JEE Mathematics 2023 Question with Solution

Given: $\lambda = \cos^2 2x - 2\sin^4 x - 2\cos x$

Find: The set of all possible values of $\lambda$.

Convert everything into terms of $\cos x$:

$$\lambda = (2\cos^2 x - 1)^2 - 2(1 - \cos^2 x)^2 - 2\cos x$$ $$= 4\cos^4 x - 4\cos^2 x + 1 - 2(1 - 2\cos^2 x + \cos^4 x) - 2\cos x$$

From the working shown, this simplifies to

$$\lambda = 2\cos^4 x - 2\cos^2 x - 1$$

So,

$$\lambda = 2\left[\cos^4 x - \cos^2 x - \frac{1}{2}\right]$$ $$= 2\left[\left(\cos^2 x - \frac{1}{2}\right)^2 - \frac{3}{4}\right]$$

Now, since $0 \le \cos^2 x \le 1$, we get

$$0 \le \left(\cos^2 x - \frac{1}{2}\right)^2 \le \frac{1}{4}$$

Hence,

$$2\left[-\frac{3}{4}\right] \le \lambda \le 2\left[-\frac{1}{2}\right]$$ $$-\frac{3}{2} \le \lambda \le -1$$

Therefore, the range is $\left[-\frac{3}{2}, -1\right]$ and the correct option is D.

The solution labels option B, but the computed interval matches option D.

Given: The expression in $\lambda$ is reduced to a polynomial in $\cos^2 x$.

Find: Its maximum and minimum values.

Write

$$\lambda = 2\left[\cos^4 x - \cos^2 x - \frac{1}{2}\right]$$

Complete the square inside the bracket:

$$\cos^4 x - \cos^2 x - \frac{1}{2} = \left(\cos^2 x - \frac{1}{2}\right)^2 - \frac{3}{4}$$

Therefore,

$$\lambda = 2\left[\left(\cos^2 x - \frac{1}{2}\right)^2 - \frac{3}{4}\right]$$

The squared term is smallest when it is $0$, giving

$$\lambda_{\min} = 2\left[-\frac{3}{4}\right] = -\frac{3}{2}$$

The squared term is largest when $\cos^2 x = 0$ or $1$, so

$$\left(\cos^2 x - \frac{1}{2}\right)^2 = \frac{1}{4}$$

Thus,

$$\lambda_{\max} = 2\left[\frac{1}{4} - \frac{3}{4}\right] = -1$$

Therefore, the set of all values is $\left[-\frac{3}{2}, -1\right]$.