Explanation

(a)

(c)(i) Position of lower quartile, \(Q_{1} = \frac{\sum f + 1}{4} = \frac{500 + 1}{4}\)

= \(\frac{501}{4} = 125.25th\) position.

\(\therefore Q_{1} = 26.0\)

Position of upper quartile, \(Q_{3} = 3(\frac{\sum f + 1}{4}) = 3(125.25) = 375.75th\) position.

\(\therefore Q_{3} = 56.3\)

Semi-interquartile range = \(\frac{Q_{3} - Q_{1}}{2} = \frac{56.3 - 26}{2}\)

= \(\frac{30.3}{2} = 15.15\)

(ii) If the pass mark is 37, then 190 students failed.

(iii) Those who scored between 0% and 20% = 100

Those who scored between 0% and 60% = 400

Those who scored between 20% and 60% = 400 - 100 = 300.

P(20% < x < 60%) = \(\frac{300}{500} = \frac{3}{5}\)

Correct Answer: B. - 2.0

Explanation

Sum of roots = \(\frac{-a}{b}\)

= \(\frac{-5m}{2}\) = 5

\(\frac{-5m}{m} = \frac{10}{-5}\)

m = -2

Explanation

We find the initial velocity using the formula \(v = u + at\)

\(33 = u + 3(6) \implies 33 = u + 18\)

\(u = 15ms^{-1}\)

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

= \(15 \times 6 + \frac{1}{2} (3)(6^{2})\)

= \(90 + 54 = 144m\)

Correct Answer: C. 0.78

Correct Answer: A. \(\frac{-1}{8}\)

Explanation

Given : \(y = x^{3} - x^{2}\)

\(\frac{\mathrm d y}{\mathrm d x} = 3x^{2} - 2x\)

\(\therefore \text{The gradient of the tangent at point (x = 2)} = 3(2^{2}) - 2(2) \)

= \(12 - 4 = 8\)

Recall, the tangent and the normal are perpendicular to each other and the product of the gradients of perpendicular lines = -1.

\(\implies \text{the gradient of the normal} = \frac{-1}{8}\)

Correct Answer: C. 56

Explanation

= \(^{8}C_{3} = \frac{8!}{(8 - 3)! 3!}\)

= \(\frac{8 \times 7 \times 6}{3 \times 2}\)

= 56 ways

Correct Answer: A. 4

Explanation

\(f(x) = \frac{1}{2 - x}, x \neq 2\)

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

\(x = \frac{1}{2 - y}\) (Let x = f(y))

\(2x - xy = 1 \implies y = \frac{2x - 1}{x}\)

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

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

= \(-2 \times -2 = 4\)

Explanation

(a)(i) Position vector of M = 3i + 7j i.e \(\overrightarrow{OM} = 3i + 7j\).

Position vector of N = 4i + 5j i.e \(\overrightarrow{ON} = 4i + 5j\)

\(\overrightarrow{MN} = \overrightarrow{ON} - \overrightarrow{OM}\)

\((4i + 5j) - (3i + 7j) = i - 2j\)

Velocity of M = 5i + 6j

Distance covered in time t = \((5i + 6j)t = 5ti + 6tj\)

When t = 3 seconds, = \(15i + 18j\)

Velocity of N = 2i + 3j

Distance covered in time t = \((2i + 3j)t = 2ti + 3tj\)

When t = 3 seconds, = \(6i + 9j\)

\(\overrightarrow{MN} = (6i + 9j) - (15i + 18j)\)

= \(-9i - 9j\)

\(|MN| = \sqrt{(-9)^{2} + (-9)^{2}} = \sqrt{162}\)

= \(9\sqrt{2} m\)

(b) \(F_{1} = 5i + pj ; F_{2} = qi + j ; F_{3} = -2pi + 3j ; F_{4} = -4i + qj\)

If the particle remains in equilibrium, then

\(F_{1} + F_{2} + F_{3} + F_{4} = 0\)

\(5i + pj + qi + j - 2pi + 3j - 4i + qj = 0\)

Equating by components,

\(i : 5 + q - 2p - 4 = 0 \implies q - 2p = -1 ...(1)\)

\(j : p + 1 + 3 + q = 0 \implies p + q = -4 ... (2)\)

\((2) - (1) : 3p = -3 \implies p = -1\)

\(-1 + q = -4 \implies q = -3\)

Correct Answer: C. 25.63

Explanation

\(y = 19.33 + 0.42x\)

\(\text{The value of y when x = 15} = 19.33 + (0.42 \times 15)\)

= \(19.33 + 6.30\)

= 25.63

Correct Answer: D. 840N

Explanation

\(\text{Net force = Upward force - weight}\)

\(ma = F - mg \implies F = ma + mg \)

\(F = (80 \times 10) + (80 \times 0.5) = 800 + 40 = 840N\)