How good are you with figures and formulas? Find out with these Mathematics past questions and answers. This Test is useful for both job aptitude test candidates and students preparing for JAMB, WAEC, NECO or Post UTME.
Integrate \(\int_{-1} ^{2} (2x^2 + x) \mathrm {d} x\)
\(4\frac{1}{2}\)
\(3\frac{1}{2}\)
\(7\frac{1}{2}\)
\(5\frac{1}{4}\)
Correct answer is C
\(\int_{-1} ^{2} (2x^2 + x) \mathrm {d} x\)
= \([\frac{2x^{2 + 1}}{3} + \frac{x^{1 + 1}}{2}]_{-1} ^{2}\)
= \([\frac{2x^{3}}{3} + \frac{x^{2}}{2}]_{-1} ^{2}\)
= \((\frac{2(2)^{3}}{3} + \frac{2^2}{2}) - (\frac{2(-1)^{3}}{3} + \frac{(-1)^{2}}{2})\)
= \((\frac{16}{3} + 2) - (\frac{-2}{3} + \frac{1}{2})\)
= \(\frac{22}{3} - (-\frac{1}{6})\)
= \(\frac{22}{3} + \frac{1}{6}\)
= \(\frac{15}{2}\)
= \(7\frac{1}{2}\)
The nth term of a sequence is given by 2\(^{2n - 1}\). Find the sum of the first four terms.
74
32
42
170
Correct answer is D
\(T_n = 2^{2n - 1}\)
\(T_1 = 2^{2(1) - 1} \)
= 2
\(T_2 = 2^{2(2) - 1}\)
= 8
\(T_3 = 2^{2(3) - 1}\)
= 32
\(T_4 = 2^{2(4) - 1}\)
= 128
\(T_1 + T_2 + T_3 + T_4 = 2 + 8 + 32 + 128\)
= 170
Solve the inequality: -7 \(\leq\) 9 - 8x < 16 - x
-1 \(\leq\) x \(\leq\) 2
-1 \(\leq\) x < 2
-1 < x < 2
-1 < x \(\leq\) 2
Correct answer is D
-7 \(\leq\) 9 - 8x < 16 - x
-7 \(\leq\) 9 - 8x and 9 - 8x < 16 - x
-7 - 9 \(\leq\) -8x and -8x + x < 16 - 9
-16 \(\leq\) -8x and -7x < 7
\(\therefore\) x \(\leq\) 2 and -1 < x
-1 < x \(\leq\) 2.
Solve for x in \(\frac{4x - 6}{3} \leq \frac{3 + 2x}{2}\)
\(x \leq 1\frac{1}{2}\)
\(x \leq \frac{21}{2}\)
\(x \geq \frac{21}{2}\)
\(x \geq 1\frac{1}{2}\)
Correct answer is B
\(\frac{4x - 6}{3} \leq \frac{3 + 2x}{2}\)
2(4x - 6) \(\leq\) 3(3 + 2x)
8x - 12 \(\leq\) 9 + 6x
8x - 6x \(\leq\) 9 + 12
2x \(\leq\) 21
\(x \leq \frac{21}{2}\)
If \(f(x) = 3x^3 + 4x^2 + x - 8\), what is the value of f(-2)?
-24
30
-18
-50
Correct answer is C
\(f(x) = 3x^3 + 4x^2 + x - 8\)
\(f(-2) = 3(-2)^3 + 4(-2)^2 + (-2) - 8\)
= \(-24 + 16 - 2 - 8\)
= -18