Explanation

(a) Number of triangles that can be formed = \(^{6}C_{3}\)

= \(\frac{6!}{(6 - 3)! 3!} = \frac{6 \times 5 \times 4}{3 \times 2}\)

= 20 ways.

(b) Tie the father and mother together so that there are 5 persons to arrange in 5! ways.

Changing the positions of the father and mother, this can be done in 2! ways.

\(\therefore \text{Ways of arranging the family = } 2!5!\)

Without restriction, they can be arranged in 6! ways.

\(\therefore \text{Father and mother not together =} 6! - 2!5! = 720 - 240 = 480 \)

Correct Answer: C. 12.06

Correct Answer: D. (1, -4)

Explanation

Equate the equation of the curve and the line in order to find their point of intersection ie The values of y in both equations.

\(y + x = 1 \implies y = 1 - x\)

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

\(x^{2} + 2x + x - 3 - 1 = 0 \implies x^{2} + 3x - 4 = 0\)

\((x - 1)(x + 4) = 0\)

\(x = (1, -4)\)

Correct Answer: D. 5x + 3y - 1 = 0

Explanation

Midpoint between P(4, 5) and Q(-6, -1) = \(\frac{4 - 6}{2}, \frac{5 - 1}{2} = (-1, 2)\)

Gradient of PQ = \(\frac{5 - (-1)}{4 - (-6)} = \frac{3}{5}\)

Gradient of line perpendicular to PQ = \(\frac{-1}{\frac{3}{5}} = -\frac{5}{3}\)

\(line = \frac{y - 2}{x + 1} = \frac{-5}{3}\)

\(3y - 6 = -5x - 5\)

\(5x + 3y - 6 + 5 = 0 \implies 5x + 3y - 1 = 0\)

Correct Answer: C. \(\frac{\sqrt{3}}{2}\)

Explanation

\(a ♦ b = \frac{ab}{4}\)

\(\sqrt{2} ♦ \sqrt{6} = \frac{\sqrt{2} \times \sqrt{6}}{4} = \frac{\sqrt{12}}{4}\)

= \(\frac{2\sqrt{3}}{4} \)

= \(\frac{\sqrt{3}}{2}\)

Correct Answer: A. \(\frac{1}{2}\) m/s

Explanation

m1u1 + m2u2 = (m1 + m2)v

m1 = 18kg, m2 = 6kg, u1 = 4ms-1, u2 = -10m/s

18(4) + 6(-10) = (18+6)v

72 - 60 = 24v

12 = 24v

v = \(\frac{1}{2}\) m/s

Correct Answer: C. 3

Explanation

\(\begin{vmatrix} k & k \\ 4 & k \end{vmatrix} + \begin{vmatrix} 2 & 3 \\ -1 & k \end{vmatrix} = 6\)

\(\begin{vmatrix} k & k \\ 4 & k \end{vmatrix} = (k^{2} - 4k)\)

\(\begin{vmatrix} 2 & 3 \\ -1 & k \end{vmatrix} = (2k + 3)\)

\(\therefore (k^{2} - 4k) + (2k + 3) = k^{2} - 2k + 3 = 6\)

\(k^{2} - 2k - 3 = 0\), factorising, we have \(k + 1 = 0\) or \(k - 3 = 0\)

Since k > 0, k = 3.

Correct Answer: D. \(x^{2} + y^{2} - 8x - 10y + 21 = 0\)

Explanation

On the y-intercept, x=0. At this point, y = 5(0) - 2y = -6; y = 3, so the y- intercept has coordinates (0,3).

The equation of the circle is given as \((x-a)^{2} + (y-b)^{2} = r^{2}\), where (a,b) is the centre and r is the radius.

The radius of the circle through the y- intercept = \(\sqrt{(4-0)^{2} + (5-3)^{2}} = \sqrt{20}\)

The equation of the circle is \((x-4)^{2} + (y-5)^{2} = 20\); expanding, we have

\(x^{2} + y^{2} + 8x - 10y + 21 = 0\)

Explanation

(a) u = 10ms\(^{-1}\), a = 0.2m\(^{-2}\), s = 100m, t = ?

using S = ut + \(\frac{1}{2} at^2\)

100 = 10t + \(\frac{1}{2}(0.2)t^2\)

100 = 10t + 0.1t\(^2\)

0.1t\(^2\) + 10t - 100 = 0

t = \(\frac{-10 \pm \sqrt{10^2 - 4(0.1)(-100)}}{2(0.1)}\)

t = \(\frac{-10\pm \sqrt{100 + 40}}{0.2}\)

t = \(\frac{-10 \pm \sqrt{140}}{0.2}\)

t = \(\frac{-10 \pm 11.83}{0.2}\)

t = \(\frac{-10 - 11.82}{2}\)

t = \(\frac{1.83}{0.2}\)

t = \(\frac{21.83}{2}\)

t = 9.15 seconds

(b)(i) S = t\(^2\) - 5 + 6

v = \(\frac{ds}{dt}\)

= 2t - 5

at t = 0, y = 2(0) - 5

= - 5ms\(^{-1}\)

The initial velocity is 5 in the negative direction

(ii) When t = 0, S = t\(^2\) - 5t + 6

S = 0\(^2\) - 5(0) + 6

S = 6

The distance is 6m

Correct Answer: D. 4

Explanation

\(\sqrt{128} = \sqrt{64\times2} = 8\sqrt{2}\)

\(\sqrt{32} = \sqrt{16\times2} = 4\sqrt{2}\)

Simplifying, we have \(\frac{8\sqrt{2}}{4\sqrt{2} - 2\sqrt{2}} = \frac{8\sqrt{2}}{2\sqrt{2}}\)

= 4