Let P(x) = x² + bx + c be a quadratic polynomial with real coefficients such that ∫₀¹ P(x)dx = 1 and

Q: Wait, why do we use the Remainder Theorem here? Can't we just perform long division?

You absolutely could perform polynomial long division, but in the JEE, time is your most precious currency! The Remainder Theorem is a powerful shortcut that bypasses the tedious process of division. It tells us that $P(2)$ is exactly the remainder when $P(x)$ is divided by $(x-2)$. It is elegant, fast, and less prone to calculation errors.

Q: I got $3b + 6c = 4$ and $2b + c = 1$. How do I know if I should solve for $b$ or $c$ first?

That is a great question! In this specific case, it doesn't matter which one you eliminate first. However, look at the target expression: $9(b + c)$. Since you need the sum of $b$ and $c$, you can solve for them individually and add them up, or use substitution. As long as your algebra is consistent, the path you choose will lead to the same beautiful result.

Q: Why is the integral of $P(x)$ from $0$ to $1$ equal to $1$? Does that have a physical meaning?

In this context, it is a mathematical constraint defining the 'area' under the curve of your polynomial. While it doesn't represent a physical force or energy here, think of it as a boundary condition. Just like a string fixed at two points, your polynomial is 'fixed' by the area it must enclose. It restricts the possible values of $b$ and $c$ to a single unique pair.

Q: I keep getting a fraction for $b$ and $c$. Did I make a mistake?

Not at all! Fractions are the bread and butter of JEE Advanced problems. Don't let the sight of $\frac{2}{9}$ or $\frac{5}{9}$ scare you. Often, these fractions are designed to cancel out perfectly in the final step, just like they do here when you multiply by $9$. Trust your process!

Q: Is there any other way to solve this without solving for $b$ and $c$ separately?

Yes! You could treat $(b+c)$ as a single variable $S$ and try to express your equations in terms of $S$. However, for a quadratic with only two variables, solving for $b$ and $c$ individually is usually the most straightforward and least confusing method. Stick to the fundamentals when you are under exam pressure!

Question

Let $P(x) = x^2 + bx + c$ be a quadratic polynomial with real coefficients such that $\int_0^1 P(x)dx = 1$ and $P(x)$ leaves remainder 5 when it is divided by $(x - 2)$. Then the value of $9(b + c)$ is equal to:

Options:

(A) 9

(B) 15

(C) 7

(D) 11

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are an architect, but instead of steel and glass, you are building with the pure, abstract logic of algebra. You are given a blueprint: a quadratic polynomial $P(x) = x^2 + bx + c$.

You know the shape—a parabola—but its exact position in the coordinate plane is hidden, masked by the unknown coefficients $b$ and $c$. To find them, we must look at the two constraints provided by the problem.

### The First Constraint: The Integral

Our first piece of information is the integral: $\int_0^1 P(x) dx = 1$. This is a geometric requirement stating that the area trapped between our parabola, the $x$-axis, and the vertical lines $x=0$ and $x=1$ must be exactly $1$ square unit.

To unlock this, we perform the integration:

$$
\int_0^1 (x^2 + bx + c) dx = \left[ \frac{x^3}{3} + \frac{bx^2}{2} + cx ight]_0^1 = 1
$$

When we plug in the limits, the $0$ vanishes, leaving us with the expression $\frac{1}{3} + \frac{b}{2} + c = 1$. To simplify, we multiply the entire equation by $6$ to clear the denominators.

This gives us our first structural pillar: $2 + 3b + 6c = 6$, or more simply:

$$
3b + 6c = 4
$$

### The Second Constraint: The Remainder Theorem

Now, we turn to the second constraint. We are told that when $P(x)$ is divided by $(x - 2)$, the remainder is $5$.

We invoke the **Remainder Theorem**, which states that the remainder of $P(x) \div (x - a)$ is simply $P(a)$. Thus, we have $P(2) = 5$.

Substituting $x = 2$ into our original polynomial, we get:

$$
P(2) = (2)^2 + b(2) + c = 5
$$

$$
4 + 2b + c = 5 \Rightarrow 2b + c = 1
$$

This is our second pillar. We now have a system of two linear equations: $3b + 6c = 4$ and $2b + c = 1$.

### The Final Synthesis

We are now at the climax of our problem. We have two equations and two unknowns. To solve this, we use the elimination method.

If we multiply our second equation by $6$, we get $12b + 6c = 6$. Now, we subtract the first equation from this new one:

$$
(12b + 6c) - (3b + 6c) = 6 - 4
$$

$$
9b = 2 \Rightarrow b = \frac{2}{9}
$$

With $b$ in hand, finding $c$ is a simple matter of substitution into $2b + c = 1$:

$$
2\left(\frac{2}{9}ight) + c = 1 \Rightarrow \frac{4}{9} + c = 1 \Rightarrow c = \frac{5}{9}
$$

Finally, we calculate the value requested: $9(b + c)$.

$$
b + c = \frac{2}{9} + \frac{5}{9} = \frac{7}{9}
$$

$$
9 	imes \left(\frac{7}{9}ight) = 7
$$

The complexity collapses into the final answer of **7**. You have successfully navigated the constraints, balanced the equations, and arrived at the truth.

Animated Solution for Mathematics - Definite Integration: Let $P (x) = x^{2} + b x + c$ be a quadratic polynomial with real coefficients such that $\int_{0}^{1} P (x) d x = 1$ and $P (x)$ leaves remainder 5 when it is divided by $(x - 2)$ . Then the value of $9 (b + c)$ is equal to:

Select Answer:

Visualized Solution

Define the Polynomial $P (x)$

The Integral Condition

Set up the Integral Equation

Integrate the Polynomial

Apply the Definite Limits

Simplify to Equation 1

Apply the Remainder Theorem

Set up the Remainder Equation

Simplify to Equation 2

Elimination Strategy

Solve for $b$

Solve for $c$

Calculate the Final Expression

Final Result

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Analyzing the Setup

The First Constraint

The Second Constraint

The Final Synthesis

Similar Questions

Let $f$ be a polynomial function such that $f (x^{2} + 1) = x^{4} + 5 x^{2} + 2$ , for all $x \in R$ . Then $\int_{0}^{3} f (x) d x$ is equal to

Let $f : R \to R$ be defined $f (x) = a e^{2 x} + b e^{x} + c x$ . If $f (0) = - 1, f^{'} (lo g_{e} 2) = 21$ and $\int_{0}^{l o g_{e} 4} (f (x) - c x) d x = \frac{39}{2}$ , then the value of $∣ a + b + c ∣$ equals:

If $\int_{0}^{1} 4 cot^{- 1} (1 - 2 x + 4 x^{2}) d x = a tan^{- 1} (2) - b lo g_{e} (5)$ , where $a, b \in N$ , then $(2 a + b)$ is equal to .........

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 $\int_{- 1}^{1} P (x) d x = 18$ , then the sum of all the coefficients of the polynomial $P (x)$ is equal to ___

Let $[t]$ denote the largest integer less than or equal to $t$ . If $\int_{0}^{3} ([x^{2}] + [\frac{x ^{2}}{2}]) d x = a + b 2 - 3 - 5 + c 6 - 7$ , where $a, b, c \in Z$ , then $a + b + c$ is equal to

If $\int_{0}^{1} \frac{1}{3 + x + 1 + x} d x = a + b 2 + c 3$ , where $a, b, c$ are rational numbers, then $2 a + 3 b - 4 c$ is equal to :

Let $f (x) + 2 f (\frac{1}{x}) = x^{2} + 5$ and $2 g (x) - 3 g (\frac{1}{2}) = x, x > 0$ . If $α = \int_{1}^{2} f (x) d x$ , and $β = \int_{1}^{2} g (x) d x$ , then the value of $9 α + β$ is:

Let the domain of the function $f (x) = lo g_{2} lo g_{4} lo g_{6} (3 + 4 x - x^{2})$ be $(a, b)$ . If $\int_{0}^{b - a} [x^{2}] d x = p - q - r, p, q, r \in N, gcd (p, q, r) = 1$ , where [.] is the greatest integer function, then $p + q + r$ is equal to

If for nonzero $x$ , $a f (x) + b f (1/ x) = \frac{1}{x} - 5$ where $a \neq = b$ , then $\int_{1}^{2} f (x) d x = \dots\dots\dots$

If $\int_{0}^{3} \frac{15 x ^{3}}{1 + x ^{2} + ( 1 + x ^{2} ) ^{3}} d x = α 2 + β 3$ , where $α, β$ are integers, then $α + β$ is equal to