Explanation

(a) \(7(x + 4) - \frac{2}{3}(x - 6) \leq 2[x - 3(x + 5)]\)

Multiply through by 3 to clear the fraction :

\(21(x + 4) - 2(x - 6) \leq 6[x - 3(x + 5)]\)

\(21x + 84 - 2x + 12 \leq 6x - 18x - 90\)

\(19x + 96 \leq -12x - 90\)

\(19x + 12x \leq -90 - 96\)

\(31x \leq - 186\)

Hence, \(x \leq -6\).

(b) Let : t represent number of tricycles; and b represent number of taxicabs

Then \(t + b = 20 ... (1)\)

\(2t + 4b = 66 .... (2)\)

From (1), b = 20 - t. Put that into eqtn (2), we have:

\(2t + 4(20 - t) = 2t + 80 - 4t = 66\)

\(80 - 66 = 14 = 2t \)

\(t = 7\)

The transport company has 7 tricycles.

Correct Answer: c. 402

Correct Answer: B. 3 minutes

Correct Answer: C. 585°

Explanation

\(180° = \pi radian\)

\(\frac{13}{4}\pi = \frac{13}{4} \times 180° = 585°\)

Correct Answer: D. TSK935AH

Explanation

(a) \(f(x) = \frac{x - 3}{2x - 1}, x \neq \frac{1}{2}\)

\(g(x) = \frac{x - 1}{x + 1}, x \neq -1\)

\(g \circ f = g(f(x)) = \frac{\frac{x - 3}{2x - 1} - 1}{\frac{x - 3}{2x - 1} + 1}\)

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

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

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

(b)(i) \(y = 9x - x^{3}\)

If x = 0, y = 0.

If y = 0, \(9x - x^{3} = 0\)

\(x(9 - x^{2}) = x (3 - x)(3 + x) = 0\)

\(\implies x = 0, -3, 3\)

Turning points : \(\frac{\mathrm d y}{\mathrm d x} = 9 - 3x^{2}\)

\(9 - 3x^{2} = 0 \implies 9 = 3x^{2}\)

\(x^{2} = 3 \implies x = \pm \sqrt{3} = \pm 1.73\)

When x = 1.73, \(y = 9(1.73) - (1.73)^{3} = 15.57 - 5.18 = 10.39\)

Point is (1.73, 10.39).

When x = -1.73, \(y = 9(-1.73) - (-1.73)^{3} = -15.57 + 5.18 = -10.39\)

Point is (-1.73, -10.39).

Maximum or minimum ?

\(\frac{\mathrm d^{2} y}{\mathrm d x^{2}} = - 6x^{2}\)

At x = 1.73, \(\frac{\mathrm d^{2} y}{\mathrm d x^{2}} < 0 \therefore (1.73, 10.39) \text{ is a maximum point}\)

At x = -1.73, \(\frac{\mathrm d^{2} y}{\mathrm d x^{2}} > 0 \therefore (-1.73, -10.39) \text{ is a maximum point}\).

Areas between x = -3 to x = 0 and x = 0 and x = 3 are equal.

\(\therefore Area = 2 \int_{0} ^{3} (9x - x^{3}) \mathrm {d} x = 2[\frac{9x^{2}}{2} - \frac{x^{4}}{4}]_{0} ^{3}\)

= \(81 - 40\frac{1}{2} = 40\frac{1}{2} sq. units\)

Correct Answer: C. 24.5

Explanation

Modal class 20 - 29

Modal class mark = \(\frac{20+29}{2}\)

= \(\frac{49}{2}\)

= 24.5

Correct Answer: d. N75,000

Correct Answer: D. 66\(\frac{2}{3}\)%

Explanation

Percentage of students with marks ranging from 35 to 50 = \(\frac{f_{35} + f{40} + f{50}}{\sum f}\)

= \(\frac{35 + 40 + 45}{20 + 35 + 40 + 45 + 25 + 15}\) x 100%

= \(\frac{120}{180}\) x 100%

= 66\(\frac{2}{3}\)%

Correct Answer: D. 8.66 N

Explanation

\(F_{1} = (5 N, 060°) = 5 \cos 60 i + 5 \sin 60 j = 2.5 i + \frac{5\sqrt{3}}{2}j\)

\(F_{2} = (10 N, 180°) = 10 \cos 180 i + 10 \sin 180 j = -10i\)

\(R = F_{1} + F_{2} = -7.5 i + \frac{5\sqrt{3}}{2}j\)

\(|R| = \sqrt{(-7.5)^{2} + (\frac{5\sqrt{3}}{2})^{2}}\)

= \(\sqrt{56.25 + 18.75} = \sqrt{75}\)

= 8.66 N