If α be a repeated root of a quadratic equation f(x) = 0 and A(x), B(x) and C(x) be polynomials of d

Q: Wait, why do we need to check the derivative $F'(\alpha)$? Isn't $F(\alpha) = 0$ enough?

Great question! If $F(\alpha) = 0$, we only know $(x - \alpha)$ is a factor. But we need to prove divisibility by $f(x)$, which is a quadratic with a repeated root, meaning $(x - \alpha)^2$ is the factor. To prove $(x - \alpha)^2$ is a factor, we absolutely need the derivative to vanish at that point too. It is the difference between a simple root and a point of tangency.

Q: Can't I just expand the determinant and simplify it? It looks like a lot of algebra.

You could, but that is the 'brute force' trap! In JEE Advanced, time is your currency. Expanding a $3 \times 3$ determinant with polynomial entries is a recipe for calculation errors. Using properties of determinants and the Factor Theorem is the elegant, 'JEE-style' path that saves you time and prevents silly mistakes.

Q: Why does the derivative of the determinant only involve the first row?

Because the second and third rows are constants! When you differentiate a determinant, you differentiate one row at a time. Since the derivatives of constant rows are zero, only the row containing the variable $x$ contributes to the derivative. It is a beautiful property that simplifies the problem significantly.

Question

If $\alpha$ be a repeated root of a quadratic equation $f(x) = 0$ and $A(x), B(x)$ and $C(x)$ be polynomials of degree 3, 4 and 5 respectively, then show that $\begin{vmatrix} A(x) & B(x) & C(x) \ A(\alpha) & B(\alpha) & C(\alpha) \ A'(\alpha) & B'(\alpha) & C'(\alpha) \end{vmatrix}$ is divisible by $f(x)$, where prime denotes the derivatives.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler of the JEE path. Today, we are going to dismantle a problem that, at first glance, looks like a terrifying, sprawling mess of algebra.

You see a $3 	imes 3$ determinant filled with polynomials $A(x)$, $B(x)$, and $C(x)$, and your instinct might be to panic, to start expanding, to drown in a sea of terms. But stop. Take a breath.

In the world of JEE Advanced, we do not fight monsters with brute force; we fight them with insight. Let us peel back the layers of this problem together.

### The Psychological Barrier

The problem asks us to show that a determinant $F(x)$ is divisible by a quadratic $f(x)$, where $f(x)$ has a repeated root $\alpha$. The moment you see a determinant, your brain might scream, "Expand it!"

But look at the structure. The first row contains the variable $x$, while the second and third rows are evaluated at a constant $\alpha$. This is not a random arrangement; it is a deliberate trap for those who rush.

By defining this determinant as a function $F(x)$, we transform a static matrix into a dynamic object that we can manipulate using the tools of calculus. We are not just solving a matrix problem; we are analyzing the behavior of a function.

### The Geometry of a Repeated Root

What does it mean for $\alpha$ to be a repeated root of $f(x) = 0$? Geometrically, it means the parabola $y = f(x)$ touches the x-axis at $x = \alpha$ without crossing it. It is a point of tangency.

Algebraically, this is profound. It means $f(x) = k(x - \alpha)^2$. Our goal is to prove that $F(x)$ is divisible by $(x - \alpha)^2$.

This is where the Factor Theorem becomes our best friend. To prove a polynomial $P(x)$ is divisible by $(x - \alpha)^2$, we do not need to perform long division. We simply need to show two things: $P(\alpha) = 0$ and $P'(\alpha) = 0$. This is the bridge between algebra and calculus.

### The Vanishing Act

Let us test the first condition: $F(\alpha) = 0$. We substitute $x = \alpha$ into our determinant $F(x)$. Look at the result:

$$
F(\alpha) = \begin{vmatrix} A(\alpha) & B(\alpha) & C(\alpha) \ A(\alpha) & B(\alpha) & C(\alpha) \ A'(\alpha) & B'(\alpha) & C'(\alpha) \end{vmatrix}
$$

Do you see it? The first row and the second row are identical! The property of determinants is absolute: if any two rows are identical, the determinant is zero. Just like that, the first condition is satisfied. The "monster" has blinked.

### The Calculus Connection

Now, for the second condition: $F'(\alpha) = 0$. How do we differentiate a determinant? We differentiate it row by row.

The rule is: $F'(x)$ is the sum of determinants where we differentiate one row at a time while keeping the others fixed. But here is the beauty of the problem: the second and third rows are constants. Their derivatives are zero.

Therefore, when we differentiate $F(x)$, only the first row changes. The other terms in the sum vanish. We are left with:

$$
F'(x) = \begin{vmatrix} A'(x) & B'(x) & C'(x) \ A(\alpha) & B(\alpha) & C(\alpha) \ A'(\alpha) & B'(\alpha) & C'(\alpha) \end{vmatrix}
$$

Now, evaluate this at $x = \alpha$:

$$
F'(\alpha) = \begin{vmatrix} A'(\alpha) & B'(\alpha) & C'(\alpha) \ A(\alpha) & B(\alpha) & C(\alpha) \ A'(\alpha) & B'(\alpha) & C'(\alpha) \end{vmatrix}
$$

Look closely. The first row and the third row are now identical! Again, the determinant vanishes. Thus, $F'(\alpha) = 0$.

### The Final Victory

We have proven that $F(\alpha) = 0$ and $F'(\alpha) = 0$. By the Factor Theorem, $(x - \alpha)^2$ must be a factor of $F(x)$.

Since $f(x) = k(x - \alpha)^2$, it follows immediately that **$F(x)$ is divisible by $f(x)$**. We did not expand a single term. We did not get lost in the algebra.

We used the properties of the system to reveal the truth. This, my friend, is the essence of JEE Advanced. It is not about how much you can calculate; it is about how much you can understand. Keep this mindset, and no problem will ever be too big for you.

Visualized Solution

Defining the Determinant $F (x)$

The Nature of a Repeated Root

Condition for Divisibility

Evaluating $F (x)$ at $x = α$

Vanishing Determinant

Differentiating the Determinant

Evaluating $F^{'} (x)$ at $x = α$

Second Vanishing Determinant

Applying the Factor Theorem

Final Conclusion

The Sigma Insight: Properties of Determinants

Analyzing the Setup

The Psychological Barrier

The Geometry of a Repeated Root

The Vanishing Act

The Calculus Connection

The Final Victory

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