Correct Answer: A. 4

Explanation

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

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

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

\(\lim \limits_{x \to 3} \frac{x^{2} - 2x - 3}{x - 3} \equiv \lim \limits_{x \to 3} (x + 1)\) (L'Hopital rule)

\(\lim \limits_{x \to 3} (x + 1) = 3 + 1 = 4\)

Correct Answer: B. 2y - x - 3< 0

Explanation

(0, 1.5), (-3, 0)

m = \(\frac{0 - 1.5}{-3, 0}\) = 0.5

0.5 = \(\frac{y - 1.5}{-3.0}\) = 0.5

y = 0.5x + 1.5

2y = x + 3

2y - x - 3 < 0

Correct Answer: C. 0.6

Explanation

\(\rho = 1 - \frac{6\sum{d^{2}}}{n(n^{2} - 1)}\)

\( 1 - \frac{6 \times 4}{4(4^{2} - 1)} = 1 - \frac{24}{60}\)

= \(1 - 0.4 = 0.6\)

Correct Answer: A. 0.58

Explanation

When x = 2

y = 2(2)\(^3\) - 2(2)\(^2\)

-5(2) + 5

y = 16 - 8 - 10 + 5

y = 3

when x = 2.05

y = 2(2.105)\(^3\) - 2(2.05)\(^2\)

- 5(2.05) + 5

y = 17.23 - 8.405 - 10.25 + 5

y = 3.575 - 3

Difference = 3.575 - 3

= 0.58

Explanation

(a) \(\overrightarrow{AB} = \begin{pmatrix} 4 \\ 5 \end{pmatrix} = 4i + 5j\)

\(\overrightarrow{AC} = \overrightarrow{AB} + \overrightarrow{BC}\)

= \(\begin{pmatrix} 4 \\ 5 \end{pmatrix} + \begin{pmatrix} -3 \\ 5 \end{pmatrix}\)

= \(\begin{pmatrix} 1 \\ 10 \end{pmatrix}\)

(i) Let \(\alpha\) be the angle between \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\).

\(\overrightarrow{AB} \cdot \overrightarrow{AC} = |AB||AC| \cos \alpha\)

\((4i + 5j) \cdot (i + 10j) = (\sqrt{4^{2} + 5^{2}})(\sqrt{1^{2} + 10^{2}}) \cos \alpha\)

\(4 + 50 = (\sqrt{41})(\sqrt{101}) \cos \alpha\)

\(\cos \alpha = \frac{54}{(\sqrt{41})(\sqrt{101})}\)

\(\cos \alpha = 0.8392\)

\(\alpha = 32.94°\)

(ii) \(\overrightarrow{AB} - \overrightarrow{BC} = \begin{pmatrix} 4 \\ 5 \end{pmatrix} - \begin{pmatrix} -3 \\ 5 \end{pmatrix}\)

=\(\begin{pmatrix} 7 \\ 0 \end{pmatrix} = 7i\)

\(\overrightarrow{AB} - \overrightarrow{BC} = 7i\)

Unit vector in the direction of 7i = \(\frac{7i}{7} = i\)

(b) \(\overrightarrow{PQ} = \begin{pmatrix} 2 \\ 8 \end{pmatrix} ; \overrightarrow{PR} = \begin{pmatrix} 11 \\ -12 \end{pmatrix}\)

\(\overrightarrow{QR} = \overrightarrow{QP} + \overrightarrow{PR} \)

= \(-\overrightarrow{PQ} + \overrightarrow{PR}\)

= \(\begin{pmatrix} -2 \\ -8 \end{pmatrix} + \begin{pmatrix} 11 \\ -12 \end{pmatrix}\)

= \(\begin{pmatrix} 9 \\ -20 \end{pmatrix}\)

\(\frac{QM}{MR} = \frac{QP + PM}{MP + PR} = \frac{-PQ + PM}{-PM + PR}\)

\(\frac{QM}{MR} = \frac{3}{7}\)

\(7QM = 3MR\)

\(7(-PQ + PM) = 3(-PM + PR)\)

\(7(-PQ) + 7PM = 3(-PM) + 3(PR)\)

\(7(\begin{pmatrix} -2 \\ -8 \end{pmatrix} - 3(\begin{pmatrix} 11 \\ -12 \end{pmatrix} = -10(PM)\)

\(\begin{pmatrix} -47 \\ -20 \end{pmatrix} = -10(PM)\)

\(PM = \begin{pmatrix} 4.7 \\ 2 \end{pmatrix}\)

Correct Answer: C. 72 N

Explanation

\(F = ma \)

\(a = \frac{v - u}{t}\)

\(\therefore F = m(\frac{v - u}{t})\)

\(F = 12(\frac{35 - 5}{5}) = 12 \times 6 = 72 N\)

Correct Answer: B. 1.54m

Explanation

for 20 students, mean = 1.67

∑fx = μf

∑fx = 20 × 1.67 = 33.4

for group 2

∑fx = 16 × 1.50 = 24

for group 3

∑fx = 14 × 1.40 = 19.6

33.4 + 24 + 19.6 = 77

∑f = 20+16+14 = 50

Correct Answer: D. \(\frac{4\sqrt{10}}{5}\)

Explanation

\(4x^{2} + 5kx + 10 = (2x + \sqrt{10})^{2}\)

Expanding the right hand side equation, we have

\(4x^{2} + 4x\sqrt{10} + 10\)

Comparing with the left hand side, we have

\(5k = 4\sqrt{10} \implies k = \frac{4}{5}\sqrt{10}\)

Correct Answer: A. y = 2x + 11

Explanation

The gradient of the line y = 2x + 5 ;

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

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

\(y - 3 = 2(x + 4) \implies y = 2x + 11\)

Correct Answer: C. \(\frac{5}{12}\)

Explanation

\(p(P) = \frac{1}{2}, p(P') = \frac{1}{2}\)

\(p(Q) = \frac{1}{3}, p(Q') = \frac{2}{3}\)

\(p(R) = \frac{3}{4}, p(R') = \frac{1}{4}\)

p(exactly two hit the target) = p(P and Q and R') + p(P and R and Q') + p(Q and R and P')

= \((\frac{1}{2} \times \frac{1}{3} \times \frac{1}{4}) + (\frac{1}{2} \times \frac{3}{4} \times \frac{2}{3}) + (\frac{1}{3} \times \frac{3}{4} \times \frac{1}{2})\)

= \(\frac{1}{24} + \frac{6}{24} + \frac{3}{24}\)

= \(\frac{5}{12}\)