Math, asked by Adyush, 1 year ago

pls answer this question​

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Answered by rishabh1894041
0

Step-by-step explanation:

y = 4 {sin}^{2}  \alpha  - cos2 \alpha  \\ y =  4 {sin}^{2}  \alpha  - 1  + 2 {sin}^{2}  \alpha  \\ y = 6 {sin}^{2}  \alpha  - 1 \\  \\ as \: we \: know \: that \\  - 1 \leqslant sin \alpha  \leqslant 1 \\ 0 \leqslant  {sin}^{2}   \alpha  \leqslant 1 \\ 0 \leqslant  6{sin}^{2}  \alpha  \leqslant 6 \\  - 1 \leqslant 6 {sin}^{2}  \alpha  \leqslant 5 \\  - 1 \leqslant y \leqslant 5 \\ Option (2) is  the  correct  answer. \\ Hope \: it \: will \: help \: you..

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