WAEC SSCE - Further Mathematics - 2015
Tambaya 1 Rahoto
In how many ways can a committee of 5 be selected from 8 students if 2 particular students are to be included?
Bayanin Amsa
Since 2 particular students must be on the committee, we only need to select 3 more students from the remaining 6 students. The number of ways to do this is the number of combinations of 3 students from a set of 6, which is given by: \(\binom{6}{3}=\frac{6!}{3!3!}=20\) Therefore, the answer is 20.
Tambaya 3 Rahoto
In calculating the mean of 8 numbers, a boy mistakenly used 17 instead of 25 as one of the numbers. If he obtained 20 as the mean, find the correct mean.
Bayanin Amsa
Let's call the sum of the 8 numbers as S. Then, the boy's calculation of the mean is: $$ \text{Mean} = \frac{S+17}{8} $$ However, the correct calculation of the mean should be: $$ \text{Correct Mean} = \frac{S+25}{8} $$ We know that the boy obtained 20 as the mean, so we can set the two expressions equal to each other and solve for S: $$ \frac{S+17}{8} = 20 \implies S+17 = 160 \implies S = 143 $$ Therefore, the correct mean is: $$ \text{Correct Mean} = \frac{S+25}{8} = \frac{143+25}{8} = \frac{168}{8} = 21 $$ So the correct mean is 21. Therefore, the answer is (3) 21.
Tambaya 4 Rahoto
Four fair coins are tossed once. Calculate the probability of having equal heads and tails.
Bayanin Amsa
When tossing a coin, there are two possible outcomes: heads or tails. Since we are tossing four coins, there are \(2^{4} = 16\) possible outcomes. We want to find the probability of having an equal number of heads and tails. The possible outcomes where we have an equal number of heads and tails are as follows: HHTT, HTHT, THHT, THTH, HTTH, TTHH. There are six such outcomes. Therefore, the probability of having an equal number of heads and tails is \(\frac{6}{16} = \frac{3}{8}\). So the answer is.
Tambaya 5 Rahoto
Given that \(y = 4 - 9x\) and \(\Delta x = 0.1\), calculate \(\Delta y\).
Bayanin Amsa
Tambaya 6 Rahoto
Find the coordinates of the point which divides the line joining P(-2, 3) and Q(4, 6) internally in the ratio 2 : 3.
Tambaya 7 Rahoto
Forces \(F_{1} = (8N, 030°)\) and \(F_{2} = (10N, 150°)\) act on a particle. Find the horizontal component of the resultant force.
Bayanin Amsa
To find the horizontal component of the resultant force, we need to resolve each force into its horizontal and vertical components. For \(F_{1} = (8N, 030°)\), the horizontal component is given by \(F_{1}\cos{30°} = 8\cos{30°} \approx 6.93N\). For \(F_{2} = (10N, 150°)\), the horizontal component is given by \(F_{2}\cos{150°} = 10\cos{150°} \approx -4.32N\). The negative sign in the horizontal component of \(F_2\) means that the force is acting in the opposite direction. To find the horizontal component of the resultant force, we sum up the horizontal components of the two forces: \[\text{Horizontal component of resultant force} = 6.93N - 4.32N = 2.61N \approx 1.7N\] Therefore, the horizontal component of the resultant force is approximately 1.7N. The option that matches this value is (a) 1.7N.
Tambaya 8 Rahoto
Forces of magnitude 8N and 5N act on a body as shown above. Calculate, correct to 2 d.p., the resultant force acting at O.
Bayanin Amsa
Tambaya 9 Rahoto
Given that \(f : x \to x^{2}\) and \(g : x \to x + 3\), where \(x \in R\), find \(f o g(2)\).
Bayanin Amsa
To find \(f \circ g(2)\), we first need to apply the function g to 2, which gives us \(g(2) = 2 + 3 = 5\). Then we apply the function f to the result, which gives us \(f(5) = 5^2 = 25\). Therefore, the answer is 25.
Tambaya 10 Rahoto
Given that \(r = 2i - j\), \(s = 3i + 5j\) and \(t = 6i - 2j\), find the magnitude of \(2r + s - t\).
Bayanin Amsa
We are given that: \begin{align*} r &= 2i - j\\ s &= 3i + 5j\\ t &= 6i - 2j \end{align*} We can now substitute these expressions into \(2r+s-t\) and simplify: \begin{align*} 2r + s - t &= 2(2i-j) + (3i+5j) - (6i - 2j)\\ &= 4i - 2j + 3i + 5j - 6i + 2j\\ &= i + 5j \end{align*} The magnitude of a vector with components \(a\) and \(b\) is given by \(\sqrt{a^2 + b^2}\). So, the magnitude of \(i+5j\) is: \begin{align*} \sqrt{i^2 + 5j^2} &= \sqrt{1^2 + 5^2} \\ &= \sqrt{26} \end{align*} Therefore, the answer is \(\sqrt{26}\).
Tambaya 11 Rahoto
If \(36, p, \frac{9}{4}, q\) are consecutive terms of an exponential sequence (G.P.). Find the sum of p and q.
Tambaya 12 Rahoto
A particle starts from rest and moves in a straight line such that its acceleration after t secs is given by \(a = (3t - 2) ms^{-2}\). Find the distance covered after 3 secs.
Bayanin Amsa
To solve this problem, we need to use the kinematic equation: $$s = ut + \frac{1}{2}at^2$$ where: - s is the distance covered - u is the initial velocity (which is zero in this case) - a is the acceleration - t is the time taken We are given that the acceleration is given by \(a = (3t-2)ms^{-2}\). Integrating this with respect to time, we get: $$v = \int a dt = \int (3t-2) dt = \frac{3}{2}t^2 - 2t + C$$ where v is the velocity of the particle at time t, and C is a constant of integration. Since the particle starts from rest, we have v = 0 when t = 0, so: $$0 = \frac{3}{2}(0)^2 - 2(0) + C$$ which gives C = 0. Therefore, the velocity of the particle at time t is given by: $$v = \frac{3}{2}t^2 - 2t$$ Now, we can use the equation for distance covered to find the distance covered by the particle after 3 seconds: $$s = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2}(3t-2)t^2 = \frac{3}{2}t^3 - t^2$$ So, when t = 3, we have: $$s = \frac{3}{2}(3)^3 - (3)^2 = \frac{27}{2} - 9 = \frac{9}{2}m$$ Therefore, the distance covered by the particle after 3 seconds is \(\frac{9}{2}m\). Thus, the correct option is: - \(\frac{9}{2}m\)
Tambaya 13 Rahoto
If \(x = i - 3j\) and \(y = 6i + j\), calculate the angle between x and y.
Bayanin Amsa
Tambaya 14 Rahoto
Given that \(f : x \to \frac{2x - 1}{x + 2}, x \neq -2\), find \(f^{-1}\), the inverse of f.
Bayanin Amsa
Tambaya 15 Rahoto
Find the minimum value of \(y = x^{2} + 6x - 12\).
Bayanin Amsa
To find the minimum value of the quadratic function, we need to determine its vertex. We can do this by completing the square or using the formula for the x-coordinate of the vertex, which is given by: $$x = -\frac{b}{2a}$$ where a and b are the coefficients of the quadratic function in standard form: \(y = ax^{2} + bx + c\). In this case, we have: $$y = x^{2} + 6x - 12$$ where a = 1 and b = 6. Therefore, the x-coordinate of the vertex is: $$x = -\frac{b}{2a} = -\frac{6}{2(1)} = -3$$ To find the minimum value of y, we substitute x = -3 into the quadratic function: $$y = (-3)^{2} + 6(-3) - 12 = -21$$ Therefore, the minimum value of y is -21. So, the correct option is (a) -21.
Tambaya 16 Rahoto
The sum, \(S_{n}\), of a sequence is given by \(S_{n} = 2n^{2} - 5\). Find the 6th term.
Bayanin Amsa
To find the 6th term of the sequence, we need to find the difference between the sum of the first 6 terms and the sum of the first 5 terms. The sum of the first 5 terms is: $S_5 = 2(5)^2 - 5 = 45$ The sum of the first 6 terms is: $S_6 = 2(6)^2 - 5 = 67$ Therefore, the 6th term of the sequence is: $S_6 - S_5 = 67 - 45 = 22$ So the answer is 22.
Tambaya 17 Rahoto
A particle starts from rest and moves in a straight line such that its acceleration after t seconds is given by \(a = (3t - 2) ms^{-2}\). Find the other time when the velocity would be zero.
Bayanin Amsa
The acceleration of the particle is given as \(a = (3t - 2) ms^{-2}\). We can integrate this to get the velocity function, \(v(t)\), of the particle: \[\int a dt = \int (3t - 2) dt\] \[v(t) = \frac{3}{2}t^2 - 2t + C\] where C is the constant of integration. Since the particle starts from rest, we have \(v(0) = 0\), which implies that C = 0. Therefore, the velocity function of the particle is given as: \[v(t) = \frac{3}{2}t^2 - 2t\] To find the other time when the velocity would be zero, we need to solve the equation \(v(t) = 0\): \[\frac{3}{2}t^2 - 2t = 0\] \[t(3t - 4) = 0\] This equation has two roots, \(t = 0\) and \(t = \frac{4}{3}\). Since the particle is starting from rest, we discard the root \(t = 0\). Therefore, the other time when the velocity would be zero is \(t = \frac{4}{3}\) seconds. Hence, the answer is \(\frac{4}{3} seconds\).
Tambaya 18 Rahoto
Given the statements:
p : the subject is difficult
q : I will do my best
Which of the following is equivalent to 'Although the subject is difficult, I will do my best'?
Bayanin Amsa
The given sentence can be rewritten as "Despite the subject being difficult, I will do my best." We can now use the logical operators to find the equivalent statement. "Despite the subject being difficult" can be represented as \(\sim p\) (not p), and "I will do my best" is equivalent to q. Then, the statement "Although the subject is difficult, I will do my best" is equivalent to \(\sim p \wedge q\). Therefore, the correct option is \( \sim p \wedge q \).
Tambaya 19 Rahoto
If \(\begin{vmatrix} 1+2x & -1 \\ 6 & 3-x \end{vmatrix} = -3 \), find the values of x.
Tambaya 20 Rahoto
Solve: \(2\cos x - 1 = 0\).
Bayanin Amsa
To solve the equation \(2\cos x - 1 = 0\), we first isolate the cosine term by adding 1 to both sides, giving: $$2\cos x = 1$$ Next, we divide both sides by 2 to obtain: $$\cos x = \frac{1}{2}$$ From the unit circle or trigonometric ratios, we know that the solutions to this equation are the angles whose cosine is equal to 1/2, which are \(\frac{\pi}{3}\) and \(\frac{5\pi}{3}\) (or their reference angles in the interval [0, 2π)). Therefore, the answer is \((\frac{\pi}{3}, \frac{5\pi}{3})\).
Tambaya 21 Rahoto
If \(\alpha\) and \(\beta\) are the roots of \(x^{2} + x - 2 = 0\), find the value of \(\frac{1}{\alpha^{2}} + \frac{1}{\beta^{2}}\).
Bayanin Amsa
To find the value of \(\frac{1}{\alpha^{2}} + \frac{1}{\beta^{2}}\), we first need to find the values of \(\alpha\) and \(\beta\). Given that \(x^{2} + x - 2 = 0\), we can factorize the quadratic equation as \((x-1)(x+2)=0\). Thus, the roots are \(\alpha = 1\) and \(\beta = -2\). Now, we can substitute these values into the expression to get: \[\frac{1}{\alpha^{2}} + \frac{1}{\beta^{2}} = \frac{1}{1^{2}} + \frac{1}{(-2)^{2}} = 1 + \frac{1}{4} = \frac{5}{4}\] Therefore, the answer is \(\frac{5}{4}\).
Tambaya 22 Rahoto
A 24N force acts on a body such that it changes its velocity from 5m/s to 9m/s in 2 secs.If the body is travelling in a straight line, calculate the distance covered in the period.
Bayanin Amsa
The problem involves finding the distance covered by a body when a force is applied to it and it changes its velocity from an initial value to a final value over a period of time. The force acting on the body is 24N and the change in velocity is (9 - 5)m/s = 4m/s. The time taken for this change is 2 seconds. Using the formula, distance = (initial velocity + final velocity) * time / 2, we can calculate the distance covered by the body. Substituting the given values, we get distance = (5m/s + 9m/s) * 2s / 2 = 14m Therefore, the distance covered by the body is 14m. Answer: 14m
Tambaya 23 Rahoto
Find the 3rd term of \((\frac{x}{2} - 1)^{8}\) in descending order of x.
Bayanin Amsa
Tambaya 24 Rahoto
Given that \(\frac{2x}{(x + 6)(x + 3)} = \frac{P}{x + 6} + \frac{Q}{x + 3}\), find P and Q.
Bayanin Amsa
We can begin by multiplying both sides of the equation by the denominator \((x+6)(x+3)\). This will eliminate the denominators on the left side of the equation, leaving us with: $$2x = P(x+3) + Q(x+6)$$ Next, we can expand the right side of the equation: $$2x = Px + 3P + Qx + 6Q$$ We can then simplify by collecting like terms: $$(P + Q)x + (3P + 6Q) = 2x$$ Since this equation must be true for all values of x, we can equate the coefficients of x and the constants on both sides of the equation: $$\begin{aligned} P + Q &= 0 \\ 3P + 6Q &= 2 \end{aligned}$$ Solving the first equation for P in terms of Q, we get: $$P = -Q$$ Substituting this expression for P into the second equation, we get: $$3(-Q) + 6Q = 2$$ Simplifying this equation, we get: $$3Q = 2$$ Thus, we have: $$Q = \frac{2}{3}$$ Substituting this value for Q into the equation P = -Q, we get: $$P = -\frac{2}{3}$$ Therefore, we have: $$P = -\frac{2}{3}\text{ and }Q = \frac{2}{3}$$ So the answer is, P = 4 and Q = -2.
Tambaya 25 Rahoto
Find \(\int \frac{x^{3} + 5x + 1}{x^{3}} \mathrm {d} x\)
Bayanin Amsa
We can write the given expression as: $$\int \frac{x^{3} + 5x + 1}{x^{3}} \mathrm {d} x = \int \left(1 + \frac{5}{x^{2}} + \frac{1}{x^{3}}\right) \mathrm {d} x$$ Now, we integrate each term separately: $$\int \mathrm {d} x + 5 \int \frac{1}{x^{2}} \mathrm {d} x + \int \frac{1}{x^{3}} \mathrm {d} x = x - \frac{5}{x} - \frac{1}{2x^{2}} + c$$ Therefore, the correct option is: $$\boxed{x - \frac{5}{x} - \frac{1}{2x^{2}} + c}$$
Tambaya 26 Rahoto
Given that \(P = \begin{pmatrix} -2 & 1 \\ 3 & 4 \end{pmatrix}\) and \(Q = \begin{pmatrix} 5 & -3 \\ 2 & -1 \end{pmatrix}\), find PQ - QP.
Tambaya 27 Rahoto
A line passes through the origin and the point \((1\frac{1}{4}, 2\frac{1}{2})\). Find the y-coordinate of the line when x = 4.
Tambaya 29 Rahoto
| Marks | 0 | 1 | 2 | 3 | 4 | 5 |
| Number of candidates | 6 | 4 | 8 | 10 | 9 | 3 |
The table above shows the distribution of marks scored by students in a test. How many candidates scored above the median score?
Bayanin Amsa
To find the median score, we need to arrange the scores in ascending or descending order. From the table, we can see that there are 40 candidates in total. To determine the median score, we need to find the middle value. Since there are an even number of candidates (40), the median will be the average of the two middle values. The two middle values are the 20th and 21st scores, which are both 3. Therefore, the median score is 3. To find how many candidates scored above the median score, we can add up the number of candidates who scored 4 and 5, which are the scores above 3. From the table, we can see that 9 candidates scored 4 and 3 candidates scored 5, making a total of 12 candidates who scored above the median score of 3. Therefore, the answer is 12.
Tambaya 30 Rahoto
Which of the following is a factor of the polynomial \(6x^{4} + 2x^{3} + 15x + 5\)?
Bayanin Amsa
To determine if a given expression is a factor of a polynomial, we can use the factor theorem which states that if a polynomial f(a) is divided by x - a, then the remainder is zero if and only if x - a is a factor of the polynomial. To use the factor theorem in this case, we can try to divide the polynomial \(6x^{4} + 2x^{3} + 15x + 5\) by each of the given expressions to see if the remainder is zero. We can use long division or synthetic division to perform the division. Using synthetic division, we can test each option as follows: - For 3x + 1: -1/3 | 6 2 0 15 5 | -2 2 -5 -------------- 6 0 2 10 0 The remainder is zero, so 3x + 1 is a factor of the polynomial. - For x + 1: -1 | 6 2 0 15 5 | -6 -8 -7 ------------ 6 -4 -8 8 The remainder is not zero, so x + 1 is not a factor of the polynomial. - For 2x + 1: -1/2 | 6 2 0 15 5 | -3 -2.5 ------------ 6 0 3 12.5 The remainder is not zero, so 2x + 1 is not a factor of the polynomial. - For x + 2: -2 | 6 2 0 15 5 | -12 -20 -30 ------------- 6 -10 -20 -15 The remainder is not zero, so x + 2 is not a factor of the polynomial. Therefore, only 3x + 1, is a factor of the polynomial \(6x^{4} + 2x^{3} + 15x + 5\). The answer is: 3x + 1.
Tambaya 31 Rahoto
A mass of 75kg is placed on a lift. Find the force exerted by the floor of the lift on the mass when the lift is moving up with constant velocity. \([g = 9.8ms^{-2}]\)
Bayanin Amsa
When the lift is moving up with a constant velocity, the net force acting on the mass must be zero since the mass is not accelerating. Therefore, the force exerted by the floor of the lift on the mass is equal in magnitude and opposite in direction to the force of gravity acting on the mass. The force of gravity acting on the mass is given by: F = m*g = 75kg * 9.8m/s^2 = 735N Therefore, the force exerted by the floor of the lift on the mass when the lift is moving up with constant velocity is 735N. Hence, the correct option is: 735N.
Tambaya 32 Rahoto
If \(2\sin^{2} \theta = 1 + \cos \theta, 0° \leq \theta \leq 90°\), find the value of \(\theta\).
Bayanin Amsa
We are given an equation in terms of trigonometric functions, and we are asked to find the value of one of the angles involved. One possible way to approach this problem is to use trigonometric identities to simplify the equation and then solve for the unknown angle. First, we can use the identity \(\sin^{2}\theta + \cos^{2}\theta = 1\) to rewrite the left-hand side of the equation as \(\sin^{2}\theta = 1 - \cos^{2}\theta\). Substituting this into the given equation, we get: \[2(1-\cos^{2}\theta) = 1 + \cos\theta\] Expanding and rearranging terms, we obtain a quadratic equation in terms of \(\cos\theta\): \[2\cos^{2}\theta + \cos\theta - 1 = 0\] We can solve this quadratic equation using the quadratic formula: \[\cos\theta = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}\] where \(a=2\), \(b=1\), and \(c=-1\). Plugging in these values, we get: \[\cos\theta = \frac{-1 \pm \sqrt{1+8}}{4} = \frac{-1 \pm 3}{4}\] Therefore, the two possible values of \(\cos\theta\) are \(\frac{1}{2}\) and \(-1\). However, we are given that \(0° \leq \theta \leq 90°\), which means that \(\cos\theta \geq 0\) in this range. Therefore, we must choose \(\cos\theta = \frac{1}{2}\), which gives: \[\cos\theta = \frac{1}{2} \implies \theta = \cos^{-1}\frac{1}{2} \approx 60°\] Therefore, the answer is (B) 60°.
Tambaya 33 Rahoto
If \(\log_{3} x = \log_{9} 3\), find the value of x.
Tambaya 34 Rahoto
Forces of magnitude 8N and 5N act on a body as shown above. Calculate, correct to 2 dp, the angle that the resultant makes with the horizontal.
Tambaya 35 Rahoto
A line passes through the origin and the point \((1\frac{1}{4}, 2\frac{1}{2})\), what is the gradient of the line?
Tambaya 36 Rahoto
| Marks | 0 | 1 | 2 | 3 | 4 | 5 |
| Number of candidates | 6 | 4 | 8 | 10 | 9 | 3 |
The table above shows the distribution of marks scored by students in a test. Find the interquartile range of the distribution.
Bayanin Amsa
Tambaya 37 Rahoto
Simplify: \(^{n}C_{r} ÷ ^{n}C_{r-1}\)
Tambaya 38 Rahoto
Solve: \(4(2^{x^2}) = 8^{x}\)
Bayanin Amsa
We can rewrite \(8^x\) as \((2^3)^x = 2^{3x}\). Substituting this into the equation, we have: $$4(2^{x^2}) = 2^{3x}$$ Dividing both sides by \(2^{x^2}\) and simplifying, we get: $$2^{x^2 - 3x + 2} = 0$$ Using the zero-product property of multiplication, we can set the exponent equal to zero: $$x^2 - 3x + 2 = 0$$ Factoring the quadratic, we get: $$(x-1)(x-2) = 0$$ Therefore, the solutions are \(x=1\) and \(x=2\). Hence, the answer is (1, 2).
Tambaya 39 Rahoto
Each of the 90 students in a class speak at least Igbo or Hausa. If 56 students speak Igbo and 50 speak Hausa, find the probability that a student selected at random from the class speaks Igbo only.
Bayanin Amsa
Tambaya 40 Rahoto
Given that \(x^{2} + 4x + k = (x + r)^{2} + 1\), find the value of k and r.
Bayanin Amsa
We know that the left side of the equation can be expanded to give: $$x^{2} + 4x + k = x^{2} + 2xr + r^{2} + 1$$ By comparing the coefficients of x and x^2, we get: $$2xr + 4x = 0 \implies x(2r+4) = 0$$ Since x cannot be equal to zero, we have: $$2r + 4 = 0 \implies r = -2$$ Substituting r = -2 in the original equation, we get: $$x^{2} + 4x + k = (x-2)^{2} + 1 = x^{2} - 4x + 5$$ Comparing the coefficients of x^2 and x, we get: $$k = 5$$ Therefore, the solution is k = 5, r = -2. Note that this matches with.
Tambaya 41 Rahoto
Points (2, 1) and (6, 7) are opposite vertices of a square which is inscribed in a circle. Find the :
(a) centre of the circle ; (b) equation of the circle.
Tambaya 42 Rahoto
Five finalists in a beauty pageant were ranked by two judges X and Y as shown in the table :
| Judges | Anne | Linda | Susan | Rose | Erica |
| X | 1 | 4 | 3 | 5 | 2 |
| Y | 3 | 2 | 4 | 5 | 1 |
Calculate the Spearman's rank correlation coefficient.
Tambaya 43 Rahoto
(a) If \(\frac{\sqrt{5} + 4}{3 - 2\sqrt{5}} - \frac{2 + \sqrt{5}}{4 - 2\sqrt{5}} = a + b\sqrt{5}\), find the values of a and b.
(b)(i) Evaluate : \(\begin{vmatrix} 2 & -1 & 2 \\ 1 & 3 & 4 \\ 1 & 2 & 1 \end{vmatrix}\)
(ii) Using the result in b(i), find, correct to two decimal places, the value of x in the system of equations.
\(2x - y + 2z + 5 = 0\)
\(x + 3y + 4z - 1 = 0\)
\(x + 2y + z + 2 = 0\)
Bayanin Amsa
None
Tambaya 44 Rahoto
There are 8 boys and 6 girls in a class. If two students are selected at random from the class, find the probability that they are of
(a) the same sex ;
(b) different sex.
Tambaya 45 Rahoto
Forces of magnitude 3N, 4N and 2N act along the vectors \(j ; -i + j\) and \(i + j\) respectively. Calculate, correct to one decimal place, the magnitude of the resultant of the forces.
Tambaya 46 Rahoto
(a) Given that \(\overrightarrow{AB} = \begin{pmatrix} 4 \\ 5 \end{pmatrix}\) and \(\overrightarrow{BC} = \begin{pmatrix} -3 \\ 5 \end{pmatrix}\); find the :
(i) angle between the vectors AB and AC ; (ii) unit vector along \(\overrightarrow{AB} - \overrightarrow{BC}\).
(b) P, Q, R and M are points in the \(O_{XY}\) plane. If \(\overrightarrow{PQ} = 2i + 8j , \overrightarrow{PR} = 11i - 12j\) and M divides QR internally in the ratio 3 : 7, find \(\overrightarrow{PM}\).
Tambaya 47 Rahoto
(a)(i) Write down the binomial expansion of \((1 + x)^{4}\).
(ii) Use the result in (a)(i) to evaluate, correct to three decimal places \((\frac{5}{4})^{4}\).
(b) The first, second and fifth terms of a linear sequence (A.P) are three consecutive terms of an exponential sequence (G.P). If the first term of the linear sequence is 7, find the common difference.
Tambaya 48 Rahoto
(a) Edem and his wife were invited to a dinner by a family of 5. They all sat in such a way in such a way that Edem sat next to his wife. Find the number of ways of seating them in a row.
(b) A bag contains 4 red and 5 black identical balls. If 5 balls are selected at random, one after the other with replacement, find the probability that :
(i) a red ball was picked 3 times ; (ii) a black ball was picked at most 2 times.
Tambaya 49 Rahoto
(a) A bucket full of water with a mass of 8kg is pulled out of a well with a light inextensible rope. Find its acceleration when tha tension in the rope is 150N. [Take \(g = 10ms^{-2}\)].
(b) A mass of 12kg is acted upon by a force F, changing its speed from 15 m/s to 25 m/s after covering a distance of 50m. Find the :
(i) value of F ; (ii) distance covered when its speed is 35 m/s.
Tambaya 50 Rahoto
The table shows the distribution of the heights of a group of people.
| Height (m) |
0.4 - 0.5 | 0.6 - 0.9 | 1.0 - 1.2 | 1.3 - 1.4 | 1.5 - 1.7 |
| Number of people | 2 | 8 | 12 | 6 | 6 |
(a) Draw a histogram to illustrate the distribution.
(b) Using an assumed mean of 1.1m, find, correct to one decimal place, the mean height of the group.
Tambaya 51 Rahoto
(a) The functions \(f : x \to x^{2} + 1\) and \(g : x \to 5 - 3x\) are defined on the set of the real numbers, R.
(i) State the domain of \(f^{-1}\), the inverse of f ; (ii) find \(g^{-1} (2)\).
(b) Evaluate : \(\int \frac{(x + 3)}{x^{2} + 6x + 9} \mathrm {d} x\)
Tambaya 52 Rahoto
A car moving with an initial velocity, u, travels in a straight line with a constant acceleration of 3ms\(^{-2}\) until it attains a velocity of 33ms\(^{-1}\) after 6 seconds. Calculate the distance travelled by the car.
Bayanin Amsa
None
Tambaya 53 Rahoto
If \(f ' '(x) = 2\), \(f ' (1) = 0\) and \(f(0) = - 8\), find f(x).
Tambaya 54 Rahoto
A binary operation A is defined on the set of real numbers, R, by \(a \Delta b = a^{3} - b^{3}\). Without using calculator, find the value of \((\sqrt{3} + \sqrt{2}) \Delta (\sqrt{3} - \sqrt{2})\) leaving the answer in surd form.
Tambaya 55 Rahoto
Solve : \(\tan (2x - 15)° - 1 = 0\), for values of x such that \(0° \leq x \leq 360°\).
