WAEC SSCE - Further Mathematics - 2013
Ajụjụ 1 Ripọtì
If \(V = \begin{pmatrix} -2 \\ 4 \end{pmatrix}\) and \(U = \begin{pmatrix} -1 \\ 5 \end{pmatrix}\), find \(|U + V|\).
Akọwa Nkọwa
We can find the sum of the vectors U and V by adding their corresponding components, i.e., $$ U + V = \begin{pmatrix} -1 \\ 5 \end{pmatrix} + \begin{pmatrix} -2 \\ 4 \end{pmatrix} = \begin{pmatrix} -3 \\ 9 \end{pmatrix}. $$ Then, the magnitude (length) of the vector U + V is given by the formula $$ |U + V| = \sqrt{(-3)^2 + 9^2} = \sqrt{90} = 3\sqrt{10}. $$ Therefore, the answer is \(3\sqrt{10}\).
Ajụjụ 2 Ripọtì
Find the equation of the straight line that passes through (2, -3) and perpendicular to the line 3x - 2y + 4 = 0.
Akọwa Nkọwa
Ajụjụ 3 Ripọtì
The equation of a circle is \(x^{2} + y^{2} - 8x + 9y + 15 = 0\). Find its radius.
Akọwa Nkọwa
To find the radius of a circle given its equation, we need to complete the square for both the x and y terms. We can do this by rearranging the equation as follows: \begin{align*} x^2 - 8x + y^2 + 9y &= -15 \\ (x^2 - 8x + 16) + (y^2 + 9y + 20.25) &= -15 + 16 + 20.25 \\ (x - 4)^2 + (y + 4.5)^2 &= 21.25 \end{align*} We can now see that the equation of the circle is in the standard form: \begin{equation*} (x - a)^2 + (y - b)^2 = r^2 \end{equation*} where the center of the circle is at point (a, b) and the radius is r. From the completed square form, we can identify that the center of the circle is at point (4, -4.5) and the radius is the square root of 21.25, which simplifies to: \begin{equation*} r = \sqrt{21.25} = \frac{1}{2} \sqrt{85} \end{equation*} Therefore, the correct answer is option (C), \(\frac{1}{2}\sqrt{85}\).
Ajụjụ 4 Ripọtì
The angle subtended by an arc of a circle at the centre is \(\frac{\pi}{3} radians\). If the radius of the circle is 12cm, calculate the perimeter of the major arc.
Akọwa Nkọwa
Ajụjụ 5 Ripọtì
If \(f(x) = x^{2}\) and \(g(x) = \sin x\), find g o f.
Akọwa Nkọwa
To find g o f, we need to substitute f(x) into g(x). Therefore, we have: g o f = g(f(x)) = g(x^2) = sin(x^2) So the answer is \(\sin x^{2}\). Option (B).
Ajụjụ 6 Ripọtì
A committee consists of 5 boys namely: Kofi, John, Ojo, Ozo and James and 3 girls namely: Rose, Ugo and Ama. In how many ways can a sub-committee consisting of 3 boys and 2 girls be chosen, if Ozo must be on the sub-committee?
Akọwa Nkọwa
The problem states that a sub-committee must be formed consisting of 3 boys and 2 girls. Also, Ozo must be on the sub-committee. This means that we already have one of the three boys chosen. We need to choose two more boys from the remaining four boys, and two girls from the three girls. The number of ways to choose two boys from the remaining four boys is given by the combination formula: C(4,2) = 6. (Alternatively, we can list all the possible combinations of two boys from the remaining four boys: Kofi and John, Kofi and Ojo, Kofi and James, John and Ojo, John and James, Ojo and James. This gives us a total of 6 combinations.) Similarly, the number of ways to choose two girls from the three girls is given by the combination formula: C(3,2) = 3. (Alternatively, we can list all the possible combinations of two girls from the three girls: Rose and Ugo, Rose and Ama, Ugo and Ama. This gives us a total of 3 combinations.) To find the total number of ways to choose the sub-committee, we multiply the number of ways to choose two boys and two girls: 6 x 3 = 18. Therefore, the answer is 18.
Ajụjụ 7 Ripọtì
If \(2\sin^{2}\theta = 1 + \cos \theta, 0° \leq \theta \leq 90°\), find \(\theta\).
Akọwa Nkọwa
We can use trigonometric identities to solve the equation. First, we know that $\cos^2 \theta + \sin^2 \theta = 1$, so we can write: \begin{align*} 2\sin^2 \theta &= 1 + \cos \theta \\ 2(1 - \cos^2 \theta) &= 1 + \cos \theta \\ 2 - 2\cos^2 \theta &= 1 + \cos \theta \\ 2\cos^2 \theta + \cos \theta - 1 &= 0 \\ (2\cos \theta - 1)(\cos \theta + 1) &= 0 \\ \end{align*} Therefore, either $2\cos \theta - 1 = 0$ or $\cos \theta + 1 = 0$. Solving for $\theta$ in each case, we get: \begin{align*} 2\cos \theta - 1 &= 0 \\ \cos \theta &= \frac{1}{2} \\ \theta &= 60^\circ \\ \\ \cos \theta + 1 &= 0 \\ \cos \theta &= -1 \\ \theta &= 180^\circ \\ \end{align*} However, we are given that $0^\circ \leq \theta \leq 90^\circ$, so the only valid solution is $\theta = 60^\circ$. Therefore, the answer is (C) 60°.
Ajụjụ 8 Ripọtì
Find the coordinates of the point which divides the line joining P(-2, 3) and Q(4, 9) internally in the ratio 2 : 3.
Akọwa Nkọwa
To solve this problem, we will use the section formula. Let the point which divides the line segment PQ internally in the ratio 2:3 be R(x,y). Then we have: \begin{align*} x = \frac{3\times(-2) + 2\times4}{2+3} = \frac{6}{5} \\ y = \frac{3\times3 + 2\times9}{2+3} = \frac{24}{5} \end{align*} Therefore, the coordinates of R are \((\frac{6}{5},\frac{24}{5})\), which corresponds to option (B).
Ajụjụ 9 Ripọtì
If \(\alpha\) and \(\beta\) are the roots of the equation \(2x^{2} - 6x + 5 = 0\), evaluate \(\frac{\beta}{\alpha} + \frac{\alpha}{\beta}\).
Akọwa Nkọwa
Ajụjụ 10 Ripọtì
Given that \(\frac{\mathrm d y}{\mathrm d x} = \sqrt{x}\), find y.
Akọwa Nkọwa
Given that \(\frac{\mathrm d y}{\mathrm d x} = \sqrt{x}\), we can integrate both sides with respect to x to obtain the original function y(x). $$ \frac{\mathrm d y}{\mathrm d x} = \sqrt{x} \\ \int \frac{\mathrm d y}{\mathrm d x} \mathrm d x= \int \sqrt{x} \mathrm d x\\ y= \int x^{\frac{1}{2}} \mathrm d x \\ y= \frac{2}{3}x^{\frac{3}{2}} + c $$ where c is an arbitrary constant of integration. Therefore, the correct option is (b) \(\frac{2}{3}x^{\frac{3}{2}} + c\).
Ajụjụ 11 Ripọtì
A body is kept at rest by three forces \(F_{1} = (10N, 030°), F_{2} = (10N, 150°)\) and \(F_{3}\). Find \(F_{3}\).
Akọwa Nkọwa
Ajụjụ 12 Ripọtì
Find the upper quartile of the following scores: 41, 29, 17, 2, 12, 33, 45, 18, 43 and 5.
Akọwa Nkọwa
To find the upper quartile, we first need to arrange the numbers in order from smallest to largest: 2, 5, 12, 17, 18, 29, 33, 41, 43, 45 The upper quartile divides the data set into quarters, with 25% of the data falling above this point. To find the upper quartile, we need to find the median of the upper half of the data set. The upper half of the data set is: 29, 33, 41, 43, 45 The median of this data set is the middle number, which is 41. Therefore, the upper quartile is 41. So the correct answer is.
Ajụjụ 13 Ripọtì
Forces 50N and 80N act on a body as shown in the diagram. Find, correct to the nearest whole number, the horizontal component of the resultant force.
Akọwa Nkọwa
Ajụjụ 14 Ripọtì
Given that \(a^{\frac{5}{6}} \times a^{\frac{-1}{n}} = 1\), solve for n.
Akọwa Nkọwa
We can simplify the left side of the equation as follows: $$a^{\frac{5}{6}} \times a^{\frac{-1}{n}} = a^{\frac{5}{6} - \frac{1}{n}}$$ Since this expression equals 1, we have: $$a^{\frac{5}{6} - \frac{1}{n}} = 1$$ We can rewrite this as: $$\frac{5}{6} - \frac{1}{n} = 0$$ Solving for n, we get: $$\frac{1}{n} = \frac{5}{6}$$ $$n = \frac{6}{5} = 1.2$$ Therefore, the value of n is 1.20, which is.
Ajụjụ 15 Ripọtì
Given that \(f(x) = 2x^{3} - 3x^{2} - 11x + 6\) and \(f(3) = 0\), factorize f(x).
Akọwa Nkọwa
To factorize the given expression, we can use synthetic division or long division to divide \(f(x)\) by \(x - 3\) which is one of its factors as given. The result of this division is a quadratic expression that can be factored easily. Using synthetic division, we get: \begin{array}{c|cccc} & 2 & -3 & -11 & 6 \\ \hline 3 & & 6 & 9 & -6 \\ & & & -6 & 27 \\ \hline & 2 & 3 & -2 & 21 \end{array} Thus, \(f(x) = (x - 3)(2x^{2} + 3x - 2)\). Now we need to factorize the quadratic expression \(2x^{2} + 3x - 2\). We can use the quadratic formula or factorization by grouping to factorize the quadratic expression. Factorization by grouping is simpler and can be done as follows: \begin{align*} 2x^{2} + 3x - 2 &= 2x^{2} + 4x - x - 2 \\ &= 2x(x + 2) - 1(x + 2) \\ &= (2x - 1)(x + 2) \end{align*} Therefore, the complete factorization of \(f(x)\) is: $$f(x) = (x - 3)(2x - 1)(x + 2)$$ Hence, the correct option is (C) (x - 3)(x + 2)(2x - 1).
Ajụjụ 16 Ripọtì
An object is thrown vertically upwards from the top of a cliff with a velocity of \(25ms^{-1}\). Find the time, in seconds, when it is 20 metres above the cliff. \([g = 10ms^{-2}]\).
Ajụjụ 17 Ripọtì
Calculate the mean deviation of 1, 2, 3, 4, 5, 5, 6, 7, 8, 9.
Akọwa Nkọwa
To calculate the mean deviation, we first need to find the mean of the given data set. Mean = (1 + 2 + 3 + 4 + 5 + 5 + 6 + 7 + 8 + 9) / 10 = 50 / 10 = 5 Next, we find the deviation of each value from the mean by subtracting the mean from each value: 1 - 5 = -4 2 - 5 = -3 3 - 5 = -2 4 - 5 = -1 5 - 5 = 0 5 - 5 = 0 6 - 5 = 1 7 - 5 = 2 8 - 5 = 3 9 - 5 = 4 To find the mean deviation, we need to find the average of the absolute values of these deviations. Mean Deviation = (|-4| + |-3| + |-2| + |-1| + |0| + |0| + |1| + |2| + |3| + |4|) / 10 = 20 / 10 = 2 Therefore, the mean deviation of the given data set is 2. The answer is (a) 2.
Ajụjụ 18 Ripọtì
Two out of ten tickets on sale for a raffle draw are winning tickets. If a guest bought two tickets, what is the probability that both tickets are winning tickets?
Akọwa Nkọwa
There are ten tickets on sale for the raffle draw, and two of them are winning tickets. If a guest buys two tickets, there are a total of \(_{10}C_2\) ways to choose two tickets out of the ten. This is because we can choose any two tickets out of ten in \(_{10}C_2\) ways, and each combination is equally likely. The number of ways to choose two winning tickets out of the two available winning tickets is \(_2C_2\), which is equal to 1. The probability of selecting two winning tickets is therefore: $$ \frac{\text{number of ways to choose two winning tickets}}{\text{number of ways to choose any two tickets}} = \frac{\binom{2}{2}}{\binom{10}{2}} = \frac{1}{\frac{10\times9}{2\times1}} = \frac{1}{45} $$ Therefore, the probability that both tickets are winning tickets is \(\frac{1}{45}\).
Ajụjụ 19 Ripọtì
A circular ink blot on a piece of paper increases its area at the rate \(4mm^{2}/s\). Find the rate of the radius of the blot when the radius is 8mm. \([\pi = \frac{22}{7}]\).
Akọwa Nkọwa
Ajụjụ 20 Ripọtì
The sales of five salesgirls on a certain day are as follows; GH¢ 26.00, GH¢ 39.00, GH¢ 33.00, GH¢ 25.00 and GH¢ 37.00. Calculate the standard deviation if the mean sale is GH¢ 32.00.
Akọwa Nkọwa
To calculate the standard deviation, we need to first find the deviations of each sale from the mean sale, square them, sum them up, divide by the total number of sales, and then take the square root of the result. The mean sale is GH¢ 32.00, so the deviations of each sale from the mean are: - GH¢ -6.00 - GH¢ 7.00 - GH¢ 1.00 - GH¢ -7.00 - GH¢ 5.00 Squaring these deviations, we get: - 36 - 49 - 1 - 49 - 25 Summing these up, we get: 36 + 49 + 1 + 49 + 25 = 160 Dividing by the total number of sales (which is 5), we get: 160/5 = 32 Taking the square root of this result, we get: sqrt(32) = 5.66 Therefore, the standard deviation is GH¢ 5.66. So the answer is.
Ajụjụ 21 Ripọtì
Two bodies of masses 3kg and 5kg moving with velocities 2 m/s and V m/s respectively in opposite directions collide. If they move together after collision with velocity 3.5 m/s in the direction of the 5kg mass, find the value of V.
Ajụjụ 22 Ripọtì
Solve: \(\sin \theta = \tan \theta\)
Akọwa Nkọwa
We can use the trigonometric identity \(\tan \theta = \dfrac{\sin \theta}{\cos \theta}\) to rewrite the given equation as \(\sin \theta = \dfrac{\sin \theta}{\cos \theta}\). Multiplying both sides by \(\cos \theta\) and simplifying, we get \(\cos \theta = 1\). Therefore, the possible solutions are those angles whose cosine is 1, which are multiples of 360 degrees. Among the given options, only 0 degrees (option D) is a multiple of 360 degrees. So, the answer is \(\theta = 0^{\circ}\).
Ajụjụ 23 Ripọtì
If \(\frac{^{n}C_{3}}{^{n}P_{2}} = 1\), find the value of n.
Ajụjụ 24 Ripọtì
Express \(\log \frac{1}{8} + \log \frac{1}{2}\) in terms of \(\log 2\).
Akọwa Nkọwa
We know that: \[\log a + \log b = \log ab\] Using this property, we can rewrite the given expression as: \[\log \left(\frac{1}{8} \cdot \frac{1}{2}\right) = \log \frac{1}{16}\] Now, we use another property of logarithms: \[\log a^b = b\log a\] to rewrite \(\log \frac{1}{16}\) in terms of \(\log 2\): \[\log \frac{1}{16} = \log 2^{-4} = -4\log 2\] Therefore, \(\log \frac{1}{8} + \log \frac{1}{2} = -4\log 2\) and the answer is: \(-4 \log 2\).
Ajụjụ 25 Ripọtì
The fourth term of an exponential sequence is 192 and its ninth term is 6. Find the common ratio of the sequence.
Akọwa Nkọwa
Let's assume that the first term of the exponential sequence is "a" and the common ratio is "r". Then, we can write the fourth and ninth terms as: Fourth term: ar^3 = 192 Ninth term: ar^8 = 6 We can then divide the two equations to eliminate "a" and obtain a relationship between the two powers of "r": (ar^8) / (ar^3) = 6/192 r^5 = 1/32 Taking the fifth root of both sides, we get: r = (1/32)^(1/5) = 1/2 Therefore, the common ratio of the exponential sequence is \(\frac{1}{2}\).
Ajụjụ 26 Ripọtì
The function \(f : F \to R\)
= \(f(x) = \begin{cases} 3x + 2 : x > 4 \\ 3x - 2 : x = 4 \\ 5x - 3 : x < 4 \end{cases}\). Find f(4) - f(-3).
Akọwa Nkọwa
The function f is defined piecewise as follows: For x > 4, f(x) = 3x + 2 For x = 4, f(x) = 3x - 2 For x < 4, f(x) = 5x - 3 To find f(4) - f(-3), we first need to find f(4) and f(-3) separately. Since f(4) is defined as 3x - 2 when x = 4, we have f(4) = 3(4) - 2 = 10. Since f(x) is defined as 5x - 3 when x < 4 and we are looking for f(-3), we substitute -3 for x to get f(-3) = 5(-3) - 3 = -18. Now we can calculate f(4) - f(-3) as: f(4) - f(-3) = 10 - (-18) = 28 Therefore, the answer is 28.
Ajụjụ 27 Ripọtì
If \(s = 3i - j\) and \(t = 2i + 3j\), find \((t - 3s).(t + 3s)\).
Akọwa Nkọwa
First, let's find the value of the expressions inside the parentheses: \begin{align*} t - 3s &= 2i + 3j - 3(3i - j)\\ &= 2i + 3j - 9i + 3j\\ &= -7i + 6j\\ \\ t + 3s &= 2i + 3j + 3(3i - j)\\ &= 2i + 3j + 9i - 3j\\ &= 11i \end{align*} Now we can substitute these expressions into the dot product: \begin{align*} (t - 3s).(t + 3s) &= (-7i + 6j).(11i)\\ &= -77i + 66j\\ \end{align*} Therefore, the answer is -77.
Ajụjụ 28 Ripọtì
Differentiate \(x^{2} + xy - 5 = 0\).
Akọwa Nkọwa
To differentiate the equation, we need to find the derivative of each term in it, using the rules of differentiation. \begin{align*} \frac{d}{dx}(x^{2} + xy - 5) &= \frac{d}{dx}(0) \\ \frac{d}{dx}(x^{2}) + \frac{d}{dx}(xy) - \frac{d}{dx}(5) &= 0 \\ 2x + y\frac{dx}{dx} + x\frac{dy}{dx} - 0 &= 0 \\ 2x + y + x\frac{dy}{dx} &= 0 \\ x\frac{dy}{dx} &= -(2x + y) \\ \frac{dy}{dx} &= -\frac{2x+y}{x} \end{align*} Therefore, the correct answer is \(\frac{-(2x + y)}{x}\).
Ajụjụ 29 Ripọtì
A particle is acted upon by two forces 6N and 3N inclined at an angle of 120° to each other. Find the magnitude of the resultant force.
Akọwa Nkọwa
To find the magnitude of the resultant force, we can use the Law of Cosines, which states that in any triangle, the square of one side is equal to the sum of the squares of the other two sides minus twice the product of those two sides and the cosine of the included angle. In this case, let the two forces be a and b, with an included angle of 120°. Then, the magnitude of the resultant force, which we'll call R, is: $$R^2 = a^2 + b^2 - 2ab\cos{120^\circ}$$ Since cosine of 120° is -1/2, we can simplify this to: $$R^2 = a^2 + b^2 + ab$$ Substituting the values given in the problem, we get: $$R^2 = (6\text{ N})^2 + (3\text{ N})^2 + (6\text{ N})(3\text{ N}) = 54\text{ N}^2$$ Taking the square root of both sides, we get: $$R = \sqrt{54\text{ N}^2} = 3\sqrt{6}\text{ N} \approx 7.746\text{ N}$$ Therefore, the correct answer is (D) \(3\sqrt{3}\) N.
Ajụjụ 30 Ripọtì
Find the third term in the expansion of \((a - b)^{6}\) in ascending powers of b.
Akọwa Nkọwa
To find the third term in the expansion of \((a-b)^6\) in ascending powers of b, we can use the binomial theorem, which states that: \[(a-b)^6 = \sum_{n=0}^{6} {6 \choose n} a^{6-n}(-b)^n\] The third term is the term where n=2, so we plug in n=2 into the above formula and simplify: \[{6 \choose 2} a^{6-2}(-b)^2 = 15a^4b^2\] Therefore, the third term in the expansion of \((a-b)^6\) in ascending powers of b is \(15a^4b^2\). So the correct answer is option B, "15a^4b^2".
Ajụjụ 32 Ripọtì
Express \(\frac{x^{2} + x + 4}{(1 - x)(x^{2} + 1)}\) in partial fractions.
Akọwa Nkọwa
Ajụjụ 33 Ripọtì
Given that \(P = \begin{pmatrix} 3 & 4 \\ 2 & x \end{pmatrix}; Q = \begin{pmatrix} 1 & 3 \\ -2 & 4 \end{pmatrix}; R = \begin{pmatrix} -5 & 25 \\ -8 & 26 \end{pmatrix}\) and PQ = R, find the value of x.
Akọwa Nkọwa
To find x such that PQ = R, we need to find the matrix product of P and Q, and then compare it to R. The matrix product of P and Q is given by: $$ PQ = \begin{pmatrix} 3 & 4 \\ 2 & x \end{pmatrix} \begin{pmatrix} 1 & 3 \\ -2 & 4 \end{pmatrix} = \begin{pmatrix} (3\times 1) + (4\times -2) & (3\times 3) + (4\times 4) \\ (2\times 1) + (x\times -2) & (2\times 3) + (x\times 4) \end{pmatrix} = \begin{pmatrix} -5 & 25 \\ 2-2x & 22+4x \end{pmatrix} $$ We are given that PQ = R, which means we can equate the corresponding entries of the two matrices. In particular, the entry in the first row and first column of PQ must be equal to the entry in the first row and first column of R, i.e. -5. This gives us the equation: $$ -5 = -5 $$ Next, we can equate the corresponding entries in the second row and first column of PQ and R, i.e. 2-2x = -8. Solving for x, we get: $$ 2-2x = -8 \Rightarrow x = 5 $$ Therefore, the value of x that satisfies the equation PQ = R is 5.
Ajụjụ 34 Ripọtì
If \(\sqrt{x} + \sqrt{x + 1} = \sqrt{2x + 1}\), find the possible values of x.
Akọwa Nkọwa
Ajụjụ 35 Ripọtì
A binary operation * is defined on the set of real numbers, R, by \(x * y = x + y - xy\). If the identity element under the operation * is 0, find the inverse of \(x \in R\).
Akọwa Nkọwa
Ajụjụ 36 Ripọtì
Evaluate \(\int_{0}^{2} (8x - 4x^{2}) \mathrm {d} x\).
Akọwa Nkọwa
To evaluate \(\int_{0}^{2} (8x - 4x^{2}) \mathrm {d} x\), we first need to find the antiderivative of the integrand with respect to x. \[\int (8x - 4x^{2}) \mathrm {d} x = 4x^{2} - \frac{4}{3} x^{3} + C\] where C is the constant of integration. Using the limits of integration, we can evaluate the definite integral as follows: \[\int_{0}^{2} (8x - 4x^{2}) \mathrm {d} x = \left[4x^{2} - \frac{4}{3} x^{3}\right]_{0}^{2}\] \[= \left[4(2)^{2} - \frac{4}{3}(2)^{3}\right] - \left[4(0)^{2} - \frac{4}{3}(0)^{3}\right]\] \[= \frac{16}{3}\] Therefore, the answer is \(\frac{16}{3}\). Option (c) is the correct answer.
Ajụjụ 37 Ripọtì
Given that \(P = \begin{pmatrix} y - 2 & y - 1 \\ y - 4 & y + 2 \end{pmatrix}\) and |P| = -23, find the value of y.
Ajụjụ 38 Ripọtì
Find the range of values of x for which \(x^{2} + 4x + 5\) is less than \(3x^{2} - x + 2\)
Akọwa Nkọwa
Ajụjụ 39 Ripọtì
Find the equation of the line that is perpendicular to \(2y + 5x - 6 = 0\) and bisects the line joining the points P(4, 3) and Q(-6, 1).
Akọwa Nkọwa
Ajụjụ 40 Ripọtì
P and Q are the points (3, 1) and (7, 4) respectively. Find the unit vector along PQ.
Akọwa Nkọwa
Ajụjụ 41 Ripọtì
(a) Find the maximum and minimum points of the curve \(y = 2x^{3} - 3x^{2} - 12x + 4\).
(b) Sketch the curve in (a) above.
Ajụjụ 42 Ripọtì
(a) The sum of the first three terms of a decreasing exponential sequence (G.P) is equal to 7 and the product of these three is equal to 8. Find the :
(i) common ratio ; (ii) first three terms of the sequence.
(b) Using the trapezium rule with the ordinates at x = 1, 2, 3, 4 and 5, calculate, correct to two decimal places, the value of \(\int_{1} ^{5} (x + \frac{2}{x^{2}}) \mathrm {d} x\).
Ajụjụ 43 Ripọtì
A stone is dropped vertically downwards from the top of a tower of height 45m with a speed of 20 ms\(^{-1}\). Find the :
(a) time it takes to reach the ground ;
(b) speed with which it hits the ground. [Take \(g = 10 ms^{-2}\)].
Ajụjụ 44 Ripọtì
Three school prefects are to be chosen from four girls and five boys. What is the probability that :
(a) only boys will be chosen ;
(b) more girls than boys will be chosen ?
Ajụjụ 45 Ripọtì
A particle is under the action of forces \(P = (4N, 030°)\) and \(R = (10N, 300°)\). Find the force that will keep the particle in equilibrium.
Ajụjụ 46 Ripọtì
(a) Differentiate \(\frac{x^{2} + 1}{(x + 1)^{2}}\) with respect to x.
(b)(i) Evaluate \(\begin{vmatrix} 1 & 2 & -1 \\ 2 & 3 & -1 \\ -1 & 1 & 3 \end{vmatrix}\).
(ii) Using the answer in (b)(i), solve the system of equations.
\(x + 2y - z = 4\)
\(2x + 3y - z = 2\)
\(-x + y + 3z = -1\).
None
Akọwa Nkọwa
None
Ajụjụ 47 Ripọtì
(a) Three vectors a, b and c are \(\begin{pmatrix} 8 \\ 3 \end{pmatrix}, \begin{pmatrix} 6 \\ -5 \end{pmatrix}\) and \(\begin{pmatrix} 2 \\ -3 \end{pmatrix}\) respectively. Find the vector d such that \(|d| = \sqrt{41}\) and d is in the direction of \(a + b - 2c\).
(b) The coordinates of A and B are (3, 4) and (3, n) respectively. If AOB = 30°, find, correct to 2 decimal places, the values of n.
Ajụjụ 48 Ripọtì
The line \(2y = x + 3\) meets the circle \(x^{2} + y^{2} - 2x + 6y - 15 = 0\) at points M and N, where N is in the first quadrant. Find the coordinates of M and N.
Ajụjụ 49 Ripọtì
(a) If the coefficient of \(x^{2}\) and \(x^{3}\) in the expansion of \((p + qx)^{7}\) are equal, express q in terms of p.
(b) A man makes a weekly contribution into a fund. In the first week, he paid N180.00, second week N260.00, third week N340.00 and so on. How much would he have contributed in 16 weeks?
Ajụjụ 50 Ripọtì
(a) The probability that a man wins a race is 0.8. In four different races, what is the probability that he wins : (i) all races ; (ii) no race ; (iii) at most 3 races ?
(b) A class consists of 5 girls and 10 boys. If a committee of 5 is chosen at random from the class, find the probability that :
(i) 3 boys are selected ; (ii) at least one girl is selected.
Ajụjụ 51 Ripọtì
The table gives the relationship between the height, in metres, of a plant and the number of days it is left to grow.
| Number of days (x) |
10 | 20 | 30 | 40 | 50 | 60 | 70 | 80 |
| Height (y) | 1.0 | 1.1 | 1.2 | 1.4 | 1.6 | 1.8 | 2.0 | 2.3 |
(a) Using a scale of 2 cm to represent 0.5 units on the y- axis and 2cm to 10 units on the x- axis, draw a scatter diagram for the information.
(b) Find \(\bar{x}\), the mean of x, and \(\bar{y}\), the mean of y, and plot \((\bar{x}, \bar{y})\) on the diagram.
(c) Draw the line of best fit to pass through \((\bar{x}, \bar{y})\) and \((10, 1)\).
(d) From graph, find the :
(i) equation of the line of best fit ; (ii) height of plant in 75 days.
Ajụjụ 52 Ripọtì
(a) Using a scale of 2 cm to 30° on the x- axis, 2 cm to 0.2 units on the y- axis, on the same graph sheet, draw the graphs of \(y = \sin 2x\) and \(y = \cos x\) for \(0° \leq x \leq 210°\) at intervals of 30°.
(b) Using the graphs in (a), find the truth set of :
(i) \(\sin 2x = 0\) ; (ii) \(\sin 2x - \cos x = 0\).
Ajụjụ 53 Ripọtì
The initial velocity of a particle of mass 0.1kg is 40 m/s in the direction of the unit vector j. The velocity of the particle changed to 30 m/s in the direction of the unit vector i. Find the change in momentum.
Ajụjụ 54 Ripọtì
Calculate the gradient of the curve \(x^{3} + y^{3} - 2xy = 11\) at (2, -1).
Ajụjụ 55 Ripọtì
A side of a rectangle is three times the other. If the perimeter increases by 2%, find the percentage increase in the area of the rectangle.
Ajụjụ 56 Ripọtì
The table shows the distribution of ages of 22 students in a school.
| Age (years) | 12-14 | 15-17 | 18-20 | 21-23 | 24-26 |
| Frequency | 6 | 10 | 3 | 2 | 1 |
Using an assumed mean of 19, calculate, correct to three significant figures, the :
(a) mean age ; (b) standard deviation ; of the distribution.
Ajụjụ 57 Ripọtì
(a) A bag contains 5 blue, 4 green and 3 yellow balls. All the balls are identical except for colour. Three balls are drawn at random without replacement. Find the probability that : (i) all three balls have the same colour ; (ii) two balls have the same colour.
(b) The table shows the ranks of the marks scored by 7 candidates in Physics and Chemistry tests.
| Physics | 6 | 5 | 4 | 3 | 2 | 7 | 1 |
| Chemistry | 7 | 6 | 2 | 4 | 1 | 5 | 3 |
Calculate the Spearman's rank correlation coefficient.
Ajụjụ 58 Ripọtì
The displacement S metres of a particle from a fixed point O at time t seconds is given by \(S = t^{2} - 6t + 5\).
(a) On a graph sheet, draw a displacement- time graph for the interval \(0 \leq x \leq 6\).
(b) From the graph, find the : (i) time at which the velocity is zero ; (ii) average velocity over the interval \(0 \leq x \leq 4\) ; (iii) total distance covered in the interval \(0 \leq x \leq 5\).
