F4 Maths

(1)(a)

∵ the graph of y = 2x² - 11x + (k + 14) intersects the x-axis at two points
∴ Δ = (-11)² - 4(2)(k + 14) > 0
=> 121 - 8k - 112 > 0 => - 8k > - 9 => k < 9/8
since k is a positive integer, k = 1

∴ y = 2x² - 11x + 15, solve 2x² - 11x + 15 = 0 => x = 3 or -5/2
thus the coordinates of A and B are (3, 0) and (-5/2, 0)

(1)(b)

the x coordinate of M is [3 + (-5/2)]/2 = 1/4
the y coordinate of M is (0 + 0)/2 = 0
thus the coordinates of M are (1/4, 0)


(2)(a)

substitute x = 0 into y = - x² + 4x + k - 4, we have y = k - 4
thus, the coordinates of C are (0, k - 4)

(2)(b)(i)

since a and b are the roots of - x² + 4x + k - 4 = 0,
a + b = sum of roots of "- x² + 4x + k - 4 = 0" = - (4)/(-1) = 4

(2)(b)(ii)

since a and b are the roots of - x² + 4x + k - 4 = 0,
ab = product of roots of "- x² + 4x + k - 4 = 0" = (k - 4)/(-1) = 4 - k

(2)(c)

(b - a)² = b² - 2ab + a² = a² + 2ab + b² - 4ab = (a + b)² - 4ab = (4)² - 4(4 - k) = 4k
the area of ΔABC = (b - a)[0 - (k - 4)]/2 (since b > a and y - int = k - 4 < 0)
= [√ (b - a)²](4 - k)/2 = [√(4k)](4 - k) / 2 = (4 - k)√k