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.
63.616
63.167
62.628
62.629
Correct answer is B
Put 1 + (0.998) = 1 + x;
x = (0.998)
Hence; 1 + 6(0.998) + 15(0.998)(0.998)\(^2\) + 20(0.998)\(^3\) + 15(0.998)\(^4\) + 6(0.998)\(^5\) + (0.998)\(^6\)
= 1 + 5.988 + 14.790 + 19.880 + 14.880 + 5.940 + 0.990
≈ 63.167
g(x) = 2x + 3 and f(x) = 3x\(^2\) - 2x + 4 find f {g (-3)}.
37
1
-3
-179
Correct answer is A
g(-3) = 2(-3) + 3 = -6 + 3 = -3
F(-3) = 3 (-3)\(^2\) - 2(-3) + 4
= 27 + 6 + 4 = 37
Solve (\(\frac{1}{9}\))\(^{x + 2}\) = 243\(^{x - 2}\)
\(\frac{7}{5}\)
\(\frac{6}{7}\)
\(\frac{-7}{6}\)
\(\frac{-6}{7}\)
Correct answer is A
(\(\frac{1}{9}\))\(^{x + 2}\) = 243\(^{x - 2}\)
3\(^{-2(x + 2)}\) = 3\(^{5(x - 2)}\);
-2(x+2) = 5(x - 2)
-2x -4 = 5x - 10
-7x = -6
x = \(\frac{6}{7}\)
Simplify \(\frac{1}{3}\) log8 + \(\frac{1}{3}\) log 64 - 2 log6
log \(\frac{2}{7}\)
log 2
log \(\frac{2}{9}\)
log 9
Correct answer is C
\(\frac{1}{3}\) log8 + \(\frac{1}{3}\) log 64 - 2 log6
= log 2 + log 4 - log 36
= log \(\frac{8}{36}\)
= log \(\frac{2}{9}\)
\(\begin{pmatrix} 1 & 6 \\ 1 & 18 \end{pmatrix}\)
\(\begin{pmatrix} -1 & -6 \\ 1 & 18 \end{pmatrix}\)
\(\begin{pmatrix} 1 & 6 \\ -1 & -18 \end{pmatrix}\)
\(\begin{pmatrix} -1 & -6 \\ -1 & -18 \end{pmatrix}\)
Correct answer is B
3M = 3 \(\begin{pmatrix} 3 & 2 \\ -1 & 4 \end{pmatrix}\)
= \(\begin{pmatrix} 9 & 6 \\ -3 & 12 \end{pmatrix}\)
2N = 2 \(\begin{pmatrix} 5 & 6 \\ -2 & -3 \end{pmatrix}\)
= \(\begin{pmatrix} 10 & 12 \\ -4 & -6 \end{pmatrix}\)
(3M - 2N) = \(\begin{pmatrix} 9 & 6 \\ -3 & 12 \end{pmatrix}\) - \(\begin{pmatrix} 10 & 12 \\ -4 & -6 \end{pmatrix}\)