Animated Solution for Mathematics - Functions: Let $f:\mathbb{R}\to \mathbb{R}$ be a function defined by $f(x)=(2+3a)x^2+(\frac{a+2}{a-1})x+b,a\ne 1$ If $f(x+y)=f(x)+f(y)+1-\frac{2}{7}xy$, then the value of $28{\sum }_{i=1}^5|f(i)|$ is

Welcome, future engineers! Today, we are not just solving a problem; we are embarking on a detective mission. Functional equations often appear as intimidating, abstract monsters in JEE Advanced papers.

They seem to hide their true nature behind layers of $f(x+y)$ and $xy$ terms. But remember, every complex equation is just a puzzle waiting for the right key. Let us peel back the layers of this quadratic mystery together.

The Great Simplification

We are given the function $f(x)=(2+3a)x^2+(\frac{a+2}{a-1})x+b$ and the functional equation $f(x+y)=f(x)+f(y)+1-\frac{2}{7}xy$.

At first glance, that $+1$ is the 'noise' in our signal. It prevents the equation from being a clean, homogeneous form. In mathematics, when you see a constant term disrupting a functional equation, your first instinct should be to shift the function.

Let us define a new function, $g(x)=f(x)+1$. This implies $f(x)=g(x)-1$.

Why do we do this? Because when we substitute this into our original equation, watch what happens:

Look at the magic! The constants on the right side are $-1-1+1$, which simplifies to $-1$. Now, we have:

The $-1$ on both sides cancels out beautifully, leaving us with the elegant, clean form:

The Detective Work

Now, we need to identify the form of $g(x)$. The presence of the $xy$ term is a massive clue. We know that the expansion of $(x+y{)}^2$ is $x^2+y^2+2xy$.

This suggests that $g(x)$ must be a quadratic polynomial. Let us assume $g(x)=A{x}^2+Cx$.

Substituting this into our simplified equation:

Expanding the left side:

By comparing the coefficients of the $xy$ term on both sides, we immediately find that $2A=-\frac{2}{7}$, which gives us $A=-\frac{1}{7}$. We have successfully cracked the quadratic coefficient!

Reconstructing the Identity

Now that we have $g(x)=-\frac{1}{7}{x}^2+Cx$, we can easily find $f(x)$ because $f(x)=g(x)-1$. Thus:

But wait! We were given the original form of $f(x)=(2+3a)x^2+(\frac{a+2}{a-1})x+b$. Since these two expressions represent the same function, their coefficients must be identical.

1. Constant term: $b=-1$. 2. $x^2$ coefficient: $2+3a=-\frac{1}{7}$. Solving this, $3a=-\frac{15}{7}$, so $a=-\frac{5}{7}$. 3. $x$ coefficient: $C=\frac{a+2}{a-1}$. Substituting $a=-\frac{5}{7}$, we get $C=\frac{-5/7+2}{-5/7-1}=\frac{9/7}{-12/7}=-\frac{3}{4}$.

Our function is fully revealed: $f(x)=-\frac{1}{7}{x}^2-\frac{3}{4}x-1$.

The Final Calculation

We are asked to find $28{\sum }_{i=1}^5|f(i)|$. Before we jump into the summation, look at the function. Since the coefficient of $x^2$ is negative, the parabola opens downwards.

For any positive $i$, $f(i)$ will be negative. Therefore, $|f(i)|=-f(i)=\frac{1}{7}{i}^2+\frac{3}{4}i+1$.

Now, we calculate the sum:

Using standard summation formulas: - $\sum i^2=\frac{5(6)(11)}{6}=55$ - $\sum i=\frac{5(6)}{2}=15$ - $\sum 1=5$

So, the sum is $\frac{55}{7}+\frac{3}{4}(15)+5=\frac{55}{7}+\frac{45}{4}+5$.

Finally, multiplying by $28$:

There you have it! Through logical substitution and careful coefficient comparison, we dismantled the problem. Keep this mindset—break the problem down, simplify the noise, and the answer is 675.