Test your knowledge of advanced level mathematics with this aptitude test. This test comprises Further Maths questions and answers from past JAMB and WAEC examinations.
Find correct to the nearest degree, the acute angle formed by the lines y = 2x + 5 and 2y = x - 6
76\(^∘\)
53\(^∘\)
37\(^∘\)
14\(^∘\)
Correct answer is C
tanθ | = | m1 - m2
1 + m1m2 |
y = 2x + 5
m1 = 2
2y = x - 6
y | = | 1
2 |
x | - | 3 |
m2 | = | 1
2 |
tanθ | = | 2 - \(\frac{1}{2}\)
1+2(\(\frac{1}{2}\)) |
tanθ = \(\frac{3}{2}\) ÷ (1+1)
tanθ = \(\frac{3}{2}\) ÷ 2
tanθ | = | 3
4 |
θ = \(tan^{-1} (\frac{3}{4})\)
θ = 36.87º
θ = 37º
1.63m
1.54m
1.52m
1.42m
Correct answer is B
for 20 students, mean = 1.67
μ | = | ∑fx
f |
∑fx = μf
∑fx = 20 × 1.67 = 33.4
for group 2
∑fx = 16 × 1.50 = 24
for group 3
∑fx = 14 × 1.40 = 19.6
33.4 + 24 + 19.6 = 77
∑f = 20+16+14 = 50
μ | = | 77
50 |
= | 1.54m |
\(\frac{1}{2}\) m/s
\(\frac{1}{3}\) m/s
2m/s
3m/s
Correct answer is A
m1u1 + m2u2 = (m1 + m2)v
m1 = 18kg, m2 = 6kg, u1 = 4ms-1, u2 = -10m/s
18(4) + 6(-10) = (18+6)v
72 - 60 = 24v
12 = 24v
v = \(\frac{1}{2}\) m/s
\(\frac{-1}{2}, 8\)
\(\frac{1}{2}, -8\)
\(\frac{-1}{2}, -8\)
\(\frac{1}{2}, 8\)
Correct answer is A
U1 = x - 4
U2 = x + 2
U3 = 3x + 1
\(\frac{u_2}{u_1} = \frac{u_3}{u_2}\)
\(\frac{x+2}{x-4} = \frac{3x+1}{x+2}\)
(x+2)(x+2) = (x-4)(3x+1)
x\(^2\) + 4x + 4 = 3x2 - 11x - 4
collecting like terms
2x\(^2\) - 15x - 8 =0
2x\(^2\) + x - 16x - 8 = 0
x(2x + 1) - 8(2x + 1) = 0
(x-8)(2x+1) = 0
x = (\(\frac{-1}{2}, 8\))
Find the coefficient of x\(^3\)y\(^2\) in the binomial expansion of (x-2y)\(^5\)
-80
10
40
90
Correct answer is C
x\(^3\)y\(^2\) in (x-2y)\(^5\)
n = 5, r = 3, p = x, q = -2y
5C\(_3\) * x\(^3\) -2y\(^2\)
5C\(_3\) = \(\frac{5!}{[5-3]!3!}\)
\(\frac{5*4*3!}{2! 3!}\) → \(\frac{5*4}{2}\)
5C\(_3\) = 10
: 5C\(_3\) * x\(^3\) -2y\(^2\) = 10 * x\(^3\) 4y\(^2\)
40x\(^3\)y\(^2\)
the coefficient is 40