2019 - JAMB Mathematics Past Questions and Answers - page 3
If given two points A(3, 12) and B(5, 22) on a x-y plane. Find the equation of the straight line with intercept at 2.
y = 5x + 2
y = 5x + 3
y = 12x + 2
y = 22x + 3
The equation of a straight line is given as: \(y = mx + b\)
Where:
m = the slope of the line
b = intercept
Given the points A(3, 12) and B(5, 22),
slope = \(\frac{22 - 12}{5 - 3}\)
= \(\frac{10}{2}\) = 5
Hence,
the equation of the line is \(y = 5x + 2\).
If P(2, m) is the midpoint of the line joining Q(m, n) and R(n, -4), find the values of m and n.
m = 0, n = 4
m = 4, n = 0
m = 2, n = 2
m = -2, n = 4
Q(m, n) and R(n, -4)
Midpoint of the line: P(2, m)
\(\implies (\frac{m + n}{2}, \frac{n - 4}{2}) = (2, m)\)
\(m + n = 2 \times 2 \implies m + n = 4 ... (i)\)
\(n - 4 = 2 \times m \implies n - 4 = 2m ... (ii)\)
From (i) and (ii),
m = 0 and n = 4.
If \(\begin{vmatrix} 2 & -4 \ x & 9 \end{vmatrix} = 58\), find the value of x.
\(\begin{vmatrix} 2 & -4 \ x & 9 \end{vmatrix} = 58\)
\(\implies (2 \times 9) - (-4 \times x) = 58\)
\(18 + 4x = 58 \implies 4x = 58 - 18 = 40\)
\(x = 10\)
If \(y = 6x^3 + 2x^{-2} - x^{-3}\), find \(\frac{\mathrm d y}{\mathrm d x}\).
\(y = 6x^3 + 2x^{-2} - x^{-3}\)
\(\frac{\mathrm d y}{\mathrm d x} = 18x^2 - 4x^{-3} + 3x^{-4}\)
\(\frac{d}{dx} [\log (4x^3 - 2x)]\) is equal to
\(\frac{d}{dx} [\log (4x^3 - 2x)]\) ... (1)
Let u = 4x\(^3\) - 2x.
\(\frac{\mathrm d}{\mathrm d x} (\log (4x^3 - 2x)) = (\frac{\mathrm d}{\mathrm d u})(\frac{\mathrm d u}{\mathrm d x})\)
\(\frac{\mathrm d}{\mathrm d u} (\log u)\) = \(\frac{1}{u}\)
\(\frac{\mathrm d u}{\mathrm d x} = 12x^2 - 2\)
\(\therefore \frac{d}{dx} [\log (4x^3 - 2x)] = \frac{12x^2 - 2}{u}\)
= \(\frac{12x^2 - 2}{4x^3 - 2x}\)
If \(f(x) = 3x^3 + 4x^2 + x - 8\), what is the value of f(-2)?
\(f(x) = 3x^3 + 4x^2 + x - 8\)
\(f(-2) = 3(-2)^3 + 4(-2)^2 + (-2) - 8\)
= \(-24 + 16 - 2 - 8\)
= -18
Solve for x in \(\frac{4x - 6}{3} \leq \frac{3 + 2x}{2}\)
\(\frac{4x - 6}{3} \leq \frac{3 + 2x}{2}\)
2(4x - 6) \(\leq\) 3(3 + 2x)
8x - 12 \(\leq\) 9 + 6x
8x - 6x \(\leq\) 9 + 12
2x \(\leq\) 21
\(x \leq \frac{21}{2}\)
Solve the inequality: -7 \(\leq\) 9 - 8x < 16 - x
-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.
The nth term of a sequence is given by 2\(^{2n - 1}\). Find the sum of the first four terms.
\(T_n = 2^{2n - 1}\)
\(T_1 = 2^{2(1) - 1} \)
= 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
Integrate \(\int_{-1} ^{2} (2x^2 + x) \mathrm {d} x\)
\(\int_{-1} ^{2} (2x^2 + x) \mathrm {d} x\)
= \([\frac{2x^{2 + 1}}{3} + \frac{x^{1 + 1}}{2}]_{-1} ^{2}\)
= \([\frac{2x^{3}}{3} + \frac{x^{2}}{2}]_{-1} ^{2}\)
= \((\frac{2(2)^{3}}{3} + \frac{2^2}{2}) - (\frac{2(-1)^{3}}{3} + \frac{(-1)^{2}}{2})\)
= \((\frac{16}{3} + 2) - (\frac{-2}{3} + \frac{1}{2})\)
= \(\frac{22}{3} - (-\frac{1}{6})\)
= \(\frac{22}{3} + \frac{1}{6}\)
= \(\frac{15}{2}\)
= \(7\frac{1}{2}\)