J's answer to Lee Meimei's Secondary 3 E Maths Singapore question.
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Another A Math technique (optional, might be good to know)
c = 3 , k = 3, a + b = 1 (from Eric's above workings)
If the point (1,4) is taken as the maximum point, then it is tangent to the line y = 4. So it only meets that line at that one point.
We have y = 4 and y = ax² + bx + 3
4 = ax² + bx + 3
ax² + bx - 1 = 0
Since line meets the curve at only 1 point, there are real and equal roots. So discriminant (b² - 4ac) = 0
b² - 4a(-1) = 0
b² + 4a = 0 ①
Since a + b = 1, b = 1 - a ②
Sub ② into ①, (1 - a)² + 4a = 0
1 - 2a + a² + 4a = 0
a² + 2a + 1 = 0
(a + 1)² = 0
a + 1 = 0
a = -1
Then b = 1 - (-1) = 1 + 1 = 2
c = 3 , k = 3, a + b = 1 (from Eric's above workings)
If the point (1,4) is taken as the maximum point, then it is tangent to the line y = 4. So it only meets that line at that one point.
We have y = 4 and y = ax² + bx + 3
4 = ax² + bx + 3
ax² + bx - 1 = 0
Since line meets the curve at only 1 point, there are real and equal roots. So discriminant (b² - 4ac) = 0
b² - 4a(-1) = 0
b² + 4a = 0 ①
Since a + b = 1, b = 1 - a ②
Sub ② into ①, (1 - a)² + 4a = 0
1 - 2a + a² + 4a = 0
a² + 2a + 1 = 0
(a + 1)² = 0
a + 1 = 0
a = -1
Then b = 1 - (-1) = 1 + 1 = 2
Date Posted:
4 years ago
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