Question

pls answer this question​

Answer 1

Step-by-step explanation:

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