Correct Answer: D. \(-\frac{3}{2}\)

Explanation

\(\log_{0.25} 8 = x\)

\(8 = 0.25^{x}\)

\(2^{3} = (2^{-2})^{x} \implies 3 = -2x\)

\(x = -\frac{3}{2}\)

Explanation

The gradient of two parallel lines are equal.

m1 = m2

6i + 8j is parallel to →OP

m = \(\frac{y_2 - y_1}{x_2 - x_1}\)

gradient(m) of 6i + 8j

m = \(\frac{8}{6}\) = \(\frac{4}{3}\)

\(\frac{y}{x}\) = \(\frac{4}{3}\)

3y - 4x =0

|OP| = 80

|OP| = √(x\(^2\) + y\(^2\)) = 80

x\(^2\) + y\(^2\) = 80\(^2\)

x\(^2\) + y\(^2\) = 6400

\(y = \frac{4x}{3}\)

\(x^2 + ({\frac{4x}{3}})^2\) = 6400

\(x^2 + \frac{16x^2}{9}\) = 6400

multiply through by 9

9x\(^2\) + 16x2 = 57600

25x\(^2\) = 57600

x\(^2\) = \(\frac{57600}{25}\)

x\(^2\) = 2304

x = √2304

x = 48

\(y = \frac{4x}{3}\) → \(y = \frac{4*48}{3}\)

y = 64

→OP =xi + yj

→OP = 48i + 64j

8i - 6j is parallel to →OQ

gradient of 8i - 6j = \(\frac{-6}{8}\)

\(\frac{y}{x}\) = \(\frac{-6}{8}\)

y = \(\frac{-6x}{8}\)

\(x^2 + ({\frac{-6x}{8}})^2\) = 14400

\(x^2 + \frac{36x^2}{64}\) = 14400

multiply through by 64

64x\(^2\) + 36x\(^2\) = 921600

100x\(^2\) = 921600

divide both sides by 100

x\(^2\) = 9216

find the square root of both sides

x = √9216

x = 96

y = \(\frac{-6x}{8}\)

y = \(\frac{-6[96]}{8}\)

y = -72

→OQ = xi + yj

→OQ = 96i - 72j

→PQ = (96i - 72j) - (48i + 64j)

→PQ = (96-48)i + (-72-64)j

→PQ = 48i - 136j

|→PQ| = √(48\(^2\) + (-136)\(^2\))

|→PQ| = √(2304 + 18496)

|→PQ| = √20800

|→PQ| = 40√13

c = 40, k = 13

Correct Answer: C. \(2\sqrt{21}\)

Explanation

Given lines \(OA\) and \(OB\) inclined at angle \(\theta\), the line \(AB\) is gotten using cosine rule.

\(|AB|^{2} = |OA|^{2} + |OB|^{2} - 2|OA||OB|\cos \theta\)

\(|AB|^{2} = 4^{2} + 5^{2} - 2(4)(5)\cos 60\)

= \(16 + 25 - 20\)

\(|AB|^{2} = 21 \implies |AB| = \sqrt{21}\)

\(\implies \text{The two particles are} \sqrt{21} m \text{apart in 1 sec}\)

In two seconds, the particles will be \(2\sqrt{21} m\) apart.

Correct Answer: D. (62 N , 020°)

Explanation

r = (10 N, 200º) and n = (16 N, 020º)

In rectangular form:

r = 10cos 200ºi + 10sin 200ºj = -9.397i - 3.420j

n = 16cos 20ºi + 16sin 20ºj = 15.035i + 5.472j

3r = -28.191i - 10.260j

2n = 30.070i + 10.945j

3r - 2n = (-28.191i - 10.260j) - (30.070i + 10.945j)

3r - 2n = -58.261i - 21.205j

|3r - 2n| = √((-58.261)\(^2\) + (-21.205)\(^2\)) = 62 N

\(tan θ =\frac{-21.205}{-58.261} = 0.3640\)

\(θ = tan^{-1} (0.3640) = 20^o\)

∴ (62 N , 020º)

Correct Answer: A. \(f : x \to 4x - 1\)

Explanation

The inverse of the inverse of a function gives the function

i.e \(f^{-1}(f^{-1}(x)) = f(x)\)

\(f^{-1}(x) = \frac{x + 1}{4}\)

Take y = x, so

\(f^{-1}(y) = \frac{y + 1}{4}\)

Let \(x = f^{-1}(y)\),

\(x = \frac{y + 1}{4} \implies 4x = y + 1\)

\(y = f(x) = 4x - 1\)

Explanation

(a) \(^{k}P_{2} = \frac{k!}{(k - 2)!}\)

= \(\frac{k (k - 1) (k - 2)!}{(k - 2)!} = 72\)

\(k(k - 1) = 72 \implies k^{2} - k - 72 = 0\)

\(k^{2} - 9k + 8k - 72 = 0 \implies (k - 9)(k + 8) = 0\)

\(k = \text{-8 or 9} \implies k = 9\) (since k cannot be negative).

(b) \(2\cos^{2} \theta - 5\cos \theta = 3\)

\(2\cos^{2} \theta - 5\cos \theta - 3 = 0\)

\(2\cos^{2} \theta - 6\cos \theta + \cos \theta - 3 = 0\)

\(2\cos \theta (\cos \theta - 3) + 1(\cos \theta - 3) = 0\)

\((2\cos \theta + 1)(\cos \theta - 3) = 0\)

\(2 \cos \theta + 1 = 0 \implies 2\cos \theta = -1\)

\(\cos \theta = -0.5 \implies \theta = \cos^{-1} (-0.5) = 120°, 240°\)

\(\cos \theta - 3 = 0 \implies \cos \theta = 3\) (has no solution).

Explanation

. Total bulbs: n(Red)=5, n(B) = 7, n(G) = 4

(5+7+4) = 16

p(R)= 5/16, p(B) = 7/16, p(G) = 4/16

(a) p(All same colour of bulbs) = p(RR) Or p(BB) or p(GG)

\(\frac{5}{16}\) * \(\frac{4}{15}\) + \(\frac{7}{16}\) * \(\frac{6}{15}\) + \(\frac{4}{16}\) * \(\frac{3}{15}\)

= \(\frac{1}{12}\) + \(\frac{7}{40}\) + \(\frac{1}{20}\)

= \(\frac{10+21+6}{120}\) = \(\frac{37}{120}\)

(b) All different colors = 1-p (AIl the same colour) =

1 - \(\frac{37}{120}\) = \(\frac{83}{120}\)

(c) p(At least one red) = p(RR) + p(RB) + p(RG)

\(\frac{5}{16}\) * \(\frac{4}{15}\) + \(\frac{5}{16}\) * \(\frac{7}{15}\) + \(\frac{5}{16}\) * \(\frac{4}{15}\)

= \(\frac{1}{12}\) + \(\frac{7}{48}\) + \(\frac{1}{12}\)

= \(\frac{8+7}{48}\) = \(\frac{15}{48}\)

= \(\frac{5}{16}\)

Explanation

(a) \(f(x + 2) = 6x^{2} + 5x - 8\)

\(f(5) \implies x + 2 = 5\)

\(\therefore x = 3\)

\(f(5) = f(3 + 2) = 6(3^{2}) + 5(3) - 8\)

= \(54 + 15 - 8\)

= \(61\)

(b) \(\frac{7\sqrt{2} + 3\sqrt{3}}{4\sqrt{2} - 2\sqrt{3}}\)

Rationalizing, we multiply the numerator and denominator by \(4\sqrt{2} + 2\sqrt{3}\)

\((\frac{7\sqrt{2} + 3\sqrt{3}}{4\sqrt{2} - 3\sqrt{3}})(\frac{4\sqrt{2} + 2\sqrt{3}}{4\sqrt{2} + 2\sqrt{3}})\)

= \(\frac{28(2) + 14\sqrt{6} + 12\sqrt{6} + 6(3)}{16(2) - 4(3)}\)

= \(\frac{74 + 26\sqrt{6}}{20}\)

= \(3.7 + 1.3\sqrt{6}\)

Explanation

(a)(i) \(F_{1} = \begin{pmatrix} 3 \cos 210° \\ 3 \sin 210° \end{pmatrix} ; F_{2} = \begin{pmatrix} 4 \cos 120° \\ 4 \sin 120° \end{pmatrix}\)

\(R = F_{1} + F_{2}\)

= \(\begin{pmatrix} 3 \cos 210° \\ 3 \sin 210° \end{pmatrix} + \begin{pmatrix} 4 \cos 120° \\ 4 \sin 120° \end{pmatrix}\)

= \(\begin{pmatrix} -3 \cos 30° \\ -3 \sin 210° \end{pmatrix} + \begin{pmatrix} -4 \cos 60° \\ 4 \sin 60° \end{pmatrix}\)

= \(\begin{pmatrix} -3 \times 0.866 \\ -3 \times 0.500 \end{pmatrix} + \begin{pmatrix} -4 \times 0.500 \\ 4 \times 0.866 \end{pmatrix}\)

= \(\begin{pmatroix} -2.598 \\ -1.500 \end{pmatrix} + \begin{pmatrix} -2.00 \\ 3.464 \end{pmatrix}\)

= \(\begin{pmatrix} -4.598 \\ 1.964 \end{pmatrix}\)

\(|R| = \sqrt{(-4.598)^{2} + (1.964)^{2}}\)

= \(\sqrt{21.14 +3.86} = \sqrt{25}\)

= 5N

Acceleration = \(\frac{force}{mass}\)

= \(\frac{5}{7} = 0.71 ms^{-2}\)

(ii) \(Velocity = acceleration \times time\)

= \(0.71 \times 3 = 2.13 ms^{-1}\).

(b) Let fourth force be \(F_{4} = \begin{pmatrix} a \\ b \end{pmatrix}\).

\(F_{1} = \begin{pmatrix} 2 \\ 3 \end{pmatrix}; F_{2} = \begin{pmatrix} 0 \\ -5 \end{pmatrix} ; F_{3} = \begin{pmatrix} 6 \\ -4 \end{pmatrix}\)

Particle is in equilibrium when the sum of all forces acting on the particle = 0

\(\begin{pmatrix} 2 \\ 3 \end{pmatrix} + \begin{pmatrix} 0 \\ -5 \end{pmatrix} + \begin{pmatrix} 6 \\ -4 \end{pmatrix} + \begin{pmatrix} a \\ b \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}\)

\(\implies 8 + a = 0 ; a = -8\)

\(-6 + b = 0 \implies b = 6\)

\(F_{4} = \begin{pmatrix} -8 \\ 6 \end{pmatrix}\)

\(|F_{4}| = \sqrt{(-8)^{2} + 6^{2}} = 10N\)

Let the angle be \(\theta\).

\(\tan \theta = \frac{6}{-8} = -0.75\)

\(\theta = -36.87°\)

= \(180° - 36.87° = 143.13°\)

The equilibriant is 10N and acts in the direction of N53.13°W.

Explanation

(a) The range of values of p must satisfy p\(^2\) - 4 x 4 x 1 \(\geq\) 0 which will simplify to (p - 4)(p + 4) \(\geq\) 0

Thus; p \(\geq\) -4 or p \(\leq\) 4

(b)(i) (1 + 3x)\(^6\) which is to be in ascending powers of x as (\(^6_0\))(3x)\(^o\) + (\(^6_1\))(3x)\(^1\) + (\(^6_2\))(3x)\(^2\) + (\(^6_3\))(3x)\(^3\) + (\(^6_4\))(3x)\(^4\) + (\(^6_5\))(3x)\(^5\) + (\(^6_6\))(3x)\(^6\)

= 1 + 18x + 135x\(^2\) + 540x\(^3\) + 1215x\(^4\) + 1458x\(^5\) + 729x\(^6\)

(b)(ii) Observe that 1 + 3x = 1.03

x = 0.01

Substituting for x in the expansion;

(1.03)\(^6\) = 1 + 18(0.01) + 135(0.01)\(^2\) + 540(0.01)\(^3\) + 1215(0.01)\(^4\) + 1458(0.01)\(^5\) + 729(0.01)\(^6\)

= 1 + 0.18 + 0.0135 + 0.00054 + 0.00001215 + ...

= 1.19405215

= 1.194 correct to four significant figures.