QJEEM19Maf07a: JEE Main 2019 Mathematics - Class 12 JEEM19Maf07a Solution: If ,where –1 x 1, – 2 y 2, x , then for all x, y, 4x 2 – 4xy…

Question

Verified Solution • 2025–26 Session moderate

If

{\cos ^{ - 1}}x - {\cos ^{ - 1}}{y \over 2} = \alpha

,where –1

\le

x

\le

1, – 2

\le

y

\le

2, x

\le

{y \over 2}

, then for all x, y, 4x² – 4xy cos

\alpha

+ y² is equal to :

Accepted Answer

{\cos ^{ - 1}}x - {\cos ^{ - 1}}{y \over 2} = \alpha

\Rightarrow \cos \left( {{{\cos }^{ - 1}}x - {{\cos }^{ - 1}}\left( {{y \over 2}} \right)} \right) = \cos \alpha

\Rightarrow x{y \over 2} + \sqrt {1 - {x^2}} \sqrt {1 - {{{y^2}} \over 4}} = \cos \alpha

\left( {\cos \alpha - {{xy} \over 2}} \right) = \sqrt {1 - {x^2}} \sqrt {1 - {{{y^2}} \over 4}}

squaring both sides

{x^2} + {{{y^2}} \over 4} - xy\cos \alpha = 1 - {\cos ^2}\alpha = {\sin ^2}\alpha

\therefore

4x² – 4xy cos

\alpha

+ y² = 4

{\sin ^2}\alpha

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