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WAEC Further Mathematics Questions and Answers 2022
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WAEC Further Mathematics
1. Evaluate log927
Solution; let’s assume x = log927
Then 9x = 27
Therefore, (32)x = 33
32x = 33
Divide it through by the coefficient of 2x
i.e. 32x /3 = 33/3
= 2x = 3 (equating indices) and x =1½
2. Solve log50.04
Solution: Let r be = log50.04, then 5r = 0.04 = 1/25 = 5-2 (equating indices)
Hence r = -2 Ans.
3. Solve 2×2 + 9x =5.
Solving solution: 2×2 + 9x =5
2×2 + 9x -5 =0
(2x – 1) (x + 5) = 0
Either 2x – 1 = 0 or x + 5 = 0
Therefore 2x = 1 or x = -5 so, x = ½ or x = -5 Ans.
4. Solve 3×2 –xy = 0 and 2y – 5x = 1 simultaneously
Solution solving
3×2 – xy = 0 ———————- equation 1
2y – 5x = 1 ———————-equation 2
From equation 2, we have 2y = 1 + 5x
Y = ½ (1 + 5x) ——————-equation 3
Substitute ½(1 + 5x) for y in equation 1
3×2 – x [½(1 + 5x)] = 0
3×2 – ½x – 2½x2 = 0
½x2 – ½x = 0
½x (x – 1) = 0
Therefore, x =0 or x =1
Substituting for x in equation 3
When x =0, y = ½(1 + 5 X 0) = ½
And, when x =1, y = ½(1 + 5 X 0) =3
Therefore, the solutions are x =0 and y =½
x =1 and y =3
Or, in ordered pairs (0, ½) and (1, 3) Ans.
5. An item is marked ‘cost N7000 VAT inclusive’. If the rate of VAT is 5%, how much VAT does the Government receive?
Solution: The price tag means that a 5% VAT surcharge is included in the total price. Thus N7000 represents 105% of the actual cost of the item.
105% of the cost price =N7000
1% of the cost of the item = N7000/105
5% of the cost (i.e. the VAT) =N7000/105 X 5
=N333 1/3 Ans.
6. Use the chain rule to differentiate (3x + 8)6
Solution: Let y = (3x + 8)6 and t = 3x + 8 then,
y = t6, dt/dx =3 and dy/dt =6t5
By the chain rule, we have; dy/dx = dy/dt X dt/dx
= 6t5 X 3
= 6 (3x + 8)5 X 3
= 18 (3x + 8)5 Ans.
7. Find the equation of the tangent to the curve y = x2 – 4x + 3 at the point (3, 1).
Solution: y = x2 – 4x + 3
= dy/dx = 2x – 4, the gradient function.
If x = 3, the gradient of the curve at this point is 2 X 3 – 4 =2
Hence, the gradient of the tangent at x=3 is 2, i.e. the same as the gradient of the curve. The equation of the tangent is, y-1/x-3 = 2
→ y – 1 = 2(x-3)
→ y = 2x – 5 Ans.
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(13ai)
Given: mass ,m =10kg
Force,F = 40N
Time, t = 0.5secs
Impulse, I = Ft = 40×0.5 = 20Ns
(13aii) Ft = m(v-u) where u= 0 (at rest)
20 = 10(v-0)
20 = 10v
V = 20/10 = 2m/s
Final speed = 2m/s
(13aiii)
Given: u=0 ; v=2m/s ; t=0.5secs
S= 1/2(u+v)t
S= 1/2(0+2)×0.5
S= 0.5 metres
Distance = 1/2 metre or 50cm
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