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

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

Statement - 2 : 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

 

Answers (2)

best_answer

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.

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