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Inverse Trigonometric Functions question

2025 · 2 Apr · Shift 2 · Q46
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Inverse Trigonometric Functions question

2025 · 2 Apr · Shift 2 · Q46

JEE MainMathematicsInverse Trigonometric FunctionsNumerical+4 / −1
 If y=cos⁡(π3+cos⁡−1x2), then (x−y)2+3y2 is equal to \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 } If y=cos(3π​+cos−12x​), then (x−y)2+3y2 is equal to 
Numerical answer
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Correct answer: 3

  1. Given expression

We have y=cos⁡(π3+cos⁡−1x2).y=\cos\left(\frac{\pi}{3}+\cos^{-1}\frac{x}{2}\right).y=cos(3π​+cos−12x​).

Let θ=cos⁡−1x2.\theta=\cos^{-1}\frac{x}{2}.θ=cos−12x​. Then cos⁡θ=x2.\cos\theta=\frac{x}{2}.cosθ=2x​. So the given becomes y=cos⁡(π3+θ).y=\cos\left(\frac{\pi}{3}+\theta\right).y=cos(3π​+θ).

  1. Use cosine addition formula

Recall, cos⁡(A+B)=cos⁡Acos⁡B−sin⁡Asin⁡B.\cos(A+B)=\cos A\cos B-\sin A\sin B.cos(A+B)=cosAcosB−sinAsinB. Hence, y=cos⁡π3cos⁡θ−sin⁡π3sin⁡θ.y=\cos\frac{\pi}{3}\cos\theta-\sin\frac{\pi}{3}\sin\theta.y=cos3π​cosθ−sin3π​sinθ. Now, cos⁡π3=12,sin⁡π3=32.\cos\frac{\pi}{3}=\frac12, \qquad \sin\frac{\pi}{3}=\frac{\sqrt3}{2}.cos3π​=21​,sin3π​=23​​. Therefore, y=12cos⁡θ−32sin⁡θ.y=\frac12\cos\theta-\frac{\sqrt3}{2}\sin\theta.y=21​cosθ−23​​sinθ.

Since cos⁡θ=x2,\cos\theta=\frac{x}{2},cosθ=2x​, we get y=x4−32sin⁡θ.y=\frac{x}{4}-\frac{\sqrt3}{2}\sin\theta.y=4x​−23​​sinθ.

  1. Express sin⁡θ\sin\thetasinθ in terms of xxx

Because θ=cos⁡−1x2,\theta=\cos^{-1}\frac{x}{2},θ=cos−12x​, we have θ∈[0,π]\theta\in[0,\pi]θ∈[0,π], so sin⁡θ≥0\sin\theta\ge 0sinθ≥0. Thus,

\sqrt{1-\left(\frac{x}{2}\right)^2}= rac{\sqrt{4-x^2}}{2}.$$ Substitute: $$y=\frac{x}{4}-\frac{\sqrt3}{2}\cdot \frac{\sqrt{4-x^2}}{2} =\frac{x-\sqrt3\sqrt{4-x^2}}{4}.$$ 4. **Compute $(x-y)^2+3y^2$ cleverly** Instead of substituting this complicated form directly, use the relation $$4y=x-\sqrt3\sqrt{4-x^2}.$$ So, $$x-4y=\sqrt3\sqrt{4-x^2}.$$ Now square both sides: $$(x-4y)^2=3(4-x^2).$$ Expand: $$x^2-8xy+16y^2=12-3x^2.$$ Thus, $$4x^2-8xy+16y^2=12.$$ Divide by $4$: $$x^2-2xy+4y^2=3.$$ But $$x^2-2xy+4y^2=(x-y)^2+3y^2.$$ Hence, $$(x-y)^2+3y^2=3.$$ 5. **Final answer** Therefore, the required integer is $$\boxed{3}.$$
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