WAEC SSCE - Further Mathematics - 2025 (Essay)

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: \( 3y - x + 1 = 0\)

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

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.

Assumed mean, Am = 24.5

Mean (\(\overline{x}\)) = Am + \(\frac{\sum fd}{\sum f}\) = 24.5 + \(\frac{(-300)}{120}\)

Mean (\(\overline{x}\)) = 24.5 - 2.5 = 22.0

Note:(x = sum of upper and lower limit divided by 2)

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

a. (x + 2) and (x -1) are factors of f(x) = 6x\(^4\) + mx\(^3\) - 13x\(^2\) + n x + 14

Then, f(-2) and f(1) are = 0

f(-2) = 6(-2)\(^4\) + m(-2)\(^3\) - 13(-2)\(^2\) + n (-2) + 14 = 0

 = 6 x 16 - 8m - 13 x 2 - 2n + 14 = 0 = 96 - 8m - 52 - 2n + 14 = 0

= 8m + 2n = 58: divide through by 2

= 4m + n = 29 - - -- - - - - - -(i)

f(1) =  6x\(^4\) + mx\(^3\) - 13x\(^2\) + n x + 14 = 0

f(1) =  6(1)\(^4\) + m(1)\(^3\) - 13(1)\(^2\) + n (1) + 14 = 0

= 6 + m - 13 + n + 14 = 0

= m + n = -7 - - - - - - - - -(ii)

Solving eqn i and ii simultaneously

from eqn i  - - - - -   n = 58 - 4m

put n = 58 - 4m into eqn ii

m + n = -7 = m + 29 - 4m = -7

- 3m = -36

m = \(\frac{36}{3}\) = 12

put m = 12  into eqn ii

m + n = -7 = 12 + n = -7

n = -7 - 12 = -19.

b. 6x\(^4\) + mx\(^3\) - 13x\(^2\) + n x + 14 = 0 becomes 6x\(^4\) + 12x\(^3\) - 13x\(^2\) - 19x + 14 = 0

f(x) = 6x\(^4\) + 12x\(^3\) - 13x\(^2\) - 19x + 14 = 0 

remainder when f(x) is divided by (x + 1)

Let (x + 1) = 0 then, x = -1

f(-1) = 6(-1)\(^4\) + 12(-1)\(^3\) - 13(-1)\(^2\) - 19(-1) + 14 

= 6 - 12 - 13 + 19 + 14 = 14

Therefore, the remainder = 14.

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.

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

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.

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.

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.

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.

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:  
\(\frac{1}{2}\overrightarrow{AB} = \begin{pmatrix} -2 \\ 3 \end{pmatrix},\)
\(-\frac{1}{2}\overrightarrow{AB} = \begin{pmatrix} 2 \\ -3 \end{pmatrix}.\) 
\(\overrightarrow{CV} = \begin{pmatrix} 2 \\ 3 \end{pmatrix} + \begin{pmatrix} 2 \\ -3 \end{pmatrix} = \begin{pmatrix} 4 \\ 0 \end{pmatrix}.\)