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Tambaya 1 Rahoto
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, \(\tan 15^\circ = \frac{8 - 4\sqrt{3}}{4} = 2 - \sqrt{3}.\)
Bayanin Amsa
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, \(\tan 15^\circ = \frac{8 - 4\sqrt{3}}{4} = 2 - \sqrt{3}.\)
Tambaya 2 Rahoto
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: \( 3y - x + 1 = 0\)
Bayanin Amsa
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: \( 3y - x + 1 = 0\)
Tambaya 3 Rahoto
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.
Bayanin Amsa
\(\frac{150}{sin 100} = \frac{T_1}{sin 140}\)
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.
Tambaya 4 Rahoto
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.
Bayanin Amsa
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.
Tambaya 5 Rahoto
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)
Bayanin Amsa
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)
Tambaya 6 Rahoto
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.
Bayanin Amsa
| 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.
Tambaya 7 Rahoto
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.
Bayanin Amsa
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.
Tambaya 8 Rahoto
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.
Bayanin Amsa
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.
Tambaya 9 Rahoto
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\(\frac{1}{4}\).
Bayanin Amsa
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\(\frac{1}{4}\).
Tambaya 10 Rahoto
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
Bayanin Amsa
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
Tambaya 11 Rahoto
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.
Bayanin Amsa
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.
Tambaya 12 Rahoto
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
Bayanin Amsa
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
Za ka so ka ci gaba da wannan aikin?