JEE Main 2025 — Functions Question with Solution

$$\begin{array}{rl}& f(x)=(3a+2)x^2+\left(\frac{a+2}{a-1}\right)x+b\\ & f\left(x+\frac{1}{2}\right)=f(x)+f(y)+1-\frac{2}{7}xy\text{.... (1)}\end{array}$$

In (1) Put $x=y=0\Rightarrow f(0)=2f(0)+1\Rightarrow f(0)=-1$

So, $\mathrm{f}(0)=0+0+\mathrm{b}=-1\Rightarrow \mathrm{b}=-1$

$$\begin{array}{rl}& \text{ In (1) Put }y=-x\Rightarrow f(0)=f(x)+f(-x)+1+\frac{2}{7}{x}^2\\ & -1=2(3a+2)x^2+2b+1+\frac{2}{7}{x}^2\end{array}$$

$$\begin{array}{rl}& -1=\left(2(3a+2)+\frac{2}{7}\right)x^2+1-2\\ & \Rightarrow 6a+4+\frac{2}{7}=0\\ & a=-\frac{5}{7}\end{array}$$

So $\mathrm{f}(\mathrm{x})=-\frac{1}{7}{\mathrm{x}}^2-\frac{3}{4}\mathrm{x}-1$

$$\Rightarrow |\mathrm{f}(\mathrm{x})|=\frac{1}{28}\left|4{\mathrm{x}}^2+21\mathrm{x}+28\right|$$

Now, $28{\sum }_{i=1}^5|f(i)|=28(|f(1)|+|f(2)|+\dots +|f(5)|)$

$28\cdot \frac{1}{28}\cdot 675=675$