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Question 1 Report
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.
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.
Question 2 Report
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, \(\tan 15^\circ = \frac{8 - 4\sqrt{3}}{4} = 2 - \sqrt{3}.\)
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, \(\tan 15^\circ = \frac{8 - 4\sqrt{3}}{4} = 2 - \sqrt{3}.\)
Question 3 Report
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.
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.
Question 4 Report
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
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
Question 5 Report
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.
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.
Question 6 Report
14a. Two pupils are chosen at random from a group of 4 boys and 5 girls. Find the probability that the two pupils chosen would be boys. Leave your answer in fraction 'a/b.'
b. Twenty percent of the total production of transistors produced by a machine are below standard. If a random sample of six transistors produced by the machine is taken, what is the probability of getting
i. Exactly 2
ii. Exactly 1. Leave your answer in six decimal places " a.bcdefg."
iii. At least 2. Leave your answer in five decimal places " a. bcdeg."
iv. At most 2 standard transistors
14a. Total number of pupils = 4 boys + 5 girls = 9.
Number of ways to choose 2 boys = \(\binom{4}{2} = 6\).
Total number of ways to choose any 2 pupils = \(\binom{9}{2} = 36\).
Probability that both chosen are boys = \(\dfrac{6}{36} = \dfrac{1}{6}\).
14b. This is a binomial distribution with n = 6 trials and p = 0.80 (probability a transistor is standard), q = 0.2 (Prb. of a transistor below standard)
Let Y = number of standard transistors.
P(Y = k) = C(6, k) × (0.8)\(^k\) × (0.2)\(^{(6−k)}\)
i: k = 2 → 15 × 0.64 × 0.0016 = 0.015360
ii: k = 1 → 6 × 0.8 × 0.00032 = 0.001536
iii: P(Y ≥ 2) = 1 − P(Y = 0) − P(Y = 1) = 1 − 0.000064 − 0.001536 = 0.99840
iv: P(Y ≤ 2) = P(Y = 0) + P(Y = 1) + P(Y = 2) = 0.000064 + 0.001536 + 0.015360 = 0.016960
14a. Total number of pupils = 4 boys + 5 girls = 9.
Number of ways to choose 2 boys = \(\binom{4}{2} = 6\).
Total number of ways to choose any 2 pupils = \(\binom{9}{2} = 36\).
Probability that both chosen are boys = \(\dfrac{6}{36} = \dfrac{1}{6}\).
14b. This is a binomial distribution with n = 6 trials and p = 0.80 (probability a transistor is standard), q = 0.2 (Prb. of a transistor below standard)
Let Y = number of standard transistors.
P(Y = k) = C(6, k) × (0.8)\(^k\) × (0.2)\(^{(6−k)}\)
i: k = 2 → 15 × 0.64 × 0.0016 = 0.015360
ii: k = 1 → 6 × 0.8 × 0.00032 = 0.001536
iii: P(Y ≥ 2) = 1 − P(Y = 0) − P(Y = 1) = 1 − 0.000064 − 0.001536 = 0.99840
iv: P(Y ≤ 2) = P(Y = 0) + P(Y = 1) + P(Y = 2) = 0.000064 + 0.001536 + 0.015360 = 0.016960
Question 7 Report
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:
\(\frac{1}{2}\overrightarrow{AB} = \begin{pmatrix} -2 \\ 3 \end{pmatrix},\)
\(-\frac{1}{2}\overrightarrow{AB} = \begin{pmatrix} 2 \\ -3 \end{pmatrix}.\)
\(\overrightarrow{CV} = \begin{pmatrix} 2 \\ 3 \end{pmatrix} + \begin{pmatrix} 2 \\ -3 \end{pmatrix} = \begin{pmatrix} 4 \\ 0 \end{pmatrix}.\)
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:
\(\frac{1}{2}\overrightarrow{AB} = \begin{pmatrix} -2 \\ 3 \end{pmatrix},\)
\(-\frac{1}{2}\overrightarrow{AB} = \begin{pmatrix} 2 \\ -3 \end{pmatrix}.\)
\(\overrightarrow{CV} = \begin{pmatrix} 2 \\ 3 \end{pmatrix} + \begin{pmatrix} 2 \\ -3 \end{pmatrix} = \begin{pmatrix} 4 \\ 0 \end{pmatrix}.\)
Question 8 Report
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.
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.
Question 9 Report
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.
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.
Question 10 Report
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
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
Question 11 Report
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: \( 3y - x + 1 = 0\)
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: \( 3y - x + 1 = 0\)
Question 12 Report
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.
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.
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