WAEC SSCE - Further Mathematics - 2025 (Essay)

Détails de la réponse

14a. Total number of pupils = 4 boys + 5 girls = 9.  
Number of ways to choose 2 boys = \(\binom{4}{2} = 6\).  
Total number of ways to choose any 2 pupils = \(\binom{9}{2} = 36\).  
Probability that both chosen are boys = \(\dfrac{6}{36} = \dfrac{1}{6}\).

14b. This is a binomial distribution with n = 6 trials and p = 0.80 (probability a transistor is standard), q = 0.2 (Prb. of a transistor below standard)

Let Y = number of standard transistors. 

P(Y = k) = C(6, k) × (0.8)\(^k\) × (0.2)\(^{(6−k)}\)

i: k = 2 → 15 × 0.64 × 0.0016 = 0.015360

ii: k = 1 → 6 × 0.8 × 0.00032 = 0.001536

iii: P(Y ≥ 2) = 1 − P(Y = 0) − P(Y = 1) = 1 − 0.000064 − 0.001536 = 0.99840

iv: P(Y ≤ 2) = P(Y = 0) + P(Y = 1) + P(Y = 2) = 0.000064 + 0.001536 + 0.015360 = 0.016960

Détails de la réponse

Maximum height, h, v = 0m/s at max height

Recall v\(^2\) = u\(^2\) − 2gh

0\(^2\)  = 80\(^2\) − 2(10)h

0 = 6400 - 20h

h = \(\frac{6400}{20}\)

h = 320 m

b. Time taken to reach maximum height, t

 V = u − gt

0 = 80−10t

10t = 80

t = \(\frac{80}{10}\) = 8s

Since it takes 8 s to reach maximum height, it will also take 8 s to return to the point of projection; hence, it is 16 s

Détails de la réponse

17a. Since the plane is smooth (frictionless), the only force acting parallel to the inclined plane is the component of the weight of the body down the plane.  

F = mg sin 30\(^\circ\)  
Given m = 5  kg and taking g = 10 m/s\(^2\).  
F = 5 \(\times 10 \times \sin 30^\circ\) = 5 \(\times 10 \times\) 0.5 = 25 N.
The magnitude of the force acting parallel to the plane is 25 N.

bi. The uniform plank PQ is 10 m long with supports at A and B such that PA = 1 m and BQ = 1 m. Thus, AB = 8 m. The plank’s centre of mass G is at its midpoint (5 m from P), so AG = 4 m. The 8 kg load is placed at C, where AC = 3.5 m.  

Weight of plank = \( 10m \) N  
Weight of load = \( 8 \times 10 = 80 \) N  
Reaction at B, \( R_B = 100 \) N (given).  

For rotational equilibrium, take moments about A (clockwise = anticlockwise):  
\((10m) \times 4 + 80 \times 3.5 = 100 \times 8\)
40m + 280 = 800 
40m = 520 \(\implies m = \frac{520}{40}\) = 13
Thus, m = 13 kg.

bii. For vertical equilibrium:   \(R_A + R_B = \text{total weight}\)
\(R_A + 100 = (10 \times\) 13) + 80
R\(_A\) + 100 = 130 + 80 = 210
R\(_A\) = 210 - 100 = 110 N.
The reaction at A is 110 N.

Détails de la réponse

11a. Using the substitution \( u = 5 - x^2 \),  
\( du = -2x \, dx \) so \( x \, dx = -\frac{1}{2} du \).  

Limits: when \( x = 1 \), \( u = 4 \); when \( x = 2 \), \( u = 1 \).  

\(\int_1^2 \frac{x}{\sqrt{5 - x^2}} \, dx = \int_4^1 \frac{1}{\sqrt{u}} \left( -\frac{1}{2} \right) du = \frac{1}{2} \int_1^4 u^{-1/2} \, du\)

\(= \frac{1}{2} \left[ 2u^{1/2} \right]_1^4 = \left[ \sqrt{u} \right]_1^4 = \sqrt{4} - \sqrt{1}\) = 2 - 1 = 1.

11b. Given \( y = p x^2 + q x \),  
\( \frac{dy}{dx} = 2px + q \).  

But it is given that \( \frac{dy}{dx} = 6x + 7 \).  

Equate coefficients of like terms:   2p = 6  ⇒  p = 3,  q = 7.  

Check: \( \frac{d^2 y}{dx^2} = 2p = 2 \times 3 = 6 \), which matches the given condition.  

Thus, p = 3, and q = 7.

Détails de la réponse

\(\frac{150}{sin 100} = \frac{T_1}{sin 140}\)

T\(_1\) = \(\frac{150 sin 140}{sin 100}\) = 97.91N

In same manner:

\(\frac{150}{sin 100} = \frac{T_2}{sin 120}\)

T\(_2\) = \(\frac{150 sin 120}{sin 100}\)

T\(_2\) = 131.91N.

i. At the highest point,  v = 0 when  t = 2 s:  
\(20 + 2k \cdot 2 = 0 \)
20 + 4k = 0
k = - 5

ii. The highest point "from the point of throw" is the upward displacement above the release point t = 0, where the variable terms begin.  
At  t = 2:  
2 + 20 x  2 + k x (2)\(^2\) = 2 + 40 + (-5) x 4 = 42 - 20 = 22 m.

Détails de la réponse

b. Median = (\(\frac{ n + 1}{2}\))\(^{th}\) data

Median = \(\frac{80 + 1}{2}\) = \(\frac{81}{2}\) = 40.5th

So, from the graph, the median is 48.8 hours.

ii. Lower quartile, Q\(_1\) = (\(\frac{n + 1}{4}\))\(^{th}\) data

Q\(_1\) = \(\frac{80 + 1}{4}\) = \(\frac{81}{4}\) = 20.25th data

Q\(_1\) = 40 hours from the graph

iii. 40th percentile

 = \(\frac{40}{100}\) x 80 = 32\(^{nd}\) data.

From the graph, 40th percentile = 45.5 hours.

iv. The number of employees who spent at least 50 hours 30 mins = 32.

Détails de la réponse

a. fifth term, T\(_5\) =? But the nth term of an A.P.  T\(_n\) = a (n - 1) d

T\(_2\) = a + d

T\(_3\) = a + 2d

T\(_5\) = a + 4d

T\(_2\) + T\(_5\) = a + d + a + 4d = 2a + 5d = 42 .....(i)

T\(_6\) = a + 5d

T\(_3\) = a + 2d

T\(_6\) - T\(_3\) = (a + 5d) - (a + 2d) = 3d = 12

3d = 12

d = 4

b. put d = 4 into equation (i)

2a + 5(4) = 42

2a = 42 - 20

2a = 22

a = 11 first term,

a = 11

c. T\(_{20}\) = a + 19d

                    = 11 + 19 x 4

T\(_{20}\) = 87.

Détails de la réponse

16a.Let the mass of ball P be \( m \) and the mass of ball Q be \( M \).  

Conservation of linear momentum (before and after collision):  
\(m \times 2u + M \times 0 = m \times \left(-\frac{1}{2}u\right) + M \times u\) 
2mu = -\(\frac{1}{2}\)mu + Mu  
Divide through by \( u \) (assuming \( u \neq 0 \)):  
2m = -\(\frac{1}{2}\)m + M
 M = 2m + \(\frac{1}{2}\)m = \(\frac{5}{2}\)m
Thus, the ratio of the mass of P to the mass of Q is  
m: M = 2: 5

bi.  Let the angle between the two forces be \( \theta \).  

By the parallelogram law (or cosine rule for vector addition):  
\(R^2 = 3^2 + 7^2 + 2 \times 3 \times 7 \times \cos\theta\)
\(5^2 = 9 + 49 + 42\cos\theta\)
\(25 = 58 + 42\cos\theta\)
\(42\cos\theta\) = 25 - 58 = -33
\(\cos\theta = -\frac{33}{42} = -\frac{11}{14} \approx -0.7857\)
\(\theta = \cos^{-1}(-0.7857) = 141.8^\circ\)

c.  Given: \(\overrightarrow{AB} = \begin{pmatrix} -4 \\ 6 \end{pmatrix}, \quad
\overrightarrow{CB} = \begin{pmatrix} 2 \\ 3 \end{pmatrix}\)

V is the midpoint of AB, so its position vector satisfies  
\(\vec{V} = \frac{\vec{A} + \vec{B}}{2}.\)
Vector \(\overrightarrow{CV}\) is  
\(\overrightarrow{CV} = \vec{V} - \vec{C} = \overrightarrow{CB} - \frac{1}{2}\overrightarrow{AB}.\)
Substitute the given vectors:  
\(\frac{1}{2}\overrightarrow{AB} = \begin{pmatrix} -2 \\ 3 \end{pmatrix},\)
\(-\frac{1}{2}\overrightarrow{AB} = \begin{pmatrix} 2 \\ -3 \end{pmatrix}.\) 
\(\overrightarrow{CV} = \begin{pmatrix} 2 \\ 3 \end{pmatrix} + \begin{pmatrix} 2 \\ -3 \end{pmatrix} = \begin{pmatrix} 4 \\ 0 \end{pmatrix}.\)

Détails de la réponse

a.  \(\frac{\sqrt{75}-3}{\sqrt{3}+1}\)

First, simplify the radical:
\(\sqrt{75} = \sqrt{25 \cdot 3} = 5\sqrt{3}\)

So the expression becomes:
\(\frac{5\sqrt{3}-3}{\sqrt{3}+1}\)

Now rationalize the denominator by multiplying top and bottom by \(\sqrt{3}\) - 1):
\(\frac{(5\sqrt{3}-3)(\sqrt{3}-1)}{(\sqrt{3}+1)(\sqrt{3}-1)}\)

Denominator: 3 - 1 = 2
Numerator: 15 - 5\(\sqrt{3} - 3\sqrt{3} + 3 = 18 - 8\sqrt{3}\)

So: \(\frac{18 - 8\sqrt{3}}{2} = 9 - 4\sqrt{3}\).

bi. The general equation of a circle is \(x^2 + y^2 + 2gx + 2fy + c = 0\)  

Substitute the three points:  Point (7, 3): 
\(49 + 9 + 14g + 6f + c = 0 \implies 14g + 6f + c = -58 \quad \text{(1)}\)

Point (2, 8): 
\(4 + 64 + 4g + 16f + c = 0 \implies 4g + 16f + c = - 68 \quad \text{(2)}\)

Point (-3, 3): 
\(9 + 9 - 6g + 6f + c = 0 \implies -6g + 6f + c = -18 \quad \text{(3)}\)

Subtract (2) from (1):  
\(10g - 10f = 10 \implies g - f = 1 \implies g = f + 1 \quad \text{(4)}\)

Subtract (3) from (2):  
\(10g + 10f = -50 \implies g + f = -5 \quad \text{(5)}\)

Solve (4) and (5):  
\((f + 1) + f = -5 \implies 2f + 1 = -5 \implies 2f = -6 \implies f = -3\)
g = -3 + 1 = -2

Substitute \( g = -2 \), \( f = -3 \) into (3):  
\(-6(-2) + 6(-3) + c = -18 \implies 12 - 18 + c = -18 \implies c = -12\)

The equation of the circle is: \(x^2 + y^2 - 4x - 6y - 12 = 0\)

bii. The centre is  -g, -f  = (2, 3).  
The radius \( r \) is given by  \(r = \sqrt{g^2 + f^2 - c} = \sqrt{(-2)^2 + (-3)^2 - (-12)} = \sqrt{4 + 9 + 12} = \sqrt{25} = 5\)

Radius of the circle = 5 units.

Détails de la réponse

10a. The equation of the curve is \( y = 4x^3 \).  
The gradient of the tangent at any point is given by the derivative:  
\(\frac{dy}{dx} = 12x^2.\)  
At points P and Q, the gradient is 108:  
\(12x^2 = 108 \implies x^2 = 9 \implies x = \pm 3.\)

When \( x = 3 \):  
\(y = 4(3)^3 = 4 \times 27 = 108.\)
So one point is 3, 108).  

When x = -3 :  
\(y = 4(-3)^3 = 4 \times (-27) = -108.\) 
So the other point is (-3, -108).  

Coordinates of P and Q:  (3, 108), and (-3, -108).

10bi.  We use the angle subtraction formulas:  
sin(A - B) = sin A cos B - cos A sin B,  
cos(A - B) = cos A cos B + sin A sin B,
where \( A = 45^\circ \) and \( B = 30^\circ \).  

Known values:  
\(\sin 45^\circ = \cos 45^\circ = \frac{\sqrt{2}}{2}, \quad \cos 30^\circ = \frac{\sqrt{3}}{2}, \quad \sin 30^\circ = \frac{1}{2}\)

Now,  
\(\sin 15^\circ = \sin(45^\circ - 30^\circ) = \left( \frac{\sqrt{2}}{2} \right) \left( \frac{\sqrt{3}}{2} \right) - \left( \frac{\sqrt{2}}{2} \right) \left( \frac{1}{2} \right) = \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4} = \frac{\sqrt{6} - \sqrt{2}}{4}.\)  

\(\cos 15^\circ = \cos(45^\circ - 30^\circ) = \left( \frac{\sqrt{2}}{2} \right) \left( \frac{\sqrt{3}}{2} \right) + \left( \frac{\sqrt{2}}{2} \right) \left( \frac{1}{2} \right) = \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4} = \frac{\sqrt{6} + \sqrt{2}}{4}.\) 

10bii.  \(\tan 15^\circ = \frac{\sin 15^\circ}{\cos 15^\circ} = \frac{\frac{\sqrt{6} - \sqrt{2}}{4}}{\frac{\sqrt{6} + \sqrt{2}}{4}} = \frac{\sqrt{6} - \sqrt{2}}{\sqrt{6} + \sqrt{2}}.\) 

Rationalise the denominator by multiplying numerator and denominator by \(\sqrt{6} - \sqrt{2}\):  
Numerator: \((\sqrt{6} - \sqrt{2})^2 = 6 - 2\sqrt{12} + 2 = 8 - 4\sqrt{3}\).  
Denominator: \((\sqrt{6})^2 - (\sqrt{2})^2 = 6 - 2 = 4 \).  

Thus,  \(\tan 15^\circ = \frac{8 - 4\sqrt{3}}{4} = 2 - \sqrt{3}.\)

Détails de la réponse

To find the equation of the tangent to the curve \( y = \frac{x - 1}{2x + 1} \) at the point (1, 0):

First, verify the point lies on the curve:  
Substitute \( x = 1 \): \( y = \frac{1 - 1}{2(1) + 1} = \frac{0}{3} = 0 \). Yes, it does.

Next, find the slope of the tangent by computing the derivative \( \frac{dy}{dx} \):  
\(y = \frac{x - 1}{2x + 1}\)
Using the quotient rule: 
\(\frac{dy}{dx} = \frac{(1)(2x + 1) - (x - 1)(2)}{(2x + 1)^2} = \frac{2x + 1 - 2x + 2}{(2x + 1)^2} = \frac{3}{(2x + 1)^2}\)

At \( x = 1 \): 
\(m = \frac{3}{(2(1) + 1)^2} = \frac{3}{9} = \frac{1}{3}\)

Equation of the tangent line at (1, 0):  
\(y - 0 = \frac{1}{3}(x - 1)\)
\(y = \frac{1}{3}x - \frac{1}{3}\)
Multiply through by 3: \( 3y = x - 1 \)  
Rearrange: \( 3y - x + 1 = 0\)

Détails de la réponse

a. Mass , m = 5 kg

Resultant force, f = (6i+2j) + (5i+4j) + (4i−j) = (6+5+4)i + (2+4−1)j  =  F = 15i + 5j

From Newton second law of motion F = \(\frac{m \Delta V}{\text{t}}\)

​F = \(\frac{m(v - u)}{\text{t}}\)

where u=0 ms\(^{−1}\)

F = \(\frac{\text{mv}}{\text{t}}\)

v = \(\frac{\text{Ft}}{\text{m}}\)

v = \(\frac{t(15i + 5j)}{5}\)

v = t(3i + j)m/s

b. Velocity, V  after 4 s

v = t(3i + j) = (4)(3i + j) = 12i + 4jm/s.