WAEC Further Mathematics Past Questions & Answers - Page 10

dy/dx = 2x - 6

y = ∫ 2x - 6

y = \(\frac{2x^2}{2} - 6 + c\)
y = x\(^2\) - 6x + c
passes through (1,2)
2 = 1\(^2\) - 6(1) + c
2 = 1 - 6 + c
c = 7
y = x\(^2\) - 6x + c

y = x\(^2\) -  6x + 7

2x\(^2\) + 2y\(^2\) - x - 3y - 41

standard equation of circle
(x-a)\(^2\) + (x-b)\(^2\) = r\(^2\)
General form of equation of a circle.
x\(^2\) + y\(^2\) + 2gx + 2fy + c = 0
a = -g, b = -f., r2 = g2 + f2 - c
the centre of the circle is (a,b)
comparing the equation with the general form of equation of circle.
2x\(^2\) + 2y\(^2\) - x - 3y - 41

= x\(^2\) + y\(^2\) + 2gx + 2fy + c
2x\(^2\) + 2y\(^2\) - x - 3y - 41 = 0
divide through by 2

g = \(\frac{-1}{4}\) ; 2g = \(\frac{-1}{2}\)

f = \(\frac{-3}{4}\) ; 2f = \(\frac{-3}{2}\)

a = -g  → - \(\frac{-1}{4}\) ; = \(\frac{1}{4}\)

b = -f → - (\frac{-3}{4}\) = (\frac{3}{4}\)

therefore the centre is (\(\frac{1}{4}\), \(\frac{3}{4}\))

No explanation has been provided for this answer.

2x\(^2\) + 7x - 15 ≥ 0

2x\(^2\) -3x + 10x - 15 ≥ 0
x(2x - 3) + 5(2x - 3) ≥ 0
(x+5)(2x-3) ≥ 0
the points on x-axis where the graph ≥ 0

x ≤ -5 or x ≥ \(\frac{3}{2}\)

4sin\(^2\)θ + 1 = 2

4sin\(^2\)θ  = 2 - 1

4sin\(^2\)θ = 1

\(\sqrt sin^2θ\) = \(\sqrt \frac{1}{4}\)

sinθ = \(\frac{1}{2}\)

θ = \(sin^{-1} \frac{1}{2}\)

θ = 30º 0r 150º