Correct Answer: D. 2

Explanation

\(S_{n} = \frac{n}{2}(2a + (n - 1)d \)

\(\frac{n}{2}(2a + (n - 1) d = n^{2} + 2n\)

\(n(2a + (n - 1) d = 2n^{2} + 4n\)

\(2an + n^{2}d - nd = 2n^{2} + 4n\)

Comparing the equations, d = 2 (coefficient of \(n^{2}\)).

Correct Answer: C. 8s

Explanation

F = ma

\(200 = 20a \implies a = 10m/s^{2}\)

\(s = ut + \frac{1}{2}at^{2}\)

\(320 = (0 \times t) + \frac{1}{2} 10t^{2}\)

\(320 = 5t^{2} \implies t^{2} = 64\)

\(t = 8secs\)

Explanation

(5, 2), (-4, k), (2, 1)

A straight line has constant gradient.

\(\frac{k - 2}{-4 - 5} = \frac{1 - k}{2 - (-4)}\)

\(\frac{k - 2}{-9} = \frac{1 - k}{6}\)

\(-9(1 - k) = 6(k - 2) \implies -9 + 9k = 6k - 12\)

\(9k - 6k = -12 + 9 \implies 3k = -3\)

\(k = -1\)

Explanation

Correct Answer: A. 2

Explanation

\(\frac{3x + 4}{(x - 2)(x + 3)}≡\frac{P}{x + 3}+\frac{Q}{x - 2}\)

\(\frac{3x + 4}{(x - 2)(x + 3)}=\frac{P(x-2)+Q(x+3)}{(x-2)(x+3)}\)

=\(3x+4=P(x-2)+Q(x+3)\)

=\(3x+4=Px-2P+Qx+3Q\)

=\(3x+4=Px+Qx-2P+3Q\)

=\(3x+4=(P+Q)x-2P+3Q\)

Equating x and the constants

P+Q=3------(i)

2P+3Q=4-------(ii)

From equation (i)

P=3-Q------(iii)

Substitute (3-Q) for P inequation (ii)

=-2(3-Q)+3Q=4

=-6+2Q+3Q=4

=-6+5Q=4

=5Q=4+6

=5Q=10

∴Q=\(\frac{10}{5}=2\)

Correct Answer: B. -4

Explanation

If (x + 1) is a factor, then f(-1) = 0.

\((-1)^{3} + p(-1)^{2} + (-1) + 6 = 0\)

\(-1 + p - 1 + 6 = 0 \implies p + 4 = 0\)

\(p = -4\)

Correct Answer: B. \(3^{\frac{1}{2}}\)

Explanation

\(\log_{3} x = \log_{9} 3 \implies \log_{3} x = \log_{9} 9^{\frac{1}{2}} = \frac{1}{2}\log_{9} 9\)

\(\log_{3} x = \frac{1}{2} \)

\(\therefore x = 3^{\frac{1}{2}}\)

Correct Answer: B. -4

Explanation

Given x = (-1, 5) for the equation \(x^{2} + kx - 5 = 0\)

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

\((x + 1)(x - 5) = 0\), expanding,

\(x^{2} - 5x + x - 5 = 0 \therefore x^{2} - 4x - 5 = 0\)

\(\therefore\) k = -4.

Explanation

(a) \(m=2kg;u=(3i+2j)ms^{-1};v=(15i-4j)ms^{-1};t=4s;a=?\)

\(a=\frac{v - u}{t}=\frac{(15i - 4j) - (3i + 2j)}{4}\)

\(a=\frac{15i - 4j - 3i - 2j}{4}=\frac{12i - 6j}{4}\)

\(∴a=3i-\frac{3}{2}j ms^{-2}\)

(b) \(m=2kg;a=3i-\frac{3}{2} j\)

\(F=ma=2(3i-\frac{3}{2}j)\)

F = 6i - 3j

\(|F| = √(6^2 + (-3)^2)\)

\(|F| = √(36 + 9) = √45\)

\(∴ |F| = 3√5 N = 6.71 N\)

(c) \(u=3i+2jms^{-1};a=3i-\frac{3}{2}j ms^{-2};t=8s\)

v = u + at

\(v=(3i+2j)+8(3i-\frac{3}{2} j)\)

\(v = 3i + 2j + 24i - 12j\)

\(v = 27i - 10j\)

\(|v| = √(27^2 + (-10)^2)\)

\(|v| = √(729 + 100) = √829\)

\(∴ |v| = 28.792 ms^{-1} (to 3d.p)\)

Explanation

\(\overline{MN}\) = 8i + 3j

\(\overline{MO}\) = 14i - 5j

Consider ?MON

\(\overline{MN}\) + \(\overline{NO}\) = \(\overline{MO}\)

\(\overline{NO}\) = \(\overline{MO}\) - \(\overline{MN}\)

\(\overline{NO}\) = 14i - 5j - (8i + 3j) = 6i - 8j

Since P is the midpoint of \(\overline{NO}\), then

\(\overline{NP} =\frac{1}{2}( \overline{NO} )\)

\(=\frac{1}{2}(6i - 8j) = 3i - 4j\)

Consider ?MNP

\(\overline{MN}\) + \(\overline{NP}\) = \(\overline{MP}\)

(8i + 3j) + (3i - 4j) = \(\overline{MP}\)

\(\overline{MP}\) = 8i + 3i + 3j - 4j

∴ \(\overline{MP}\) = 11i - j