Explanation

(a)

Take moments about P

Clockwise moment = \(200 \times 1.5 + 100 \times 4\)

= \(300 + 400 = 700N\)

Anti-clockwise moment = \(5R_{2} N\)

\(5R_{2} = 700 ; R_{2} = 140N\)

For vertical equilibrium, \(R_{1} + R_{2} = 300\)

\(R_{1} = 300 - 140 = 160N\)

(b)

Suppose he walked x m from P

Let each reaction be R N

Equating forces, 2R = 300

\(R = 150N\)

We take moments about P.

Clockwise moments =\( 200x + 100 \times 4\)

= \(200x + 400 Nm\)

Anti-clockwise moments = 5R = 750 Nm

\(200x + 400 = 750\)

\(200x = 350 ; x =1.75m\)

(c) Plank tips over when he is between S and Q.

Let boy be y m from S.

We take moments about S.

\(200y = 100(5 - 4) = 100\)

\(y = 0.5m\)

The boy must be 5.5m from P.

Correct Answer: C. 3

Explanation

\(\sin \theta \leq 1\)

i.e Maximum value of \(\sin \theta \forall \theta = 1\).

Therefore, \(2 + \sin \theta \leq 2 + 1 = 3\)

Correct Answer: C. \(\frac{1}{3}\)

Explanation

\(\cos (A + B) = \cos A \cos B - \sin A \sin B\)

\(\sin A = \frac{3}{5} ; \cos B = \frac{15}{17}\)

\(\cos A = -\frac{4}{5}\) (A is obtuse)

\(\sin B = \frac{8}{17}\)

\(\cos (A + B) = \cos A \cos B - \sin A \sin B\)

= \((\frac{-4}{5} \times \frac{15}{17}) - (\frac{3}{5} \times \frac{8}{17})\)

= \(\frac{-60}{85} - \frac{24}{85}\)

= \(\frac{-84}{85}\)

Correct Answer: A. 1

Explanation

\(\lim \limits_{x \to 1} \frac{1 - x}{x^{2} - 3x + 2}\)

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

= \(\frac{-1}{x - 2}\)

\(\lim \limits_{x \to 1} \frac{1 - x}{x^{2} - 3x + 2} = \lim \limits_{x \to 1} \frac{-1}{x - 2}\)

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

Explanation

4 red, 6 blue, 8 green = 18 marbles.

p(red) = \(\frac{4}{18} = \frac{2}{9}\); p(not red) = \(\frac{7}{9}\)

p(blue) = \(\frac{6}{18} = \frac{1}{3}\) ; p(not blue) = \(\frac{2}{3}\)

p(green) = \(\frac{8}{18} = \frac{4}{9}\) ; p(not green) = \(\frac{5}{9}\)

(a) p( 3 green) = \(\frac{8}{18} \times \frac{7}{17} \times \frac{6}{16} = \frac{7}{102}\)

(b) p(all same colour) = p(all red) or p(all blue) or p(all green)

= \(\frac{4}{18} \times \frac{4}{18} \times \frac{4}{18} + \frac{6}{18} \times \frac{6}{18} \times \frac{6}{18} + \frac{8}{18} \times \frac{8}{18} \times \frac{8}{18}\)

= \(\frac{792}{5832}\)

= \(\frac{11}{81}\)

Explanation

With the 6N force acting in the direction 090°, the diagram is redrawn as :

Resultant = \(\begin{pmatrix} 4 \cos 65° \\ 4 \sin 65° \end{pmatrix} + \begin{pmatrix} 6 \cos 90° \\ 6 \sin 90° \end{pmatrix} + \begin{pmatrix} 8 \cos 145° \\ 8 \sin 145° \end{pmatrix} + \begin{pmatrix} 4 \cos 245° \\ 4 \sin 245° \end{pmatrix} + \begin{pmatrix} 5 \cos 295° \\ 5 \sin 295° \end{pmatrix}\)

= \(\begin{pmatrix} 4 \cos 65 \\ 4 \sin 65 \end{pmatrix} + \begin{pmatrix} 6 \cos 90 \\ 6 \sin 90 \end{pmatrix} + \begin{pmatrix} -8 \cos 35 \\ 8 \sin 35 \end{pmatrix} + \begin{pmatrix} -4 \cos 65 \\ -4 \sin 65 \end{pmatrix} + \begin{pmatrix} 5 \cos 65 \\ -5 \sin 65 \end{pmatrix}\)

= \(\begin{pmatrix} 4 \times 0.4226 \\ 4 \times 0.9063 \end{pmatrix} + \begin{pmatrix} 0 \\ 6 \end{pmatrix} + \begin{pmatrix} -8 \times 0.8192 \\ 8 \times 0.5736 \end{pmatrix} + \begin{pmatrix} -4 \times 0.4226 \\ -4 \times 0.9063 \end{pmatrix} + \begin{pmatrix} 5 \times 0.4226 \\ -5 \times 0.9063 \end{pmatrix}\)

= \(\begin{pmatrix} 1.6904 \\ 3.6252 \end{pmatrix} + \begin{pmatrix} 0 \\ 6 \end{pmatrix} + \begin{pmatrix} -6.5536 \\ 4.5888 \end{pmatrix} + \begin{pmatrix} -1.6904 \\ -3.6252 \end{pmatrix} + \begin{pmatrix} 2.113 \\ -4.5315 \end{pmatrix}\)

= \(\begin{pmatrix} -4.4406 \\ 6.0573 \end{pmatrix}\)

\(|Resultant| = \sqrt{(-4.4406)^{2} + (6.0573)^{2}} = \sqrt{56.41}\)

= 7.5107N.

The magnitude of the resultant = 7.5N.

(b) Let \(\theta\) be the direction of the resultant.

\(\tan \theta = \frac{6.0573}{-4.4406} = - 1.364\)

\(\theta = -53.75° \)

= \(180° - 53.75° = 126.25°\) with the positive x- axis.

Correct Answer: C. \(\frac{2x^{\frac{7}{2}}}{7} + c\)

Explanation

\(x^{2}\sqrt{x} \equiv x^{2}. x^{\frac{1}{2}} = x^{\frac{5}{2}}\)

\(\implies \frac{\mathrm d y}{\mathrm d x} = x^{\frac{5}{2}}\)

\(y = \int x^{\frac{5}{2}} \mathrm d x\)

= \(\frac{x^{\frac{5}{2} + 1}}{\frac{5}{2} + 1} + c\)

= \(\frac{2x^{\frac{7}{2}}}{7} + c\)

Correct Answer: A. (7, -2)

Explanation

Let (x1, y1) be the image of the point (x, y) under the given transformations.

x1 = 3x - y

y1 = x + 4y

\(\begin{vmatrix} 3 & -1 \\ 1 & 4 \end{vmatrix} \begin{vmatrix} x \\ y = \end{vmatrix} \begin{vmatrix} x_1 \\ y_1 \end{vmatrix}\)

\(\begin{vmatrix} 3 & -1 \\ 1 & 4 \end{vmatrix} \begin{vmatrix} 2 \\ 1 = \end{vmatrix} \begin{vmatrix} 7 \\ -2 \end{vmatrix}\)

Correct Answer: B. 12

Explanation

6(2x\(^2\))\(^2\) (\(\frac{1}{x^2}\))\(^2\)

= 6 x 2

= 12