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1.Relation and Function
hard
ધારો કે $f: R \rightarrow R$ એ $f(x)=(2+3 a) x^2+\left(\frac{a+2}{a-1}\right) x+ b , a \neq 1$ પ્રમાણે વ્યાખ્યાયિત એક વિધેય છ. જો $f(x+ y )=f(x)+f( y )+1-\frac{2}{7} x y$ હોય, તો $28 \sum_{i=1}^5|f(i)|$ નું મૂલ્ય _________ છે.
A$715$
B$735$
C$545$
D$675$
(JEE MAIN-2025)
Solution
$f(x)=(3 a+2) x^2+\left(\frac{a+2}{a-1}\right) x+b$
$f\left(x+y\right)=f(x)+f(y)+1-\frac{2}{7} x y$
In (1) Put $x=y=0 \Rightarrow f(0)=2 f(0)+1 \Rightarrow f(0)=-1$
So, $f(0)=0+0+b=-1 \Rightarrow b=-1$
$\text { In (1) Put } y=-x \Rightarrow f(0)=f(x)+f(-x)+1+\frac{2}{7} x^2$
$-1=2(3 a+2) x^2+2 b+1+\frac{2}{7} x^2$
$-1=\left(2(3 a+2)+\frac{2}{7}\right) x^2+1-2$
$\Rightarrow 6 a+4+\frac{2}{7}=0$
$a=-\frac{5}{7}$
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, $28 \sum_{i=1}^5|f(6)|=28(|f(1)|+f(2)+\ldots+f(5))$
$28 \cdot \frac{1}{28} \cdot 675=675$
$f\left(x+y\right)=f(x)+f(y)+1-\frac{2}{7} x y$
In (1) Put $x=y=0 \Rightarrow f(0)=2 f(0)+1 \Rightarrow f(0)=-1$
So, $f(0)=0+0+b=-1 \Rightarrow b=-1$
$\text { In (1) Put } y=-x \Rightarrow f(0)=f(x)+f(-x)+1+\frac{2}{7} x^2$
$-1=2(3 a+2) x^2+2 b+1+\frac{2}{7} x^2$
$-1=\left(2(3 a+2)+\frac{2}{7}\right) x^2+1-2$
$\Rightarrow 6 a+4+\frac{2}{7}=0$
$a=-\frac{5}{7}$
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, $28 \sum_{i=1}^5|f(6)|=28(|f(1)|+f(2)+\ldots+f(5))$
$28 \cdot \frac{1}{28} \cdot 675=675$
Standard 12
Mathematics
