If 2a + 3b + 6c = 0, then at least one root of the equation ax² + bx + c = 0 lies in the interval (A

Q: Why do we integrate? Can't we just use the discriminant or the quadratic formula?

The discriminant tells you about the nature of the roots (real, imaginary, equal), but it doesn't help you locate them within a specific interval. Integration is a powerful 'bridge' technique. By integrating, we transform the algebraic condition on the coefficients into a statement about the values of a function at two points. This allows us to use the existence theorems of calculus, like Rolle's Theorem, which are specifically designed to prove that a root exists in a certain range.

Q: Why does Rolle's Theorem apply here?

Rolle's Theorem requires three things: the function must be continuous on the closed interval, differentiable on the open interval, and the function values at the endpoints must be equal. Since our auxiliary function $f(x) = \frac{ax^3}{3} + \frac{bx^2}{2} + cx$ is a polynomial, it is continuous and differentiable everywhere. We calculated $f(0) = 0$ and $f(1) = 0$. Because $f(0) = f(1)$, the conditions are perfectly met, guaranteeing a point where the derivative is zero.

Q: What if the interval was different? How do I know which interval to pick?

The interval is determined by the coefficients in the given condition. If the condition was $a + b + c = 0$, you would evaluate the integral at $x=1$ to get $f(1) = \frac{a}{3} + \frac{b}{2} + c$, which doesn't match. You have to look at the denominators produced by integration. In this problem, the denominators were $3$ and $2$, which led us to evaluate at $x=1$ because $\frac{1}{3}$ and $\frac{1}{2}$ are the coefficients of the integral. The problem is designed so that the evaluation at $x=1$ yields the given linear condition.

Question

If $2a + 3b + 6c = 0$, then at least one root of the equation $ax^2 + bx + c = 0$ lies in the interval

Options:

(A) (1, 3)

(B) (1, 2)

(C) (2, 3)

(D) (0, 1)

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Have you ever looked at a problem and felt like the numbers were whispering a secret? That is exactly what is happening here. We are given a quadratic equation $ax^2 + bx + c = 0$ and a strange, seemingly unrelated constraint: $2a + 3b + 6c = 0$.

At first glance, you might be tempted to reach for the quadratic formula or start testing values. But stop. Take a breath. In JEE Advanced, when you see a linear relation between coefficients, it is rarely just algebra; it is a call to use calculus.

### The Calculus Bridge

Why would we integrate a quadratic? Think about the power rule of integration. When you integrate $ax^2 + bx + c$, the powers of $x$ increase, and the coefficients get divided by the new powers.

Specifically, the integral of $ax^2$ is $\frac{ax^3}{3}$, the integral of $bx$ is $\frac{bx^2}{2}$, and the integral of $c$ is $cx$. Look at those denominators: $3$ and $2$.

Now look at the given condition $2a + 3b + 6c = 0$. If we divide this entire equation by $6$, we get:

$$\frac{2a}{6} + \frac{3b}{6} + \frac{6c}{6} = 0$$

This simplifies to:

$$\frac{a}{3} + \frac{b}{2} + c = 0$$

Do you see the magic? This is exactly what we get if we evaluate the integral function $f(x) = \frac{ax^3}{3} + \frac{bx^2}{2} + cx$ at $x = 1$. We have built a bridge between the algebra of the coefficients and the geometry of a function.

### The Rolle's Theorem Revelation

Now that we have our auxiliary function $f(x) = \frac{ax^3}{3} + \frac{bx^2}{2} + cx$, let's test its behavior. At $x = 0$, every term vanishes, so $f(0) = 0$.

We just showed that at $x = 1$, $f(1) = \frac{a}{3} + \frac{b}{2} + c$, which is $\frac{2a + 3b + 6c}{6}$. Since the problem tells us $2a + 3b + 6c = 0$, it follows that $f(1) = 0$.

We have a polynomial function that starts at zero and ends at zero on the interval $[0, 1]$. This is the classic setup for **Rolle's Theorem**.

Rolle's Theorem tells us that if a continuous and differentiable function has the same value at two points, there must be at least one point $\alpha$ between them where the slope of the tangent is zero. In our case, that means there exists some $\alpha \in (0, 1)$ such that $f'(\alpha) = 0$.

### The Final Connection

What is $f'(\alpha)$? It is simply the original quadratic expression evaluated at $\alpha$:

$$f'(\alpha) = a\alpha^2 + b\alpha + c$$

Since we proved $f'(\alpha) = 0$, it means $\alpha$ is a root of the equation $ax^2 + bx + c = 0$. And because $\alpha$ was guaranteed to be in the interval $(0, 1)$, we have proven that **at least one root of the quadratic equation must lie between $0$ and $1$**.

It is elegant, it is powerful, and it is the kind of thinking that separates the good from the great. Keep looking for these hidden connections, and you will find that mathematics is not just a subject to study, but a story to be told.

Animated Solution for Mathematics - Differentiation: If $2 a + 3 b + 6 c = 0$ , then at least one root of the equation $a x^{2} + b x + c = 0$ lies in the interval

Select Answer:

Visualized Solution

Analyze the Given Condition

The Logic Bridge: Integration

Define the Auxiliary Function $f (x)$

Evaluate $f (x)$ at $x = 0$

Evaluate $f (x)$ at $x = 1$ (Raw Setup)

Evaluate $f (x)$ at $x = 1$ (Compute)

Apply the Given Condition

Verify Rolle's Theorem Conditions

Apply Rolle's Theorem Conclusion

Connect Back to the Original Equation

Final Conclusion

The Sigma Insight: Mean Value Theorems

Analyzing the Setup

The Calculus Bridge

The Rolle's Theorem Revelation

The Final Connection

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Animated Solution for Mathematics - Differentiation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If 2a+3b+6c=0, then at least one root of the equation ax2+bx+c=0 lies in the interval