Given that f(x) = ax^2 + b x + c and f(x + y) = f(x) + f(y) + xy for

The detailed step-by-step solution, refer to the video

Refer screenshot for a quick understanding of the procedure.

from given function we can determine f(1) ; a+b+c =3
f(2) ; f(1+1)= f(1)+f(1)+1*1 ; 3+3+1 ; 7
f(3) ; f(2+1) ; f(2)+f(1)+2*1 ; 7+3+2 ; 12
f(4) ; f(3+1) ; f(3)+f(1)+3*1 ; 12+3+3; 18
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we see that for every f ∆ ; the value difference increases by 1 more than earlier
so for f(10) ; we can determine that the value f(10) = 75
IMO D


Given that f(x)=ax2+bx+cf(x)=ax2+bx+c and f(x+y)=f(x)+f(y)+xyf(x+y)=f(x)+f(y)+xy for all numbers x and y, where a, b, and c are constants. If a+b+c=3a+b+c=3, what is the value of f(10) ?

A. 60
B. 63
C. 72
D. 75
E. 88

Given that f(x)=ax2+bx+cf(x)=ax2+bx+c and f(x+y)=f(x)+f(y)+xyf(x+y)=f(x)+f(y)+xy for all numbers x and y, where a, b, and c are constants. If a+b+c=3a+b+c=3, what is the value of f(10) ?

Ans:
1. f(x)=ax2+bx+cf(x)=ax2+bx+c ---- Yields ---- f(0) = c
2. f(x+y)=f(x)+f(y)+xyf(x+y)=f(x)+f(y)+xy ---- Yields ---- f(0) = f(0+0) = f(0) + f(0) + 0*0 = 2c
From 1 and 2, c = 0

3. f(x)=ax2+bx+cf(x)=ax2+bx+c --- Yields f(2) = 4a+2b+c (from qn)
4. f(x+y)=f(x)+f(y)+xyf(x+y)=f(x)+f(y)+xy ---- Yields ---- f(2)=f(1+1)=2(a+b+c) + 1
5. With c = 0 and from 3 and 4, 4a+2b+c = 2(a+b+c) + 1 ==> 4a+2b=2a+2b+1 ==> 2a=1 ==> a = 0.5
6. With c=0 , a = 0.5 (from 5) and a+b+c =3, we get b=2.5

f(10) = 100a+10b+c = 100*0.5+10*2.5+0 = 75
Ans: Choice D

ALTERNATIVELY (I think the following is a shorter method)

We know that f(1)=3=a+b+c
f(2) = f(1+1)=f(1)+f(1)+1=3+3+1=7
f(3)=f(2+1)=7+3+2=12
f(5)=f(3+2)=12+7+6=25
f(10)=f(5+5)=f(5)+f(5)+5*5=25+25+25=75
Ans: D

Given that \(f(x)=ax^2+bx+c\) and \(f(x+y)=f(x)+f(y)+xy\) for all numbers x and y, where a, b, and c are constants.
If \(a+b+c=3\), what is the value of \(f(10)\) ?

\(f(x)=ax^2+bx+c \)--> \(f(1)=a+b+c =3\)

\(f(2)= f(1+1)= f(1) +f(1) +1*1= 3+3+1= 7\)

\(f(4)= f(2+2)= f(2)+f(2)+ 2*2= 7+7+4= 18\)

\(f(8)= f(4+4)= f(4)+ f(4) +4*4= 18+18+16= 52\)

\(f(10)= f(8+2)= f(8)+f(2)+ 2*8= 52+7+16= 75 \)

The answer is D.

Hi chondro48
f(10) = f(10+0) = f(10) + c. --> c = 0. We have a + b + c = 3 --> a + b = 3 (1)
f(2) = 4a + 2b + c = 4a + 2b. Also f(2) = f(1+1) = 2f(1) + 1. We have f(1) = a + b + c = 3 --> f(2) = 2.3 + 1 = 7 --> 4a + 2b = 7 (2)
Solve (1) and (2) we get a = 1/2 and b = 5/2
So f(10) = (1/2).100 + (5/2).10 + 0 = 50 + 25 = 75 (D)
Thank you for introducing this question to me.

I understand the solution, now that I see it. However, I don't have enough practice to be able to easily apply it on a different problem. How likely are we to run into a problem like this on the GMAT?

yes, these type of questions is quite frequent. For more practice, look for the bottom of the page for "SIMILAR TOPICS".

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