JEE Main · 2025 · 28 Jan 2025 · Shift 1 · Sets Relations and Functions

$f(x)=(2+3a)x^2+\left(\frac{a+2}{a-1}\right)x+b$, and $f(x+y)=f(x)+f(y)+1-\frac{2}{7}xy$

Put $x=y=0$: $f(0)=2f(0)+1\Rightarrow f(0)=-1$, so $b=-1$.

Put $y=-x$: $f(0)=f(x)+f(-x)+1+\frac{2}{7}{x}^2$. Since the linear terms cancel in $f(x)+f(-x)$, this gives

$-1=2(2+3a)x^2+2b+1+\frac{2}{7}{x}^2$

which must hold for every $x$, so the coefficient of $x^2$ vanishes:

$2(2+3a)+\frac{2}{7}=0\Rightarrow 2+3a=-\frac{1}{7}\Rightarrow a=-\frac{5}{7}$

Then the quadratic coefficient is $2+3a=-\frac{1}{7}$, and the linear coefficient is

$\frac{a+2}{a-1}=\frac{9/7}{-12/7}=-\frac{3}{4}$

So $f(x)=-\frac{1}{7}{x}^2-\frac{3}{4}x-1=-\frac{1}{28}\left(4x^2+21x+28\right)$

Since $4x^2+21x+28>0$ for $x=1,\dots ,5$:

$28|f(i)|=4i^2+21i+28$

$28\sum _{i=1}^5|f(i)|=4\sum i^2+21\sum i+28(5)=4(55)+21(15)+140=220+315+140=675$