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Question 1 Report
Question 2 Report
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Question 3 Report
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Question 4 Report
Find the area bounded by the curve \(y = 3x^2 - 2x + 1\), the coordinates \(x = 1\) and \(x = 3\) and the x-axis
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Question 7 Report
If \(x = \begin{pmatrix}1 & 2 \\ 0 & 3\end{pmatrix}\) and \(y = \begin{pmatrix}2 & 1 \\ 4 & 3\end{pmatrix}\). Find \(xy\).
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Question 8 Report
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Question 9 Report
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Question 10 Report
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Question 11 Report
| Age in years | 13 | 14 | 15 | 16 | 17 |
| No. of students | 3 | 10 | 30 | 42 | 15 |
The frequency distribution above shows the ages of students in a secondary school. In a pie chart constructed to represent the data, the angles corresponding to the 15 years old is
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Question 12 Report
If \(a \ast b = +\sqrt{ab}\), evaluate \(2 \ast (12 \ast 27)\)
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Question 13 Report
Solve the inequality \( (x - 3)(x - 4) \le 0 \)
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Question 15 Report
Find the distance between two towns p(45\(^{o}\)N, 30\(^{o}\)W) and Q(15\(^{o}\)S, 30\(^{o}\)W) if the radius of the earth is 7000km. [\(\pi = \frac{22}{7}\)]
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Question 16 Report
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Question 18 Report
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Question 19 Report
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Question 20 Report
| Class Interval | Frequency | Class boundaries | Class Mid-point |
| \(1.5 - 1.9\) | \(2\) | \(1.45 - 1.95\) | \(1.7\) |
| \(2.0 - 2.4\) | \(21\) | \(1.95 - 2.45\) | \(2.2\) |
| \(2.5 - 2.9\) | \(4\) | \(2.45 - 2.95\) | \(2.7\) |
| \(3.0 - 2.9\) | \(15\) | \(2.95 - 3.45\) | \(3.2\) |
| \(3.5 - 3.9\) | \(10\) | \(3.45 - 3.95\) | \(3.7\) |
| \(4.0 - 4.4\) | \(5\) | \(3.95 - 4.45\) | \(4.2\) |
| \(4.5 - 4.9\) | \(3\) | \(4.45 - 4.95\) | \(4.7\) |
The median of the distribution above is
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Question 21 Report
In the diagram, find PQ if the area of triangle PQR is 35cm2
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Question 23 Report
Find T in terms of K, Q and S if S = 2r\( \pi \)QT + K)
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T = s24Qπr2 - k
Question 24 Report
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Question 25 Report
| class | 1−3 | 4−6 | 7−9 |
| Frequency | 5 | 8 | 5 |
Find the standard deviation of the data using the table above
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Question 27 Report
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Question 28 Report
Express \( \frac{5x-12}{(x-2)(x-3)} \) in partial fractions
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5x−12(x−2)(x−3)=Ax−2+Bx−3
= A(x−3)+B(x−2)(x−2)(x−3)
⟹5x−12=Ax−3A+Bx−2B
A+B=5...(i)
−(3A+2B)=−12⟹3A+2B=12...(ii)
From (i), A=5−B
3(5−B)+2B=12
15−3B+2B=12⟹B=3
A+3=5⟹A=2
5x−12(x−2)(x−3)=2x−2+3x−3
Question 29 Report
For what value of x is the tangent to the curve \(y = x^2 - 4x + 3\) parallel to the x-axis?
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Question 30 Report
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m = 1.05, r = 0.6
m + r = 1.05 + 0.5
= 1.65
Question 31 Report
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Question 32 Report
Let \(p\) be a probability function on set \(S\), where \(S = (a_1, a_2, a_3, a_4)\). Find \(P(a_1)\) if \(P(a_2) = \frac{1}{3}\), \(p(a_3) = \frac{1}{6}\) and \(p(a_4) = \frac{1}{5}\)
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Question 33 Report
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Question 34 Report
If \(x - 1\) and \(x + 1\) are both factors of the equation \(x^3 + px^2 + qx + 6 = 0\), evaluate \(p\) and \(q\)
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Question 35 Report
The graph of f(x) = x2 - 5x + 6 crosses the x-axis at the points
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Question 36 Report
Simplify \( \frac{x^2 - 1}{x^3 + 2x^2 - x - 2} \)
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Question 37 Report
Find the sum to infinity of the following sequence 1. \( \frac{9}{10} \), 2. (\( \frac{9}{10} \)), 3. (\( \frac{9}{10} \))
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Question 38 Report
| Class Interval | Frequency | Class boundaries | Class Mid−point |
| \(1.5-1.9\) | \(2\) | \(1.45-1.95\) | \(1.7\) |
| \(2.0-2.4\) | \(21\) | \(1.95-2.45\) | \(2.2\) |
| \(2.5-2.9\) | \(4\) | \(2.45-2.95\) | \(2.7\) |
| \(3.0-2.9\) | \(15\) | \(2.95-3.45\) | \(3.2\) |
| \(3.5-3.9\) | \(10\) | \(3.45-3.95\) | \(3.7\) |
| \(4.0-4.4\) | \(5\) | \(3.95-4.45\) | \(4.2\) |
| \(4.5-4.9\) | \(3\) | \(4.45-4.95\) | \(4.7\) |
Find the mode of the distribution above to find the mode of the distribution.
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Mode = a + (b - a)(fm - Fb)
2Fm - Fa - Fb
= 3.0 + (3.4?3)(15?4)2(15)?4?10
= 3 + (6.4)(11)30?14
= 3 + 4.416
= 3 + 0.275
= 3.275
= 3.3cm
Question 39 Report
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Question 40 Report
In the diagram, the base diameter is 14cm while the height is 12cm. Calculate the total surface area if the cylinder has both a base and a top.[\( \pi \frac{22}{7} \)]
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Question 41 Report
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Question 42 Report
Two variables x and y are such that \( \frac{dy}{dx} = 4x - 3 \) and y = 5 when x = 2. Find y in terms of x
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Question 43 Report
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Question 44 Report
Use the graph of the curve \(y = f(x)\)to solve the inequality \(f(x) \le 0\)
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Combining solutions
= x ≤ 1; 1 ≥ x ≥ 2
Question 45 Report
PT is a tangent to the circle TYZX. YT = YX and < PTX = 50o. Calculate < TZY
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Question 47 Report
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