Correct Answer: D. 6

Explanation

The mode is the occurrence with the highest frequency which, from the table, is 45 (the occurrence of obtaining a 6).

Correct Answer: D. x\(^2\) - y\(^2\) + 3x - 5y = 2

Correct Answer: C. 192

Explanation

Using the sum of an AP, S\(_n\) = \(\fra{n}{2}\) [ 2a + (n - 1)d]

S\(_3\) = \(\fra{3}{2}\) [ 2a + (3 - 1)d]

18 = \(\fra{3}{2}\) [ 2a + 2d]

2a + 2d = 12

a = 4

2(4) + 2d = 18 --> 8 + 2d = 12

2d = 4; d = 2

a = 4: a + d

= 4 + 2 = 6

a + 2d = 4 + 2]2]

= 8

product of the terms = 4 * 6 * 8 = 192

Explanation

Since the class intervals are not equal, we draw the histogram in a way that the area of the bars, rather than the height must be proportional to the frequency. On the vertical scale, we plot frequency density instead of frequency where frequency density = frequency/ class width.

Correct Answer: B. \(\frac{3x - 7}{2x - 4}, x \neq 2\)

Explanation

\(h : x \to 2 - \frac{1}{2x - 3}\)

\(h(x) = \frac{2(2x - 3) - 1}{2x - 3} = \frac{4x - 7}{2x - 3}\)

Let x = h(y)

\(x = \frac{4y - 7}{2y - 3}\)

\(x(2y - 3) = 4y - 7 \implies 2xy - 4y = 3x - 7\)

\(y = \frac{3x - 7}{2x - 4}\)

\(h^{-1}(x) = \frac{3x - 7}{2x - 4}\)

Explanation

(a) \(a = (-3i + 4j) ; b = (-8i - 15j)\)

Let the angle between them be \(\theta\).

\(a \cdot b = |a||b| \cos \theta\)

\(24 - 60 = (\sqrt{(-3)^{2} + 4^{2}})(\sqrt{(-8)^{2} + (-15)^{2}}) \cos \theta\)

\(-36 = 5 \times 17 \times \cos \theta\)

\(\cos \theta = \frac{-36}{85} = -0.4235 \implies \theta = \cos^{-1} (-0.4235)\)

\(\theta = 115.06°\)

(b) \(a = 4 \cos 60° \\ 4 \sin 60° \end{pmatrix} = \begin{pmatrix} 2 \\ 3.464 \end{pmatrix}\)

\(b = \begin{pmatrix} 3 \cos 120° \\ 3 \sin 120° \end{pmatrix} = \begin{pmatrix} -1.5 \\ 2.598 \end{pmatrix}\)

\(a - b = \begin{pmatrix} 2 - (-1.5) \\ 3.464 - 2.598 \end{pmatrix} = \begin{pmatrix} 3.5 \\ 0.866 \end{pmatrix}\)

\(|a - b| = \sqrt{(3.5)^{2} + (0.866)^{2}} = \sqrt{12.25 + 0.75}\)

= \(\sqrt{13}\)

Unit vector along a - b = \(\frac{3.5i + 0.866j}{\sqrt{13}}\).

Correct Answer: B. \(\frac{140}{221}\)

Explanation

\(\cos(x+y) = \cos x\cos y - \sin x\sin y\)

Given \(\sin\) of an angle implies we have the value of the opposite and hypotenuse of the right-angled triangle. We find the adjacent side using Pythagoras' theorem.

\(Adj^{2} = Hyp^{2} - Opp^{2}\)

For triangle with angle x, \(adj = \sqrt{13^{2} - 5^{2}} = \sqrt{144} = 12\)

For triangle with angle y, \(adj = \sqrt{17^{2} - 8^{2}} = \sqrt{225} = 15\)

\(\therefore \cos x = \frac{12}{13}; \cos y = \frac{15}{17}\)

\(\cos(x+y) = (\frac{12}{13}\times\frac{15}{17}) - (\frac{5}{13}\times\frac{8}{17}) = \frac{180}{221} - \frac{40}{221}\)

= \(\frac{140}{221}\)

Explanation

M = \(\begin {pmatrix} 0 & 7 \\ 3 & -1 \end {pmatrix}\)

N = \(\begin {pmatrix} 2 & -1 \\ 5 & 3 \end {pmatrix}\)

(b) p(2,-3) = (2(2) + 3, 5(2) + 3(-3))

= (7, 10 - 9) = (7, 1)

p(7, 1) = (7(1), 2(7) - 1)

= (7, 21, -1) = (7, 20)

(c) Q`(2, 4) = (2(2) - 4, 5(2) + 14)

= (4 - 4, 10 + 12)

= (0, 22)

Correct Answer: B. 9

Explanation

Rearrange; 2, 2, 5, 7, 9, 9, 9, 10, 10, 11, 12, 18 Median = 9

Correct Answer: A. -i - 5j

Explanation

\(\overline{PQ}\) = 5i - 2j; \(\overline{QR}\) = 4i + 3j

\(\overline{RP}\) = \(\overline{PQ}\) - \(\overline{QR}\)

= 5i - 2j - [4i + 3j] = 5i - 2j - 4i - 3j

= i - 5j