Correct Answer: C. {x : 2 ≤ x ≤ 4}

Explanation

P = {x : 2 ≤ x ≤ 8} = {2, 3, 4, 5, 6, 7, 8}

Q = {x : 4 < x ≤ 12} = {5, 6, 7, 8, 9, 10, 11, 12}

Q1 = R - {5, 6, 7, 8, 9, 10, 11, 12} = {..., 2, 3, 4, 13, 14, 15, ...}

∴ P ? Q\(^1\) = {2, 3, 4} = {x : 2 ≤ x ≤ 4}

Correct Answer: D. -3

Explanation

\(\begin{pmatrix} 1 & -3 \\ 1 & 4 \end{pmatrix} \begin{pmatrix} -6 \\ P \end{pmatrix} = \begin{pmatrix} 3 \\ -26 \end{pmatrix}\)

\(\implies (1 \times -6) + (-3 \times P) = 3\)

\(-6 - 3P = 3 \implies -3P = 9\)

\(P = -3\)

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

Explanation

cos150° = -cos30°

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

Correct Answer: D. {-9, -5, -1, 3, 7}

Explanation

The elements of x are {-2, -1, 0, 1, 2}

\(y = 4x - 1\) = 4(-2) - 1 = -9; 4(-1) - 1 = -5; 4(0) - 1 = -1; 4(1) - 1 = 3; 4(2) - 1 = 7.

The range of x is {-9, -5, -1, 3 7}.

Correct Answer: B. 73.40°

Explanation

\(F_{1} = 7i + 0j\)

\(F_{2} = (4\cos\theta)i + (4\sin\theta)j\)

\(9 = \sqrt{(7 + 4\cos\theta)^{2} + (4\sin\theta)^{2}}\)

\(9^{2} = (7 + 4\cos\theta)^{2} + (4\sin\theta)^{2} \implies 81 = 49 + 56\cos\theta + 16\cos^{2}\theta + 16\sin^{2}\theta\)

\(81 = 49 + 56\cos\theta + 16(\cos^{2}\theta + \sin^{2}\theta)\)

Recall, \(\cos^{2}\theta + \sin^{2}\theta = 1\)

\(81 = 49 + 56\cos\theta + 16 \implies 81 - 49 -16 = 56\cos\theta\)

\(16 = 56\cos\theta \implies \cos\theta = \frac{16}{56} = 0.2857\)

\(\theta = \cos^{-1} 0.2857 = 73.40°\)

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: A. \(\frac{\sqrt{2}}{5}\)

Explanation

\(F = F_{1} + F_{2}\)

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

\(F = ma \implies (-1, -1) = 5a\)

\(a = (-\frac{1}{5}, -\frac{1}{5})\)

\(|a| = \sqrt{(\frac{-1}{5})^2 + (\frac{-1}{5})^2} = \sqrt{2}{25}\)

\(|a| = \frac{\sqrt{2}}{5} ms^{-2}\)

Explanation

P(a sum of 7) = \(\frac{6}{36}\) and P(a sum of 4) = \(\frac{3}{36}\)

P(a sum of 4) = \(\frac{3}{36}\)

P(a sum of 7 and a sum of 4) = \(\frac{6}{36} \times \frac{3}{36}\) = \(\frac{1}{72}\)

Explanation

\(\frac{\sqrt{75} - 3}{\sqrt{3} + 1}\)

Rationalizing, \((\frac{\sqrt{75} - 3}{\sqrt{3} + 1})(\frac{\sqrt{3} - 1}{\sqrt{3} - 1}\)

\(\frac{\sqrt{3 \times 75} - \sqrt{75} - 3\sqrt{3} + 3}{3 - \sqrt{3} + \sqrt{3} - 1}\)

= \(\frac{15 - 5\sqrt{3} - 3\sqrt{3} + 3}{2}\)

= \(\frac{18 - 8\sqrt{3}}{2}\)

= \(9 - 4\sqrt{3}\)

(b)(i) The general form of the circle equation is \(x^{2} + y^{2} + 2gx + 2fy + c = 0\) where (x, y) is a point on the circle.

\(\therefore A(7, 3) = 7^{2} + 3^{2} + 7(2g) + 3(2f) + c = 0\)

\(14g + 6f + c = -58 .... (1)\)

\(B(2, 8) = 2^{2} + 8^{2} + 2(2g) + 2(2f) + c = 0\)

\(4g + 4f + c = -68 ..... (2)\)

\(C(-3, 3) = -3^{2} + 3^{2} - 3(2g) + 3(2f) + c = 0\)

\(6f - 6g + c = -18 ....(3)\)

Solving the equations, we get \(g = -2; f = -3; c = -12\)

\(\text{The equation of the circle is} x^{2} + y^{2} - 4x - 6y - 12 = 0\).

(b)(ii) Radius = \(\sqrt{g^{2} + f^{2} - c}\)

= \(\sqrt{-2^{2} + (-3^{2}) - (-12)}\)

\(\sqrt{25} = 5 units\)

Explanation

\(2u m/s\) \(0 m/s\) : \(\frac{1}{2} u m/s\) \(u m/s\)

P Q P Q

Before collision : After collision

Momentum before collision = \(M_{P} (2u) kg m/s\)

Momentum after collision = \(-\frac{1}{2}u (M_{P}) + M_{Q} (u)\)

\(2u (M_{P}) + \frac{1}{2} u (M_{P}) = M_{Q} (u)\)

\(\frac{5}{2} M_{P} = M_{Q}\)

\(\frac{M_{P}}{M_{Q}} = \frac{2}{5}\)

The ratio = 2:5

(b)

From cosine rule, \(5^{2} = 3^{2} + 7^{2} - 2 \times 3 \times 7 \cos (180 - \alpha)\)

\(25 = 9 + 49 - 42(-\cos \alpha)\)

\(25 = 58 + 42\cos \alpha\)

\(\cos \alpha = \frac{-33}{42} = -0.7857\)

\(\alpha = \cos^{-1} (-0.7857) = 141.79°\)

(c)

\(\overrightarrow{VB} = \frac{1}{2}\overrightarrow{AB} = \frac{1}{2} \begin{pmatrix} -4 \\ 6 \end{pmatrix} = \begin{pmatrix} -2 \\ 3 \end{pmatrix}\)

\(\overrightarrow{CV} + \overrightarrow{VB} = \overrightarrow{CB}\)

\(\overrightarrow{CV} = \overrightarrow{CB} - \overrightarrow{VB}\)

= \(\begin{pmatrix} 2 \\ -3 \end{pmatrix} - \begin{pmatrix} -2 \\ 3 \end{pmatrix}\)

= \(\begin{pmatrix} 2 + 2 \\ -3 - 3 \end{pmatrix}\)

= \(\begin{pmatrix} 4 \\ -6 \end{pmatrix}\)