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Ibeere 1 Ìròyìn
In an examination, 70% of the candidates passed. If 12 candidates are selected at random, find the probability that:
(a) at least two of them failed;
(b) exactly half of them passed;
(c) not more than one - six of them failed.
70% passed = p
30% failed = q
Using binomial theorem
P[x = x] = \(\begin{pmatrix} n \\ x \end{pmatrix}\)P\(^x\)q\(^{n - x}\)
(a) Probability that at least two of them failed:
P(X \(\geq\) 2) = 1 − P(X=0) − P(X=1)
When P(x = 0) = \(\begin{pmatrix} 12 \\ 12 \end{pmatrix}\)\(\begin{pmatrix} 7 \\ 10 \end{pmatrix}\)\(^2\)\(\begin{pmatrix} 3 \\ 10 \end{pmatrix}\)\(^0\)
= 0.01384129
When P(x = 1) = \(\begin{pmatrix} 12 \\ 11 \end{pmatrix}\)\(\begin{pmatrix} 7 \\ 10 \end{pmatrix}\)\(^{11}\)\(\begin{pmatrix} 3 \\ 10 \end{pmatrix}\)\(^1\)
= 0.07118376
P(X \(\geq\) 2) = 1 - 0.01384129 - 0.0711837 = 0.9149751 ≈ 0.915
(b) half of twelve = 6
P(x = 6) = \(\begin{pmatrix} 12 \\ 6 \end{pmatrix}\) \(\begin{pmatrix} 7 \\ 10 \end{pmatrix}\)\(^6\) \(\begin{pmatrix} 3 \\ 10 \end{pmatrix}\)\(^6\)
= 924 x 0.117649 x 0.000729 = 0.079245
(c) \(\frac{1}{6}\) x 12 = 2, i.e. P[x \(\leq\) 2) = P[x = 0] + P[ x = 1] + P[x = 2]
= 0.01384129 + 0.07118376 + \(\begin{pmatrix} 12 \\ 10 \end{pmatrix}\) \(\begin{pmatrix} 7 \\ 10 \end{pmatrix}\)\(^{10}\) \(\begin{pmatrix} 3 \\ 10 \end{pmatrix}\)\(^2\)
= 0.01384129 + 0.07118376 + 0.167790298 = 0.252815348.
Awọn alaye Idahun
70% passed = p
30% failed = q
Using binomial theorem
P[x = x] = \(\begin{pmatrix} n \\ x \end{pmatrix}\)P\(^x\)q\(^{n - x}\)
(a) Probability that at least two of them failed:
P(X \(\geq\) 2) = 1 − P(X=0) − P(X=1)
When P(x = 0) = \(\begin{pmatrix} 12 \\ 12 \end{pmatrix}\)\(\begin{pmatrix} 7 \\ 10 \end{pmatrix}\)\(^2\)\(\begin{pmatrix} 3 \\ 10 \end{pmatrix}\)\(^0\)
= 0.01384129
When P(x = 1) = \(\begin{pmatrix} 12 \\ 11 \end{pmatrix}\)\(\begin{pmatrix} 7 \\ 10 \end{pmatrix}\)\(^{11}\)\(\begin{pmatrix} 3 \\ 10 \end{pmatrix}\)\(^1\)
= 0.07118376
P(X \(\geq\) 2) = 1 - 0.01384129 - 0.0711837 = 0.9149751 ≈ 0.915
(b) half of twelve = 6
P(x = 6) = \(\begin{pmatrix} 12 \\ 6 \end{pmatrix}\) \(\begin{pmatrix} 7 \\ 10 \end{pmatrix}\)\(^6\) \(\begin{pmatrix} 3 \\ 10 \end{pmatrix}\)\(^6\)
= 924 x 0.117649 x 0.000729 = 0.079245
(c) \(\frac{1}{6}\) x 12 = 2, i.e. P[x \(\leq\) 2) = P[x = 0] + P[ x = 1] + P[x = 2]
= 0.01384129 + 0.07118376 + \(\begin{pmatrix} 12 \\ 10 \end{pmatrix}\) \(\begin{pmatrix} 7 \\ 10 \end{pmatrix}\)\(^{10}\) \(\begin{pmatrix} 3 \\ 10 \end{pmatrix}\)\(^2\)
= 0.01384129 + 0.07118376 + 0.167790298 = 0.252815348.
Ibeere 2 Ìròyìn
The data shows the ordered marks scored by students in a test: 11, 12, (2x + y), (x + 2y), 14, and ((y\(^2\) - 2x). Given that the median is 13\(\frac{1}{2}\) and y is greater than x by 1, find:
(a) the values of x and y
(b) correct to three significant figures, the standard deviation of the distribution.
Given: 11, 12, (2x + y), (x + 2y), 14, and ((y\(^2\) - 2x). median = 13\(\frac{1}{2}\) = \(\frac{27}{2}\)
Median = \(\frac{(2x + y) + (x + 2y)}{2}\) = \(\frac{27}{2}\)
\(\frac{3x + 3y}{2}\) = \(\frac{27}{2}\)
3x + 3y = 27
x + y = 9 - - - - - - - -(i)
also, y = x + 1 - - - - -(ii)
put y = x + 1 into eqn (ii)
x + x + 1 = 9
2x = 9 - 1 = 8
x = \(\frac{8}{2}\) = 4
y = x + 1 = 4 + 1 = 5
(b) Since x = 4 and y = 5
11, 12, (2(3) + 5), (4 + (2(5)), 14, (5\(^2\) - 2(4))
11, 12, 13, 14, 14, 17
Mean(\(\overline{x}\)) = \(\frac{\sum{x}}{n}\) = \(\frac{11 + 12 + 13 + 14 + 14 + 17}{6}\) = 13.5
| x | f | fx | (x - \(\overline{x}\)) | (x - \(\overline{x}\))\(^2\) | f(x - \(\overline{x}\))\(^2\) |
| 11 | 1 | 11 | - 2.5 | 6.25 | 6.25 |
| 12 | 1 | 12 | - 1.5 | 2.25 | 2.25 |
| 13 | 1 | 13 | - 0.5 | 0.25 | 0.25 |
| 14 | 2 | 28 | 0.5 | 0.25 | 0.50 |
| 17 | 1 | 17 | 3.5 | 12.5 | 12.25 |
\(\sum f(x - \overline{x})^2\) = 21.50
S.D = \(\sqrt{\frac{\sum f(x - \overline{x})^2}{\sum {f}}}\) = \(\sqrt{\frac{21.50}{6}}\) = 1.89
Awọn alaye Idahun
Given: 11, 12, (2x + y), (x + 2y), 14, and ((y\(^2\) - 2x). median = 13\(\frac{1}{2}\) = \(\frac{27}{2}\)
Median = \(\frac{(2x + y) + (x + 2y)}{2}\) = \(\frac{27}{2}\)
\(\frac{3x + 3y}{2}\) = \(\frac{27}{2}\)
3x + 3y = 27
x + y = 9 - - - - - - - -(i)
also, y = x + 1 - - - - -(ii)
put y = x + 1 into eqn (ii)
x + x + 1 = 9
2x = 9 - 1 = 8
x = \(\frac{8}{2}\) = 4
y = x + 1 = 4 + 1 = 5
(b) Since x = 4 and y = 5
11, 12, (2(3) + 5), (4 + (2(5)), 14, (5\(^2\) - 2(4))
11, 12, 13, 14, 14, 17
Mean(\(\overline{x}\)) = \(\frac{\sum{x}}{n}\) = \(\frac{11 + 12 + 13 + 14 + 14 + 17}{6}\) = 13.5
| x | f | fx | (x - \(\overline{x}\)) | (x - \(\overline{x}\))\(^2\) | f(x - \(\overline{x}\))\(^2\) |
| 11 | 1 | 11 | - 2.5 | 6.25 | 6.25 |
| 12 | 1 | 12 | - 1.5 | 2.25 | 2.25 |
| 13 | 1 | 13 | - 0.5 | 0.25 | 0.25 |
| 14 | 2 | 28 | 0.5 | 0.25 | 0.50 |
| 17 | 1 | 17 | 3.5 | 12.5 | 12.25 |
\(\sum f(x - \overline{x})^2\) = 21.50
S.D = \(\sqrt{\frac{\sum f(x - \overline{x})^2}{\sum {f}}}\) = \(\sqrt{\frac{21.50}{6}}\) = 1.89
Ibeere 3 Ìròyìn
PART II
A particle of weight 12 N lying on a horizontal ground is acted by forces F\(_1\) = (10 N, 090º), F\(_2\) = (16 N, 180º), F\(_3\) = (7 N, 300º) and F\(_4\) = (12N, 030º)
(a) Express all the forces acting on the particle as column vectors
(b) Find, correct to two decimal places, the magnitude of the:
(i) resultant forces;
(ii) acceleration with which the particle starts to move.[Take g = 10 ms\(^{-2}\)]
W = 12 N
From the diagram above,
| F(N) | F\(_x\) | F\(_y\) |
| 10 | 10cos90 | 10sin90 |
| 16 | 16cos180 | 16sin180 |
| 7 | 7cos30 | 7sin30 |
| 12 | 12cos300 | 12sin300 |
\(\sum{F_x}\) = - 3.938 N, \(\sum{F_y}\) = 3.108 N
Expressing in column vector
\(\begin{pmatrix} i & j \\ 10cos 90º & 10sin90º \\ 16cos180º & 16sin180º \\ 7cos30º & 7sin30º \\ 12cos 300º & 12sin300º \end{pmatrix}\)
Resultant R = \(\sqrt{(F_x)^2 + (F_y)^2}\)
R = \(\sqrt{( - 3.938)^2 + (3.108)^2}\) = 5.018N ≈ 5.02 N
(ii) R = ma (from Newton's law)
But, W = mg
m = \(\frac{\text{W}}{\text{g}}\) = \(\frac{12}{10}\) = 1.2 kg
From, R = ma, then, a = \(\frac{\text{R}}{\text{m}}\) = \(\frac{5.018}{1.2}\) = 4.182ms\(^{-2}\) ≈ 4.18 ms\(^{-2}\)
Awọn alaye Idahun
W = 12 N
From the diagram above,
| F(N) | F\(_x\) | F\(_y\) |
| 10 | 10cos90 | 10sin90 |
| 16 | 16cos180 | 16sin180 |
| 7 | 7cos30 | 7sin30 |
| 12 | 12cos300 | 12sin300 |
\(\sum{F_x}\) = - 3.938 N, \(\sum{F_y}\) = 3.108 N
Expressing in column vector
\(\begin{pmatrix} i & j \\ 10cos 90º & 10sin90º \\ 16cos180º & 16sin180º \\ 7cos30º & 7sin30º \\ 12cos 300º & 12sin300º \end{pmatrix}\)
Resultant R = \(\sqrt{(F_x)^2 + (F_y)^2}\)
R = \(\sqrt{( - 3.938)^2 + (3.108)^2}\) = 5.018N ≈ 5.02 N
(ii) R = ma (from Newton's law)
But, W = mg
m = \(\frac{\text{W}}{\text{g}}\) = \(\frac{12}{10}\) = 1.2 kg
From, R = ma, then, a = \(\frac{\text{R}}{\text{m}}\) = \(\frac{5.018}{1.2}\) = 4.182ms\(^{-2}\) ≈ 4.18 ms\(^{-2}\)
Ibeere 4 Ìròyìn
The magnitude of two vectors u and v are 10N and 12N respectively. If the magnitude of their resultant is 15N, calculate the angle between them.
From the diagram above, using cosine's law
Cos \(\theta\) = \(\frac{10^2 + 12^2 - 15^2}{2 \times 10 \times 12}\)
Cos \(\theta\) = \(\frac{100 + 144 - 225}{2 \times 10 \times 12}\) = \(\frac{244 - 225}{240}\)
= \(\frac{19}{240}\) = 0.0791667
\(\theta\) = Cos\(^{-1}\)(0.0791667) = 85.459º
But the angle between the two vectors = 180º - \(\theta\) = 180º - 85.459º = 94.54º
Awọn alaye Idahun
From the diagram above, using cosine's law
Cos \(\theta\) = \(\frac{10^2 + 12^2 - 15^2}{2 \times 10 \times 12}\)
Cos \(\theta\) = \(\frac{100 + 144 - 225}{2 \times 10 \times 12}\) = \(\frac{244 - 225}{240}\)
= \(\frac{19}{240}\) = 0.0791667
\(\theta\) = Cos\(^{-1}\)(0.0791667) = 85.459º
But the angle between the two vectors = 180º - \(\theta\) = 180º - 85.459º = 94.54º
Ibeere 5 Ìròyìn
Two events M and N are such that P(M) = \(\frac{1}{2}\), P(N) = \(\frac{9}{20}\) and P(M ∩ N) = \(\frac{11}{50}\)
(a) P(M ∩ N'): (b) P(M' ∩ N)
Given that: P(M) = \(\frac{1}{2}\), P(N) = \(\frac{9}{20}\) and P(M ∩ N) = \(\frac{11}{50}\)
Using the diagram above
(a) P(M ∩ N') = \(\frac{1}{2} - \frac{11}{50}\) = \(\frac{25 - 11}{50}\) = \(\frac{14}{50}\) = \(\frac{7}{25}\)
(b) P(M' ∩ N) = \(\frac{9}{20} - \frac{11}{50}\) = \(\frac{45 - 22}{100}\) = \(\frac{23}{100}\).
Awọn alaye Idahun
Given that: P(M) = \(\frac{1}{2}\), P(N) = \(\frac{9}{20}\) and P(M ∩ N) = \(\frac{11}{50}\)
Using the diagram above
(a) P(M ∩ N') = \(\frac{1}{2} - \frac{11}{50}\) = \(\frac{25 - 11}{50}\) = \(\frac{14}{50}\) = \(\frac{7}{25}\)
(b) P(M' ∩ N) = \(\frac{9}{20} - \frac{11}{50}\) = \(\frac{45 - 22}{100}\) = \(\frac{23}{100}\).
Ibeere 6 Ìròyìn
If tan x = \(\frac{1}{3}\), where 180º < x < 270º
evaluate \(\frac{sin2 x - cos x}{2 tan x + sin 2x}\), leaving the answer in surd form (radicals)
hyp\(^2\) = 3\(^2\) + 1\(^2\) (Pythagoras's theorem)
hyp = \(\sqrt{10}\)
Given: tan x = \(\frac{1}{3}\) ( sin and cos will be in the third quadrant and are both negative where 180º < x < 270º)
sin x = \(\frac{-1}{\sqrt{10}}\), cos x = \(\frac{-3}{\sqrt{10}}\)
\(\frac{sin2 x - cos x}{2 tan x + sin 2x}\) = \(\frac{2sinx cos x - cos x}{2 tan x + 2sin x cos x}\) but 2snx cox = \(\frac{3}{5}\)
\(\frac{sin2 x - cos x}{2 tan x + sin 2x}\) = \(\frac{\frac{3}{5} + \frac{3}{\sqrt{10}}}{2(\frac{1}{3}) + \frac{3}{5}}\) = \(\frac{\frac{3}{5} + \frac{3}{\sqrt{10}}}{\frac{19}{15}}\)
= (\(\frac{3}{5}\) + \(\frac{3\sqrt{10}}{10}\)) \(\div\) \(\frac{19}{15}\) = \(\frac{15}{19}\)[\(\frac{3}{5}\) + \(\frac{3}{\sqrt{10}}\)]
= \(\frac{9}{19}\) + \(\frac{45}{19\sqrt{10}}\)
= \(\frac{9}{19}\) + \(\frac{45 \sqrt{10}}{19(10)}\)
= \(\frac{9}{19}\) + \(\frac{9 \sqrt{10}}{19(2)}\)
= \(\frac{9}{19}\)[1 + \(\frac{\sqrt{10}}{2}\)]
Awọn alaye Idahun
hyp\(^2\) = 3\(^2\) + 1\(^2\) (Pythagoras's theorem)
hyp = \(\sqrt{10}\)
Given: tan x = \(\frac{1}{3}\) ( sin and cos will be in the third quadrant and are both negative where 180º < x < 270º)
sin x = \(\frac{-1}{\sqrt{10}}\), cos x = \(\frac{-3}{\sqrt{10}}\)
\(\frac{sin2 x - cos x}{2 tan x + sin 2x}\) = \(\frac{2sinx cos x - cos x}{2 tan x + 2sin x cos x}\) but 2snx cox = \(\frac{3}{5}\)
\(\frac{sin2 x - cos x}{2 tan x + sin 2x}\) = \(\frac{\frac{3}{5} + \frac{3}{\sqrt{10}}}{2(\frac{1}{3}) + \frac{3}{5}}\) = \(\frac{\frac{3}{5} + \frac{3}{\sqrt{10}}}{\frac{19}{15}}\)
= (\(\frac{3}{5}\) + \(\frac{3\sqrt{10}}{10}\)) \(\div\) \(\frac{19}{15}\) = \(\frac{15}{19}\)[\(\frac{3}{5}\) + \(\frac{3}{\sqrt{10}}\)]
= \(\frac{9}{19}\) + \(\frac{45}{19\sqrt{10}}\)
= \(\frac{9}{19}\) + \(\frac{45 \sqrt{10}}{19(10)}\)
= \(\frac{9}{19}\) + \(\frac{9 \sqrt{10}}{19(2)}\)
= \(\frac{9}{19}\)[1 + \(\frac{\sqrt{10}}{2}\)]
Ibeere 7 Ìròyìn
Find the sum of all natural numbers between 403 and 603 which are divisible by 7
Given: 403, 404, 405, 406, . . . . 602
Numbers divisible by 7 are 406, 413, 420, . . . . . 602.
S\(_n\) = \(\frac{n}{2}\)[2a + (n - 1)d]
a = 406, d = 7
T\(_n\) = a + (n - 1)d
602 = 406 + (n - 1)7
602 - 406 = 7n - 7
7n = 602 - 406 + 7 = 203
n = \(\frac{203}{7}\) = 29.
S\(_29\) = \(\frac{29}{2}\)[2(406) + (29 - 1)7]
= \(\frac{29}{2}\)[812+ (28 x 7)]
= \(\frac{29}{2}\)[812+ (196)]
= \(\frac{29}{2}\)[1008] = 14616
Awọn alaye Idahun
Given: 403, 404, 405, 406, . . . . 602
Numbers divisible by 7 are 406, 413, 420, . . . . . 602.
S\(_n\) = \(\frac{n}{2}\)[2a + (n - 1)d]
a = 406, d = 7
T\(_n\) = a + (n - 1)d
602 = 406 + (n - 1)7
602 - 406 = 7n - 7
7n = 602 - 406 + 7 = 203
n = \(\frac{203}{7}\) = 29.
S\(_29\) = \(\frac{29}{2}\)[2(406) + (29 - 1)7]
= \(\frac{29}{2}\)[812+ (28 x 7)]
= \(\frac{29}{2}\)[812+ (196)]
= \(\frac{29}{2}\)[1008] = 14616
Ibeere 8 Ìròyìn
SECTION B (PART 1)
A curve is given by y = 8x + \(\frac{27}{2x^2}\),
FIND:
(a) an expression for \(\frac{dy}{dx}\),
(b) the coordinates of the stationary point on the curve and the nature of the stationary point;
(c) the equation of the normal to the curve at (2, 2).
Given: y = 8x + \(\frac{27}{2x^2}\) = y = 8x + \(\frac{27}{2}\)x\(^{-2}\),
(a) \(\frac{dy}{dx}\) = 8x\(^0\) + \(\frac{27}{2}\)(-2)x\(^{-3}\)
= 8 - 27x\(^{-3}\) = 8 - \(\frac{27}{x^3}\)
(b) \(\frac{dy}{dx}\) = 8 - \(\frac{27}{x^3}\) = 0
= 8 = \(\frac{27}{x^3}\)
= 8x\(^3\) = 27
x\(^3\) = \(\frac{27}{8}\)
x = \( \sqrt[3]{\frac{27}{8}}\) = \(\frac{3}{2}\)
x = \(\frac{3}{2}\)
y = 8x + \(\frac{27}{2x^2}\) = 8(\(\frac{3}{2}\)) + \(\frac{27}{2}\)(\(\frac{3}{2}\))\(^{-2}\)
y = 4(3) + \(\frac{27}{2}\)(\(\frac{4}{9}\))
y = 12 + 6 = 18
Thus, stationary points are (\(\frac{3}{2}\), 18)
\(\frac{d^2y}{dx^2}\) = - \(\frac{27}{x^3}\) = - 27(-3)x\(^{-4}\) = \(\frac{81}{x^4}\) = \(\frac{81}{(\frac{3}{2})^4}\) = 81 x \(\frac{16}{81}\) = 16 > 0
For all x, the stationary point is a minimum.
(c) gradient = 8 - \(\frac{27}{x^3}\) = 8 - \(\frac{27}{2^3}\) = 8 - \(\frac{27}{8}\) = \(\frac{37}{8}\)
Equation of normal, at (2, 2)
but m\(_2\) = \(\frac{- 1}{m_1}\) = \(\frac{- 1}{\frac{37}{8}}\) = \(\frac{- 8}{37}\)
y - y\(_1\) = m\(_2\)[x - x\(_1\)]
y - 2 = \(\frac{- 8}{37}\)[x - 2]
37y - 74 = - 8x + 16
37y + 8x - 90 = 0
Awọn alaye Idahun
Given: y = 8x + \(\frac{27}{2x^2}\) = y = 8x + \(\frac{27}{2}\)x\(^{-2}\),
(a) \(\frac{dy}{dx}\) = 8x\(^0\) + \(\frac{27}{2}\)(-2)x\(^{-3}\)
= 8 - 27x\(^{-3}\) = 8 - \(\frac{27}{x^3}\)
(b) \(\frac{dy}{dx}\) = 8 - \(\frac{27}{x^3}\) = 0
= 8 = \(\frac{27}{x^3}\)
= 8x\(^3\) = 27
x\(^3\) = \(\frac{27}{8}\)
x = \( \sqrt[3]{\frac{27}{8}}\) = \(\frac{3}{2}\)
x = \(\frac{3}{2}\)
y = 8x + \(\frac{27}{2x^2}\) = 8(\(\frac{3}{2}\)) + \(\frac{27}{2}\)(\(\frac{3}{2}\))\(^{-2}\)
y = 4(3) + \(\frac{27}{2}\)(\(\frac{4}{9}\))
y = 12 + 6 = 18
Thus, stationary points are (\(\frac{3}{2}\), 18)
\(\frac{d^2y}{dx^2}\) = - \(\frac{27}{x^3}\) = - 27(-3)x\(^{-4}\) = \(\frac{81}{x^4}\) = \(\frac{81}{(\frac{3}{2})^4}\) = 81 x \(\frac{16}{81}\) = 16 > 0
For all x, the stationary point is a minimum.
(c) gradient = 8 - \(\frac{27}{x^3}\) = 8 - \(\frac{27}{2^3}\) = 8 - \(\frac{27}{8}\) = \(\frac{37}{8}\)
Equation of normal, at (2, 2)
but m\(_2\) = \(\frac{- 1}{m_1}\) = \(\frac{- 1}{\frac{37}{8}}\) = \(\frac{- 8}{37}\)
y - y\(_1\) = m\(_2\)[x - x\(_1\)]
y - 2 = \(\frac{- 8}{37}\)[x - 2]
37y - 74 = - 8x + 16
37y + 8x - 90 = 0
Ibeere 9 Ìròyìn
(a) Express \(\frac{9x}{(2x + 1)(x^2 + 1)}\) in partial fraction
(b) If \(^{2m}P_2\) - 10 = \(^m P_2\), find the positive value of m.
\(\frac{9x}{(2x + 1)(x^2 + 1)}\) = \(\frac{A}{2x + 1}\) + \(\frac{Bx + C}{x^2 + 1}\) = \(\frac{A(x^2 + 1) + (Bx + C)(x + 1)}{(2x + 1)(x^2 + 1)}\) - - - - -- - - -(i)
Using the cover-up method, put x = \(\frac{-1}{2}\)
A = \(\frac{9x}{x^2 + 1}\) = \(\frac{9(\frac{-1}{2})}{(\frac{-1}{2})^2 + 1}\) = \(\frac{\frac{-9}{2}}{\frac{5}{4}}\) = \(\frac{-18}{5}\)
From equation (i)
9x = A(\(x^2\) + 1) + (Bx + C)(x + 1)
9x = A\(x^2\) + A + 2B\(x^2\) + Bx + 2x + C
9x = (A + B)\(x^2\) + (B + 2C)x + (A + C)
comparing coefficients(x\(^2\)
A + 2B = 0
but A = \(\frac{-18}{5}\)
- A = 2B
-(\(\frac{-18}{5}\)) = 2B
18 = 10B
B = \(\frac{18}{10}\) = \(\frac{9}{5}\).
comparing coefficients (constant terms)
A + C = 0
C = - A = -(\(\frac{-18}{5}\))
C = \(\frac{18}{5}\)
\(\frac{9x}{(2x + 1)(x^2 + 1)}\) = \(\frac{\frac{-18}{5}}{2x + 1}\) + \(\frac{\frac{9x}{5} + \frac{18}{5}}{x^2 + 1}\)
= \(\frac{-18}{5(2x + 1)}\) + \(\frac{9x}{5(x^2 + 1)}\) + \(\frac{18}{5(x^2 + 1)}\)
= \(\frac{18}{5}\)[\(\frac{1}{x^2 + 1}\) + \(\frac{1}{2(x^2 + 1)}\) - \(\frac{1}{2x + 1}\)]
2(b) \(^{2m}P_2\) - 10 = \(^m P_2\)
\(\frac{(2m)!}{(2m - 2)!}\) - 10 = \(\frac{m!}{(m - 2)!}\)
\(\frac{(2m)(2m - 1) \times (2m - 2)!}{(2m - 2)!}\) - 10 = \(\frac{m \times (m - 1) \times (m - 2)!}{(m - 2)!}\)
⇒ 2m(2m - 1) - 10 = m(m - 1)
4m\(^2\) - 2m - 10 = m\(^2\) - m
4m\(^2\) - m\(^2\) - 2m + m - 10 = 0
3 m\(^2\) - m - 10 = 0
(3m + 5)(m - 2) = 0
m = 2( +ve value only)
Awọn alaye Idahun
\(\frac{9x}{(2x + 1)(x^2 + 1)}\) = \(\frac{A}{2x + 1}\) + \(\frac{Bx + C}{x^2 + 1}\) = \(\frac{A(x^2 + 1) + (Bx + C)(x + 1)}{(2x + 1)(x^2 + 1)}\) - - - - -- - - -(i)
Using the cover-up method, put x = \(\frac{-1}{2}\)
A = \(\frac{9x}{x^2 + 1}\) = \(\frac{9(\frac{-1}{2})}{(\frac{-1}{2})^2 + 1}\) = \(\frac{\frac{-9}{2}}{\frac{5}{4}}\) = \(\frac{-18}{5}\)
From equation (i)
9x = A(\(x^2\) + 1) + (Bx + C)(x + 1)
9x = A\(x^2\) + A + 2B\(x^2\) + Bx + 2x + C
9x = (A + B)\(x^2\) + (B + 2C)x + (A + C)
comparing coefficients(x\(^2\)
A + 2B = 0
but A = \(\frac{-18}{5}\)
- A = 2B
-(\(\frac{-18}{5}\)) = 2B
18 = 10B
B = \(\frac{18}{10}\) = \(\frac{9}{5}\).
comparing coefficients (constant terms)
A + C = 0
C = - A = -(\(\frac{-18}{5}\))
C = \(\frac{18}{5}\)
\(\frac{9x}{(2x + 1)(x^2 + 1)}\) = \(\frac{\frac{-18}{5}}{2x + 1}\) + \(\frac{\frac{9x}{5} + \frac{18}{5}}{x^2 + 1}\)
= \(\frac{-18}{5(2x + 1)}\) + \(\frac{9x}{5(x^2 + 1)}\) + \(\frac{18}{5(x^2 + 1)}\)
= \(\frac{18}{5}\)[\(\frac{1}{x^2 + 1}\) + \(\frac{1}{2(x^2 + 1)}\) - \(\frac{1}{2x + 1}\)]
2(b) \(^{2m}P_2\) - 10 = \(^m P_2\)
\(\frac{(2m)!}{(2m - 2)!}\) - 10 = \(\frac{m!}{(m - 2)!}\)
\(\frac{(2m)(2m - 1) \times (2m - 2)!}{(2m - 2)!}\) - 10 = \(\frac{m \times (m - 1) \times (m - 2)!}{(m - 2)!}\)
⇒ 2m(2m - 1) - 10 = m(m - 1)
4m\(^2\) - 2m - 10 = m\(^2\) - m
4m\(^2\) - m\(^2\) - 2m + m - 10 = 0
3 m\(^2\) - m - 10 = 0
(3m + 5)(m - 2) = 0
m = 2( +ve value only)
Ibeere 10 Ìròyìn
A body of mass 40 kg is placed on a rough inclined plane which makes an angle of 30\(^0\) with the horizontal. If a force of 420 N is applied upwards parallel to the plane. find the:
(a) maximum friction force that will keep the body in equilibrium;
(b) coefficient of friction.[Take g = 10ms\(^{-1}\)
From the diagram above;
fr + mg sin \(\theta\) = 420
fr = 420 - 40 x 10 sin 30º
fr = 420 - 200 = 220N
(b) coefficient of friction(μ) = \(\frac{\text{frictional force}}{\text{normal reaction}}\) = \(\frac{fr}{R}\)
μ = \(\frac{fr}{mg cos \theta}\) = \(\frac{220}{400 cos 30}\) = 0.63508
Therefore, μ = 0.635.
Awọn alaye Idahun
From the diagram above;
fr + mg sin \(\theta\) = 420
fr = 420 - 40 x 10 sin 30º
fr = 420 - 200 = 220N
(b) coefficient of friction(μ) = \(\frac{\text{frictional force}}{\text{normal reaction}}\) = \(\frac{fr}{R}\)
μ = \(\frac{fr}{mg cos \theta}\) = \(\frac{220}{400 cos 30}\) = 0.63508
Therefore, μ = 0.635.
Ibeere 11 Ìròyìn
Three linear transformations, P, Q, and R in the oxy plane are defined by
P: (x, y) → (-4x - y, 2x)
Q: (x, y) → (y, 6x - 9y)
R: (x, y) → (x - 2y, 3x + 5y)
(a) write down the matrices of P, Q, and R
(b) Find:
(i) 2P - 3R + Q;
(ii) QR;
(iii) the inverse of the matrix R.
Given:
P: (x, y) → (-4x - y, 2x)
Q: (x, y) → (y, 6x - 9y)
R: (x, y) → (x - 2y, 3x + 5y)
(a) the matrix of P, Q, and R
P = \(\begin{pmatrix} - 4 & -1 \\ 2 & 0 \end{pmatrix}\), Q = \(\begin{pmatrix} 0 & 1 \\ 6 & -9 \end{pmatrix}\), R = \(\begin{pmatrix} 1 & -2 \\ 3 & 5 \end{pmatrix}\),
(b) (i) 2P - 3R + Q
= 2\(\begin{pmatrix} - 4 & -1 \\ 2 & 0 \end{pmatrix}\) - 3\(\begin{pmatrix} 1 & -2 \\ 3 & 5 \end{pmatrix}\) + \(\begin{pmatrix} 0 & 1 \\ 6 & -9 \end{pmatrix}\)
= \(\begin{pmatrix} -8 & -2 \\ 4 & 0 \end{pmatrix}\) - \(\begin{pmatrix} 3 & -6 \\ 9 & 15 \end{pmatrix}\) + \(\begin{pmatrix} 0 & 1 \\ 6 & -9 \end{pmatrix}\)
= \(\begin{pmatrix} -11 & 5 \\ 1 & -24 \end{pmatrix}\)
(ii) QR = \(\begin{pmatrix} 0 & 1 \\ 6 & -9 \end{pmatrix}\)\(\begin{pmatrix} 1 & -2 \\ 3 & 5 \end{pmatrix}\)
= \(\begin{pmatrix} 0 + 3 & 0+ 5 \\ 6 - 27 & -12 - 45 \end{pmatrix}\)
= \(\begin{pmatrix} 3 & 5 \\ -21 & -57 \end{pmatrix}\)
(iii) the inverse of the matrix R. \(\begin{pmatrix} 1 & -2 \\ 3 & 5 \end{pmatrix}\),
|R| = (1 x 5) - ( -2 x 3) = 5 + 6 = 11
R\(^{-1}\) = \(\frac{Adj (R)}{|R|}\) = \(\frac{1}{11}\)\(\begin{pmatrix} 5 & 2 \\ -3 & 1 \end{pmatrix}\)
= \(\begin{pmatrix} \frac{5}{11} & \frac{2}{11} \\ \frac{-3}{11} & \frac{1}{11} \end{pmatrix}\)
Awọn alaye Idahun
Given:
P: (x, y) → (-4x - y, 2x)
Q: (x, y) → (y, 6x - 9y)
R: (x, y) → (x - 2y, 3x + 5y)
(a) the matrix of P, Q, and R
P = \(\begin{pmatrix} - 4 & -1 \\ 2 & 0 \end{pmatrix}\), Q = \(\begin{pmatrix} 0 & 1 \\ 6 & -9 \end{pmatrix}\), R = \(\begin{pmatrix} 1 & -2 \\ 3 & 5 \end{pmatrix}\),
(b) (i) 2P - 3R + Q
= 2\(\begin{pmatrix} - 4 & -1 \\ 2 & 0 \end{pmatrix}\) - 3\(\begin{pmatrix} 1 & -2 \\ 3 & 5 \end{pmatrix}\) + \(\begin{pmatrix} 0 & 1 \\ 6 & -9 \end{pmatrix}\)
= \(\begin{pmatrix} -8 & -2 \\ 4 & 0 \end{pmatrix}\) - \(\begin{pmatrix} 3 & -6 \\ 9 & 15 \end{pmatrix}\) + \(\begin{pmatrix} 0 & 1 \\ 6 & -9 \end{pmatrix}\)
= \(\begin{pmatrix} -11 & 5 \\ 1 & -24 \end{pmatrix}\)
(ii) QR = \(\begin{pmatrix} 0 & 1 \\ 6 & -9 \end{pmatrix}\)\(\begin{pmatrix} 1 & -2 \\ 3 & 5 \end{pmatrix}\)
= \(\begin{pmatrix} 0 + 3 & 0+ 5 \\ 6 - 27 & -12 - 45 \end{pmatrix}\)
= \(\begin{pmatrix} 3 & 5 \\ -21 & -57 \end{pmatrix}\)
(iii) the inverse of the matrix R. \(\begin{pmatrix} 1 & -2 \\ 3 & 5 \end{pmatrix}\),
|R| = (1 x 5) - ( -2 x 3) = 5 + 6 = 11
R\(^{-1}\) = \(\frac{Adj (R)}{|R|}\) = \(\frac{1}{11}\)\(\begin{pmatrix} 5 & 2 \\ -3 & 1 \end{pmatrix}\)
= \(\begin{pmatrix} \frac{5}{11} & \frac{2}{11} \\ \frac{-3}{11} & \frac{1}{11} \end{pmatrix}\)
Ibeere 12 Ìròyìn
Using the trapezium rule with seven ordinates, evaluate, correct to three decimal places, \(\int_{2.4}^{3.6} \frac{1}{\sqrt{x^2 - 2}}\)dx
\(\int_{2.4}^{3.6} \frac{1}{\sqrt{x^2 - 2}}\)dx using trapezium rule
| x | 2.4 | 2.6 | 2.8 | 3.0 | 3.2 | 3.4 | 3.6 |
| x\(^2\) - 2 | 3.76 | 4.76 | 5.84 | 7.00 | 8.24 | 9.56 | 10.96 |
| \(\sqrt{x^2 - 2}\) |
1.9391 | 2.1817 | 2.4166 | 2.6458 | 2.8705 | 3.0919 | 3.3106 |
| \(\frac{1}{\sqrt{x^2 - 2}}\) | 0.5157 | 0.4583 | 0.4138 | 0.3780 | 0.3484 | 0.3234 | 0.3021 |
| \(y_1\) | \(y_2\) | \(y_3\) | \(y_4\) | \(y_5\) | \(y_6\) | \(y_7\) |
= \(\frac{1}{2}\)(h)[[(\(y_1\) + \(y_7\)] + 2[\(y_2\) + \(y_3\) + \(y_4\) + \(y_5\) + \(y_6\)]]
= \(\frac{1}{2}\)(0.2)[0.5157 + 0.3021 ] + 2[0.4583 + 0.4138 + 0.3780 + 0.3484 + 0.3234]
= (0.1)[0.8178 + 3.8440] = 0.46618 ≈ 0.466 to 3 dp.
Awọn alaye Idahun
\(\int_{2.4}^{3.6} \frac{1}{\sqrt{x^2 - 2}}\)dx using trapezium rule
| x | 2.4 | 2.6 | 2.8 | 3.0 | 3.2 | 3.4 | 3.6 |
| x\(^2\) - 2 | 3.76 | 4.76 | 5.84 | 7.00 | 8.24 | 9.56 | 10.96 |
| \(\sqrt{x^2 - 2}\) |
1.9391 | 2.1817 | 2.4166 | 2.6458 | 2.8705 | 3.0919 | 3.3106 |
| \(\frac{1}{\sqrt{x^2 - 2}}\) | 0.5157 | 0.4583 | 0.4138 | 0.3780 | 0.3484 | 0.3234 | 0.3021 |
| \(y_1\) | \(y_2\) | \(y_3\) | \(y_4\) | \(y_5\) | \(y_6\) | \(y_7\) |
= \(\frac{1}{2}\)(h)[[(\(y_1\) + \(y_7\)] + 2[\(y_2\) + \(y_3\) + \(y_4\) + \(y_5\) + \(y_6\)]]
= \(\frac{1}{2}\)(0.2)[0.5157 + 0.3021 ] + 2[0.4583 + 0.4138 + 0.3780 + 0.3484 + 0.3234]
= (0.1)[0.8178 + 3.8440] = 0.46618 ≈ 0.466 to 3 dp.
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