The 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). ???

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Answer 1 · 2017-05-29T13:11:18

# -49.#

Given that, #(ast) f(x)+f(2x+y)+5xy=f(3x-y)+2x^2+1;x,y in RR.#

#(ast), &, x=y=0 rArr f(0)+f(0)+0=f(0)+0+1.#

# rArr f(0)=1.................(1)#

#(ast), &, x=0, y=y rArr f(0)+f(y)+0=f(-y)+0+1.#

#rArr 1+f(y)=f(-y)+1,......[because, (1)]#

#rArr f(y)=f(-y)," &, since, this holds "AA y in RR, #

# f" is even...........................(2)."#

Finally, #(ast), &, x=2y rArr f(2y)+f(5y)+10y^2=f(5y)+8y^2+1.#

#rArr f(2y)=-2y^2+1," or, switching to x from y, we have,"#

# f(x)=-2(x/2)^2+1=-1/2x^2+1.#

Accordingly, the Desired Value, #f(10)=-49.#

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