Question

A real-valued function f satisfies the relation
f(x)f(y)= f(2xy +3) +3f(x+y)-3f(y)+6y, for all real numbers x and y.
Then the value of f(8) is___​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

A real-valued function f satisfies the relation

f(x) f(y) = f(2xy +3) + 3 f(x+y) - 3f(y) + 6y, for all real numbers x and y.

Now,

$\rm :\longmapsto\:f(x)f(y) = f(2xy +3) +3f(x+y)-3f(y)+6y$

On substituting x = 0 and y = 0, we get

$\rm :\longmapsto\:f(0)f(0) = f(3) +3f(0)-3f(0)+0$

$\bf :\longmapsto\:f(0) {}^{2} = f(3) - - - (1)$

Now, again Consider,

$\rm :\longmapsto\:f(x)f(y) = f(2xy +3) +3f(x+y)-3f(y)+6y$

On substituting y = 0, we get

$\rm :\longmapsto\:f(x)f(0) = f(3) +3f(x)-3f(0)+0$

$\rm :\longmapsto\:f(x)f(0) = {f(0)}^{2} +3f(x)-3f(0) \: \: \: \: \{ \: using \: (1) \: \}$

$\rm :\longmapsto\:f(x)f(0) - {f(0)}^{2} = 3f(x)-3f(0)$

$\rm :\longmapsto\:[f(x) - f(0)] f(0) = 3(f(x)-f(0))$

So,

$\bf\implies \:f(0) = 3 - - - (2) \: \: \: as \: f(x) \: \ne \: f(0)$

Now, again Consider,

$\rm :\longmapsto\:f(x)f(y) = f(2xy +3) +3f(x+y)-3f(y)+6y$

On substituting, x = 0 and y = 3, we get

$\rm :\longmapsto\:f(0)f(3) = f(3) +3f(3)-3f(3)+18$

$\rm :\longmapsto\:3f(3) = f(3) + 18 \: \: \: \: \: \: \{using \: 2) \}$

$\rm :\longmapsto\:3f(3)-f(3) = 18 \: \: \: \: \: \:$

$\rm :\longmapsto\:2f(3) = 18 \: \: \: \: \: \:$

$\bf :\longmapsto\:f(3) =9 - - - - (3)$

Now, Again Consider,

$\rm :\longmapsto\:f(x)f(y) = f(2xy +3) +3f(x+y)-3f(y)+6y$

On substituting x = 0 and y = 8, we get

$\rm :\longmapsto\:f(0)f(8) = f(3) +3f(8)-3f(8)+48$

$\rm :\longmapsto\:3f(8) = 9 +48$

$\rm :\longmapsto\:3f(8) = 57$

$\bf :\longmapsto\:f(8) = 19$

Hence,

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \purple{ \underbrace{ \boxed{ \blue{ \bf \: f(8) = 19 \: \: }}}}$

Answer 2

$f(x)f(y)=f(2xy+3)+3f(x+y)-3f(y)+6y$

$Putting\ x=y=0$

$We\ get,\ f(0)f(0)=f(3)+3f(0)-3f(0)+6y$

$Or,\ \ \ \ \ \ f^{2} (0)=f(3)$

$Putting\ y = 0$

$We\ get,\ f(x)f(0)=f(3)+3f(x)-3f(0)$

$Or,\ \ \ \ \ \ f(x)f(0)=f^{2}(0)+3f(x)-3f(0)$

$Either,\ f(x)-f(0)=0\ or,\ f(0)-3=0$

$Since\ all\ functions\ cannot\ have\ the\ same\ value, f(0)=3.$

$Putting \ x=0, y=3$

$We\ get,\ f(0)f(3)=f(3)+3f(3)-3f(3)+18$

$Or,\ \ \ \ \ \ f(0) f(3) = f(3) + 3f(3) -3f(3) +18 = f(3) +18$

$Or,\ \ \ \ \ \ 3 f(3) = f(3) + 18$

$Therefore, f(3) =9$

$Putting\ x=0, y=8$

$We\ get,\ f (0) f (8) = f (3) + 3f (8) - 3f (8) + 48$

$Or,\ \ \ \ \ \ 3f(8) = f(3) + f(8) -f(8) + 48$

$Or, \ \ \ \ \ \ 3 f(8) = 9 + 48.$

$Or,\ \ \ \ \ \ f(8) = 19$

$The\ question\ is\\''A\ real-valued\ function\ f\ satisfies\ the\ relation\\f(x)f(y) = f(2xy + 3) + 3f(x + y) - 3f(y) + 6y,\\ for\ all\ real\ numbers\ x\ and\ y, then\ the \ value\ of\ f(8)\ is''$

$Hence, the\ answer\ is\ 19$

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