Let α and β be the distinct roots of 2x^2 + (cos θ)x - 1 = 0, θ ∈ (0, 2π). If m and M are the minimu

Q: Wait, why can't we just solve for $x$ using the quadratic formula and then raise it to the power of 4?

That is a classic trap! If you solve for $x$, you will get expressions involving square roots of trigonometric functions. Raising those to the power of 4 will lead to an algebraic mess that is almost impossible to simplify. Using Vieta's formulas allows you to work with the symmetric sums of the roots, which is the elegant and intended path.

Q: How do I know the range of $t = \cos^{2} \theta$ is $[0, 1]$?

Great question! Since $\theta$ is in the interval $(0, 2\pi)$, the value of $\cos \theta$ ranges from $-1$ to $1$. When you square any value in the interval $[-1, 1]$, the result must be non-negative, and the maximum value is $1^{2} = 1$. Thus, $t = \cos^{2} \theta$ is strictly confined to the interval $[0, 1]$.

Q: Why do we use $t = \cos^{2} \theta$ instead of $\cos \theta$?

Because the expression for $\alpha^{4} + \beta^{4}$ only contains $\cos^{2} \theta$ terms. By substituting $t = \cos^{2} \theta$, we transform a trigonometric expression into a simple quadratic function in $t$. This makes finding the maximum and minimum values much easier, as we are just analyzing a parabola over a closed interval.

Question

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

Options:

(A) 24

(B) 25

(C) 17

(D) 27

Sigma Percentile · Accepted Answer

### The Algebraic Symphony: Unlocking the Quadratic

Welcome, fellow traveler on the path to JEE mastery. Today, we are not just solving a problem; we are peeling back the layers of a beautiful algebraic structure.

We are given the quadratic equation $2x^{2} + (\cos 	heta)x - 1 = 0$, where $	heta \in (0, 2\pi)$. Our mission is to find the maximum and minimum values of $\alpha^{4} + \beta^{4}$, where $\alpha$ and $\beta$ are the roots.

### Phase 1: The Power of Vieta's Formulas

Many students see a quadratic equation and immediately reach for the quadratic formula. But stop! In competitive exams, the path of least resistance is often the most elegant.

We know that for any quadratic equation $ax^{2} + bx + c = 0$, the sum of the roots $\alpha + \beta = -\frac{b}{a}$ and the product $\alpha\beta = \frac{c}{a}$. Applying this to our equation, we find:

$$\alpha + \beta = -\frac{\cos 	heta}{2}$$

$$\alpha\beta = -\frac{1}{2}$$

These are our fundamental building blocks. They are the DNA of the roots, and they will carry us to the finish line without ever needing to know the individual values of $\alpha$ or $\beta$.

### Phase 2: The Algebraic Ladder

We need to reach $\alpha^{4} + \beta^{4}$. We cannot jump there in one leap, so we climb the ladder. First, we find $\alpha^{2} + \beta^{2}$ using the identity:

$$\alpha^{2} + \beta^{2} = (\alpha + \beta)^{2} - 2\alpha\beta$$

Substituting our values, we get:

$$\alpha^{2} + \beta^{2} = \left(-\frac{\cos 	heta}{2}ight)^{2} - 2\left(-\frac{1}{2}ight) = \frac{\cos^{2} 	heta}{4} + 1$$

Now, we apply the same logic to reach the fourth power. We know that $(\alpha^{2} + \beta^{2})^{2} = \alpha^{4} + \beta^{4} + 2(\alpha\beta)^{2}$. Rearranging this, we get:

$$\alpha^{4} + \beta^{4} = (\alpha^{2} + \beta^{2})^{2} - 2(\alpha\beta)^{2}$$

Substitute our previous result and the product of the roots:

$$\alpha^{4} + \beta^{4} = \left(\frac{\cos^{2} 	heta}{4} + 1ight)^{2} - 2\left(-\frac{1}{2}ight)^{2} = \left(\frac{\cos^{2} 	heta}{4} + 1ight)^{2} - \frac{1}{2}$$

### Phase 3: The Trigonometric Bridge

This expression looks daunting, but let us simplify it. Let $t = \cos^{2} 	heta$. Our expression becomes a function $f(t) = (\frac{t}{4} + 1)^{2} - \frac{1}{2}$.

Here is where the JEE trap lies: the domain. Since $	heta \in (0, 2\pi)$, $\cos 	heta$ varies between $-1$ and $1$. Therefore, $t = \cos^{2} 	heta$ must lie in the interval $[0, 1]$.

Now, we analyze $f(t) = (\frac{t}{4} + 1)^{2} - \frac{1}{2}$ on the interval $t \in [0, 1]$. This is an increasing function. The minimum value $m$ occurs at the smallest $t$, and the maximum value $M$ occurs at the largest $t$.

For $m$ (at $t = 0$):

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

For $M$ (at $t = 1$):

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

### Phase 4: The Final Victory

We are almost there. The question asks for $16(M + m)$. Let us calculate the sum first:

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

Finally, multiplying by 16:

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

And there it is! The elegance of the cancellation at the very end is the hallmark of a well-crafted JEE problem. You have navigated the algebra, respected the trigonometric domain, and arrived at the **final answer of 25**.

Animated Solution for Mathematics - Quadratic Equations: Let $α$ and $β$ be the distinct roots of $2 x^{2} + (cos θ) x - 1 = 0,$ $θ \in (0, 2 π)$ . If m and M are the minimum and the maximum values of $α_{θ}^{4} + β_{θ}^{4}$ , then $16 (M + m)$ equals:

Select Answer:

Visualized Solution

Analyze the Quadratic Equation

Sum and Product of Roots

Calculate $α^{2} + β^{2}$

Substitute Values

Express $α^{4} + β^{4}$

Substitute into Power 4

Simplify Power 4

Define Function $f (t)$

Determine Range of $t$

Find Maximum Value $M$

Find Minimum Value $m$

Calculate $M + m$

Final Computation

The Sigma Insight: Maximum and Minimum Values of Quadratic Expressions

The Algebraic Symphony

Phase 1

Phase 2

Phase 3

Phase 4

Similar Questions

The values of $λ$ such that sum of the squares of the roots of the quadratic equation $x^{2} + (3 - λ) x + 2 = λ$ has the least value is :

The number of integral values of $m$ for which the quadratic expression $(1 + 2 m) x^{2} - 2 (1 + 3 m) x + 4 (1 + m), x \in R$ , is always positive, is :

Let the sum of the maximum and the minimum values of the function $f (x) = \frac{2 x ^{2} - 3 x + 8}{2 x ^{2} + 3 x + 8}$ be $\frac{m}{n}$ , where $gcd (m, n) = 1$ . Then $m + n$ is equal to :

Let $f (x) = (1 + b^{2}) x^{2} + 2 b x + 1$ and let $m (b)$ be the minimum value of $f (x)$ . As $b$ varies, the range of $m (b)$ is

If the roots of the equation $b x^{2} + c x + a = 0$ be imaginary, then for all real values of $x$ , the expression $3 b^{2} x^{2} + 6 b c x + 2 c^{2}$ is

If $a \in R$ and the equation $- 3 (x - [x])^{2} + 2 (x - [x]) + a^{2} = 0$ (where $[x]$ denotes the greatest integer $\leq x$ ) has no integral solution, then all possible values of $a$ lie in the interval:

Find the range of values of $t$ for which $2 sin t = \frac{1 - 2 x + 5 x ^{2}}{3 x ^{2} - 2 x - 1}$ , $t \in [- π /2, π /2]$ .

If $f (x) = x^{2} + 2 b x + 2 c^{2}$ and $g (x) = - x^{2} - 2 c x + b^{2}$ such that $min f (x) > max g (x)$ , then the relation between $b$ and $c$ , is

The probability of selecting integers $a \in [- 5, 30]$ such that $x^{2} + 2 (a + 4) x - 5 a + 64 > 0$ , for all $x \in R$ , is:

Select Answer:

Visualized Solution

Analyze the Quadratic Equation

Sum and Product of Roots

Calculate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α2+β2

Substitute Values

Express .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α4+β4

Substitute into Power 4

Simplify Power 4

Define Function .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(t)

Determine Range of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t

Find Maximum Value .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } M

Find Minimum Value .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m

Calculate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } M+m