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Swali 1 Ripoti
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\(\frac{1}{4}\).
Maelezo ya Majibu
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\(\frac{1}{4}\).
Swali 2 Ripoti
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.
Maelezo ya Majibu
\(\frac{150}{sin 100} = \frac{T_1}{sin 140}\)
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.
Swali 3 Ripoti
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: \( 3y - x + 1 = 0\)
Maelezo ya Majibu
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: \( 3y - x + 1 = 0\)
Swali 4 Ripoti
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.
Maelezo ya Majibu
| 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.
Swali 5 Ripoti
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)
Maelezo ya Majibu
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)
Swali 6 Ripoti
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
Maelezo ya Majibu
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
Swali 7 Ripoti
17a. A body of mass 5 kg is placed on a smooth plane inclined at an angle of 30º to the horizontal. Find: the magnitude of the force acting parallel to the plane.
bi. A uniform plank PQ of length 10m and mass m kg rests on two support A and B. Where \PA\ = \BQ\ = 1m. A load of mass 8kg is placed on the plank at point C such that \AC\ = 3.5m, if the reaction at B is 100N. Calculate the value of m
bii. the reaction at A [ take g = 10m/s\(^2\)].
17a. Since the plane is smooth (frictionless), the only force acting parallel to the inclined plane is the component of the weight of the body down the plane.
F = mg sin 30\(^\circ\)
Given m = 5 kg and taking g = 10 m/s\(^2\).
F = 5 \(\times 10 \times \sin 30^\circ\) = 5 \(\times 10 \times\) 0.5 = 25 N.
The magnitude of the force acting parallel to the plane is 25 N.
bi. The uniform plank PQ is 10 m long with supports at A and B such that PA = 1 m and BQ = 1 m. Thus, AB = 8 m. The plank’s centre of mass G is at its midpoint (5 m from P), so AG = 4 m. The 8 kg load is placed at C, where AC = 3.5 m.
Weight of plank = \( 10m \) N
Weight of load = \( 8 \times 10 = 80 \) N
Reaction at B, \( R_B = 100 \) N (given).
For rotational equilibrium, take moments about A (clockwise = anticlockwise):
\((10m) \times 4 + 80 \times 3.5 = 100 \times 8\)
40m + 280 = 800
40m = 520 \(\implies m = \frac{520}{40}\) = 13
Thus, m = 13 kg.
bii. For vertical equilibrium: \(R_A + R_B = \text{total weight}\)
\(R_A + 100 = (10 \times\) 13) + 80
R\(_A\) + 100 = 130 + 80 = 210
R\(_A\) = 210 - 100 = 110 N.
The reaction at A is 110 N.
Maelezo ya Majibu
17a. Since the plane is smooth (frictionless), the only force acting parallel to the inclined plane is the component of the weight of the body down the plane.
F = mg sin 30\(^\circ\)
Given m = 5 kg and taking g = 10 m/s\(^2\).
F = 5 \(\times 10 \times \sin 30^\circ\) = 5 \(\times 10 \times\) 0.5 = 25 N.
The magnitude of the force acting parallel to the plane is 25 N.
bi. The uniform plank PQ is 10 m long with supports at A and B such that PA = 1 m and BQ = 1 m. Thus, AB = 8 m. The plank’s centre of mass G is at its midpoint (5 m from P), so AG = 4 m. The 8 kg load is placed at C, where AC = 3.5 m.
Weight of plank = \( 10m \) N
Weight of load = \( 8 \times 10 = 80 \) N
Reaction at B, \( R_B = 100 \) N (given).
For rotational equilibrium, take moments about A (clockwise = anticlockwise):
\((10m) \times 4 + 80 \times 3.5 = 100 \times 8\)
40m + 280 = 800
40m = 520 \(\implies m = \frac{520}{40}\) = 13
Thus, m = 13 kg.
bii. For vertical equilibrium: \(R_A + R_B = \text{total weight}\)
\(R_A + 100 = (10 \times\) 13) + 80
R\(_A\) + 100 = 130 + 80 = 210
R\(_A\) = 210 - 100 = 110 N.
The reaction at A is 110 N.
Swali 8 Ripoti
1. The sum of the 2nd and 5th terms of an arithmetic progression (A.P) is 42. If the difference between the 6th and 3rd terms is 12, find:
a. the common difference
b. the first term
c. the 20th term.
a. fifth term, T\(_5\) =? But the nth term of an A.P. T\(_n\) = a (n - 1) d
T\(_2\) = a + d
T\(_3\) = a + 2d
T\(_5\) = a + 4d
T\(_2\) + T\(_5\) = a + d + a + 4d = 2a + 5d = 42 .....(i)
T\(_6\) = a + 5d
T\(_3\) = a + 2d
T\(_6\) - T\(_3\) = (a + 5d) - (a + 2d) = 3d = 12
3d = 12
d = 4
b. put d = 4 into equation (i)
2a + 5(4) = 42
2a = 42 - 20
2a = 22
a = 11 first term,
a = 11
c. T\(_{20}\) = a + 19d
= 11 + 19 x 4
T\(_{20}\) = 87.
Maelezo ya Majibu
a. fifth term, T\(_5\) =? But the nth term of an A.P. T\(_n\) = a (n - 1) d
T\(_2\) = a + d
T\(_3\) = a + 2d
T\(_5\) = a + 4d
T\(_2\) + T\(_5\) = a + d + a + 4d = 2a + 5d = 42 .....(i)
T\(_6\) = a + 5d
T\(_3\) = a + 2d
T\(_6\) - T\(_3\) = (a + 5d) - (a + 2d) = 3d = 12
3d = 12
d = 4
b. put d = 4 into equation (i)
2a + 5(4) = 42
2a = 42 - 20
2a = 22
a = 11 first term,
a = 11
c. T\(_{20}\) = a + 19d
= 11 + 19 x 4
T\(_{20}\) = 87.
Swali 9 Ripoti
3. If (x + 2) and (x -1) are factors of f(x) = 6x\(^4\) + mx\(^3\) - 13x\(^2\) + n x + 14, find the:
a. value of m and n. Leave your answer in this format 'm,n.'
b. remainder when f(x) is divided by (x + 1)
a. (x + 2) and (x -1) are factors of f(x) = 6x\(^4\) + mx\(^3\) - 13x\(^2\) + n x + 14
Then, f(-2) and f(1) are = 0
f(-2) = 6(-2)\(^4\) + m(-2)\(^3\) - 13(-2)\(^2\) + n (-2) + 14 = 0
= 6 x 16 - 8m - 13 x 2 - 2n + 14 = 0 = 96 - 8m - 52 - 2n + 14 = 0
= 8m + 2n = 58: divide through by 2
= 4m + n = 29 - - -- - - - - - -(i)
f(1) = 6x\(^4\) + mx\(^3\) - 13x\(^2\) + n x + 14 = 0
f(1) = 6(1)\(^4\) + m(1)\(^3\) - 13(1)\(^2\) + n (1) + 14 = 0
= 6 + m - 13 + n + 14 = 0
= m + n = -7 - - - - - - - - -(ii)
Solving eqn i and ii simultaneously
from eqn i - - - - - n = 58 - 4m
put n = 58 - 4m into eqn ii
m + n = -7 = m + 29 - 4m = -7
- 3m = -36
m = \(\frac{36}{3}\) = 12
put m = 12 into eqn ii
m + n = -7 = 12 + n = -7
n = -7 - 12 = -19.
b. 6x\(^4\) + mx\(^3\) - 13x\(^2\) + n x + 14 = 0 becomes 6x\(^4\) + 12x\(^3\) - 13x\(^2\) - 19x + 14 = 0
f(x) = 6x\(^4\) + 12x\(^3\) - 13x\(^2\) - 19x + 14 = 0
remainder when f(x) is divided by (x + 1)
Let (x + 1) = 0 then, x = -1
f(-1) = 6(-1)\(^4\) + 12(-1)\(^3\) - 13(-1)\(^2\) - 19(-1) + 14
= 6 - 12 - 13 + 19 + 14 = 14
Therefore, the remainder = 14.
Maelezo ya Majibu
a. (x + 2) and (x -1) are factors of f(x) = 6x\(^4\) + mx\(^3\) - 13x\(^2\) + n x + 14
Then, f(-2) and f(1) are = 0
f(-2) = 6(-2)\(^4\) + m(-2)\(^3\) - 13(-2)\(^2\) + n (-2) + 14 = 0
= 6 x 16 - 8m - 13 x 2 - 2n + 14 = 0 = 96 - 8m - 52 - 2n + 14 = 0
= 8m + 2n = 58: divide through by 2
= 4m + n = 29 - - -- - - - - - -(i)
f(1) = 6x\(^4\) + mx\(^3\) - 13x\(^2\) + n x + 14 = 0
f(1) = 6(1)\(^4\) + m(1)\(^3\) - 13(1)\(^2\) + n (1) + 14 = 0
= 6 + m - 13 + n + 14 = 0
= m + n = -7 - - - - - - - - -(ii)
Solving eqn i and ii simultaneously
from eqn i - - - - - n = 58 - 4m
put n = 58 - 4m into eqn ii
m + n = -7 = m + 29 - 4m = -7
- 3m = -36
m = \(\frac{36}{3}\) = 12
put m = 12 into eqn ii
m + n = -7 = 12 + n = -7
n = -7 - 12 = -19.
b. 6x\(^4\) + mx\(^3\) - 13x\(^2\) + n x + 14 = 0 becomes 6x\(^4\) + 12x\(^3\) - 13x\(^2\) - 19x + 14 = 0
f(x) = 6x\(^4\) + 12x\(^3\) - 13x\(^2\) - 19x + 14 = 0
remainder when f(x) is divided by (x + 1)
Let (x + 1) = 0 then, x = -1
f(-1) = 6(-1)\(^4\) + 12(-1)\(^3\) - 13(-1)\(^2\) - 19(-1) + 14
= 6 - 12 - 13 + 19 + 14 = 14
Therefore, the remainder = 14.
Swali 10 Ripoti
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.
Maelezo ya Majibu
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.
Swali 11 Ripoti
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.
Maelezo ya Majibu
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.
Swali 12 Ripoti
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
Maelezo ya Majibu
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
Je, ungependa kuendelea na hatua hii?