If x=2\sin^{2}\alpha - \cos2\alpha, then \:\: x \:\: lies \:\: in\:\: the \:\: interval

  • Option 1)

    \left [-1, 3 \right ]

  • Option 2)

    \left [1, 2 \right ]

  • Option 3)

    \left [-2, 4 \right ]

  • Option 4)

    None of these

 

Answers (1)

 

Results from General Solution -

 If \sin ^{2}\Theta = \sin ^{2}\alpha

    \cos ^{2}\Theta = \cos ^{2}\alpha

    \tan ^{2}\Theta = \tan ^{2}\alpha

then \Theta = n\pi \pm \alpha

- wherein

\alpha is the given angle

 

 x= \sin ^{2}\alpha -\cos ^{2}\alpha

\Rightarrow 2\sin ^{2}\alpha - (1-2\sin ^{2}\alpha )

\Rightarrow 2\sin ^{2}\alpha -1+2\sin ^{2}\alpha

\Rightarrow 4\sin ^{2}\alpha -1+2\sin ^{2}\alpha

4\sin ^{2}\alpha -1                                                      \therefore 0\leq \sin ^{2}\alpha \leq 1

0-1, 4-1

\left [ -1,3 \right ]


Option 1)

\left [-1, 3 \right ]

This solution is correct.

Option 2)

\left [1, 2 \right ]

This solution is incorrect.

Option 3)

\left [-2, 4 \right ]

This solution is incorrect.

Option 4)

None of these

This solution is incorrect.

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