Let f: R rightarrow R be a function defined by f(x)=(2+3 a) x 2 + (a+2/a-1 ) x+b, a neq 1 . If f(x+y)=f(x)+f(y)+1-2/7 x y , then the value of 28 sum i=1 5 f(i) is:

Mathematics · Functions

JEE Main 2025 — 28 January, Morning Shift — Question 5

Let $f: R \rightarrow R$ be a function defined by $f(x)=(2+3 a) x^{2}+\left(\frac{a+2}{a-1}\right) x+b, a \neq 1$ . If

$f(x+y)=f(x)+f(y)+1-\frac{2}{7} x y$ , then the value of $28 \sum_{i=1}^{5}|f(i)|$ is:

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

$f(x+y)=f(x)+f(y)+1-\frac{2}{7} x y$

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

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

In (1) Put $y=-x \Rightarrow f(0)=f(x)+f(-x)+1+\frac{2}{7} x^{2}$

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

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

$\Rightarrow 6 \mathrm{a}+4+\frac{2}{7}=0$

$a=-\frac{5}{7}$ So $f(x)=-\frac{1}{7} x^{2}-\frac{3}{4} x-1$

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

Now, $28 \sum_{i=1}^{5}|f(6)|=28(|f(1)|+f(2)+\ldots+f(5))$

Answer key and solution verified before publishing.

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