Question

he function $f$ satisfies\[f(x) + f(2x + y) + 5xy = f(3x - y) + 2x^2 + 1\]for all real numbers $x$, $y$. Determine the value of $f(10)$.

Answer 1

Answer:

We have f(10) = -49.

Explanation:

We have

$f(x)+f(2x+y)+5xy=f(3x-y)+2x^{2}+1$

Putting x = 0 and y = 0 we get,

$f(0)+f(0)+0=f(0)+0+1\\\\\therefore f(0)=1$

Now put x = 0 in the function equation we get

$f(0)+f(y)+0=f(-y)+1\\\\1+f(y)=f(-y)+1(\because f(0)=1)\\\\\Rightarrow f(-y)=f(y)$

Hence the given function is an even function

Now put x = 2y in the functional equation we get

$f(2y)+f(5y)+10y^{2}=f(5y)+8y^{2}+1\\\\f(2y)=-2y^{2}+1$

Now put y = x/2 we get

$f(x)=-2(\frac{x^{2}}{4})+1\\\\f(x)=\frac{-x^{2}}{2}+1$

Thus f(10) equals

$f(10)=\frac{-100}{2}+1$

thus f(10) = -49