JEE Main 2025 Jan 28 Shift 1, Mathematics Q5: Some Important Properties Of Functions

Use the functional equation to pin down the coefficients.

Step 1. Set $x=y=0$: $f(0)=2f(0)+1$, so $f(0)=-1$. From the formula $f(0)=b$, hence $b=-1$.

Step 2. Set $y=-x$: $f(0)=f(x)+f(-x)+1+\frac{2}{7}{x}^2$.

Now $f(x)+f(-x)=2(2+3a)x^2+2b$ (the linear term cancels). So

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

For this to hold for all $x$, the $x^2$ coefficient must vanish: $2(2+3a)+\frac{2}{7}=0\Rightarrow 6a+4+\frac{2}{7}=0\Rightarrow a=-\frac{5}{7}$.

Step 3. Now compute the coefficients of $f$:

$2+3a=2-\frac{15}{7}=-\frac{1}{7}$, and $\frac{a+2}{a-1}=\frac{9/7}{-12/7}=-\frac{9}{12}=-\frac{3}{4}$.

Hence $f(x)=-\frac{1}{7}{x}^2-\frac{3}{4}x-1$, and $|f(x)|=\frac{1}{28}|4x^2+21x+28|$.

Step 4. The quadratic $4x^2+21x+28$ is positive for all real $x$ (discriminant $=441-448<0$, leading coefficient positive), so the absolute value is unnecessary on positive integers.

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