ધારો કે f: R rightarrow R એ f(x)=(2+3 a) x^2+ | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

$f(x)=(3 a+2) x^2+\left(\frac{a+2}{a-1}\right) x+b$
$f\left(x+y\right)=f(x)+f(y)+1-\frac{2}{7} x y$
In (1) Put $x=y=0 \Rightarrow f(0)=2 f(0)+1 \Rig

Vedclass · Accepted Answer

$675$

Vedclass · Answer

$f(x)=(3 a+2) x^2+\left(\frac{a+2}{a-1}ight) x+b$
$f\left(x+yight)=f(x)+f(y)+1-\frac{2}{7} x y$
In (1) Put $x=y=0 \Rightarrow f(0)=2 f(0)+1 \Rightarrow f(0)=-1$
So, $f(0)=0+0+b=-1 \Rightarrow b=-1$
$	ext { In (1) Put } y=-x \Rightarrow f(0)=f(x)+f(-x)+1+\frac{2}{7} x^2$
$-1=2(3 a+2) x^2+2 b+1+\frac{2}{7} x^2$
$-1=\left(2(3 a+2)+\frac{2}{7}ight) x^2+1-2$
$\Rightarrow 6 a+4+\frac{2}{7}=0$
$a=-\frac{5}{7}$
So $f(x)=-\frac{1}{7} x^2-\frac{3}{4} x-1$
$\Rightarrow|f(x)|=\frac{1}{28}\left|4 x^2+21 x+28ight|$
Now, $28 \sum_{i=1}^5|f(6)|=28(|f(1)|+f(2)+\ldots+f(5))$
$28 \cdot \frac{1}{28} \cdot 675=675$

Similar Questions

જો વિધેય $f(x)=\sec ^{-1}\left(\frac{2 x}{5 x+3}\right)$ નો પ્રદેશ $[\alpha, \beta) U (\gamma, \delta]$ હોય, તો $|3 \alpha+10(\beta+\gamma)+21 \delta|=..........$

જો $f(x + ay,\;x - ay) = axy$, તો $f(x,\;y) =$

વિધેય $f(x)={\left( {1 + \frac{1}{x}} \right)^x}$ હોય તો f (x) નો વિસ્તાર મેળવો

જો $f(x) = {\cos ^{ - 1}}\left( {\frac{{2x}}{{1 + {x^2}}}} \right) + {\sin ^{ - 1}}\left( {\frac{{1 - {x^2}}}{{1 + {x^2}}}} \right)$ તો $f(1) + f(2)$ ની કિમંત મેળવો.