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 the fourth term of the binomial expansion of (x−k)5 in descending powers of x.<
10x3k2
5x3k2
−5x2k3
−10x2k3
Correct answer is D
(x−k)5=5C0x5(−k)0+5C1x4(−k)1+...
The fourth term in the expansion = ^{5}C_{4 - 1}(x)^{5 - 3}(-k)^{3 = 10 \times x^{2} \times -k^{3}
= −10x2k3
2xx−3,x≠3
2xx+3,x≠−3
3xx−3,x≠3
3xx+3,x≠−3
Correct answer is A
From the rules of binary operation, x∗e=x
⟹x∗e=3x+3e−xe=x
3e−xe=x−3x=−2x
e=2xx−3,x≠3
Simplify \frac{\tan 80° - \tan 20°}{1 + \tan 80° \tan 20°}
3\sqrt{2}
2\sqrt{3}
\sqrt{3}
\sqrt{2}
Correct answer is C
\tan (x - y) = \frac{\tan x - \tan y}{1 + \tan x \tan y}
\implies \frac{\tan 80 - \tan 20}{1 + \tan 80 \tan 20} = \tan (80 - 20) = \tan 60°
\tan 60 = \frac{\sin 60}{\cos 60} = \frac{\sqrt{3}}{2} ÷ \frac{1}{2}
= \sqrt{3}
Simplify \frac{\sqrt{3}}{\sqrt{3} - 1} + \frac{\sqrt{3}}{\sqrt{3} +1}
\frac{1}{2}
\frac{1}{2}\sqrt{3}
3
2\sqrt{3}
Correct answer is C
\frac{\sqrt{3}}{\sqrt{3} - 1} + \frac{\sqrt{3}}{\sqrt{3} + 1}
= \frac{\sqrt{3}(\sqrt{3} + 1) + \sqrt{3}(\sqrt{3} - 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)}
= \frac{6}{3 - 1}
= 3
-4
0
4
12
Correct answer is D
Gradient = \frac{y_{1} - y_{2}}{x_{1} - x_{2}}
\frac{1}{2} = \frac{5 - 9}{4 - x}
\frac{1}{2} = \frac{-4}{4 - x} \implies -8 = 4 - x
x = 4 + 8 = 12