2007 - JAMB Mathematics Past Questions and Answers - page 8

If the line 3y = 4x – 1 is parallel t[ line qy = x + 3
Implies gradient of 3y = 4x – 1
Y = ⁴/₃x - ¹/₃
∴Gradient = ⁴/₃
Gradient of line qy = x + 3
Y = ¹/_qx + ³/_q
∴ Gradient = ¹/_q
⁴/₃ = ¹/_q
4q = 3
Q = ³/₄

x = 30^o (alternate ∠s b/c PQ//Sr)
Y = 50^o (vertical opposite ∠s)
r = 30^o (alternate ∠s b/c PQ//SR)
But = y + q + t = 180^o (∠s on a straight line)
50 + q + 30 = 180
q + 80 = 180
q = 180 -80
q = 100^o

Volume of bowl \(\frac{2}{3}\pi r^2\\ 718\frac{2}{3}=\frac{2}{3}\pi r^2\\ \frac{2156}{3}=\frac{2}{3} \times \frac{22}{7} \times r^3
∴r^3 = \frac{2156 \times 3 \times 7}{3 \times 2 \times 22}\\ r^3 = 343\\ r = \sqrt[3]{343}\\ r= 7.0cm\)

If the point of intersection is (1, -2), it implies that x = 1 and y = -2 when the two equation are solved simultaneously.
∴ substitute x = 1 and y = -2 in any of the equations
2y - k x + 2 = 0
2(-2) - k(1) + 2 = 0
-4 - k + 2 = 0
-4 + 2 = k
-2 = k

\(Tan 30 = \frac{h}{40}\\ \frac{1}{\sqrt{3}}=\frac{h}{40}\\ h\sqrt{3}=40\\ h = \frac{40}{\sqrt{3}}\\ h = \frac{40}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}\\ h = \frac{40\sqrt{3}}{3}\)

Locus of point P equidistant from y - 5 = 0 and y - 3 = 0 is y = 4 i.e y - 4 = 0

(1-k+2) / 2 = - k and -4 + k + 1 = k
3-k = -2k and -3 + k = 2k
K = -3 and k = -3

Each Exterior ∠ = ³⁶⁰/_n
= ³⁶⁰/₈
= 45^o

\(\frac{tan 60^o - tan 30^o}{tan 60^o + tan 30^o}= \frac{\sqrt{3}-\frac{1}{\sqrt{3}}}{\sqrt{3}+\frac{1}{\sqrt{3}}}\\ =\left(\frac{\sqrt{3}\sqrt{3}-1}{\sqrt{3}}\right)\div \left(\frac{\sqrt{3}\sqrt{3}+1}{\sqrt{3}}\right)\\ =\frac{(3-1)}{(3+1)}\\ =\frac{2}{4}=\frac{1}{2}\)