Correct Answer: B. \(\begin{pmatrix} 2 & -5 \\ -1 & 3 \end{pmatrix}\)

Explanation

Let A = \(\begin{pmatrix} 3 & 5 \\ 1 & 2 \end{pmatrix}\)

|A| = (3 x 2 - 5 x 1)

= 6 - 5

= 1

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

Correct Answer: B. 3

Explanation

When you have two lines, \(y_{1}, y_{2}\), perpendicular to each other, the product of their slopes = -1.

\(3x + 4y + 6 = 0 \implies 4y = -6 - 3x\)

\(\therefore y = \frac{-6}{4} - \frac{3}{4}x\)

\(\frac{\mathrm d y}{\mathrm d x} = \frac{-3}{4}\)

Also, \(4x - by + 3 = 0 \implies by = 4x + 3\)

\(y = \frac{4}{b}x + \frac{3}{b}\)

\(\frac{\mathrm d y}{\mathrm d x} = \frac{4}{b}\)

\(\frac{-3}{4} \times \frac{4}{b} = -1 \implies \frac{4}{b} = \frac{4}{3}\)

\(b = 3\)

Correct Answer: C. 0.53 square units

Explanation

Area = \(\frac{1}{2} \times 1.33 \times 0.8\)

= 0.53 square unit

Correct Answer: D. 2n - 3

Explanation

\(\frac{^{n}P_{3}}{^{n}C_{2}} + ^{n}P_{0}\)

\(\frac{^{n}P_{3}}{^{n}C_{2}} = \frac{n!}{(n - 3)!} ÷ \frac{n!}{(n - 2)! 2!}\)

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

\(^{n}P_{0} = \frac{n!}{(n - 0)!} = 1\)

\(\frac{^{n}P_{3}}{^{n}C_{2}} + ^{n}P_{0} = 2n - 4 + 1 = 2n - 3\)

Explanation

They were expected to resolve the forces vertically and horizontally to have; (\(^{5 \cos 60^o}_{5 \sin 60^o}\)) + (\(^{P \cos 30^o}_{P \sin 30^o}\)) + (\(^{Q \cos 60^o}_{-Q \sin 60^o}\)) + (\(^{3 \cos 90^o}_{-3 \sin 90^o}\)) + (\(^{-5 \cos 0^o}_{5 \sin 0^o}\)) = (\(^0_0\))

which simplifies to;

(\(^{0.866P + 0.5Q}_{0.5P - 0.866Q}\)) = (\(^{2.5}_{-1.33}\))

This will yield two equations to be solved, the equations are 0.866P + 0.5Q = 2.5 and 0.5P - 0.866Q = -1.33

When these equations are solved, they will give P = 1.5 and Q = 2.4019

Correct Answer: A. -5.5

Explanation

\(\int_{\frac{1}{2}}^{1} \frac{x^{3} - 4}{x^{3}} \mathrm {d} x \equiv \int_{\frac{1}{2}}^{1} 1 - \frac{4}{x^{3}} \mathrm {d} x\).

= \(\int_{\frac{1}{2}}^{1} 1 - 4x^{-3} \mathrm {d} x\).

= \(\left. x - \frac{4x^{-2}}{-2}\right |_{\frac{1}{2}}^{1} = \left. x + \frac{2}{x^{2}}\right |_{\frac{1}{2}}^{1}\)

= \((1 + 2) - (\frac{1}{2} + 8) = 3 - 8.5 = -5.5\)

Explanation

When f(x) is divided by (x - a), the remainder is f(a). When (x - a) is a factor, then f(a) = 0.

\(f(x) = 6x^{4} + mx^{3} - 13x^{2} + nx + 14\)

(a) When divided by (x + 2), \(f(-2) = 6(-2^{4}) + m(-2^{3}) - 13(-2^{2}) + n(-2) + 14 = 0\)

= \(96 - 8m - 52 - 2n + 14 = 0\)

\(58 = 8m + 2n .... (1)\)

When divided by (x - 1), \(f(1) = 6(1^{4}) + m(1^{3}) - 13(1^{2}) + n(1) + 14 = 0\)

\(6 + m - 13 + n + 14 = 0\)

\(m + n = -7 ... (2)\)

\(m = -7 - n \) (from equation 2)

\(\therefore 58 = 8(-7 - n) + 2n \)

\(-56 - 8n + 2n = 58\)

\(-6n = 58 + 56 = 114 \implies n = -19\)

\(m = -7 - (-19) = -7 + 19 = 12\)

\(\therefore \text{m and n = 12 and -19}\)

\(\therefore f(x) = 6x^{4} + 12x^{3} - 13x^{2} - 19x + 14\)

(b) When divided by (x + 1)

\(f(-1) = 6(-1^{4}) + 12(-1^{3}) - 13(-1^{2}) - 19(-1) + 14\)

= \(6 - 12 - 13 + 19 + 14 = 14\).

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

Explanation

\(\frac{-1 + 0 + 9 + 3 + k + 5}{6} = 2 \implies 16 + k = 12\)

\(k = -4\)

Arranging -1, 0, 9, 3, -4, 5 in order: -4, -1, 0, 3, 5, 9

Median = \(\frac{0 + 3}{2} = \frac{3}{2}\)

Correct Answer: D. 97.2N

Explanation

Horizontal component = 230 cos 64 = 97.2N

Correct Answer: C. 3

Explanation

\(\frac{^{8}P_{x}}{^{8}C_{x}} = \frac{8!}{(8 - x)!} ÷ \frac{8!}{(8 - x)! x!} = 6\)

\(\frac{8!}{(8 - x)!} \times \frac{(8 - x)! x!}{8!} = x! = 6\)

\(x = 3\)