2019 - JAMB Mathematics Past Questions and Answers - page 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 \(y = 5x + 2\).

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.

\(\begin{vmatrix} 2 & -4 \ x & 9 \end{vmatrix} = 58\)

\(\implies (2 \times 9) - (-4 \times x) = 58\)

\(18 + 4x = 58 \implies 4x = 58 - 18 = 40\)

\(x = 10\)

\(y = 6x^3 + 2x^{-2} - x^{-3}\)

\(\frac{\mathrm d y}{\mathrm d x} = 18x^2 - 4x^{-3} + 3x^{-4}\)

\(\frac{d}{dx} [\log (4x^3 - 2x)]\) ... (1)

Let u = 4x\(^3\) - 2x.

\(\frac{\mathrm d}{\mathrm d x} (\log (4x^3 - 2x)) = (\frac{\mathrm d}{\mathrm d u})(\frac{\mathrm d u}{\mathrm d x})\)

\(\frac{\mathrm d}{\mathrm d u} (\log u)\) = \(\frac{1}{u}\)

\(\frac{\mathrm d u}{\mathrm d x} = 12x^2 - 2\)

\(\therefore \frac{d}{dx} [\log (4x^3 - 2x)] = \frac{12x^2 - 2}{u}\)

= \(\frac{12x^2 - 2}{4x^3 - 2x}\)

\(f(x) = 3x^3 + 4x^2 + x - 8\)

\(f(-2) = 3(-2)^3 + 4(-2)^2 + (-2) - 8\)

= \(-24 + 16 - 2 - 8\)

= -18

\(\frac{4x - 6}{3} \leq \frac{3 + 2x}{2}\)

2(4x - 6) \(\leq\) 3(3 + 2x)

8x - 12 \(\leq\) 9 + 6x

8x - 6x \(\leq\) 9 + 12

2x \(\leq\) 21

\(x \leq \frac{21}{2}\)

-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.

\(T_n = 2^{2n - 1}\)

\(T_1 = 2^{2(1) - 1} \)

= 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

\(\int_{-1} ^{2} (2x^2 + x) \mathrm {d} x\)

= \([\frac{2x^{2 + 1}}{3} + \frac{x^{1 + 1}}{2}]_{-1} ^{2}\)

= \([\frac{2x^{3}}{3} + \frac{x^{2}}{2}]_{-1} ^{2}\)

= \((\frac{2(2)^{3}}{3} + \frac{2^2}{2}) - (\frac{2(-1)^{3}}{3} + \frac{(-1)^{2}}{2})\)

= \((\frac{16}{3} + 2) - (\frac{-2}{3} + \frac{1}{2})\)

= \(\frac{22}{3} - (-\frac{1}{6})\)

= \(\frac{22}{3} + \frac{1}{6}\)

= \(\frac{15}{2}\)

= \(7\frac{1}{2}\)