WAEC SSCE - Further Mathematics - 2012

Akọwa Nkọwa

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.

Akọwa Nkọwa

Akọwa Nkọwa

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.

Akọwa Nkọwa

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.

Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

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.

Akọwa Nkọwa

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 \(3 ms^{-2}\).

Akọwa Nkọwa

Akọwa Nkọwa

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.

Akọwa Nkọwa

Akọwa Nkọwa

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.

Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

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."

Akọwa Nkọwa

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 \(\begin{pmatrix} 0 & 10 \\ 9 & -4 \end{pmatrix}\).

Akọwa Nkọwa

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.

Akọwa Nkọwa

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

Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

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 \(\frac{1}{(x + 1)^2}\).

Akọwa Nkọwa

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).

Akọwa Nkọwa

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: - \(\frac{1}{36}\)

Akọwa Nkọwa

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.

Akọwa Nkọwa

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 \(\frac{-12i}{13} + \frac{5j}{13}\).

Akọwa Nkọwa

Akọwa Nkọwa

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.

Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

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".

Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

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.

Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

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.

Akọwa Nkọwa

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.

Akọwa Nkọwa

Akọwa Nkọwa

(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 \(a - b\):

Akọwa Nkọwa

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).

Akọwa Nkọwa

None

Akọwa Nkọwa

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.

Akọwa Nkọwa

(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.

Akọwa Nkọwa

(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 \(C = A^{-1}\).

Akọwa Nkọwa

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 \(x=1\).

Akọwa Nkọwa

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

Akọwa Nkọwa

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

Akọwa Nkọwa

(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 \(\dfrac{5}{24}\ \text{cm s}^{-1}\).

Akọwa Nkọwa

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 \(y = 4x^2 + 4x + 2\).

Akọwa Nkọwa

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.)

Akọwa Nkọwa

(a)(i)(1) With replacement

Since an item is replaced after the first selection, there are \(4 \times 4 = 16\) possible ordered outcomes:

\(S=\{pp,pq,pr,ps,qp,qq,qr,qs,rp,rq,rr,rs,sp,sq,sr,ss\}\).

(2) Without replacement

There are \(4 \times 3=12\) possible ordered outcomes:

\(S'=\{pq,pr,ps,qp,qr,qs,rp,rq,rs,sp,sq,sr\}\).

(a)(ii)(1) With replacement

The outcomes containing at least one \(r\) are:

\(\{pr,qr,rp,rq,rr,rs,sr\}\).

Therefore,

(2) Without replacement

The outcomes containing \(r\) are:

\(\{pr,qr,rp,rq,rs,sr\}\).

Therefore,

(b)(i) Numbers from 100 to 999 divisible by 4

The multiples of 4 are:

\(100,104,108,\ldots,996\).

\[ 996=100+4(n-1) \] \[ 996=100+4n-4 \] \[ 996=96+4n

Akọwa Nkọwa

(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 \(9\,\text{ms}^{-1}\):

Akọwa Nkọwa

(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

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

Akọwa Nkọwa

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 \(13.5\ \text{m s}^{-2}\).

Akọwa Nkọwa

All six classes have the same width. Each class spans five marks (\(1\text{-}5,\ 6\text{-}10,\ \dots,\ 26\text{-}30\)), so the class widths are equal and an ordinary frequency histogram may be drawn: the bar heights are simply the frequencies, and there is no need to adjust for frequency density. First convert the class limits to continuous class boundaries by subtracting \(0.5\) from each lower limit and adding \(0.5\) to each upper limit.

The bars are drawn touching (no gaps), because the class boundaries are continuous. The frequencies total \(2+7+12+8+5+1 = 35\), matching the 35 students.

The tallest bar, over the class \(11\text{-}15\), is the modal class. Reading the peak of the histogram gives an estimate of the mode within that interval.

Examination note: do not merge or re-band the classes. The intervals are already equal in width, so each bar height equals its frequency; changing the boundaries to unequal widths would distort the distribution and lose marks.

Akọwa Nkọwa

\(g:x\to x^{2}-2\) 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 \((g\circ h)\left(-\tfrac12\right)=-\dfrac{14}{9}\).