WAEC SSCE - Further Mathematics - 2018

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The given equation is \(\log_{3}a - 2 = 3\log_{3}b\). We can use the logarithmic property that \(\log_{a}b^{c} = c\log_{a}b\) to simplify the equation: \(\log_{3}a - 2 = \log_{3}b^{3}\) \(\log_{3}a = \log_{3}b^{3} + 2\) \(\log_{3}a = \log_{3}(b^{3}\cdot 3^{2})\) Using the property that if \(\log_{a}b = \log_{a}c\) then \(b = c\), we get: \(a = b^{3}\cdot 3^{2}\) \(a = 9b^{3}\) Therefore, the value of \(a\) in terms of \(b\) is \(a = 9b^{3}\).

Maelezo ya Majibu

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Maelezo ya Majibu

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Maelezo ya Majibu

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Maelezo ya Majibu

Maelezo ya Majibu

The constant term in a binomial expansion is the term that has no variable or a variable raised to the power of zero. To find the constant term in \((2x^{2} + \frac{1}{x})^{9}\), we need to look for the term that has no \(x\) or a \(x\) raised to the power of zero. To obtain the constant term, we need to choose the constant terms from each of the factors in the expansion, which are \(2x^2\) and \(\frac{1}{x}\), in such a way that their product is raised to a power that adds up to 9. In other words, we need to choose the constant terms that multiply to give a coefficient in the expansion of \((2x^2)^m (\frac{1}{x})^{9-m}\) that is independent of \(x\). The constant term in the expansion of \((2x^2)^m (\frac{1}{x})^{9-m}\) is \(\binom{9}{m}(2x^2)^m (\frac{1}{x})^{9-m}\), where \(\binom{9}{m}\) is the binomial coefficient that represents the number of ways to choose \(m\) items out of 9. For the constant term, we need to choose \(m\) such that the powers of \(x\) in the two factors add up to zero. That is, \begin{align*} 2m - (9-m) &= 0 \\ \Rightarrow m &= \frac{9}{3} \\ \Rightarrow m &= 3 \end{align*} Therefore, the constant term in the expansion of \((2x^2 + \frac{1}{x})^9\) is \(\binom{9}{3}(2x^2)^3 (\frac{1}{x})^6\). Plugging in the values, we get: \begin{align*} \binom{9}{3}(2x^2)^3 (\frac{1}{x})^6 &= \frac{9!}{3!6!}(2^3x^6)(\frac{1}{x^6}) \\ &= 84(8) \\ &= 672 \end{align*} Therefore, the constant term in the binomial expansion of \((2x^2 + \frac{1}{x})^9\) is \boxed{672}.

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To convert from kilometers per hour to meters per second, we need to use the following formula: 1 kilometer per hour = 0.277778 meters per second Therefore, to find the speed of the car in meters per second, we can multiply its speed in kilometers per hour by 0.277778. Speed in meters per second = 120 km/h x 0.277778 = 33.33336 m/s (rounded to 5 decimal places) Therefore, the correct answer is 33.3\(ms^{-1}\).

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If \(\alpha\) and \(\beta\) are the roots of the equation \(2x^{2} + 5x + n = 0\), then by the quadratic formula, we have: \[\alpha, \beta = \frac{-5 \pm \sqrt{5^{2} - 4(2)(n)}}{2(2)}\] Simplifying, we get: \[\alpha, \beta = \frac{-5 \pm \sqrt{25 - 8n}}{4}\] We are also given that \(\alpha\beta = 2\). Therefore, we have: \[\alpha\beta = \frac{-5 + \sqrt{25 - 8n}}{4} \cdot \frac{-5 - \sqrt{25 - 8n}}{4} = 2\] Expanding the left-hand side, we get: \[\frac{25 - (25 - 8n)}{16} = 2\] Simplifying, we get: \[\frac{8n}{16} = 2\] \[n = 4\] Therefore, the answer is the fourth option, 4.

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We know that the midpoint of a line segment is the point that is exactly halfway between the endpoints of the segment. So, to find the coordinates of point N, we need to use the midpoint formula. The midpoint formula is: Midpoint = [(x1 + x2)/2, (y1 + y2)/2] where (x1, y1) and (x2, y2) are the coordinates of the two endpoints of the segment. In this problem, we are given that the midpoint of segment MN is P(3, -4), and one endpoint is M(4, -1). Let's call the coordinates of the other endpoint N(x, y). Using the midpoint formula, we can write: [(4 + x)/2, (-1 + y)/2] = (3, -4) Now, we can solve for x and y: (4 + x)/2 = 3 => 4 + x = 6 => x = 2 (-1 + y)/2 = -4 => -1 + y = -8 => y = -7 Therefore, the coordinates of point N are (2, -7). So, the correct option is (2, -7).

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The distance moved by a particle is given by the equation \(S = 12t^{2} - 2t^{3}\), where S is the distance travelled in metres, and t is the time taken in seconds. To find the acceleration of the particle in 1 second, we need to differentiate the equation for distance with respect to time twice to obtain the equation for acceleration. First, we differentiate S with respect to time to get the velocity equation: $$\frac{dS}{dt} = \frac{d}{dt}(12t^{2} - 2t^{3}) = 24t - 6t^{2}$$ Next, we differentiate the velocity equation with respect to time to get the acceleration equation: $$\frac{d^2S}{dt^2} = \frac{d}{dt}(24t - 6t^{2}) = 24 - 12t$$ To find the acceleration in 1 second, we substitute t = 1 into the acceleration equation: $$\frac{d^2S}{dt^2} \bigg\rvert_{t=1} = 24 - 12(1) = 12ms^{-2}$$ Therefore, the acceleration of the particle in 1 second is 12\(ms^{-2}\).

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To find the values of x where the function f(x) is equal to 3, we need to solve the equation:

3x^2 - 12x + 12 = 3

We can start by subtracting 3 from both sides:

3x^2 - 12x + 12 - 3 = 3 - 3
3x^2 - 12x + 9 = 0

Next, we can use the quadratic formula to find the solutions for x:

x = (-b ± √(b^2 - 4ac)) / 2a

where a = 3, b = -12, c = 9. Plugging in these values, we get:

x = (-(-12) ± √((-12)^2 - 4 * 3 * 9)) / 2 * 3
x = (12 ± √(144 - 108)) / 6
x = (12 ± √36) / 6
x = (12 ± 6) / 6

So, the two values of x that solve the equation are:

x = (12 + 6) / 6 = 18 / 6 = 3
x = (12 - 6) / 6 = 6 / 6 = 1

Therefore, the two values of x that make f(x) equal to 3 are 1 and 3.

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To find the domain of the given function, we need to identify any values of x that would make the denominator of the fraction equal to zero or negative, as these values would result in an undefined function. Additionally, we need to consider any other restrictions on x that may arise from the function's algebraic properties. In this case, the denominator of the function is the square root of \(9x^{2} + 1\). Since the square root of a negative number is not defined in the real number system, we know that the expression inside the square root must be non-negative. This means that: \(9x^{2} + 1 \geq 0\) Solving this inequality, we get: \(9x^{2} \geq -1\) Dividing both sides by 9, we get: \(x^{2} \geq -\frac{1}{9}\) Since the square of any real number is non-negative, we know that this inequality is satisfied for all real values of x. Therefore, there are no values of x that would make the denominator of the function equal to zero or negative, and the domain of the function is all real numbers. In other words, the correct answer is: \(x: x \in R\)

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To find the equation whose roots are \((\alpha + 1)\) and \((\beta + 1)\), we can use the relationship between the roots and coefficients of a quadratic equation. Let's start by finding the sum and product of the roots of the original equation, \(2x^{2} - 5x + 6 = 0\). We can use the formulae: Sum of roots, \(S = -\frac{b}{a} = -\frac{-5}{2} = \frac{5}{2}\) Product of roots, \(P = \frac{c}{a} = \frac{6}{2} = 3\) Now, let's add 1 to both roots: \(\alpha + 1\) and \(\beta + 1\) The sum of these roots would be: \((\alpha + 1) + (\beta + 1) = \alpha + \beta + 2\) And the product would be: \((\alpha + 1)(\beta + 1) = \alpha \beta + \alpha + \beta + 1\) Now, we want to find the equation whose roots are \((\alpha + 1)\) and \((\beta + 1)\). Let's call this equation \(ax^{2} + bx + c = 0\). According to the relationship between roots and coefficients, we know that: \(\frac{-b}{a} = \frac{5}{2}\) and \(\frac{c}{a} = 3\) Solving for \(b\) and \(c\) in terms of \(a\), we get: \(b = -\frac{5a}{2}\) and \(c = 3a\) Now, we can use the sum and product of the new roots to form two equations: \(\frac{-b}{a} = \frac{5}{2} \implies -\frac{5a}{2a} = \frac{5}{2} \implies -5 = 5\) This equation is clearly not true, which means that the first option, \(2x^{2} - 9x + 15 = 0\), is not the correct answer. Let's move on to the second option: \((\alpha + \beta) + 2 = \frac{5}{2} + 2 = \frac{9}{2}\) \((\alpha \beta + \alpha + \beta + 1) = 3 + \frac{5}{2} + \frac{5}{2} + 1 = 8\) Substituting the values of \(b\) and \(c\) in terms of \(a\), we get: \(a(x^{2} - \frac{9}{2}x + \frac{13}{2}) = 0\) Simplifying, we get: \(2x^{2} - 9x + 13 = 0\) Therefore, the answer is the second option, \(2x^{2} - 9x + 13 = 0\).

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We know that a perfect square trinomial has the form of \((ax + b)^2\), where a and b are constants. So, if we have a trinomial of the form \(4x^2 + 5kx + 10\), we can write it as \((2x + c)^2\), where c is another constant. Expanding the square, we get: \begin{align*} (2x + c)^2 &= 4x^2 + 4cx + c^2 \\ &= 4x^2 + (4c)x + c^2 \\ \end{align*} Comparing the coefficients of the two trinomials, we get: \begin{align*} 4 &= 4 \\ 5k &= 4c \\ 10 &= c^2 \end{align*} Dividing the second equation by 4, we get: \begin{align*} \frac{5k}{4} &= c \end{align*} Since c = \( \sqrt{10} \), we get: \begin{align*} \frac{5k}{4} &= \sqrt{10} \\ \Rightarrow k &= \frac{4\sqrt{10}}{5} \end{align*} So the value of k is \( \frac{4\sqrt{10}}{5} \).

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To convert radians to degrees, we use the following formula:

degrees = radians × 180°/π

where π is approximately equal to 3.14.

Substituting the given value of radians, we get:

degrees = (13/4)π × 180°/π

The π cancels out, leaving us with:

degrees = (13/4) × 180°

Simplifying the expression, we get:

degrees = 585°

Therefore, the answer is 585°.

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We can use the formula \(F = ma\) to solve the problem, where \(F\) is the force in Newtons, \(m\) is the mass in kilograms, and \(a\) is the acceleration in meters per second squared. We need to convert the mass from grams to kilograms and find the acceleration first. Given that the initial velocity, \(u\) = 0 \(ms^{-1}\), final velocity, \(v\) = \(5.4ms^{-1}\), and time, \(t\) = 18 seconds. We can use the formula \(v = u + at\) to find the acceleration: \(v = u + at\) \(5.4 = 0 + a\times 18\) \(a = \frac{5.4}{18} = 0.3ms^{-2}\) Now, we can substitute the values of mass and acceleration in the formula to find the force: \(F = ma\) \(F = 0.028\times 0.3\) \(F = 0.0084N\) Therefore, the value of force is 0.0084N.

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The values given by X and Y are already ranks (1 to 8), so \(n = 8\). Form \(d = X - Y\) and \(d^2\).

\(r_s \approx 0.57\) indicates a moderate positive agreement between the two panels of judges in ranking the brands of cooking oil.

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None

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Note that \(\log_{x}3=\dfrac{1}{\log_{3}x}\). Let \(y=\log_{3}x\). The equation becomes

\[y-\frac{3}{y}+2=0\]

Multiply through by \(y\):

\[y^{2}+2y-3=0\Rightarrow(y+3)(y-1)=0\Rightarrow y=-3\ \text{or}\ y=1\]

Convert back with \(y=\log_{3}x\):

\[y=1\Rightarrow x=3^{1}=3,\qquad y=-3\Rightarrow x=3^{-3}=\frac{1}{27}\]

\[x=3\quad\text{or}\quad x=\frac{1}{27}\]

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Let \(x\) = number of heads and \(f\) = frequency. First find \(N = \sum f\).

(a) Mean number of heads.

(b) Probability of an even number of heads. Even outcomes are \(x = 0,2,4,6,8,10\) with frequencies \(2+23+11+100+8+3 = 147\).

(c) Probability of an odd number of heads. Odd outcomes are \(x = 1,3,5,7,9\) with frequencies \(7+36+61+12+5 = 121\).

Check: \(0.549 + 0.451 = 1.000\).

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Measure positions from \(X\): \(X=0,\ P=0.8\text{m},\ Q=0.8+2.2=3.0\text{m},\ Y=4.0\text{m}\). The uniform weight \(350\text{N}\) acts at the centre, \(2.0\text{m}\). The \(80\text{N}\) acts at \(X=0\); the \(1000\text{N}\) is \(0.6\text{m}\) from \(Q\), i.e. at \(2.4\text{m}\).

(a) Vertical equilibrium: \(R_P+R_Q=80+350+1000=1430\).

Take moments about \(P\) (anticlockwise positive):

\[80(0.8)-350(1.2)-1000(1.6)+R_Q(2.2)=0\]

\[64-420-1600+2.2R_Q=0\Rightarrow R_Q=\frac{1956}{2.2}=889.09\text{N}\]

Then \(R_P=1430-889.09=540.91\text{N}\).

(b) Replace with \(1200\text{N}\) at distance \(s\) from \(Q\) (position \(3.0-s\)). Keeping the reaction at \(P\) unchanged at \(540.91\text{N}\), take moments about \(Q\):

\[R_P(2.2)=80(3.0)+350(1.0)+1200\,s\]

\[540.91(2.2)=240+350+1200s\Rightarrow 1190=590+1200s\Rightarrow s=0.5\text{m}\]

So the \(1200\text{N}\) weight should be placed \(0.5\text{m}\) from \(Q\) (between \(P\) and \(Q\)).

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Given \(m=3i-2j,\ n=2i-3j,\ p=-i+6j\).

\[4m=12i-8j,\qquad 2n=4i-6j,\qquad 3p=-3i+18j\]

\[4m+2n-3p=(12i-8j)+(4i-6j)-(-3i+18j)\]

Collect \(i\) terms: \(12+4+3=19\). Collect \(j\) terms: \(-8-6-18=-32\).

\[4m+2n-3p=19i-32j\]

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(a) \(P(K)=\tfrac{1}{3}\Rightarrow P(K')=\tfrac{2}{3}\); \(P(T)=\tfrac{1}{5}\Rightarrow P(T')=\tfrac{4}{5}\). The events are independent.

Only one solves = (Kunle solves, Tayo fails) OR (Kunle fails, Tayo solves):

\[P=\left(\tfrac{1}{3}\times\tfrac{4}{5}\right)+\left(\tfrac{2}{3}\times\tfrac{1}{5}\right)=\frac{4}{15}+\frac{2}{15}=\frac{6}{15}=\frac{2}{5}\]

(b) Choosing 8 from 10 with no restriction:

\[\binom{10}{8}=\binom{10}{2}=\frac{10\times9}{2}=45\ \text{ways}\]

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(a) With \(u=x-2\), we have \(x=u+2\), so

\[x^{3}+5=(u+2)^{3}+5=u^{3}+6u^{2}+12u+8+5=u^{3}+6u^{2}+12u+13\]

Therefore

\[\frac{x^{3}+5}{(x-2)^{4}}=\frac{u^{3}+6u^{2}+12u+13}{u^{4}}=\frac{1}{u}+\frac{6}{u^{2}}+\frac{12}{u^{3}}+\frac{13}{u^{4}}\]

(b) Replace \(u\) by \(x-2\):

\[\frac{x^{3}+5}{(x-2)^{4}}=\frac{1}{x-2}+\frac{6}{(x-2)^{2}}+\frac{12}{(x-2)^{3}}+\frac{13}{(x-2)^{4}}\]

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None

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By conservation of linear momentum, total momentum before = total momentum after. After collision they move together with common velocity \(v\); total mass \(=20+30=50\,\text{kg}\).

(a) Same direction before collision. Take both velocities as positive:

\[20(80)+30(50)=50v\]

\[1600+1500=50v\Rightarrow3100=50v\Rightarrow v=62\ \text{m/s}\]

(b) Opposite directions before collision. Take the second body's velocity as negative:

\[20(80)+30(-50)=50v\]

\[1600-1500=50v\Rightarrow100=50v\Rightarrow v=2\ \text{m/s}\]

The positive result shows the combined body moves in the direction of the \(20\,\text{kg}\) body.

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Let the first term be \(a\) and common difference \(d\).

Sum of first 12 terms is 168:

\[S_{12}=\frac{12}{2}\big(2a+11d\big)=168\Rightarrow6(2a+11d)=168\Rightarrow2a+11d=28\quad(1)\]

Third term is 7:

\[a+2d=7\quad(2)\]

From (2), \(a=7-2d\). Substitute into (1):

\[2(7-2d)+11d=28\Rightarrow14+7d=28\Rightarrow7d=14\Rightarrow d=2\]

Then \(a=7-2(2)=3\).

\[a=3,\qquad d=2\]

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Expand the determinant along the first row.

\[\Delta=(x-3)\begin{vmatrix}2&2\\-4&6-x\end{vmatrix}-(-4)\begin{vmatrix}5&2\\2&6-x\end{vmatrix}+3\begin{vmatrix}5&2\\2&-4\end{vmatrix}\]

The \(2\times2\) minors:

\(\begin{vmatrix}2&2\\-4&6-x\end{vmatrix}=2(6-x)+8=20-2x\).

\(\begin{vmatrix}5&2\\2&6-x\end{vmatrix}=5(6-x)-4=26-5x\).

\(\begin{vmatrix}5&2\\2&-4\end{vmatrix}=-20-4=-24\).

So

\[\Delta=(x-3)(20-2x)+4(26-5x)+3(-24)\]

\[=(-2x^{2}+26x-60)+(104-20x)-72=-2x^{2}+6x-28\]

Set \(\Delta=-24\):

\[-2x^{2}+6x-28=-24\Rightarrow-2x^{2}+6x-4=0\Rightarrow x^{2}-3x+2=0\]

\[(x-1)(x-2)=0\Rightarrow x=1\ \text{or}\ x=2\]

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Position vectors: \(A=(8,-2),\ B=(2,6),\ C=(-10,4)\).

(a) In parallelogram \(ABCN\) the diagonals \(AC\) and \(BN\) bisect each other, so they share a midpoint.

Midpoint of \(AC=\left(\tfrac{8-10}{2},\tfrac{-2+4}{2}\right)=(-1,1)\).

Set midpoint of \(BN=(-1,1)\): \(\tfrac{2+N_x}{2}=-1\Rightarrow N_x=-4\); \(\tfrac{6+N_y}{2}=1\Rightarrow N_y=-4\).

\[N=-4\mathbf{i}-4\mathbf{j}\]

(b) \(\overrightarrow{AN}=N-A=(-4-8,\,-4+2)=(-12,-2)\); \(\overrightarrow{AB}=B-A=(2-8,\,6+2)=(-6,8)\).

(c) \(\overrightarrow{AN}\cdot\overrightarrow{AB}=(-12)(-6)+(-2)(8)=72-16=56\).

\(|\overrightarrow{AN}|=\sqrt{144+4}=\sqrt{148}\approx12.17\); \(|\overrightarrow{AB}|=\sqrt{36+64}=10\).

\[\cos\theta=\frac{56}{10\sqrt{148}}=0.4603\Rightarrow \theta\approx62.59^{\circ}\]

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Admission probabilities: Ali \(P(A)=\tfrac23\), Baba \(P(B)=\tfrac34\), Katty \(P(K)=\tfrac45\). Failures: \(P(A')=\tfrac13,\ P(B')=\tfrac14,\ P(K')=\tfrac15\). The three events are independent.

(a) Only Katty and Baba (so Ali fails):

\[P(A')\,P(B)\,P(K)=\tfrac13\times\tfrac34\times\tfrac45=\frac{12}{60}=\frac15\]

(b) None gains admission:

\[P(A')\,P(B')\,P(K')=\tfrac13\times\tfrac14\times\tfrac15=\frac{1}{60}\]

(c) At most two = not all three. So subtract the probability that all three gain admission:

\[1-P(A)\,P(B)\,P(K)=1-\left(\tfrac23\times\tfrac34\times\tfrac45\right)=1-\frac{24}{60}=1-\frac25=\frac35\]

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(a)(i) Expand \(\left(2-\tfrac12 x\right)^{5}\) using \(\binom{5}{k}2^{5-k}\left(-\tfrac12 x\right)^{k}\):

\[32-40x+20x^{2}-5x^{3}+\tfrac58 x^{4}-\tfrac{1}{32}x^{5}\]

(ii) Put \(2-\tfrac12 x=1.99\Rightarrow \tfrac12 x=0.01\Rightarrow x=0.02\). Substitute:

\(32-40(0.02)+20(0.02)^{2}-5(0.02)^{3}+\cdots\)

\(=32-0.8+0.008-0.00004+\cdots\approx31.20796\).

\[(1.99)^{5}\approx31.21\ \text{(2 d.p.)}\]

(b) Let \(P(x)=x^{3}+qx^{2}+rx+9\).

\((x+1)\) is a factor \(\Rightarrow P(-1)=0\): \(-1+q-r+9=0\Rightarrow q-r=-8\).

Remainder \(-17\) on division by \((x+2)\Rightarrow P(-2)=-17\): \(-8+4q-2r+9=-17\Rightarrow 4q-2r=-18\Rightarrow 2q-r=-9\).

Subtracting: \((2q-r)-(q-r)=-9-(-8)\Rightarrow q=-1\), then \(r=q+8=7\).

\[\boxed{q=-1,\ r=7}\]