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18 Feb 2020, 02:16
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
64% (02:53) correct
36%
(02:59)
wrong
based on 203
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Given that \(f(x) = ax^2 + b x + c\) 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) ?
A. 60
B. 63
C. 72
D. 75
E. 88
Are You Up For the Challenge: 700 Level Questions
Updated on: 20 Feb 2020, 05:29
Originally posted by freedom128 on 18 Feb 2020, 04:27.
Last edited by freedom128 on 20 Feb 2020, 05:29, edited 2 times in total.
Last edited by freedom128 on 20 Feb 2020, 05:29, edited 2 times in total.
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*f(1) = a + b + c = 3
*f(0) = c
*f(x + y) = f(x) + f(y) + xy
f(1 + 0) = f(1) + f(0)
3 = 3 + c --> c = 0 = f(0)
*f(1) = a+b=3 ...[1]
*f(x + y) = f(x) + f(y) + xy
f(1 - 1) = f(1) + f(-1) - 1
0 = 3 + a - b - 1
a-b=-2 ...[2]
[1]+[2]
a=0.5 and b=2.5
f(10) = a(100) + b(10) = 50+25 = 75
FINAL ANSWER IS (D)
Kindly give kudo to this explanation
*f(0) = c
*f(x + y) = f(x) + f(y) + xy
f(1 + 0) = f(1) + f(0)
3 = 3 + c --> c = 0 = f(0)
*f(1) = a+b=3 ...[1]
*f(x + y) = f(x) + f(y) + xy
f(1 - 1) = f(1) + f(-1) - 1
0 = 3 + a - b - 1
a-b=-2 ...[2]
[1]+[2]
a=0.5 and b=2.5
f(10) = a(100) + b(10) = 50+25 = 75
FINAL ANSWER IS (D)
Kindly give kudo to this explanation
18 Feb 2020, 02:35
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Given that \(f(x) = ax^2 + b x + c\) 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(1)= \(a*1^2 +b*1+c \)= a+b+c = 3
also f(2) = f(1+1) = f(1) +f(1) +1*1 =3+3+1= 7
f(3) = f(1+2)= f(1) + f(2) +1*2 = 3+7+2= 12
f(5) = f(3+2)= f(3) + f(2) +3*2 = 12+7+6= 25
f(10) = f(5+5)= f(5) + f(5) +5*5 = 25+25+25= 75
Answer = D
f(1)= \(a*1^2 +b*1+c \)= a+b+c = 3
also f(2) = f(1+1) = f(1) +f(1) +1*1 =3+3+1= 7
f(3) = f(1+2)= f(1) + f(2) +1*2 = 3+7+2= 12
f(5) = f(3+2)= f(3) + f(2) +3*2 = 12+7+6= 25
f(10) = f(5+5)= f(5) + f(5) +5*5 = 25+25+25= 75
Answer = D
