Explanation

(a) \(2x^{2} + x - 6 < 0 \implies 2x^{2} + 4x - 3x - 6 < 0\)

\(2x(x + 2) - 3(x + 2) < 0 \implies (2x - 3)(x + 2) < 0\)

If \((2x - 3)(x + 2) < 0\), then either (2x - 3) < 0 or (x + 2) < 0.

\((2x - 3) < 0 \implies x < \frac{3}{2}\), then \((x + 2) > 0 \implies x > -2\)

\((x + 2) < 0 \implies x < -2\), then \((2x - 3) > 0 \implies x > \frac{3}{2}\)

Check: Let x = -3,

\((2x - 3)(x + 2) = (-9)(-1) = 9 > 0\)

Let x = 1,

\((2x - 3)(x + 2) = (-1)(3) = -3 < 0\)

\(\therefore -2 < x < \frac{3}{2}\).

(b) \(\frac{5 - 2\sqrt{10}}{3\sqrt{5} + \sqrt{2}}\)

= \((\frac{5 - 2\sqrt{10}}{3\sqrt{5} + \sqrt{2}})(\frac{3\sqrt{5} - \sqrt{2}}{3\sqrt{5} - \sqrt{2}})\)

= \(\frac{15\sqrt{5} - 5\sqrt{2} - 30\sqrt{2} + 4\sqrt{5}}{9(5) - 3\sqrt{10} + 3\sqrt{10} - 2}\)

= \(\frac{19\sqrt{5} - 35\sqrt{2}}{45 - 2}\)

= \(\frac{19\sqrt{5} - 35\sqrt{2}}{43}\)

= \(\frac{-35}{43} \sqrt{2} + \frac{19}{43} \sqrt{5}\)

\(\therefore m = -\frac{-35}{43} ; n = \frac{19}{43}\).

Correct Answer: C. (6x² + 3) cos (2x³ + 3x - 4)

Correct Answer: D. -3

Explanation

\(\frac{2x - 1}{x + 3}\) not defined

x + 3 = 0

x = -3

Correct Answer: C. \(\frac{1}{12}\)

Correct Answer: A. (2,3)

Explanation

(a) \(3\sqrt{75} = 3 \times \sqrt{25 \times 3} = 3(5\sqrt{3}) = 15\sqrt{3}\)

\(\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}\)

\(\sqrt{108} = \sqrt{36 \times 3} = 6\sqrt{3}\)

\(\therefore 3\sqrt{75} - \sqrt{12} + \sqrt{108} \equiv 15\sqrt{3} - 2\sqrt{3} + 6\sqrt{3}\)

= \((15 - 2 + 6)\sqrt{3}\)

= \(19\sqrt{3}\).

(b) \(124_{n} = 232_{five}\)

\(124_{n} = (1 \times n^{2}) + (2 \times n^{1}) + (4 \times n^{0}) \)

= \(n^{2} + 2n + 4\)

\(232_{five} = (2 \times 5^{2}) + (3 \times 5^{1}) + (2 \times 5^{0})\)

= \(50 + 15 + 2\)

= \(67\)

\(\implies n^{2} + 2n + 4 = 67\)

\(n^{2} + 2n + 4 - 67 = 0\)

\(n^{2} + 2n - 63 = 0\)

\(n^{2} - 7n + 9n - 63 = 0\)

\(n(n - 7) + 9(n - 7) = 0\)

\((n + 9)(n - 7) = 0\)

\(\text{n = -9 or 7}\)

Hence, \(n = 7\)

\(124_{seven} = 232_{five}\)

Explanation

Put the value of d in (1), so we have

\(a + 0.75 = 2 \implies a = 1.25\)

Therefore, the length of the poles in metres = \(1.25, 2, 2.75, 3.5, 4.25, 5, 5.75, 6.5\)

Correct Answer: B. -4

Explanation

If (x + 1) is a factor, then f(-1) = 0.

\((-1)^{3} + p(-1)^{2} + (-1) + 6 = 0\)

\(-1 + p - 1 + 6 = 0 \implies p + 4 = 0\)

\(p = -4\)

Correct Answer: C. 7

Correct Answer: A. 6/35