Explanation

(a) Tie Edem and his wife. There are six people to arrange in 6! ways, Edem and his wife can be arranged in 2! ways.

\(\therefore\) Total number of ways = 2!6! = 1440 ways.

(b) 4 red, 5 black balls

Total = 9 balls.

\(p(red) = p = \frac{4}{9} ; p(black) = q = \frac{5}{9}\)

5 balls are selected. The binomial probability distribution function is

\((p + q)^{5} = p^{5} + 5p^{4}q + 10p^{3}q^{2} + 10p^{2}q^{3} + 5pq^{4} + q^{5}\)

(i) \(p(\text{red ball picked 3 times}) = 10p^{3}q^{2}\)

= \(10 \times (\frac{4}{9})^{3} \times (\frac{5}{9})^{2}\)

= \(10 \times \frac{64}{729} \times \frac{25}{81}\)

= \(0.271\)

(ii) \(p(\text{black ball at most 2 times}) = p(none) + p(once) + p(twice)\)

= \(10p^{3}q^{2} + 5p^{4}q + p^{5}\)

= \(\frac{16000}{59049} \times 5(\frac{4}{9})^{4} (\frac{5}{9}) + (\frac{4}{9})^{5}\)

= \(\frac{16000 + 6400 + 1024}{59049}\)

= \(\frac{23424}{59049}\)

= \(0.3967\)

Explanation

\(\int_{1}^{3} (\frac{x - 1}{(x + 1)^{2}}) \mathrm {d} x\)

Let \(x + 1 = u, x - 1 = u - 2\)

When x = 3, u = 3 + 1 = 4

x = 1, u = 1 + 1 = 2

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

= \(\int_{2}^{4} (\frac{u}{u^{2}} - \frac{2}{u^{2}}) \mathrm {d} u\)

= \(\int_{2}^{4} (\frac{1}{u} - 2u^{-2}) \mathrm {d} u\)

= \([\ln u - \frac{2u^{-1}}{-1}]|_{2}^{4}\)

= \([\ln u + 2u^{-1}]|_{2}^{4}\)

= \([\ln u + \frac{2}{u}]|_{2}^{4}\)

= \((\ln 4 + \frac{2}{4}) - (\ln 2 + \frac{2}{2})\)

= \(\ln 4 - \ln 2 - \frac{1}{2}\)

= \(\ln (\frac{4}{2}) - \frac{1}{2}\)

= \(\ln 2 - \frac{1}{2}\).

Correct Answer: B. 2n + 1

Explanation

\(S_{n} = \frac{n}{2}(2a + (n - 1) d = n^{2} + 2n\)

\(n(2a + (n - 1) d = 2n^{2} + 4n\)

\(2an + n^{2}d - nd = 2n^{2} + 4n\)

\(n^{2}d = 2n^{2}\)

\(d = 2\)

\((2a - d) n = 4n\)

\(2a - d = 4 \implies 2a = 4 + d = 4 + 2 = 6\)

\(a = 3\)

\(T_{n} = a + (n - 1)d\)

= \(3 + (n - 1)2 = 3 + 2n - 2 = 2n + 1\)

Correct Answer: C. \(-\frac{896x^5}{27}\)

Explanation

rth term of a binomial expansion =\(^nC_r-1 a^{n-(r-1)}b^{r-1}\)

\(n = 10,r = 6 \therefore r-1=5\)

6th term =\(^{10}C_5 1^{10} - 5 (-\frac{2}{3}x)^5\)

\(=252*1*-\frac{32x^5}{243}=-\frac{896x^5}{27}\)

Explanation

(a) Resultant of forces, \(R = -63j + (32.14i + 38.3j) + (14i - 24.25j)\)

= \(46.14i - 48.95j\)

Magnitude |R| = \(\sqrt{(46.14)^{2} + (-48.95)^{2}}\)

= \(\sqrt{2128.9 + 2396.1}\)

= \(\sqrt{4525}\)

= \(67.268 \approxeq 67.3N\) (to 1 d.p)

(b) \(a = \frac{F}{m}\)

= \(\frac{67.268}{5} = 13.454 ms^{-2}\)

\(\approxeq 13.5 ms^{-2}\)

Correct Answer: A. \(\frac{6}{5}\)

Explanation

\(\sqrt{4x^2 + 1}\) = \(\frac{13x}{6}\)

4x\(^2\) + 1 = \(\frac{169x^2}{36}\)

4 + x\(^2\) = \(\frac{169x^2}{36}\)

cross multiply

169x\(^2\) - 144x\(^2\) = 36

25x\(^2\) = 36

x\(^2\) = \(\frac{36}{25}\)

: x = \(\pm\frac{6}{5}\)

Explanation

(a) \(x = 3i - j ; y = 2i + kj\)

\(x \cdot y = |x||y| \cos \theta\)

\((3i - j) \cdot (2i + kj) = (\sqrt{3^{2} + (-1)^{2}})(\sqrt{2^{2} + k^{2}}) \cos \theta\)

\(6 - k = (\sqrt{10})(\sqrt{4 + k^{2}})(\frac{\sqrt{5}}{5})\)

\(5(6 - k) = \sqrt{10(4 + k^{2})(5)}\)

\(30 - 5k = \sqrt{200 + 50k^{2}}\)

\((30 - 5k)^{2} = 200 + 50k^{2}\)

\(900 - 300k + 25k^{2} = 200 + 50k^{2}\)

\(900 - 200 - 300k + 25k^{2} - 50k^{2} = 0\)

\(25k^{2} + 300k - 700 = 0 \implies k^{2} + 12k - 28 = 0\)

\(k^{2} - 2k + 14k - 28 = 0 \implies k(k - 2) + 14(k - 2) = 0\)

\(\text{k = 2 or -14}\).

(b)

ABCD is a parallelogram if AD = BC.

\(\overrightarrow{AD} = \overrightarrow{AB} + \overrightarrow{BD}\)

= \(\begin{pmatrix} -5 \\ -1 \end{pmatrix} + \begin{pmatrix} 4 \\ -7 \end{pmatrix}\)

= \(\begin{pmatrix} -1 \\ -8 \end{pmatrix}\)

\(\overrightarrow{BC} = \overrightarrow{BA} + \overrightarrow{AC}\)

= \(\begin{pmatrix} 5 \\ 1 \end{pmatrix} + \begin{pmatrix} -6 \\ -9 \end{pmatrix}\)

= \(\begin{pmatrix} -1 \\ -8 \end{pmatrix}\).

\(\overrightarrow{AD} = \overrightarrow{BC}\). ABCD is a parallelogram.

Correct Answer: C. 291

Explanation

Note that every 4- digit and 5- digit numbers formed is greater than 150. Therefore, \(^{5}P_{5} + ^{5}P_{4} = \frac{5!}{(5 - 5)!} + \frac{5!}{(5 - 4)!} = 120 + 120 = 240\) numbers are greater than 150 already.

Among the 3- digit numbers, the numbers 123, 124, 125, 132, 134, 135, 142, 143 and 145 are removed from the ones that meet the criterion so we have:

\(^{5}P_{3} - 9 = \frac{5!}{(5 - 3)!} = 60 - 9 = 51 \implies 240 + 51 = 291\) numbers are greater than 150.

Explanation

(i) Force parallel to plane = \(mg \sin 30\)

= \(5 \times 10 \times 0.5 \)

= \(25 N\)

(ii) Force required to keep the body in equilibrium

= \(mg \sin 30 - \mu R\)

= \(25 - \mu \times 5 \times \frac{5\sqrt{3}}{2}\)

= \(25(1 - \mu) \frac{\sqrt{3}}{2}\)

b)

Since the plank does not move, all forces and torques are balanced. Forces:

RA + RB - mCg - mpg=0,

torques (consider A as the pivot):

\(\frac{mp}{10}\)g ⋅ 0.5 − \(\frac{9}{10}\)mp?g⋅4.5 −mC?g⋅3.5+100⋅8=0

mp?=13 kg,

RA?=110 N.

Explanation

(p + 1/2√3)(1 - √3)\(^2\) = 3- √3

(1- √3)\(^2\) = 4 -2√3

(p + 1/2√3) (4 -2√3) = 3 - √3

4p - 2p√3 - 2√3 - 3 = 3 - √3

4p - 2p√3 = 3 - √3 + 3 + 2√3

= 2p(2 - √3) = 6 + √3

2p= \(\frac{6+ √3}{2-√3}\)

p = \(\frac{2(6+ √3)}{2-√3}\)

= \(\frac{12+ 2√3}{2-√3}\)

= \(\frac{(12+ 2√3)(2+√3)}{(2-√3)(2+√3)}\)

: 30 + 16√3