Question

Say the ans immediately ​

Answer 1

$\huge\mathfrak\red{Answer}$

$x = {cos}^{2} \alpha + {sin}^{4} \alpha \\ \\ x = (1 - {sin}^{2} \alpha ) + {sin}^{4} \alpha \\ x = 1 + ( {sin}^{ 4 } \alpha - {sin}^{2} \alpha ) \\ \\ so \: from \: this \: we \: can \: conclude \\ that \: maximum \: value \: of \: x \: is \: \\ only \: when \: ({sin}^{4} \alpha - {sin}^{2} \alpha ) \: \\ is \: minimum \\ \\ minimum \: value \: of \: sin \alpha \: is \: zero \\ at \: \alpha = 0 \\ \\ maximum \: value \: of \: x \\ 1 + ( {sin}^{4} \alpha - {sin}^{2} \alpha ) \\ 1 + ( {sin}^{4} 0 - {sin}^{2} 0) = 1.....(1) \\ \\ minimum \: value \: of \: x \: is \: when \\ {sin}^{4} \alpha - {sin}^{2} \alpha \: is \: minimum \\ we \: can \: find \: its \: minimum \: by \\ maxima \\ \\ let \: m = {sin}^{4} \alpha - {sin}^{2} \alpha \\ differentiate \: this \: equation \\ \\ \frac{dm}{d \alpha } = 4 {sin}^{3}.cos \alpha \alpha - 2 \sin \alpha . \cos \alpha \\ \\ 0 = \sin \alpha \cos \alpha (4 {sin}^{2} \alpha - 2) \\ \\ we \: get \: \sin \alpha = \frac{1}{ \sqrt{2} } \\ \\ minimum \: value \: of \: x \\ \\ x = 1 + ( ({ \frac{1}{ \sqrt{2} } )}^{4} - \frac{1}{ {( \sqrt{2}) }^{2} } ) \\ \\ x = 1 + \frac{1}{4} - \frac{1}{2} \\ \\ x = \frac{3}{4}....(2) \\ \\ from \: (1) \: and \: (2) \: we \: can \: say \: that \\ \frac{3}{4} \leqslant x \leqslant 1$