statement- 2 is false.

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