WAEC SSCE - Further Mathematics - 2012

Maelezo ya Majibu

Variance is not a measure of central tendency. Measures of central tendency are used to describe the typical or central value of a set of data, while variance is a measure of how spread out the data is from the mean. Variance is a measure of variability, not centrality. Therefore, the answer is "Variance."

Maelezo ya Majibu

To solve this problem, we can use the formula: \[ s = ut + \frac{1}{2}at^2 \] where s is the distance, u is the initial velocity, t is the time, and a is the acceleration due to gravity. In this case, the initial velocity is zero because the stone is dropped from rest. We also know that the distance is 45m and the acceleration due to gravity is 10 \(ms^{-2}\). Thus, we have: \begin{align*} s &= ut + \frac{1}{2}at^2 \\ 45 &= 0t + \frac{1}{2}(10)t^2 \\ 45 &= 5t^2 \\ t^2 &= 9 \\ t &= 3 \end{align*} Therefore, the time it takes for the stone to hit the ground is 3 seconds. Hence, the answer is (a) 3.0 seconds.

Maelezo ya Majibu

Maelezo ya Majibu

Maelezo ya Majibu

The correct answer is: \(-1 \leq r \leq 1\). The correlation coefficient (r) measures the degree of linear association between two variables. The value of r ranges from -1 to 1, where -1 indicates a perfect negative correlation, 0 indicates no correlation, and 1 indicates a perfect positive correlation. Since r can take on any value between -1 and 1 (including -1 and 1), the correct statement is \(-1 \leq r \leq 1\), which means that the correlation coefficient is always between -1 and 1, inclusive. Therefore, options (A), (B), and (C) are incorrect.

Maelezo ya Majibu

To find the unit vector in the direction of the vector \(-12i + 5j\), we need to divide the vector by its magnitude. The magnitude of a vector with components \(a\) and \(b\) is given by the formula \(\sqrt{a^2+b^2}\). So, the magnitude of the vector \(-12i + 5j\) is \(\sqrt{(-12)^2+5^2} = 13\). Now, to get the unit vector, we divide each component of the vector by its magnitude: \[\frac{-12}{13}i + \frac{5}{13}j\] This is the unit vector in the direction of the vector \(-12i + 5j\). Therefore, the correct option is \(\frac{-12i}{13} + \frac{5j}{13}\).

Maelezo ya Majibu

Maelezo ya Majibu

To find the point of intersection of the curve and the line, we need to solve the system of equations formed by equating the two equations: \begin{align*} y &= x^2 + 2x - 3 \\ y &= -x + 1 \end{align*} Setting the right-hand sides equal to each other, we get: \begin{align*} x^2 + 2x - 3 &= -x + 1 \\ x^2 + 3x - 4 &= 0 \\ (x + 4)(x - 1) &= 0 \end{align*} Thus, the values of x at the points of intersection are x = -4 and x = 1. To find the corresponding y-values, we substitute these values of x back into either equation. Using the equation y = x^2 + 2x - 3, we get: \begin{align*} y &= (-4)^2 + 2(-4) - 3 \\ &= 7 \end{align*} and \begin{align*} y &= (1)^2 + 2(1) - 3 \\ &= 0 \end{align*} Therefore, the points of intersection are (-4, 7) and (1, 0). So, the correct answer is (1, -4).

Maelezo ya Majibu

Let's call the sum of the 8 correct numbers "S" and the incorrect number that was added "x". The mean of the 8 numbers is: S/8 But the boy used 17 instead of 25, so the sum he used was: S + 17 - 25 = S - 8 And he got a mean of 20, so we can set up the equation: (S - 8)/8 = 20 Solving for S, we get: S = 168 So the correct mean is: S/8 = 168/8 = 21 Therefore, the correct mean is 21, which is.

Maelezo ya Majibu

Maelezo ya Majibu

To find the acceleration, we need to differentiate the distance formula with respect to time (t): \begin{align*} s &= \frac{3}{2}t^{2} - 3t \\ \frac{d}{dt}s &= \frac{d}{dt}\left(\frac{3}{2}t^{2}\right) - \frac{d}{dt}(3t) \\ \frac{d}{dt}s &= 3t - 3 \\ \end{align*} Therefore, the acceleration is the second derivative of the distance formula with respect to time: \begin{align*} \frac{d^{2}}{dt^{2}}s &= \frac{d}{dt}(3t - 3) \\ \frac{d^{2}}{dt^{2}}s &= 3 \\ \end{align*} Thus, the acceleration of the particle is a constant value of 3 \(ms^{-2}\). Therefore, the answer is \(3 ms^{-2}\).

Maelezo ya Majibu

To find the radius of the circle, we need to use the standard form of the equation of a circle, which is \((x - a)^{2} + (y - b)^{2} = r^{2}\), where \((a, b)\) is the center of the circle and \(r\) is the radius. To convert the given equation to standard form, we can complete the square for both \(x\) and \(y\): \begin{align*} 3x^{2} + 3y^{2} + 24x - 12y &= 15 \\ 3(x^{2} + 8x) + 3(y^{2} - 4y) &= 15 \\ 3(x^{2} + 8x + 16) + 3(y^{2} - 4y + 4) &= 15 + 3(16) + 3(4) \\ 3(x + 4)^{2} + 3(y - 2)^{2} &= 72 \\ (x + 4)^{2} + (y - 2)^{2} &= 8^{2} \end{align*} Comparing this with the standard form, we see that the center of the circle is \((-4, 2)\) and the radius is \(8\). Therefore, the answer is (d) 5.

Maelezo ya Majibu

Maelezo ya Majibu

Maelezo ya Majibu

To simplify this expression, we can first use the identity: $$\log_{a}(b) + \log_{a}(c) = \log_{a}(bc)$$ Using this identity, we can simplify the given expression as follows: \begin{align*} \log_{10}\left(\frac{1}{3}+\frac{1}{4}\right) + 2\log_{10}(2) + \log_{10}\left(\frac{3}{7}\right) &= \log_{10}\left(\frac{7}{12}\right) + \log_{10}(2^2) + \log_{10}\left(\frac{3}{7}\right) \\ &= \log_{10}\left(\frac{7}{12} \cdot 2^2 \cdot \frac{3}{7}\right) \\ &= \log_{10}(1) \\ &= 0 \end{align*} Therefore, the answer is 0.

Maelezo ya Majibu

Maelezo ya Majibu

An identity element in a binary operation is an element such that when it operates with any other element of the set, it does not change the other element. In this case, we need to find an element, say "x," such that for any real number "a," $$a \Delta x = a$$ Substituting the given definition of the operation, we get: $$a + x + 4 = a$$ Solving for x, we get: $$x = -4$$ Thus, -4 is the identity element for the given binary operation.

Maelezo ya Majibu

To find the mean mark, we add up all the marks and divide by the number of students. Adding up all the marks, we get: 3 + 7 + 6 + 2 + 8 + 5 + 9 + 1 + 4 + 10 = 55 There are 10 students, so we divide the sum of the marks by 10: 55 / 10 = 5.50 Therefore, the mean mark is 5.50. Hence, the correct option is: 5.50.

Maelezo ya Majibu

If (x + 1) is a factor of the polynomial, it means that if we substitute -1 in the polynomial, it should give us zero. Therefore: \((-1)^3 + p(-1)^2 -1 + 6 = 0\) Simplifying the above equation, we get: \(-1 + p + 5 = 0\) \(p = -4\) Hence, the value of p is -4. So the correct answer is.

Maelezo ya Majibu

To find \(\overrightarrow{SQ}\), we can use the fact that \(\overrightarrow{SR} = \overrightarrow{SQ} + \overrightarrow{QR}\). Rearranging this equation to solve for \(\overrightarrow{SQ}\) gives us: $$\overrightarrow{SQ} = \overrightarrow{SR} - \overrightarrow{QR} = (-5i + 3j) - (3i + 2j) = -8i + j$$ Therefore, the value of \(\overrightarrow{SQ}\) is -8i + j. So, the correct option is (A) 8i + j.

Maelezo ya Majibu

When a die is tossed, there are six possible outcomes, each with equal probability. Therefore, the probability of getting a 3 on the first toss is 1/6, and the probability of getting a 5 on the second toss is also 1/6. Since we want both events to occur, we need to multiply their probabilities: \[\frac{1}{6} \times \frac{1}{6} = \frac{1}{36}\] So, the probability of obtaining a 3 and a 5 is 1/36. Therefore, the correct option is: - \(\frac{1}{36}\)

Maelezo ya Majibu

To differentiate \(\frac{x}{x + 1}\) with respect to x, we can use the quotient rule of differentiation, which states that for functions u(x) and v(x), the derivative of \(\frac{u(x)}{v(x)}\) is given by: \[\frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x) v(x) - u(x) v'(x)}{(v(x))^2}\] Applying this rule to the given function, we have: \[u(x) = x\] \[v(x) = x + 1\] So, we need to find u'(x) and v'(x): \[u'(x) = 1\] \[v'(x) = 1\] Substituting these values into the quotient rule formula, we get: \[\frac{d}{dx} \left( \frac{x}{x + 1} \right) = \frac{1(x + 1) - x(1)}{(x + 1)^2}\] Simplifying the numerator and denominator, we get: \[\frac{d}{dx} \left( \frac{x}{x + 1} \right) = \frac{1}{(x + 1)^2}\] Therefore, the correct answer is \(\frac{1}{(x + 1)^2}\).

Maelezo ya Majibu

The formula for finding the length of an arc of a sector is given by L = rθ, where L is the length of the arc, r is the radius of the circle, and θ is the angle in radians. Using this formula and the given values, we have: L = 4 x 0.9 L = 3.6 cm Therefore, the length of the arc of the sector is 3.6 cm. Answer: 3.6 cm.

Maelezo ya Majibu

Maelezo ya Majibu

Maelezo ya Majibu

To find the acute angle between two lines, we need to find the angle between their respective direction vectors. The direction vector of the line 2x + y = 4 is v = i + 2j and the direction vector of the line -3x + y + 7 = 0 is w = -3i + j. The acute angle θ between two vectors a and b is given by the formula cos(θ) = (a.b) / (|a|.|b|), where a.b is the dot product of the vectors and |a| and |b| are their respective magnitudes. Using this formula, we can find cos(θ) for the given direction vectors as follows: cos(θ) = (v.w) / (|v|.|w|) = (-5) / (√5.√10) = -1/√2 Since we are looking for the acute angle, we need to take the inverse cosine of -1/√2 in the range [0, π/2]: θ = cos^-1(-1/√2) ≈ 45° Therefore, the acute angle between the given lines is approximately 45°, which is.

Maelezo ya Majibu

Maelezo ya Majibu

Maelezo ya Majibu

Maelezo ya Majibu

To find the value of (P + Q), we need to first expand the left-hand side of the equation using distributive property. P(x - 3) + Q(x + 1) = Px - 3P + Qx + Q Then we can simplify it by combining the like terms. Px + Qx - 3P + Q + 2x + 3 = 0 Now, we can group the like terms together: (P + Q)x - 3P + Q + 3 = 2x + 3 Since the coefficients of x on both sides of the equation are equal, we can equate the corresponding coefficients of x: (P + Q) = 2 Therefore, the value of (P + Q) is 2. So the correct answer is: - 2

Maelezo ya Majibu

Maelezo ya Majibu

Maelezo ya Majibu

To find the value of (2P - Q), we first need to compute 2P and Q, and then subtract Q from 2P. To compute 2P, we multiply each element of matrix P by 2: 2P = \(\begin{pmatrix} 4 & 2 \\ 10 & -6 \end{pmatrix}\) To subtract Q from 2P, we subtract each corresponding element of matrix Q from matrix 2P: 2P - Q = \(\begin{pmatrix} 4-4 & 2+8 \\ 10-1 & -6+2 \end{pmatrix}\) = \(\begin{pmatrix} 0 & 10 \\ 9 & -4 \end{pmatrix}\) Therefore, the answer is \(\begin{pmatrix} 0 & 10 \\ 9 & -4 \end{pmatrix}\).

Maelezo ya Majibu

Maelezo ya Majibu

Maelezo ya Majibu

Maelezo ya Majibu

Maelezo ya Majibu

We can start by using the midpoint formula which states that the midpoint of a line joining two points \((x_1, y_1)\) and \((x_2, y_2)\) is \((\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})\). So, the midpoint of the line joining (1-k, -4) and (2, k+1) is: \begin{align*} &\left(\frac{(1-k)+2}{2},\frac{(-4)+(k+1)}{2}\right) \\ &\Rightarrow \left(\frac{3-k}{2},\frac{k-3}{2}\right) \end{align*} We are given that the midpoint is (-k, k), so we can equate the x and y coordinates: \begin{align*} \frac{3-k}{2} &=-k \\ \frac{k-3}{2} &=k \end{align*} Solving for k, we get: \begin{align*} k &= 2 \\ \end{align*} Therefore, the value of k is 2, which is the answer option labeled as "-2".

Maelezo ya Majibu

We know that: $$^{n}P_{r} = \frac{n!}{(n-r)!} = 90$$ and $$^{n}C_{r} = \binom{n}{r} = \frac{n!}{r!(n-r)!} = 15$$ To find the value of r, we can use the formula: $$^{n}C_{r} = \frac{^{n}P_{r}}{r!}$$ Substituting the given values, we get: $$15 = \frac{90}{r!}$$ Simplifying the equation, we get: $$r! = 6$$ The only integer value of r that satisfies this equation is 3, since 3! = 6. Therefore, the answer is r = 3.

Maelezo ya Majibu

Maelezo ya Majibu

Each class has width 3; the mid-values are \(1.41,1.44,\dots,1.62\). Use a working-mean coding \(u=\dfrac{x-A}{c}\) with \(A=1.50\) and \(c=0.03\).

(a)(i) Mean height.\[\bar{x}=A+\frac{\sum fu}{\sum f}\,c=1.50+\frac{41}{100}(0.03)=1.50+0.0123=1.5123\approx 1.51\text{ m}\]

(a)(ii) Mean deviation. \(\text{MD}=\dfrac{\sum f\,|x-\bar{x}|}{\sum f}\) with \(\bar{x}=1.5123\).

\[\text{MD}=\frac{3.333}{100}=0.0333\approx 0.03\text{ m}\]

(b) Probability of height greater than the mean. The mean \(1.5123\) m lies below the upper boundary \(1.515\) m of the class \(1.49-1.51\), so every student in the classes \(1.52-1.54\) and above exceeds the mean:\[24+14+6+1=45\]\[P(\text{height}>\bar{x})=\frac{45}{100}=0.45\]

Mean = 1.51 m, mean deviation \(\approx\) 0.03 m, and the required probability = 0.45.

Maelezo ya Majibu

The gradient function is the derivative. For \(y = ax^2 + bx + c\),

\[\frac{dy}{dx} = 2ax + b.\]

We are told this equals \(8x + 4\), so comparing coefficients:

\[2a = 8 \Rightarrow a = 4, \qquad b = 4.\]

Using the minimum value. At the minimum the gradient is zero:

\[8x + 4 = 0 \Rightarrow x = -\tfrac{1}{2}.\]

The minimum value of \(y\) there is \(1\). Substitute \(a=4,\ b=4,\ x=-\tfrac12\):

\[y = 4\left(-\tfrac12\right)^2 + 4\left(-\tfrac12\right) + c = 4\cdot\tfrac14 - 2 + c = 1 - 2 + c = c - 1.\]

Set \(c - 1 = 1 \Rightarrow c = 2\).

Answer: \(a = 4,\; b = 4,\; c = 2\), giving \(y = 4x^2 + 4x + 2\).

Maelezo ya Majibu

(a) Since \(x^2 + x - 6 = (x+3)(x-2)\), the polynomial \(f(x) = x^3 + px^2 - 10x + q\) is zero at \(x = -3\) and \(x = 2\).

(i) Using \(f(2) = 0\): \(8 + 4p - 20 + q = 0 \Rightarrow 4p + q = 12\).

Using \(f(-3) = 0\): \(-27 + 9p + 30 + q = 0 \Rightarrow 9p + q = -3\).

Subtracting: \(5p = -15 \Rightarrow p = -3\), and \(q = 12 - 4(-3) = 24\).

So \(p = -3,\; q = 24\).

(ii) With \(f(x) = x^3 - 3x^2 - 10x + 24\), divide by \(x^2 + x - 6\):

\[x^3 - 3x^2 - 10x + 24 = (x^2 + x - 6)(x - 4).\]

The third factor is \((x - 4)\).

(b) For a cube of side \(s\), volume \(V = s^3\), so \(\dfrac{dV}{dt} = 3s^2\dfrac{ds}{dt}\).

Given \(\dfrac{dV}{dt} = 2\tfrac12 = 2.5\ \text{cm}^3\text{s}^{-1}\) and \(s = 2\ \text{cm}\):

\[2.5 = 3(2)^2\frac{ds}{dt} = 12\frac{ds}{dt} \Rightarrow \frac{ds}{dt} = \frac{2.5}{12} = \frac{5}{24} \approx 0.21\ \text{cm s}^{-1}.\]

The side is increasing at \(\dfrac{5}{24}\ \text{cm s}^{-1}\).

Maelezo ya Majibu

(a)(i) The series \(A(1+r) + A(1+r)^2 + \dots + A(1+r)^n\) is a geometric progression with first term \(a = A(1+r)\), common ratio \(R = (1+r)\) and \(n\) terms. Its sum is

\[S_n = \frac{a(R^n - 1)}{R - 1} = \frac{A(1+r)\big[(1+r)^n - 1\big]}{(1+r) - 1} = \frac{A(1+r)\big[(1+r)^n - 1\big]}{r}.\]

(ii) With \(r = 8\% = 0.08\) and \(A = 40\), the \(k\)-th term is \(T_k = 40(1.08)^k\). The sum of the 6th to 10th terms is

\[\sum_{k=6}^{10} 40(1.08)^k = 40\big[(1.08)^6 + (1.08)^7 + (1.08)^8 + (1.08)^9 + (1.08)^{10}\big].\]

Computing the powers: \(1.586874,\ 1.713824,\ 1.850930,\ 1.999005,\ 2.158925\); their sum is \(9.309558\). Hence

\[S = 40 \times 9.309558 \approx \text{GH}\phi\,372.38.\]

The sum of the 6th to 10th terms is about GH\(\phi\)372.38.

(b) For \(y = \dfrac{1}{x} = x^{-1}\), \(\dfrac{dy}{dx} = -x^{-2} = -\dfrac{1}{x^2}\).

At \(x = 2\): gradient \(= -\dfrac{1}{4}\), and the point is \(\left(2, \tfrac12\right)\). The tangent is

\[y - \tfrac{1}{2} = -\tfrac{1}{4}(x - 2) \;\Rightarrow\; y = 1 - \tfrac{1}{4}x, \quad\text{or}\quad x + 4y = 4.\]

Maelezo ya Majibu

Express \(3x^{2}-6x+10\) as \(a(x-b)^{2}+c\).

Factor 3 from the x-terms and complete the square:

So \(a=3,\ b=1,\ c=7\).

Minimum value. Since \(3(x-1)^{2}\ge 0\) and is zero when \(x=1\), the least value of the expression is

The minimum value is \(7\), occurring at \(x=1\).

Maelezo ya Majibu

(a) Simplify the integrand first. With \(\sqrt{x} = x^{1/2}\),

\[\frac{x(3x-2)}{2\sqrt{x}} = \frac{3x^2 - 2x}{2x^{1/2}} = \frac{3}{2}x^{3/2} - x^{1/2}.\]

Integrate:

\[\int\left(\frac{3}{2}x^{3/2} - x^{1/2}\right)dx = \frac{3}{2}\cdot\frac{x^{5/2}}{5/2} - \frac{x^{3/2}}{3/2} = \frac{3}{5}x^{5/2} - \frac{2}{3}x^{3/2}.\]

Evaluate from \(1\) to \(4\) (using \(4^{5/2}=32,\ 4^{3/2}=8\)):

\[\left[\frac{3}{5}(32) - \frac{2}{3}(8)\right] - \left[\frac{3}{5} - \frac{2}{3}\right] = \left(\frac{96}{5} - \frac{16}{3}\right) - \left(-\frac{1}{15}\right) = \frac{208}{15} + \frac{1}{15} = \frac{209}{15}.\]

So the integral is \(\dfrac{209}{15} = 13\tfrac{14}{15} \approx 13.93\).

(b) Divide the circle equation by 2 to get unit leading coefficients:

\[x^2 + y^2 - 4x + \tfrac{5}{2}y - 5 = 0.\]

Comparing with \(x^2 + y^2 + 2gx + 2fy + c = 0\): \(2g=-4\Rightarrow g=-2\), \(2f=\tfrac52\Rightarrow f=\tfrac54\), \(c=-5\).

(i) Centre \((-g, -f) = \left(2,\, -\tfrac{5}{4}\right)\).

(ii) Radius \(r = \sqrt{g^2 + f^2 - c} = \sqrt{4 + \tfrac{25}{16} + 5} = \sqrt{\tfrac{169}{16}} = \tfrac{13}{4} = 3.25\).

Maelezo ya Majibu

\(g:x\to x^{2}-2\) and \(h:x\to\dfrac{1}{x+2}\).

(a) Inverse of h. Let \(y=\dfrac{1}{x+2}\). Then

So, replacing \(y\) by \(x\),

(b) \(g\circ h\) at \(x=-\dfrac12\). First evaluate \(h\):

Then apply \(g\):

Hence \((g\circ h)\left(-\tfrac12\right)=-\dfrac{14}{9}\).

Maelezo ya Majibu

(a) The transformation \(A(x,y)\to(2x - y,\, -5x + 3y)\) has matrix

\[A = \begin{pmatrix} 2 & -1 \\ -5 & 3 \end{pmatrix}.\]

(b) First compute the squares.

\[A^2 = \begin{pmatrix} 2 & -1 \\ -5 & 3 \end{pmatrix}\begin{pmatrix} 2 & -1 \\ -5 & 3 \end{pmatrix} = \begin{pmatrix} 9 & -5 \\ -25 & 14 \end{pmatrix},\qquad B^2 = \begin{pmatrix} 3 & 1 \\ 5 & 2 \end{pmatrix}^2 = \begin{pmatrix} 14 & 5 \\ 25 & 9 \end{pmatrix}.\]

(i) \(\displaystyle A^2 - B^2 = \begin{pmatrix} 9-14 & -5-5 \\ -25-25 & 14-9 \end{pmatrix} = \begin{pmatrix} -5 & -10 \\ -50 & 5 \end{pmatrix}.\)

(ii) \(\displaystyle C = B^2 A = \begin{pmatrix} 14 & 5 \\ 25 & 9 \end{pmatrix}\begin{pmatrix} 2 & -1 \\ -5 & 3 \end{pmatrix} = \begin{pmatrix} 3 & 1 \\ 5 & 2 \end{pmatrix}.\)

(iii) If \(C\begin{pmatrix}x\\y\end{pmatrix} = \begin{pmatrix}10\\18\end{pmatrix}\), then \(3x + y = 10\) and \(5x + 2y = 18\).

From the first, \(y = 10 - 3x\); substituting: \(5x + 2(10-3x) = 18 \Rightarrow -x + 20 = 18 \Rightarrow x = 2\), then \(y = 4\). So \(M(2, 4)\).

(c) Note \(\det A = (2)(3)-(-1)(-5) = 1\), and \(AC = \begin{pmatrix}1&0\\0&1\end{pmatrix} = I\). Hence \(C\) is the inverse of \(A\), that is \(C = A^{-1}\).

Maelezo ya Majibu

Both parts use the binomial distribution \(P(X=r) = \binom{n}{r}p^r(1-p)^{n-r}\).

(a) Die thrown \(n = 6\) times, \(p = \tfrac16\) for a six, \(q = \tfrac56\).

(i) Exactly three sixes:

\[\binom{6}{3}\left(\tfrac16\right)^3\left(\tfrac56\right)^3 = 20\cdot\frac{1}{216}\cdot\frac{125}{216} = \frac{2500}{46656} \approx 0.054.\]

(ii) At most three sixes \(= P(0)+P(1)+P(2)+P(3)\):

\[P(0)=0.334898,\ P(1)=0.401878,\ P(2)=0.200939,\ P(3)=0.053584.\]

Sum \(= 0.991\) (to three decimal places).

(b) Screws: \(n = 10\), \(p = 0.08\) defective, \(q = 0.92\).

(i) Exactly two defective:

\[\binom{10}{2}(0.08)^2(0.92)^8 = 45(0.0064)(0.513219) \approx 0.148.\]

(ii) Not more than two \(= P(0)+P(1)+P(2)\):

\[P(0)=(0.92)^{10}=0.434390,\ P(1)=10(0.08)(0.92)^9=0.377730,\ P(2)=0.147807.\]

Sum \(= 0.960\) (to three decimal places).

Maelezo ya Majibu

The marks are grouped into equal class intervals of width 5, so the frequencies can be plotted directly against the true class boundaries. First convert each class limit to its class boundary by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit.

Because every class has the same width, the height of each bar equals its frequency. The bars are drawn adjacent to one another (no gaps), with the class boundaries marked on the horizontal axis and frequency on the vertical axis.

The tallest bar corresponds to the class 11 - 15 (boundaries 10.5 - 15.5) with a frequency of 12, which is therefore the modal class of the distribution. The completed histogram above is the required diagram.

Maelezo ya Majibu

(a) With \(p=4i-3j\) and \(q=-i+5j\):

\[2p+3q=2(4i-3j)+3(-i+5j)=(8i-6j)+(-3i+15j)=5i+9j.\]

Its magnitude is

\[|2p+3q|=\sqrt{5^2+9^2}=\sqrt{25+81}=\sqrt{106}.\]

The unit vector in the direction of \(2p+3q\) is \(\dfrac{5i+9j}{\sqrt{106}}\). Since \(|r|=15\) and \(r\) is in this direction,

\[r=15\cdot\frac{5i+9j}{\sqrt{106}}=\frac{75}{\sqrt{106}}i+\frac{135}{\sqrt{106}}j\approx 7.28i+13.11j.\]

(b) From the diagram, taking angles anticlockwise from the positive \(x\)-axis (horizontal): the \(6\ \text{N}\) is vertical (\(90^\circ\)); the \(8\ \text{N}\) is \(30^\circ\) to the left of the \(6\ \text{N}\), so at \(120^\circ\); the \(4\ \text{N}\) is \(60^\circ\) to the right of the \(6\ \text{N}\), so at \(30^\circ\).

Resolve horizontally:

\[\sum F_x=8\cos120^\circ+6\cos90^\circ+4\cos30^\circ=-4+0+3.464=-0.536\ \text{N}.\]

Resolve vertically:

\[\sum F_y=8\sin120^\circ+6\sin90^\circ+4\sin30^\circ=6.928+6+2=14.928\ \text{N}.\]

(i) Magnitude:

\[R=\sqrt{(-0.536)^2+14.928^2}=\sqrt{0.287+222.85}=\sqrt{223.14}\approx \mathbf{14.94\ \text{N}}.\]

(ii) Direction. \(\sum F_x\) is negative and \(\sum F_y\) positive, so the resultant lies in the second quadrant (just left of the upward vertical). The acute angle to the vertical is

\[\tan^{-1}\!\left(\frac{|\sum F_x|}{\sum F_y}\right)=\tan^{-1}\!\left(\frac{0.536}{14.928}\right)\approx 2.1^\circ.\]

Measured anticlockwise from the positive \(x\)-axis, the direction is \(180^\circ-\tan^{-1}(14.928/0.536)\approx \mathbf{92.1^\circ}\); that is, the resultant of about \(14.94\ \text{N}\) acts almost vertically, inclined \(2.1^\circ\) to the left of the \(6\ \text{N}\) force.

Maelezo ya Majibu

Add the forces component by component. The three forces are

\[F_1 = -63\mathbf{j},\quad F_2 = 32.14\mathbf{i} + 38.3\mathbf{j},\quad F_3 = 14\mathbf{i} - 24.25\mathbf{j}.\]

Resultant components.

\[\sum F_x = 0 + 32.14 + 14 = 46.14,\qquad \sum F_y = -63 + 38.3 - 24.25 = -48.95.\]

So \(\mathbf{R} = 46.14\mathbf{i} - 48.95\mathbf{j}\).

(a) Magnitude of the resultant force.

\[|\mathbf{R}| = \sqrt{46.14^2 + 48.95^2} = \sqrt{2128.90 + 2396.10} = \sqrt{4525.00} \approx 67.3\ \text{N}.\]

(b) Acceleration. By Newton's second law \(\mathbf{R} = m\mathbf{a}\), with \(m = 5\ \text{kg}\):

\[a = \frac{|\mathbf{R}|}{m} = \frac{67.3}{5} \approx 13.5\ \text{m s}^{-2}.\]

The acceleration is directed along the resultant force, with magnitude \(13.5\ \text{m s}^{-2}\).

Maelezo ya Majibu

None

Maelezo ya Majibu

(a) Weight \(W = mg = 5\times10 = 50\,\text{N}\). Let \(T_1\) (string \(AP\)) and \(T_2\) (string \(BP\)) act at \(40^{o}\) and \(30^{o}\) to the downward vertical on opposite sides.

Horizontal: \(T_1\sin40^{o} = T_2\sin30^{o}\).

Vertical: \(T_1\cos40^{o} + T_2\cos30^{o} = 50\).

Using Lami's theorem is neat here. The angles at \(P\) are: between \(W\) and \(T_1 = 140^{o}\), between \(W\) and \(T_2 = 150^{o}\), between \(T_1\) and \(T_2 = 70^{o}\).

Tension in \(AP \approx \mathbf{26.6\,\text{N}}\); tension in \(BP \approx \mathbf{34.2\,\text{N}}\).

(b)(i) \(a = \dfrac{9 - 5}{2} = 2\,\text{ms}^{-2}\), so \(F = ma = 5\times2 = \mathbf{10\,\text{N}}\).

(ii) Continuing at \(a = 2\,\text{ms}^{-2}\) for a further \(3\,\text{s}\) from \(9\,\text{ms}^{-1}\):

Maelezo ya Majibu

None

Maelezo ya Majibu

A.P.: \(T_{21}=5\tfrac12\) and \(S_{21}=94\tfrac12\).

Write the two facts algebraically:

Subtract (2) from (1):

From (2), \(a=\dfrac92-10\left(\tfrac{1}{10}\right)=\dfrac92-1=\dfrac72\).

(a) First term \(a=\dfrac72=3\tfrac12\).

(b) Common difference \(d=\dfrac{1}{10}=0.1\).

(c) Sum of the first 30 terms:

So \(S_{30}=148\tfrac12\).

Maelezo ya Majibu

The binomial expansion of \((1+ax)^n\) in ascending powers of \(x\) is:

\[(1+ax)^n = 1 + n(ax) + \frac{n(n-1)}{2!}(ax)^2 + \dots = 1 + nax + \frac{n(n-1)}{2}a^2x^2 + \dots\]

So the first three terms are \(1,\; nax,\; \dfrac{n(n-1)}{2}a^2x^2\).

Forming the equations. The coefficient of \(x\) is \(2\) and the coefficient of \(x^2\) is \(\tfrac{3}{2}\):

\[na = 2 \quad\text{(1)}, \qquad \frac{n(n-1)}{2}a^2 = \frac{3}{2} \quad\text{(2)}.\]

From (1), \(a = \dfrac{2}{n}\). Substituting into (2):

\[\frac{n(n-1)}{2}\cdot\frac{4}{n^2} = \frac{3}{2} \;\Rightarrow\; \frac{2(n-1)}{n} = \frac{3}{2}.\]

Cross-multiplying: \(4(n-1) = 3n \Rightarrow 4n-4 = 3n \Rightarrow n = 4\).

Then \(a = \dfrac{2}{n} = \dfrac{2}{4} = \dfrac{1}{2}\).

Answer: \(a = \tfrac{1}{2},\; n = 4\). (Check: with these values the terms are \(1 + 2x + \tfrac{3}{2}x^2\), as required.)

Maelezo ya Majibu

(a) With \(a = \begin{pmatrix}-3\\4\end{pmatrix},\ b = \begin{pmatrix}-8\\-15\end{pmatrix}\):

(b) \(a = (4\cos60^{o},\ 4\sin60^{o}) = (2,\ 2\sqrt3)\); \(b = (3\cos120^{o},\ 3\sin120^{o}) = (-1.5,\ 1.5\sqrt3)\).

Unit vector along \(a - b\):