Q62 / 125 · BITSAT 2020 · Mathematics

Let $x+y=3-\cos 4 \theta$ and $x-y=4 \sin 2 \theta$ then the greatest of $x y$ is

  • A
    $\frac{3}{4}$
  • Correct answer
    B
    1
  • C
    $\frac{1}{2}$
  • D
    2

Solution

$x=\frac{3-\cos 4 \theta+4 \sin 2 \theta}{2}$ $=\frac{3-\left(1-\sin ^{2} 2 \theta\right)+4 \sin 2 \theta}{2}=(1+\sin 2 \theta)^{2}$ $y=\frac{3-\cos 4 \theta-4 \sin 2 \theta}{2}$ $=\frac{3-\left(1-\sin ^{2} 2 \theta\right)+4 \sin 2 \theta}{2}=(1-\sin 2 \theta)^{2}$ $\therefore x y=\left(1-\sin ^{2} 2 \theta\right)^{2}=\cos ^{4} 2 \theta \leq 1$

Hence, the correct answer is option: 'B'.

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