Let c, k ∈ R. If f(x) = (c+1)x² + (1-c²)x + 2k and f(x+y) = f(x) + f(y) - xy, for all x, y ∈ R, then

Q: Why can we just assume $f(0) = 0$? Is it always true for functional equations?

Not always, but in this specific case, substituting $x=0$ and $y=0$ into $f(x+y) = f(x) + f(y) - xy$ yields $f(0) = f(0) + f(0) - 0$, which simplifies to $f(0) = 0$. It is a rigorous deduction, not an assumption! Always test the simplest inputs first.

Q: Why did we use the derivative definition? Can't we just solve it using algebra?

Algebra alone is difficult here because the functional equation relates $f(x+y)$ to $f(x)$ and $f(y)$ in a way that suggests a continuous, differentiable function. Using the derivative definition is the standard 'JEE Advanced' way to convert a functional equation into a differential equation, which is much easier to solve.

Q: Why do we compare coefficients at the end?

Because $f(x)$ is a polynomial. If two polynomials are equal for all values of $x$, their coefficients must be identical. This is the Principle of Identity for polynomials, and it allows us to bridge the gap between our derived form and the given form.

Question

Let $c, k \in \mathbb{R}$. If $f(x) = (c+1)x^2 + (1-c^2)x + 2k$ and $f(x+y) = f(x) + f(y) - xy$, for all $x, y \in \mathbb{R}$, then the value of $|2(f(1) + f(2) + f(3) + \dots + f(20))|$ is equal to ______.

Sigma Percentile · Accepted Answer

### The Symphony of Functional Equations and Calculus

Welcome, future engineers. Today, we are not just solving a problem; we are decoding the mathematical DNA of a function.

When you encounter a problem that blends functional equations with polynomial structures, it is easy to feel overwhelmed. But remember, every complex problem is just a collection of simple, elegant truths waiting to be uncovered. Let us embark on this journey together.

### Phase 1: The Anchor Point

We are given the functional equation $f(x+y) = f(x) + f(y) - xy$. This equation is a constraint that dictates the behavior of $f(x)$ for every real number $x$ and $y$.

Our first instinct should always be to find an anchor point. By substituting $x=0$ and $y=0$ into our equation, we get:

$$f(0) = f(0) + f(0) - 0$$

Subtracting $f(0)$ from both sides, we find that $f(0) = 0$. This is our first victory.

Now, looking at the given quadratic form $f(x) = (c+1)x^2 + (1-c^2)x + 2k$, we know that $f(0) = 2k$. Since $f(0) = 0$, it follows immediately that $2k = 0$, or simply $k = 0$.

The constant term vanishes, and our function simplifies to $f(x) = (c+1)x^2 + (1-c^2)x$.

### Phase 2: The Calculus Bridge

Now, how do we find $c$? We need to understand the rate of change of this function. This is where the beauty of calculus shines.

We use the first principle of derivatives:

$$f'(x) = \lim_{h 	o 0} \frac{f(x+h) - f(x)}{h}$$

Using our functional equation, we replace $f(x+h)$ with $f(x) + f(h) - xh$. Substituting this into our limit, we get:

$$f'(x) = \lim_{h 	o 0} \frac{f(x) + f(h) - xh - f(x)}{h}$$

The $f(x)$ terms cancel out, leaving us with $f'(x) = \lim_{h 	o 0} \frac{f(h)}{h} - x$. Since $f(0) = 0$, the term $\lim_{h 	o 0} \frac{f(h)}{h}$ is exactly the definition of $f'(0)$.

Thus, we have derived a beautiful differential equation: $f'(x) = f'(0) - x$.

### Phase 3: The Integration

We have a differential equation that describes the slope of our function at any point $x$. To recover the function $f(x)$, we integrate both sides with respect to $x$:

$$\int f'(x) dx = \int (f'(0) - x) dx$$

This gives us $f(x) = f'(0)x - \frac{x^2}{2} + C$. Since we already established $f(0) = 0$, the constant of integration $C$ must be zero.

Our function is now $f(x) = f'(0)x - \frac{1}{2}x^2$.

### Phase 4: The Comparison

We now have two expressions for the same function: the one given in the problem, $f(x) = (c+1)x^2 + (1-c^2)x$, and our derived form, $f(x) = -\frac{1}{2}x^2 + f'(0)x$.

By the Principle of Identity, the coefficients of corresponding powers of $x$ must be equal. Comparing the $x^2$ terms, we get $c+1 = -\frac{1}{2}$, which implies $c = -\frac{3}{2}$.

Comparing the $x$ terms, we get $f'(0) = 1 - c^2$. Substituting $c = -\frac{3}{2}$, we find:

$$f'(0) = 1 - \left(-\frac{3}{2}ight)^2 = 1 - \frac{9}{4} = -\frac{5}{4}$$

### Phase 5: The Final Summation

With $f(x) = -\frac{1}{2}x^2 - \frac{5}{4}x$, we are ready for the final act. We need to calculate $|2 \sum_{x=1}^{20} f(x)|$.

Substituting our function, we get:

$$\left| 2 \sum_{x=1}^{20} \left(-\frac{1}{2}x^2 - \frac{5}{4}xight) ight| = \left| \sum_{x=1}^{20} (-x^2 - \frac{5}{2}x) ight|$$

Using the standard summation formulas $\sum_{x=1}^{n} x^2 = \frac{n(n+1)(2n+1)}{6}$ and $\sum_{x=1}^{n} x = \frac{n(n+1)}{2}$, we calculate:

$$\sum_{x=1}^{20} x^2 = \frac{20 	imes 21 	imes 41}{6} = 2870$$

$$\sum_{x=1}^{20} x = \frac{20 	imes 21}{2} = 210$$

The final value is:

$$| -2870 - \frac{5}{2}(210) | = | -2870 - 525 | = | -3395 | = \mathbf{3395}$$

We have arrived at the destination. The path was rigorous, but the logic was sound. Keep practicing, and you will master these patterns.

Animated Solution for Mathematics - Differentiation: Let $c, k \in R$ . If $f (x) = (c + 1) x^{2} + (1 - c^{2}) x + 2 k$ and $f (x + y) = f (x) + f (y) - x y$ , for all $x, y \in R$ , then the value of $∣2 (f (1) + f (2) + f (3) + \dots + f (20)) ∣$ is equal to ______.

Enter Numerical Value:

Visualized Solution

Understanding the Problem

Finding $f (0)$

Determining the Constant $k$

Applying the Derivative Definition

Simplifying the Limit

The Differential Equation

Integrating to find $f (x)$

Comparing with the Quadratic Form

Solving for $c$ and $f^{'} (0)$

The Final Form of $f (x)$

Setting up the Summation

Calculating the Sum of Squares

Calculating the Linear Sum

Final Calculation

The Sigma Insight: Techniques of Differentiation

The Symphony of Functional Equations and Calculus

Phase 1

Phase 2

Phase 3

Phase 4

Phase 5

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