Correct Answer: D. 4

Explanation

An equation can be written as \(x^{2} - (\alpha + \beta)x + (\alpha\beta) = 0\)

Making the coefficient of \(x^{2}\) = 1 in the given equation, we have

\(x^{2} + \frac{5}{2}x + \frac{n}{2} = 0\)

Comparing, we have \(\alpha\beta = \frac{n}{2} = 2 \implies n = 4\)

Explanation

Acceleration, \(a = (4 - 2t) ms^{-2}\)

\(\therefore v = \int a \mathrm {d} t \)

= \(\int (4 - 2t) \mathrm {d} t = (4t - t^{2} + c)\)

When t = 0, v = 5 m/s

\(5 = 0 - 0 + c \implies c = 5\)

\(\therefore v = (4t - t^{2} + 5) ms^{-1}\)

Distance, \(s = \int (4t - t^{2} + 5) \mathrm {d} t\)

= \(2t^{2} - \frac{t^{3}}{3} + 5t + c\)

When t = 0, s = 18 m.

\(0 - 0 + 0 + c = 18 \implies c = 18\)

\(\therefore s = 2t^{2} - \frac{t^{3}}{3} + 5t + 18\)

(a) Velocity is greatest when a = 0.

\(4 - 2t = 0 \implies t = 2s\)

(b)(i) Particle is momentarily at rest when v = 0 m/s.

\(4t - t^{2} + 5 = 0 \implies t^{2} - 4t - 5 = 0\)

\(t(t - 5) + 1(t - 5) = 0 \implies t = -1 ; t = 5\)

Since time cannot be negative, t = 5s.

(ii) Distance when t = 5s.

\(s = 2(5^{2}) - \frac{1}{3} (5^{3}) + 5(5) + 18\)

= \(50 - \frac{125}{3} + 25 + 18\)

= \(\frac{154}{3}\)

= \(51\frac{1}{3} m\)

Explanation

\(f(x) = 6x^{3} + 13x^{2} + 2x - 5\)

f(-1) = 0 implies (x + 1) is a factor.

Using the method of long division, you find the other factors.

\(\frac{6x^{3} + 13x^{2} + 2x - 5}{x + 1} = 6x^{2} + 7x - 5\) (check).

\(6x^{2} + 7x - 5 = 6x^{2} - 3x + 10x - 5\)

\(3x(2x - 1) + 5(2x - 1)\)

Hence, the factors of f(x) are (x + 1), (3x + 5) and (2x - 1).

Correct Answer: C. Z

Explanation

From the table, x * z = x, y * z = y, z * z = z and w * z = w

∴ z is the identity element

Explanation

3cos2x - sinx = 0

cos 2A = 1 – 2 sin\(^2\)A

3(1 - 2sin\(^2\)x) - sinx = 0

3 - 6sin\(^2\)x -sinx = 0

let sinx = p

3 - 6p\(^2\) -p = 0

-6p2 - p +3 = 0

factors of p in the eqn:

p = 0.7953 or 0.6287

when p = 0.7953

p = sinx

x = sin\(^{-1}\) p

x = sin\(^{-1}\)(0.7953)

x = 52.68º

when p = 0.6287

p = sinx

x = sin\(^{-1}\) p

x = sin\(^{-1}\)(0.6287)

x = 38.95º

Explanation

Gradient of BC = \(\frac{8 + 2}{1 - 4} = -\frac{10}{3}\)

Gradient of DC = \(\frac{8 - y}{1 - x}\)

BDC is a straight line , so

Gradient of BC = Gradient of DC.

\(\frac{10}{-3} = \frac{8 - y}{1 - x}\)

\(-3(8 - y) = 10(1 - x) \implies 3y - 24 = 10 - 10x\)

\(3y + 10x = 34\)

(b) \(\overrightarrow{BC} = \overrightarrow{OC} - \overrightarrow{OB}\)

= \((i + 8j) - (4i - 2j)\)

= \(-3i + 10j\)

\(|\overrightarrow{BC}| = \sqrt{(-3)^{2} + (10)^{2}}\)

= \(\sqrt{109}\)

Unit vector in the direction of BC = \(\frac{(-3i + 10j)}{\sqrt{109}}\)

Correct Answer: D. 30

Explanation

54 x 40 = 35 x 10 + 60F

160 = 350 + 60F

F = \(\frac{1810}{60}\)

= 30N

Correct Answer: C. 25 - 29

Explanation

The upper quartile occurs at the \(\frac{3N}{4}\) position.

= \(\frac{3 \times 50}{4} = 37.5\)

This occurs in the class 25 - 29.

Correct Answer: D. -4

Explanation

\(\begin{vmatrix} 4 & x \\ 5 & 3 \end{vmatrix} = 12 - 5x = 32\)

\(5x = 12 - 32 = -20\)

\(x = -4\)

Correct Answer: B. \(x< \frac{-1}{2}, x > 3\)