Functions – Functional Equation – JEE Main 28 Jan 2025 Shift 1

Solution:

Given the functional equation $f(x+y) = f(x) + f(y) + 1 – \frac{2}{7}xy$.

First, substitute $x = 0$ and $y = 0$:
$$ f(0) = f(0) + f(0) + 1 \Rightarrow f(0) = -1 $$
From the polynomial definition $f(x) = (2+3a)x^2 + \left(\frac{a+2}{a-1}\right)x + b$, we have $f(0) = b$.

Therefore, $b = -1$.

Now, substitute $y = -x$ in the functional equation:
$$ f(0) = f(x) + f(-x) + 1 – \frac{2}{7}x(-x) $$
$$ -1 = f(x) + f(-x) + 1 + \frac{2}{7}x^2 $$
$$ f(x) + f(-x) = -2 – \frac{2}{7}x^2 $$

Using the algebraic definition of $f(x)$, we calculate $f(x) + f(-x)$:
$$ f(x) + f(-x) = \left[(2+3a)x^2 + \dots + b\right] + \left[(2+3a)(-x)^2 + \dots + b\right] $$
The terms with odd powers of $x$ cancel out:
$$ f(x) + f(-x) = 2(2+3a)x^2 + 2b $$
Substitute $b = -1$:
$$ f(x) + f(-x) = 2(2+3a)x^2 – 2 $$

Comparing the coefficient of $x^2$ in both expressions for $f(x) + f(-x)$:
$$ 2(2+3a) = -\frac{2}{7} $$
$$ 2+3a = -\frac{1}{7} \Rightarrow 3a = -2 – \frac{1}{7} = -\frac{15}{7} $$
$$ a = -\frac{5}{7} $$

Now, we find the coefficient of $x$ in $f(x)$:
$$ \frac{a+2}{a-1} = \frac{-5/7 + 2}{-5/7 – 1} = \frac{9/7}{-12/7} = -\frac{3}{4} $$
Thus, the function is:
$$ f(x) = -\frac{1}{7}x^2 – \frac{3}{4}x – 1 $$
Taking the common denominator $-28$:
$$ f(x) = -\frac{1}{28}(4x^2 + 21x + 28) $$

We need to find $28\sum_{i=1}^{5} |f(i)|$.
Since $x > 0$ for $i=1$ to $5$, the term $(4x^2 + 21x + 28)$ is always positive.
$$ |f(i)| = \frac{1}{28}(4i^2 + 21i + 28) $$
$$ 28 |f(i)| = 4i^2 + 21i + 28 $$

Now summing from $i=1$ to $5$:
$$ \sum_{i=1}^{5} (4i^2 + 21i + 28) = 4\sum i^2 + 21\sum i + \sum 28 $$
$$ = 4\left(\frac{5 \cdot 6 \cdot 11}{6}\right) + 21\left(\frac{5 \cdot 6}{2}\right) + (28 \times 5) $$
$$ = 4(55) + 21(15) + 140 $$
$$ = 220 + 315 + 140 = 675 $$

Ans. (4)