# Let $\dpi{100} f(x)=(x+1)^{2}-1,x\geq -1$Statement - 1 : The set $\dpi{100} \left \{ x:f(x)=f^{-1}(x) \right \}=\left \{ 0,-1 \right \}$Statement - 2 : $\dpi{100} f$ is a bijection.Statement - 1 (Assertion) and Statement - 2 (Reason). Option 1) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1 Option 2) Statement- 1 is true, Statement-2 is false Option 3) Statement-1 is false, Statement-2 is true Option 4) Statement-1 is true, Statement-2 is true; Statement-2 is correct explanation for Statement-1

As we learnt in

Bijective Function -

The function which is both one-one and onto is Bijective Function.

-

f(x) = (x + 1)2 - 1     $x\geq-1$

Let  y = (x + 1)2 - 1

$\therefore\ \; x+1=\pm\sqrt{y+1}$

$\therefore\ \; x =\pm\sqrt{y+1}-1$

Correct option is 2.

Option 1)

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

This is an incorrect option.

Option 2)

Statement- 1 is true, Statement-2 is false

This is the correct option.

Option 3)

Statement-1 is false, Statement-2 is true

This is an incorrect option.

Option 4)

Statement-1 is true, Statement-2 is true; Statement-2 is correct explanation for Statement-1

This is an incorrect option.

N

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