If y = ( 3 + ^-1 x2), then (x - y)^2 + 3y^2 is equal to. — Solution | JEE Mathematics

MCQMediumJEE 2025Inverse Trigonometric Functions

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

Answer

Correct answer:C

Standard Method

Given: $y = \cos\!\left(\frac{\pi}{3} + \cos^{-1}\!\frac{x}{2}\right)$

Find: The value of $(x-y)^2 + 3y^2$ .

Let

\theta = \cos^{-1}\!\left(\frac{x}{2}\right)

Then

\cos\theta = \frac{x}{2}, \qquad \sin\theta = \sqrt{1-\left(\frac{x}{2}\right)^2} = \frac{\sqrt{4-x^2}}{2}

Now use the cosine addition formula:

y = \cos\left(\frac{\pi}{3}+\theta\right) = \cos\frac{\pi}{3}\cos\theta - \sin\frac{\pi}{3}\sin\theta

So,

y = \frac{1}{2}\cdot\frac{x}{2} - \frac{\sqrt{3}}{2}\cdot\frac{\sqrt{4-x^2}}{2} = \frac{x}{4} - \frac{\sqrt{3}}{4}\sqrt{4-x^2}

Set

a = \frac{x}{4}, \qquad b = \frac{\sqrt{3}}{4}\sqrt{4-x^2}

Then

y = a-b

and

x-y = x-(a-b) = \frac{3x}{4}+b

Now compute:

(x-y)^2 + 3y^2 = \left(\frac{3x}{4}+b\right)^2 + 3(a-b)^2

Expanding,

\left(\frac{3x}{4}\right)^2 + \frac{3x}{2}b + b^2 + 3a^2 - 6ab + 3b^2

= \frac{9x^2}{16} + 3a^2 + 4b^2 + \left(\frac{3x}{2}b - 6ab\right)

Since $a = \frac{x}{4}$ , the mixed terms cancel:

\frac{3x}{2}b - 6ab = \frac{3x}{2}b - 6\left(\frac{x}{4}\right)b = 0

Also,

3a^2 = 3\left(\frac{x}{4}\right)^2 = \frac{3x^2}{16}

and

4b^2 = 4\left(\frac{3}{16}(4-x^2)\right) = \frac{12}{16}(4-x^2) = 3 - \frac{3x^2}{4}

Therefore,

(x-y)^2 + 3y^2 = \left(\frac{9x^2}{16} + \frac{3x^2}{16}\right) + \left(3 - \frac{3x^2}{4}\right)

= \frac{12x^2}{16} + 3 - \frac{3x^2}{4} = \frac{3x^2}{4} + 3 - \frac{3x^2}{4} = 3

So the expression is constant, and the correct option is C.

Therefore, $(x-y)^2 + 3y^2 = 3$ .

Identity-Based Expansion

Given: $y = \cos \left( \frac{\pi}{3} + \cos^{-1} \frac{x}{2} \right)$

Find: The value of $(x-y)^2 + 3y^2$ .

Let

\theta = \cos^{-1}\frac{x}{2}

Then

y = \cos\left(\frac{\pi}{3}+\theta\right)

Using

\cos(A+B)=\cos A\cos B-\sin A\sin B

we get

y = \frac{1}{2}\cos\theta - \frac{\sqrt{3}}{2}\sin\theta

Since $\cos\theta = \frac{x}{2}$ ,

y = \frac{x}{4} - \frac{\sqrt{3}}{2}\sin\theta

the solution concludes that on simplifying the required expression,

(x-y)^2 + 3y^2 = 3

Thus, the correct option is C and the required value is $3$ .

A common mistake is using the wrong addition identity for cosine, such as writing $\cos(A+B)=\cos A\cos B+\sin A\sin B$ . This changes the sign of the second term and gives an incorrect expression for $y$ . Use $\cos(A+B)=\cos A\cos B-\sin A\sin B$ instead.
Students often forget that if $\theta=\cos^{-1}\left(\frac{x}{2}\right)$ , then $\cos\theta=\frac{x}{2}$ and $\sin\theta=\sqrt{1-\left(\frac{x}{2}\right)^2}$ . Replacing $\sin\theta$ incorrectly leads to wrong simplification. Express both sine and cosine carefully from the inverse trigonometric definition.
Another mistake is expanding $\left(\frac{3x}{4}+b\right)^2 + 3(a-b)^2$ incorrectly and missing cancellation of the mixed terms. The cross terms cancel only after substituting $a=\frac{x}{4}$ correctly. Expand systematically before combining like terms.

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step — free to start.