If and

A and B are respectively the maximum and the minimum values of   f(\Theta ) , then (A, B) is equal to :

  • Option 1)

    \left ( 3,-1 \right )

  • Option 2)

    \left ( 4,2-\sqrt{2} \right )

  • Option 3)

    \left ( 2+\sqrt{2},2-\sqrt{2} \right )

  • Option 4)

    \left ( 2+\sqrt{2},-1)

 

Answers (2)

As we learnt in 

Value of determinants of order 3 -

-

 

 f\left ( \theta \right )=\begin{vmatrix} 1 &cos\theta &1 \\ -sin\theta & 1 &-cos\theta \\ -1& sin\theta & 1 \end{vmatrix}

f\left ( \theta \right )= 1\left ( 1+sin\theta cos\theta \right ) - cos\theta \left ( -sin\theta-cos\theta \right ) +1 \left ( -sin^{2}\theta +1 \right )

=1+sin\theta cos\theta +sin\theta cos\theta +cos^{2}\theta -sin^{2}\theta +1

2+sin2\theta +cos2\theta

= 2\pm \sqrt{2}

So, min value = 2-\sqrt{2}

max value = 2+\sqrt{2}

 

 


Option 1)

\left ( 3,-1 \right )

This option is incorrect.

Option 2)

\left ( 4,2-\sqrt{2} \right )

This option is incorrect.

Option 3)

\left ( 2+\sqrt{2},2-\sqrt{2} \right )

This option is correct.

Option 4)

\left ( 2+\sqrt{2},-1)

This option is incorrect.

N neha

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