If 2a + 3b + 6c = 0, then the quadratic equation ax² + bx + c = 0 has (A) at least one root in [0, 1

Q: Wait, why do we choose $f(x)$ to be the integral of the quadratic? How would I know to do that in an exam?

That is a fantastic question! In competitive exams like JEE, when you see a linear relation among coefficients, it is a huge hint to look for a connection to calculus. By integrating the quadratic, we create a function whose derivative is the quadratic itself. This allows us to use Rolle's Theorem, which is specifically designed to prove the existence of roots for derivatives. It is a classic 'reverse-engineering' technique in calculus.

Q: Can we just use the discriminant method instead?

You could try, but it would be incredibly messy! The discriminant $D = b^2 - 4ac$ would involve substituting $c = -\frac{2a + 3b}{6}$ into the expression, which would lead to a complex quadratic in terms of $a$ and $b$. It might work, but it is prone to algebraic errors and is far less elegant than the calculus approach. Rolle's Theorem is the 'surgical' tool here.

Q: Why does Rolle's Theorem guarantee a root in $(0, 1)$ specifically?

Rolle's Theorem states that if $f(x)$ is continuous on $[p, q]$, differentiable on $(p, q)$, and $f(p) = f(q)$, then there exists at least one $k \in (p, q)$ such that $f'(k) = 0$. We found $f(0) = 0$ and $f(1) = 0$. Since $p=0$ and $q=1$, the theorem guarantees a root $k$ in the open interval $(0, 1)$.

Question

If $2a + 3b + 6c = 0$, then the quadratic equation $ax^2 + bx + c = 0$ has

Options:

(A) at least one root in $[0, 1]$

(B) at least one root in $[2, 3]$

(C) at least one root in $[4, 5]$

(D) none of these

Sigma Percentile · Accepted Answer

### The Hidden Geometry of Coefficients

Imagine you are standing on the precipice of a challenging JEE Advanced problem. You are given a quadratic equation $ax^2 + bx + c = 0$ and a seemingly arbitrary constraint: $2a + 3b + 6c = 0$.

At first glance, this looks like a pure algebra problem. You might be tempted to start manipulating the coefficients, perhaps trying to express one in terms of the others or diving into the discriminant.

But pause for a moment. In the world of JEE, when you see a linear relationship between coefficients, it is often a siren call for calculus. Let us embark on a journey to uncover the hidden geometric reality behind this problem.

### The Calculus Bridge: Rolle's Theorem

To solve this, we need a powerful tool from our mathematical arsenal: Rolle's Theorem. This theorem is a cornerstone of calculus, providing a bridge between the values of a function and the behavior of its derivative.

It states that if a function $f(x)$ is continuous on a closed interval $[p, q]$, differentiable on the open interval $(p, q)$, and if the function values at the endpoints are equal, meaning $f(p) = f(q)$, then there must exist at least one point $k$ strictly between $p$ and $q$ where the derivative $f'(k) = 0$.

Geometrically, this is the moment where the tangent to the curve is perfectly horizontal. It is a beautiful, intuitive concept that we are about to wield with precision.

### Constructing the Auxiliary Function

We want to find a root of the quadratic equation $ax^2 + bx + c = 0$. This is equivalent to finding a point where the derivative of some function $f(x)$ is zero.

If we want $f'(x) = ax^2 + bx + c$, we simply need to integrate the quadratic expression with respect to $x$. Let us perform this integration:

$$f(x) = \int (ax^2 + bx + c) dx = \frac{ax^3}{3} + \frac{bx^2}{2} + cx$$

We can safely set the constant of integration to zero, as it does not affect the derivative. Now, we have our auxiliary function $f(x) = \frac{ax^3}{3} + \frac{bx^2}{2} + cx$.

This function is a polynomial, which means it is continuous and differentiable everywhere—perfect for Rolle's Theorem.

### The Moment of Truth: Evaluation

Now, let us test the endpoints of our interval, $x = 0$ and $x = 1$. First, evaluating at $x = 0$ is straightforward:

$$f(0) = \frac{a(0)^3}{3} + \frac{b(0)^2}{2} + c(0) = 0$$

This is a great start. Now, let us evaluate at $x = 1$:

$$f(1) = \frac{a(1)^3}{3} + \frac{b(1)^2}{2} + c(1) = \frac{a}{3} + \frac{b}{2} + c$$

To simplify this, we take a common denominator of $6$:

$$f(1) = \frac{2a + 3b + 6c}{6}$$

Look at the numerator! It is exactly the condition given in the problem statement: $2a + 3b + 6c = 0$. Substituting this in, we get:

$$f(1) = \frac{0}{6} = 0$$

### The Conclusion: A Root Emerges

We have found that $f(0) = 0$ and $f(1) = 0$. The conditions for Rolle's Theorem are perfectly satisfied on the interval $[0, 1]$.

Therefore, there must exist at least one $k \in (0, 1)$ such that $f'(k) = 0$. Since $f'(x) = ax^2 + bx + c$, this means $ak^2 + bk + c = 0$.

We have proven that the quadratic equation has at least one real root in the interval $(0, 1)$. This problem is a masterclass in recognizing patterns. It teaches us that algebra and calculus are not separate silos but interconnected languages.

When you see coefficients tied together in a linear dance, remember the integral, remember Rolle's Theorem, and watch as the complexity dissolves into elegance. Keep this strategy in your toolkit, and you will be ready for whatever the JEE throws your way!

Animated Solution for Mathematics - Differentiation: If $2 a + 3 b + 6 c = 0$ , then the quadratic equation $a x^{2} + b x + c = 0$ has

Select Answer:

Visualized Solution

Analyze the Given Condition

The Power of Rolle's Theorem

Defining the Auxiliary Function $f (x)$

Verifying the Derivative

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

Evaluating $f (x)$ at $x = 1$

Utilizing the Given Condition

Verifying Rolle's Theorem Conditions

Applying Rolle's Theorem

Conclusion and Key Takeaway

The Sigma Insight: Mean Value Theorems

The Hidden Geometry of Coefficients

The Calculus Bridge

Constructing the Auxiliary Function

The Moment of Truth

The Conclusion

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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 the quadratic equation ax2+bx+c=0 has