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Ajụjụ 1 Ripọtì
8. An object is projected vertically upward with a velocity of 80 ms\(^{-1}\). Find the;
a. Maximum height reached (Leave your answer in whole number 'abc.')
b. Time taken to return to the point of projection [ g = 10m/s\(^2\)]
Maximum height, h, v = 0m/s at max height
Recall v\(^2\) = u\(^2\) − 2gh
0\(^2\) = 80\(^2\) − 2(10)h
0 = 6400 - 20h
h = \(\frac{6400}{20}\)
h = 320 m
b. Time taken to reach maximum height, t
V = u − gt
0 = 80−10t
10t = 80
t = \(\frac{80}{10}\) = 8s
Since it takes 8 s to reach maximum height, it will also take 8 s to return to the point of projection; hence, it is 16 s
Akọwa Nkọwa
Maximum height, h, v = 0m/s at max height
Recall v\(^2\) = u\(^2\) − 2gh
0\(^2\) = 80\(^2\) − 2(10)h
0 = 6400 - 20h
h = \(\frac{6400}{20}\)
h = 320 m
b. Time taken to reach maximum height, t
V = u − gt
0 = 80−10t
10t = 80
t = \(\frac{80}{10}\) = 8s
Since it takes 8 s to reach maximum height, it will also take 8 s to return to the point of projection; hence, it is 16 s
Ajụjụ 2 Ripọtì
4. Find the equation of a tangent to the curve y = \(\frac{x - 1}{2x + 1}\), x \(\pm\) \(\frac{-1}{2}\) at the point(1, 0)
Leave your answer in this format: ay - bx + c = 0
To find the equation of the tangent to the curve \( y = \frac{x - 1}{2x + 1} \) at the point (1, 0):
First, verify the point lies on the curve:
Substitute \( x = 1 \): \( y = \frac{1 - 1}{2(1) + 1} = \frac{0}{3} = 0 \). Yes, it does.
Next, find the slope of the tangent by computing the derivative \( \frac{dy}{dx} \):
\(y = \frac{x - 1}{2x + 1}\)
Using the quotient rule:
\(\frac{dy}{dx} = \frac{(1)(2x + 1) - (x - 1)(2)}{(2x + 1)^2} = \frac{2x + 1 - 2x + 2}{(2x + 1)^2} = \frac{3}{(2x + 1)^2}\)
At \( x = 1 \):
\(m = \frac{3}{(2(1) + 1)^2} = \frac{3}{9} = \frac{1}{3}\)
Equation of the tangent line at (1, 0):
\(y - 0 = \frac{1}{3}(x - 1)\)
\(y = \frac{1}{3}x - \frac{1}{3}\)
Multiply through by 3: \( 3y = x - 1 \)
Rearrange: \( 3y - x + 1 = 0\)
Akọwa Nkọwa
To find the equation of the tangent to the curve \( y = \frac{x - 1}{2x + 1} \) at the point (1, 0):
First, verify the point lies on the curve:
Substitute \( x = 1 \): \( y = \frac{1 - 1}{2(1) + 1} = \frac{0}{3} = 0 \). Yes, it does.
Next, find the slope of the tangent by computing the derivative \( \frac{dy}{dx} \):
\(y = \frac{x - 1}{2x + 1}\)
Using the quotient rule:
\(\frac{dy}{dx} = \frac{(1)(2x + 1) - (x - 1)(2)}{(2x + 1)^2} = \frac{2x + 1 - 2x + 2}{(2x + 1)^2} = \frac{3}{(2x + 1)^2}\)
At \( x = 1 \):
\(m = \frac{3}{(2(1) + 1)^2} = \frac{3}{9} = \frac{1}{3}\)
Equation of the tangent line at (1, 0):
\(y - 0 = \frac{1}{3}(x - 1)\)
\(y = \frac{1}{3}x - \frac{1}{3}\)
Multiply through by 3: \( 3y = x - 1 \)
Rearrange: \( 3y - x + 1 = 0\)
Ajụjụ 3 Ripọtì
SECTION B
9a. Simplify \(\frac{\sqrt{75} - 3}{\sqrt{3} + 1}\), leaving your answers in the form a + b\(\sqrt{c}\), where a, b, and c are rational numbers.
bi. The points (7,3), (2,8), and (-3,3) lie on a circle. Find the equation
bii. Find the radius of the circle.
a. \(\frac{\sqrt{75}-3}{\sqrt{3}+1}\)
First, simplify the radical:
\(\sqrt{75} = \sqrt{25 \cdot 3} = 5\sqrt{3}\)
So the expression becomes:
\(\frac{5\sqrt{3}-3}{\sqrt{3}+1}\)
Now rationalize the denominator by multiplying top and bottom by \(\sqrt{3}\) - 1):
\(\frac{(5\sqrt{3}-3)(\sqrt{3}-1)}{(\sqrt{3}+1)(\sqrt{3}-1)}\)
Denominator: 3 - 1 = 2
Numerator: 15 - 5\(\sqrt{3} - 3\sqrt{3} + 3 = 18 - 8\sqrt{3}\)
So: \(\frac{18 - 8\sqrt{3}}{2} = 9 - 4\sqrt{3}\).
bi. The general equation of a circle is \(x^2 + y^2 + 2gx + 2fy + c = 0\)
Substitute the three points: Point (7, 3):
\(49 + 9 + 14g + 6f + c = 0 \implies 14g + 6f + c = -58 \quad \text{(1)}\)
Point (2, 8):
\(4 + 64 + 4g + 16f + c = 0 \implies 4g + 16f + c = - 68 \quad \text{(2)}\)
Point (-3, 3):
\(9 + 9 - 6g + 6f + c = 0 \implies -6g + 6f + c = -18 \quad \text{(3)}\)
Subtract (2) from (1):
\(10g - 10f = 10 \implies g - f = 1 \implies g = f + 1 \quad \text{(4)}\)
Subtract (3) from (2):
\(10g + 10f = -50 \implies g + f = -5 \quad \text{(5)}\)
Solve (4) and (5):
\((f + 1) + f = -5 \implies 2f + 1 = -5 \implies 2f = -6 \implies f = -3\)
g = -3 + 1 = -2
Substitute \( g = -2 \), \( f = -3 \) into (3):
\(-6(-2) + 6(-3) + c = -18 \implies 12 - 18 + c = -18 \implies c = -12\)
The equation of the circle is: \(x^2 + y^2 - 4x - 6y - 12 = 0\)
bii. The centre is -g, -f = (2, 3).
The radius \( r \) is given by \(r = \sqrt{g^2 + f^2 - c} = \sqrt{(-2)^2 + (-3)^2 - (-12)} = \sqrt{4 + 9 + 12} = \sqrt{25} = 5\)
Radius of the circle = 5 units.
Akọwa Nkọwa
a. \(\frac{\sqrt{75}-3}{\sqrt{3}+1}\)
First, simplify the radical:
\(\sqrt{75} = \sqrt{25 \cdot 3} = 5\sqrt{3}\)
So the expression becomes:
\(\frac{5\sqrt{3}-3}{\sqrt{3}+1}\)
Now rationalize the denominator by multiplying top and bottom by \(\sqrt{3}\) - 1):
\(\frac{(5\sqrt{3}-3)(\sqrt{3}-1)}{(\sqrt{3}+1)(\sqrt{3}-1)}\)
Denominator: 3 - 1 = 2
Numerator: 15 - 5\(\sqrt{3} - 3\sqrt{3} + 3 = 18 - 8\sqrt{3}\)
So: \(\frac{18 - 8\sqrt{3}}{2} = 9 - 4\sqrt{3}\).
bi. The general equation of a circle is \(x^2 + y^2 + 2gx + 2fy + c = 0\)
Substitute the three points: Point (7, 3):
\(49 + 9 + 14g + 6f + c = 0 \implies 14g + 6f + c = -58 \quad \text{(1)}\)
Point (2, 8):
\(4 + 64 + 4g + 16f + c = 0 \implies 4g + 16f + c = - 68 \quad \text{(2)}\)
Point (-3, 3):
\(9 + 9 - 6g + 6f + c = 0 \implies -6g + 6f + c = -18 \quad \text{(3)}\)
Subtract (2) from (1):
\(10g - 10f = 10 \implies g - f = 1 \implies g = f + 1 \quad \text{(4)}\)
Subtract (3) from (2):
\(10g + 10f = -50 \implies g + f = -5 \quad \text{(5)}\)
Solve (4) and (5):
\((f + 1) + f = -5 \implies 2f + 1 = -5 \implies 2f = -6 \implies f = -3\)
g = -3 + 1 = -2
Substitute \( g = -2 \), \( f = -3 \) into (3):
\(-6(-2) + 6(-3) + c = -18 \implies 12 - 18 + c = -18 \implies c = -12\)
The equation of the circle is: \(x^2 + y^2 - 4x - 6y - 12 = 0\)
bii. The centre is -g, -f = (2, 3).
The radius \( r \) is given by \(r = \sqrt{g^2 + f^2 - c} = \sqrt{(-2)^2 + (-3)^2 - (-12)} = \sqrt{4 + 9 + 12} = \sqrt{25} = 5\)
Radius of the circle = 5 units.
Ajụjụ 4 Ripọtì
5a. There are 6 points in a plane. How many triangles can be formed with the points?
b. A family of 6 is to be seated in a row. In how many ways can this be done if the father and mother are not to sit together?
Leave your answer in whole numbers " abc."
a. Assuming no three points are collinear (as is standard unless specified otherwise), the number of triangles is the number of ways to choose 3 points out of 6, which form a triangle.
I.e \(^6C_3\) = \(\frac{6!}{(6-3)!3!}\) = \(\frac{6!}{3!3!}\)
= \(\frac{6 \times 5 \times 4 \times 3!}{3! \times 3 \times 2 \times 1}\) = \(\frac{6 \times 5 \times 4}{3 \times 2 \times 1}\) = 5 x 4 = 20 triangles.
b. Total number of ways to seat 6 people in a row (no restrictions): 6! =720
Number of ways where father and mother sit together: Treat father and mother as a single unit (they can switch places within the unit: father-mother or mother-father). This gives 5 units to arrange: 5! x 2 = 120 x 2 = 240ways.
Number of ways where they are not together: Total - Together = 720 − 240 = 480 ways
Akọwa Nkọwa
a. Assuming no three points are collinear (as is standard unless specified otherwise), the number of triangles is the number of ways to choose 3 points out of 6, which form a triangle.
I.e \(^6C_3\) = \(\frac{6!}{(6-3)!3!}\) = \(\frac{6!}{3!3!}\)
= \(\frac{6 \times 5 \times 4 \times 3!}{3! \times 3 \times 2 \times 1}\) = \(\frac{6 \times 5 \times 4}{3 \times 2 \times 1}\) = 5 x 4 = 20 triangles.
b. Total number of ways to seat 6 people in a row (no restrictions): 6! =720
Number of ways where father and mother sit together: Treat father and mother as a single unit (they can switch places within the unit: father-mother or mother-father). This gives 5 units to arrange: 5! x 2 = 120 x 2 = 240ways.
Number of ways where they are not together: Total - Together = 720 − 240 = 480 ways
Ajụjụ 5 Ripọtì
16a. A ball P moving with velocity 2 u m/s, collides with a similar ball Q, of different mass, which is at rest. After the collision, Q moves with u m/s and P with velocity \(\frac{1}{2}\) u m/s in the opposite direction. Find the ratio of the mass of P and Q.
b. Two forces of magnitude 3N and 7N have a resultant of magnitude 5N. Calculate, correct to one decimal place, the angle between the two forces.
c. AB\(^→\) \(\left| \begin{array}{cc} -4 \\ 6 \end{array} \right|\) and CB\(^→\) \(\left| \begin{array}{cc} 2 \\ 3 \end{array} \right|\) are two vectors in the XY plane. If V is the midpoint AB\(^→\). Find CV\(^→\)
16a.Let the mass of ball P be \( m \) and the mass of ball Q be \( M \).
Conservation of linear momentum (before and after collision):
\(m \times 2u + M \times 0 = m \times \left(-\frac{1}{2}u\right) + M \times u\)
2mu = -\(\frac{1}{2}\)mu + Mu
Divide through by \( u \) (assuming \( u \neq 0 \)):
2m = -\(\frac{1}{2}\)m + M
M = 2m + \(\frac{1}{2}\)m = \(\frac{5}{2}\)m
Thus, the ratio of the mass of P to the mass of Q is
m: M = 2: 5
bi. Let the angle between the two forces be \( \theta \).
By the parallelogram law (or cosine rule for vector addition):
\(R^2 = 3^2 + 7^2 + 2 \times 3 \times 7 \times \cos\theta\)
\(5^2 = 9 + 49 + 42\cos\theta\)
\(25 = 58 + 42\cos\theta\)
\(42\cos\theta\) = 25 - 58 = -33
\(\cos\theta = -\frac{33}{42} = -\frac{11}{14} \approx -0.7857\)
\(\theta = \cos^{-1}(-0.7857) = 141.8^\circ\)
c. Given: \(\overrightarrow{AB} = \begin{pmatrix} -4 \\ 6 \end{pmatrix}, \quad
\overrightarrow{CB} = \begin{pmatrix} 2 \\ 3 \end{pmatrix}\)
V is the midpoint of AB, so its position vector satisfies
\(\vec{V} = \frac{\vec{A} + \vec{B}}{2}.\)
Vector \(\overrightarrow{CV}\) is
\(\overrightarrow{CV} = \vec{V} - \vec{C} = \overrightarrow{CB} - \frac{1}{2}\overrightarrow{AB}.\)
Substitute the given vectors:
\(\frac{1}{2}\overrightarrow{AB} = \begin{pmatrix} -2 \\ 3 \end{pmatrix},\)
\(-\frac{1}{2}\overrightarrow{AB} = \begin{pmatrix} 2 \\ -3 \end{pmatrix}.\)
\(\overrightarrow{CV} = \begin{pmatrix} 2 \\ 3 \end{pmatrix} + \begin{pmatrix} 2 \\ -3 \end{pmatrix} = \begin{pmatrix} 4 \\ 0 \end{pmatrix}.\)
Akọwa Nkọwa
16a.Let the mass of ball P be \( m \) and the mass of ball Q be \( M \).
Conservation of linear momentum (before and after collision):
\(m \times 2u + M \times 0 = m \times \left(-\frac{1}{2}u\right) + M \times u\)
2mu = -\(\frac{1}{2}\)mu + Mu
Divide through by \( u \) (assuming \( u \neq 0 \)):
2m = -\(\frac{1}{2}\)m + M
M = 2m + \(\frac{1}{2}\)m = \(\frac{5}{2}\)m
Thus, the ratio of the mass of P to the mass of Q is
m: M = 2: 5
bi. Let the angle between the two forces be \( \theta \).
By the parallelogram law (or cosine rule for vector addition):
\(R^2 = 3^2 + 7^2 + 2 \times 3 \times 7 \times \cos\theta\)
\(5^2 = 9 + 49 + 42\cos\theta\)
\(25 = 58 + 42\cos\theta\)
\(42\cos\theta\) = 25 - 58 = -33
\(\cos\theta = -\frac{33}{42} = -\frac{11}{14} \approx -0.7857\)
\(\theta = \cos^{-1}(-0.7857) = 141.8^\circ\)
c. Given: \(\overrightarrow{AB} = \begin{pmatrix} -4 \\ 6 \end{pmatrix}, \quad
\overrightarrow{CB} = \begin{pmatrix} 2 \\ 3 \end{pmatrix}\)
V is the midpoint of AB, so its position vector satisfies
\(\vec{V} = \frac{\vec{A} + \vec{B}}{2}.\)
Vector \(\overrightarrow{CV}\) is
\(\overrightarrow{CV} = \vec{V} - \vec{C} = \overrightarrow{CB} - \frac{1}{2}\overrightarrow{AB}.\)
Substitute the given vectors:
\(\frac{1}{2}\overrightarrow{AB} = \begin{pmatrix} -2 \\ 3 \end{pmatrix},\)
\(-\frac{1}{2}\overrightarrow{AB} = \begin{pmatrix} 2 \\ -3 \end{pmatrix}.\)
\(\overrightarrow{CV} = \begin{pmatrix} 2 \\ 3 \end{pmatrix} + \begin{pmatrix} 2 \\ -3 \end{pmatrix} = \begin{pmatrix} 4 \\ 0 \end{pmatrix}.\)
Ajụjụ 6 Ripọtì
6. The table shows the distribution of the ages of a group of people in a village.
| Ages(in years) | 15-18 | 19-22 | 23-26 | 27-30 | 31-34 | 35-38 |
| Frequency | 40 | 33 | 25 | 10 | 8 | 4 |
Using an assumed mean of 24.5. Calculate the mean distribution.
Assumed mean, Am = 24.5
| Ages | f | x | d = x - Am | fd |
| 15 - 18 | 40 | 16.5 | -8.0 | -320 |
| 19 - 22 | 33 | 20.5 | -4.0 | -132 |
| 23 - 26 | 25 | 24.5 | 0.0 | 0 |
| 27 - 30 | 10 | 28.5 | 4.0 | 40 |
| 31 - 34 | 8 | 32.5 | 8.0 | 64 |
| 35 - 38 | 4 | 36.5 | 12.0 | 48 |
| \(\sum\)f = 120 | \(\sum\)fd = - 300 |
Mean (\(\overline{x}\)) = Am + \(\frac{\sum fd}{\sum f}\) = 24.5 + \(\frac{(-300)}{120}\)
Mean (\(\overline{x}\)) = 24.5 - 2.5 = 22.0
Note:(x = sum of upper and lower limit divided by 2)
Akọwa Nkọwa
Assumed mean, Am = 24.5
| Ages | f | x | d = x - Am | fd |
| 15 - 18 | 40 | 16.5 | -8.0 | -320 |
| 19 - 22 | 33 | 20.5 | -4.0 | -132 |
| 23 - 26 | 25 | 24.5 | 0.0 | 0 |
| 27 - 30 | 10 | 28.5 | 4.0 | 40 |
| 31 - 34 | 8 | 32.5 | 8.0 | 64 |
| 35 - 38 | 4 | 36.5 | 12.0 | 48 |
| \(\sum\)f = 120 | \(\sum\)fd = - 300 |
Mean (\(\overline{x}\)) = Am + \(\frac{\sum fd}{\sum f}\) = 24.5 + \(\frac{(-300)}{120}\)
Mean (\(\overline{x}\)) = 24.5 - 2.5 = 22.0
Note:(x = sum of upper and lower limit divided by 2)
Ajụjụ 7 Ripọtì
13a. The table below shows the distribution of hours spent at work by the employees of a factory in a week
| Time(hours) | 20 - 29 | 30 - 39 | 40 - 49 | 50 - 59 | 60 - 69 | 70 - 79 |
| No. of persons | 8 | 11 | 23 | 25 | 8 | 5 |
Draw an Ogive for the distribution
b. Using your graph, estimate
i. the median.
ii. estimate the lower quartile
iii. 40th percentile
iv. number of employees that spent at least 50 hours 30 mins.
| Time(hours) | Frequency | Cumulative frequency | Upper boundary |
| 20-29 | 8 | 8 | 29.5 |
| 30-39 | 11 | 19 | 39.5 |
| 40-49 | 23 | 42 | 49.5 |
| 50-59 | 25 | 67 | 59.5 |
| 60-69 | 8 | 75 | 69.5 |
| 70-79 | 5 | 80 | 79.5 |
b. Median = (\(\frac{ n + 1}{2}\))\(^{th}\) data
Median = \(\frac{80 + 1}{2}\) = \(\frac{81}{2}\) = 40.5th
So, from the graph, the median is 48.8 hours.
ii. Lower quartile, Q\(_1\) = (\(\frac{n + 1}{4}\))\(^{th}\) data
Q\(_1\) = \(\frac{80 + 1}{4}\) = \(\frac{81}{4}\) = 20.25th data
Q\(_1\) = 40 hours from the graph
iii. 40th percentile
= \(\frac{40}{100}\) x 80 = 32\(^{nd}\) data.
From the graph, 40th percentile = 45.5 hours.
iv. The number of employees who spent at least 50 hours 30 mins = 32.
Akọwa Nkọwa
| Time(hours) | Frequency | Cumulative frequency | Upper boundary |
| 20-29 | 8 | 8 | 29.5 |
| 30-39 | 11 | 19 | 39.5 |
| 40-49 | 23 | 42 | 49.5 |
| 50-59 | 25 | 67 | 59.5 |
| 60-69 | 8 | 75 | 69.5 |
| 70-79 | 5 | 80 | 79.5 |
b. Median = (\(\frac{ n + 1}{2}\))\(^{th}\) data
Median = \(\frac{80 + 1}{2}\) = \(\frac{81}{2}\) = 40.5th
So, from the graph, the median is 48.8 hours.
ii. Lower quartile, Q\(_1\) = (\(\frac{n + 1}{4}\))\(^{th}\) data
Q\(_1\) = \(\frac{80 + 1}{4}\) = \(\frac{81}{4}\) = 20.25th data
Q\(_1\) = 40 hours from the graph
iii. 40th percentile
= \(\frac{40}{100}\) x 80 = 32\(^{nd}\) data.
From the graph, 40th percentile = 45.5 hours.
iv. The number of employees who spent at least 50 hours 30 mins = 32.
Ajụjụ 8 Ripọtì
11a. Using the substitution U = 5 - x\(^2\)
evaluate \(\int _1^2 \frac{\text{x}}{\sqrt{5 - x^2}}\) dx
b. If y = px\(^2\) + qx, \(\frac{\text{dy}}{\text{dx}}\) = 7 and \(\frac{d^2y}{dx^2}\) = 6. Find the values of p and q.
11a. Using the substitution \( u = 5 - x^2 \),
\( du = -2x \, dx \) so \( x \, dx = -\frac{1}{2} du \).
Limits: when \( x = 1 \), \( u = 4 \); when \( x = 2 \), \( u = 1 \).
\(\int_1^2 \frac{x}{\sqrt{5 - x^2}} \, dx = \int_4^1 \frac{1}{\sqrt{u}} \left( -\frac{1}{2} \right) du = \frac{1}{2} \int_1^4 u^{-1/2} \, du\)
\(= \frac{1}{2} \left[ 2u^{1/2} \right]_1^4 = \left[ \sqrt{u} \right]_1^4 = \sqrt{4} - \sqrt{1}\) = 2 - 1 = 1.
11b. Given \( y = p x^2 + q x \),
\( \frac{dy}{dx} = 2px + q \).
But it is given that \( \frac{dy}{dx} = 6x + 7 \).
Equate coefficients of like terms: 2p = 6 ⇒ p = 3, q = 7.
Check: \( \frac{d^2 y}{dx^2} = 2p = 2 \times 3 = 6 \), which matches the given condition.
Thus, p = 3, and q = 7.
Akọwa Nkọwa
11a. Using the substitution \( u = 5 - x^2 \),
\( du = -2x \, dx \) so \( x \, dx = -\frac{1}{2} du \).
Limits: when \( x = 1 \), \( u = 4 \); when \( x = 2 \), \( u = 1 \).
\(\int_1^2 \frac{x}{\sqrt{5 - x^2}} \, dx = \int_4^1 \frac{1}{\sqrt{u}} \left( -\frac{1}{2} \right) du = \frac{1}{2} \int_1^4 u^{-1/2} \, du\)
\(= \frac{1}{2} \left[ 2u^{1/2} \right]_1^4 = \left[ \sqrt{u} \right]_1^4 = \sqrt{4} - \sqrt{1}\) = 2 - 1 = 1.
11b. Given \( y = p x^2 + q x \),
\( \frac{dy}{dx} = 2px + q \).
But it is given that \( \frac{dy}{dx} = 6x + 7 \).
Equate coefficients of like terms: 2p = 6 ⇒ p = 3, q = 7.
Check: \( \frac{d^2 y}{dx^2} = 2p = 2 \times 3 = 6 \), which matches the given condition.
Thus, p = 3, and q = 7.
Ajụjụ 9 Ripọtì
10a. The gradient of a tangent to the curve y = 4x\(^3\) at points P and Q is 108. Find the coordinates of P and Q
bi. Given \(\hat{A}\) = 45º, \(\hat{B}\) = 30º, sin(A + B) = sinA sinB + sinB sinA and cos(A + B) = cosA cosB - sinA sinB. Show that sin 15º = \(\frac{\sqrt{6} - \sqrt{2}}{4}\)
and cos15º = \(\frac{\sqrt{6} + \sqrt{2}}{4}\)
ii. Hence, find tan 15º
10a. The equation of the curve is \( y = 4x^3 \).
The gradient of the tangent at any point is given by the derivative:
\(\frac{dy}{dx} = 12x^2.\)
At points P and Q, the gradient is 108:
\(12x^2 = 108 \implies x^2 = 9 \implies x = \pm 3.\)
When \( x = 3 \):
\(y = 4(3)^3 = 4 \times 27 = 108.\)
So one point is 3, 108).
When x = -3 :
\(y = 4(-3)^3 = 4 \times (-27) = -108.\)
So the other point is (-3, -108).
Coordinates of P and Q: (3, 108), and (-3, -108).
10bi. We use the angle subtraction formulas:
sin(A - B) = sin A cos B - cos A sin B,
cos(A - B) = cos A cos B + sin A sin B,
where \( A = 45^\circ \) and \( B = 30^\circ \).
Known values:
\(\sin 45^\circ = \cos 45^\circ = \frac{\sqrt{2}}{2}, \quad \cos 30^\circ = \frac{\sqrt{3}}{2}, \quad \sin 30^\circ = \frac{1}{2}\)
Now,
\(\sin 15^\circ = \sin(45^\circ - 30^\circ) = \left( \frac{\sqrt{2}}{2} \right) \left( \frac{\sqrt{3}}{2} \right) - \left( \frac{\sqrt{2}}{2} \right) \left( \frac{1}{2} \right) = \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4} = \frac{\sqrt{6} - \sqrt{2}}{4}.\)
\(\cos 15^\circ = \cos(45^\circ - 30^\circ) = \left( \frac{\sqrt{2}}{2} \right) \left( \frac{\sqrt{3}}{2} \right) + \left( \frac{\sqrt{2}}{2} \right) \left( \frac{1}{2} \right) = \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4} = \frac{\sqrt{6} + \sqrt{2}}{4}.\)
10bii. \(\tan 15^\circ = \frac{\sin 15^\circ}{\cos 15^\circ} = \frac{\frac{\sqrt{6} - \sqrt{2}}{4}}{\frac{\sqrt{6} + \sqrt{2}}{4}} = \frac{\sqrt{6} - \sqrt{2}}{\sqrt{6} + \sqrt{2}}.\)
Rationalise the denominator by multiplying numerator and denominator by \(\sqrt{6} - \sqrt{2}\):
Numerator: \((\sqrt{6} - \sqrt{2})^2 = 6 - 2\sqrt{12} + 2 = 8 - 4\sqrt{3}\).
Denominator: \((\sqrt{6})^2 - (\sqrt{2})^2 = 6 - 2 = 4 \).
Thus, \(\tan 15^\circ = \frac{8 - 4\sqrt{3}}{4} = 2 - \sqrt{3}.\)
Akọwa Nkọwa
10a. The equation of the curve is \( y = 4x^3 \).
The gradient of the tangent at any point is given by the derivative:
\(\frac{dy}{dx} = 12x^2.\)
At points P and Q, the gradient is 108:
\(12x^2 = 108 \implies x^2 = 9 \implies x = \pm 3.\)
When \( x = 3 \):
\(y = 4(3)^3 = 4 \times 27 = 108.\)
So one point is 3, 108).
When x = -3 :
\(y = 4(-3)^3 = 4 \times (-27) = -108.\)
So the other point is (-3, -108).
Coordinates of P and Q: (3, 108), and (-3, -108).
10bi. We use the angle subtraction formulas:
sin(A - B) = sin A cos B - cos A sin B,
cos(A - B) = cos A cos B + sin A sin B,
where \( A = 45^\circ \) and \( B = 30^\circ \).
Known values:
\(\sin 45^\circ = \cos 45^\circ = \frac{\sqrt{2}}{2}, \quad \cos 30^\circ = \frac{\sqrt{3}}{2}, \quad \sin 30^\circ = \frac{1}{2}\)
Now,
\(\sin 15^\circ = \sin(45^\circ - 30^\circ) = \left( \frac{\sqrt{2}}{2} \right) \left( \frac{\sqrt{3}}{2} \right) - \left( \frac{\sqrt{2}}{2} \right) \left( \frac{1}{2} \right) = \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4} = \frac{\sqrt{6} - \sqrt{2}}{4}.\)
\(\cos 15^\circ = \cos(45^\circ - 30^\circ) = \left( \frac{\sqrt{2}}{2} \right) \left( \frac{\sqrt{3}}{2} \right) + \left( \frac{\sqrt{2}}{2} \right) \left( \frac{1}{2} \right) = \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4} = \frac{\sqrt{6} + \sqrt{2}}{4}.\)
10bii. \(\tan 15^\circ = \frac{\sin 15^\circ}{\cos 15^\circ} = \frac{\frac{\sqrt{6} - \sqrt{2}}{4}}{\frac{\sqrt{6} + \sqrt{2}}{4}} = \frac{\sqrt{6} - \sqrt{2}}{\sqrt{6} + \sqrt{2}}.\)
Rationalise the denominator by multiplying numerator and denominator by \(\sqrt{6} - \sqrt{2}\):
Numerator: \((\sqrt{6} - \sqrt{2})^2 = 6 - 2\sqrt{12} + 2 = 8 - 4\sqrt{3}\).
Denominator: \((\sqrt{6})^2 - (\sqrt{2})^2 = 6 - 2 = 4 \).
Thus, \(\tan 15^\circ = \frac{8 - 4\sqrt{3}}{4} = 2 - \sqrt{3}.\)
Ajụjụ 10 Ripọtì
15a. A body of mass 15kg is suspended at a point P by two light inextensible strings XP\(^→\) and YP\(^→\). The strings are inclined at 60º and 40º, respectively, to the downward vertical. Find, correct to two decimal places, the tension in the strings (take g = 10m/s\(^2\))
b. The height h metres, of a ball thrown into the air is 2 + 20t + kt\(^2\), after t seconds. If its takes 2 seconds for the ball to reach its height point, Find:
i. the value of k
ii. its highest point from the point of throw.
\(\frac{150}{sin 100} = \frac{T_1}{sin 140}\)
T\(_1\) = \(\frac{150 sin 140}{sin 100}\) = 97.91N
In same manner:
\(\frac{150}{sin 100} = \frac{T_2}{sin 120}\)
T\(_2\) = \(\frac{150 sin 120}{sin 100}\)
T\(_2\) = 131.91N.
i. At the highest point, v = 0 when t = 2 s:
\(20 + 2k \cdot 2 = 0 \)
20 + 4k = 0
k = - 5
ii. The highest point "from the point of throw" is the upward displacement above the release point t = 0, where the variable terms begin.
At t = 2:
2 + 20 x 2 + k x (2)\(^2\) = 2 + 40 + (-5) x 4 = 42 - 20 = 22 m.
Akọwa Nkọwa
\(\frac{150}{sin 100} = \frac{T_1}{sin 140}\)
T\(_1\) = \(\frac{150 sin 140}{sin 100}\) = 97.91N
In same manner:
\(\frac{150}{sin 100} = \frac{T_2}{sin 120}\)
T\(_2\) = \(\frac{150 sin 120}{sin 100}\)
T\(_2\) = 131.91N.
i. At the highest point, v = 0 when t = 2 s:
\(20 + 2k \cdot 2 = 0 \)
20 + 4k = 0
k = - 5
ii. The highest point "from the point of throw" is the upward displacement above the release point t = 0, where the variable terms begin.
At t = 2:
2 + 20 x 2 + k x (2)\(^2\) = 2 + 40 + (-5) x 4 = 42 - 20 = 22 m.
Ajụjụ 11 Ripọtì
7a. A body of mass 5 kg resting on a smooth horizontal plane is acted upon by forces 6i + 2j, 5i + 4j, and 4i − j. Calculate: the velocity of the body
b. the magnitude of its velocity, after 4 seconds
a. Mass , m = 5 kg
Resultant force, f = (6i+2j) + (5i+4j) + (4i−j) = (6+5+4)i + (2+4−1)j = F = 15i + 5j
From Newton second law of motion F = \(\frac{m \Delta V}{\text{t}}\)
F = \(\frac{m(v - u)}{\text{t}}\)
where u=0 ms\(^{−1}\)
F = \(\frac{\text{mv}}{\text{t}}\)
v = \(\frac{\text{Ft}}{\text{m}}\)
v = \(\frac{t(15i + 5j)}{5}\)
v = t(3i + j)m/s
b. Velocity, V after 4 s
v = t(3i + j) = (4)(3i + j) = 12i + 4jm/s.
Akọwa Nkọwa
a. Mass , m = 5 kg
Resultant force, f = (6i+2j) + (5i+4j) + (4i−j) = (6+5+4)i + (2+4−1)j = F = 15i + 5j
From Newton second law of motion F = \(\frac{m \Delta V}{\text{t}}\)
F = \(\frac{m(v - u)}{\text{t}}\)
where u=0 ms\(^{−1}\)
F = \(\frac{\text{mv}}{\text{t}}\)
v = \(\frac{\text{Ft}}{\text{m}}\)
v = \(\frac{t(15i + 5j)}{5}\)
v = t(3i + j)m/s
b. Velocity, V after 4 s
v = t(3i + j) = (4)(3i + j) = 12i + 4jm/s.
Ajụjụ 12 Ripọtì
2. If 2\(^{2x -2y}\) = 32 and log\(_y\) x = 2, find the values of x and y
Leave your answer in this format "+ x,- y"
2\(^{2x -2y}\) = 32 = 2\(^{2x -2y}\) = 2\(^5\)
2x - 2y = 5 - - - - - - - -- (1)
log\(_y\) x = 2,
x = y\(^2\) - - - - - - - -(2)
Put x = y\(^2\) into equation (i) 2(y\(^2\)) - 3y = 5
2\(y^2\) - 3y - 5 = 0
2y\(^2\) + 2y - 5y - 5 = 0
2y(y + 1) - 5(y + 1) = 0
(2y - 5)(y + 1) = 0
2y - 5 = 0 or y + 1 = 0
2y = 5 or y = -1
y = \(\frac{5}{2}\) or y = -1
y = \(\frac{5}{2}\) or y = -1
When y = -1
x = (-1)\(^2\)
x = 1
When y = \(\frac{5}{2}\)
x = (\(\frac{5}{2}\))\(^2\) = \(\frac{25}{4}\) = 6\(\frac{1}{4}\).
Akọwa Nkọwa
2\(^{2x -2y}\) = 32 = 2\(^{2x -2y}\) = 2\(^5\)
2x - 2y = 5 - - - - - - - -- (1)
log\(_y\) x = 2,
x = y\(^2\) - - - - - - - -(2)
Put x = y\(^2\) into equation (i) 2(y\(^2\)) - 3y = 5
2\(y^2\) - 3y - 5 = 0
2y\(^2\) + 2y - 5y - 5 = 0
2y(y + 1) - 5(y + 1) = 0
(2y - 5)(y + 1) = 0
2y - 5 = 0 or y + 1 = 0
2y = 5 or y = -1
y = \(\frac{5}{2}\) or y = -1
y = \(\frac{5}{2}\) or y = -1
When y = -1
x = (-1)\(^2\)
x = 1
When y = \(\frac{5}{2}\)
x = (\(\frac{5}{2}\))\(^2\) = \(\frac{25}{4}\) = 6\(\frac{1}{4}\).
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