QJEEM09Ma2482: JEE Main 2009 Mathematics - Class 11 JEEM09Ma2482 Solution: Let Statement - 1 : The set . Statement - 2 : is a bijection.

Question

Verified Solution • 2025–26 Session moderate

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.

A Statement - 1 is true, Statement - 2 is true; Statement - 2 is a correct explanation for Statement - 1
B Statement - 1 is true, Statement - 2 is true; Statement - 2 is not a correct explanation for Statement - 1
C Statement - 1 is true, Statement - 2 is false
D Statement - 1 is false, Statement - 2 is true

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

is false.

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