JEE MAIN - Mathematics (2025 - 2nd April Evening Shift - No. 21)

$$ \text { If } y=\cos \left(\frac{\pi}{3}+\cos ^{-1} \frac{x}{2}\right) \text {, then }(x-y)^2+3 y^2 \text { is equal to } $$
Answer
3

Explanation

$$\begin{aligned} & y=\cos \left(\frac{\pi}{3}+\cos ^{-1} \frac{x}{2}\right) \\ & =\cos \left(\frac{\pi}{3}\right) \cos \left(\cos ^{-1}\left(\frac{x}{2}\right)\right)-\sin \left(\frac{\pi}{3 .}\right) \sin \left(\cos ^{-1}\left(\frac{x}{2}\right)\right) \\ & =\frac{1}{2} \cdot \frac{x}{2}-\frac{\sqrt{3}}{2} \cdot \sqrt{1-\frac{x^2}{4}} \\ & \Rightarrow 4 y=x-\sqrt{3} \sqrt{4-x^2} \\ & \Rightarrow(4 y-x)^2=3\left(4-x^2\right) \\ & \Rightarrow 16 y^2+x^2-8 x y=12-3 x^2 \\ & x^2+4 y^2-2 x y=3 \\ & (x-y)^2+3 y^2=3 \end{aligned}$$

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