Let P(x) be a real polynomial of degree 3 which vanishes at x = -3. Let P(x) have local minima at x

Q: Wait, why is the sum of coefficients of a polynomial always equal to P(1)?

Think about the general form of a polynomial: P(x) = a_n x^n + a_{n-1} x^{n-1} + ... + a_1 x + a_0. If you substitute x = 1, every term x^k becomes 1^k, which is just 1. So, P(1) = a_n(1)^n + a_{n-1}(1)^{n-1} + ... + a_1(1) + a_0, which simplifies exactly to the sum of the coefficients a_n + a_{n-1} + ... + a_0. It is a beautiful property that saves us from expanding everything!

Q: Can I just write P(x) = ax^3 + bx^2 + cx + d and solve for the four variables using the given conditions?

You absolutely could, and it would work! However, in JEE Advanced, time is your most precious currency. Using the derivative properties (P'(1)=0, P'(-1)=0) allows you to reduce the problem to just two variables (a and C) immediately. Always look for the path that minimizes algebraic complexity.

Q: Why did the integral of the odd part vanish? I'm still a bit confused about the property.

An odd function satisfies f(-x) = -f(x). Geometrically, the area under the curve from -a to 0 is the exact negative of the area from 0 to a. When you sum them up over the interval [-a, a], they cancel each other out perfectly. In our case, x^3/3 - x is an odd function, so its integral from -1 to 1 is guaranteed to be zero.

Question

Let $P(x)$ be a real polynomial of degree 3 which vanishes at $x = -3$. Let $P(x)$ have local minima at $x = 1$, local maxima at $x = -1$ and $\int_{-1}^1 P(x) dx = 18$, then the sum of all the coefficients of the polynomial $P(x)$ is equal to ___

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are given a cubic polynomial $P(x)$ with local extrema at $x = 1$ and $x = -1$. In the language of calculus, this is a massive hint.

At any local maximum or minimum, the tangent to the curve is perfectly horizontal. This means the derivative, $P'(x)$, must be zero at these points.

Since $P(x)$ is a cubic, its derivative $P'(x)$ must be a quadratic. If we know the roots are $1$ and $-1$, we can immediately write the derivative as:

$$P'(x) = a(x - 1)(x + 1) = a(x^2 - 1)$$

This is the skeleton of our polynomial.

### Reconstructing the Polynomial

Now, we must reverse the process. To find $P(x)$, we integrate $P'(x)$ with respect to $x$:

$$P(x) = \int a(x^2 - 1) dx = a\left(\frac{x^3}{3} - x\right) + C$$

Here, $C$ is our constant of integration. We have two unknowns, $a$ and $C$, and we have two conditions to find them.

The problem states that $P(-3) = 0$. Substituting $x = -3$ into our expression:

$$a\left(\frac{(-3)^3}{3} - (-3)\right) + C = 0 \Rightarrow -6a + C = 0$$

Keep this relationship close; it is the bridge between our two variables.

### The Power of Symmetry

Next, we tackle the definite integral:

$$\int_{-1}^{1} P(x) dx = 18$$

This is where many students get bogged down in algebra, but you are a JEE warrior—you look for the elegant path. Notice the term $a(\frac{x^3}{3} - x)$. This is an **odd function**.

In the world of integration, the integral of an odd function over a symmetric interval $[-k, k]$ is always zero. This is a gift! The entire complex part of the polynomial vanishes, leaving us with:

$$\int_{-1}^{1} C dx = 18$$

This simplifies to $2C = 18$, giving us **$C = 9$**. With $C$ in hand, we return to our bridge equation:

$$-6a + 9 = 0 \Rightarrow a = \frac{3}{2}$$

### The Final Reveal

We have our polynomial:

$$P(x) = \frac{3}{2}\left(\frac{x^3}{3} - x\right) + 9 = \frac{x^3}{2} - \frac{3x}{2} + 9$$

The question asks for the sum of the coefficients. A common trap is to try and sum them manually, but there is a beautiful property here: the sum of the coefficients of any polynomial is simply $P(1)$.

Substituting $x = 1$ into our expression:

$$P(1) = \frac{1}{2} - \frac{3}{2} + 9 = -1 + 9 = 8$$

And there it is—the elegance of mathematics in full display. The sum of the coefficients is **$8$**.

Enter Numerical Value:

Visualized Solution

Introduction to $P (x)$

Analyzing the Derivative $P^{'} (x)$

Integrating to find $P (x)$

Using the Condition $P (- 3) = 0$

The Definite Integral Condition

Property of Odd Functions

Solving for $C$ and $a$

The Final Polynomial $P (x)$

Sum of Coefficients Concept

Final Calculation and Result

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Analyzing the Setup

Reconstructing the Polynomial

The Power of Symmetry

The Final Reveal

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Enter Numerical Value:

Visualized Solution

Introduction to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(x)

Analyzing the Derivative .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P′(x)

Integrating to find .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(x)

Using the Condition .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(−3)=0