Let α and β Be the Distinct Roots of 2x² + (cos θ)x − 1 = 0, θ ∈ (0, 2π). If m and M Are the Minimum and Maximum Values of α⁴ + β⁴, Then 16(M + m) Equals

JEE Main 2025 — 22 January, Evening Shift. Previous Year Question.

Problem

Let $\alpha$ and $\beta$ be the distinct roots of $2x^2+(\cos\theta)x-1=0$, $\theta\in(0,2\pi)$. If $m$ and $M$ are the minimum and the maximum values of $\alpha^4+\beta^4$ (as $\theta$ varies), then $16(M+m)$ equals:

Key insight. Rather than treating $\alpha^4+\beta^4$ as a function of $\theta$ directly, express it in terms of $\alpha+\beta$ and $\alpha\beta$ (both known from Vieta’s formulas) — this reduces the whole expression to a function of $\cos^2\theta$ alone. Since $\cos^2\theta$ ranges over the fixed interval $[0,1]$ regardless of the messy trigonometric details, the min and max become a simple two-point check.

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Approach

Use Vieta’s formulas to write $\alpha+\beta$ and $\alpha\beta$ in terms of $\cos\theta$. Express $\alpha^4+\beta^4$ using the standard identities $\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta$ and $\alpha^4+\beta^4=(\alpha^2+\beta^2)^2-2(\alpha\beta)^2$, reducing everything to a function of $u=\cos^2\theta \in [0,1]$. Since this function is increasing in $u$, evaluate it at $u=0$ and $u=1$ to get $m$ and $M$.

Solution

Step 1 — Apply Vieta’s formulas

$$\alpha+\beta = -\frac{\cos\theta}{2}, \qquad \alpha\beta = -\frac{1}{2}$$

Step 2 — Find α² + β²

$$\alpha^2+\beta^2 = (\alpha+\beta)^2-2\alpha\beta = \frac{\cos^2\theta}{4} – 2\left(-\frac{1}{2}\right) = \frac{\cos^2\theta}{4}+1$$

Step 3 — Find α⁴ + β⁴

$$\alpha^4+\beta^4 = (\alpha^2+\beta^2)^2 – 2(\alpha\beta)^2 = \left(\frac{\cos^2\theta}{4}+1\right)^2 – 2\left(\frac{1}{4}\right)$$

$$= \left(\frac{\cos^2\theta}{4}+1\right)^2 – \frac{1}{2}$$

Step 4 — Substitute u = cos²θ

Let $f(u) = \left(\dfrac{u}{4}+1\right)^2 – \dfrac{1}{2}$, where $u\in[0,1]$ (since $\theta$ ranges over $(0,2\pi)$, $\cos\theta$ covers $[-1,1]$, so $\cos^2\theta$ covers $[0,1]$). Since $\dfrac{u}{4}+1$ is increasing in $u$ and always positive, $f(u)$ is increasing on $[0,1]$.

Step 5 — Find the minimum (at u = 0)

$$m = f(0) = (0+1)^2-\frac{1}{2} = 1-\frac{1}{2} = \frac{1}{2}$$

Step 6 — Find the maximum (at u = 1)

$$M = f(1) = \left(\frac{1}{4}+1\right)^2-\frac{1}{2} = \left(\frac{5}{4}\right)^2-\frac{1}{2} = \frac{25}{16}-\frac{8}{16} = \frac{17}{16}$$

Step 7 — Compute 16(M + m)

$$M+m = \frac{17}{16}+\frac{1}{2} = \frac{17}{16}+\frac{8}{16} = \frac{25}{16}$$

$$16(M+m) = 16\times\frac{25}{16} = 25$$

Answer

Common mistakes

• Forgetting to check whether $f(u)$ is increasing or decreasing before assuming where the min/max occur. Here $f$ happens to be increasing on $[0,1]$, so the minimum is at $u=0$ and maximum at $u=1$ — but this needs to be verified, not assumed, since a different coefficient structure could reverse it.
• Using the wrong range for $\cos^2\theta$. Since $\theta\in(0,2\pi)$ (a full period, open interval), $\cos\theta$ still attains every value in $[-1,1]$ arbitrarily closely, so $\cos^2\theta$ effectively ranges over $[0,1]$ — treating the open interval as excluding the endpoints of $\cos^2\theta$’s range would be incorrect here.

Practise next

• Let $\gamma,\delta$ be the distinct roots of $3x^2+(\sin\phi)x-2=0$, $\phi\in(0,2\pi)$. If $p,P$ are the minimum and maximum values of $\gamma^2+\delta^2$, find $9(P+p)$, using the same substitution approach.