Let [f( x ) = ( x + 1 )^2 - 1,x - 1] Statement - 1 : The set [ x:f( x ) = f^ - 1( x ) = 0,…

Question

Let $$f\left( x \right) = {\left( {x + 1} \right)^2} - 1,x \ge  - 1$$

Statement - 1 : The set $$\left\{ {x:f\left( x \right) = {f^{ - 1}}\left( x \right)} \right\} = \left\{ {0, - 1} \right\}$$.

Statement - 2 : $f$ is a bijection.

Accepted Answer

Given that $$f\left( x \right) = {\left( {x + 1} \right)^2} - 1,\,x \ge  - 1$$

Clearly $${D_f} = \left[ { -1 ,\infty } \right)$$ but co-domain is not given

$$\therefore$$ $f(x)$ need not be necessarily onto.

But if $f(x)$ is onto then as $$f\left( x \right)$$ is one one also,

$(x+1)$ being something $+ve,$ $${f^{ - 1}}\left( x \right)$$ will exist where

$${\left( {x + 1} \right)^2} - 1 = y$$

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

$$\,\,\,\,\,$$ $$\left( { + ve} \right.$$ square root as $$x + 1 \ge 0$$ $\left. {} \right) (

\] \Rightarrow x =  - 1 + \sqrt {y + 1} $$ \Rightarrow {f^{ - 1}}\left( x \right) = \sqrt {x + 1}  - 1$Then$\,\,\,\,\,\,\,\,$$f\left( x \right) = {f^{ - 1}}\left( x \right) (

\] \Rightarrow {\left( {x + 1} \right)^2} - 1 = \sqrt {x + 1}  - 1$$ \Rightarrow {\left( {x + 1} \right)^2} = \sqrt {x + 1} $$ \Rightarrow {\left( {x + 1} \right)^4} = \left( {x + 1} \right) (

\] \Rightarrow \left( {x + 1} \right)\left[ {{{\left( {x + 1} \right)}^3} - 1} \right] = 0$$ \Rightarrow x =  - 1,0$$\therefore$The statement -$1$is correct but statement-$2$

Solutions · 1