**WAEC Further Mathematics Questions and Answers 2022 is now release for the May/June 2022. WAEC Further Mathematics Theory and Objective Answers (100%legit) Further Mathematics 2 Essay verified Free (Expo) for West African Examinations Council. WAEC Further Mathematics Questions For you to have good WAEC result in Further Mathematics as well as repeated questions for free in this post.** You will also understand how WAEC Further Mathematics questions are set and how to answer them. The West African Examinations Council is an examination board established by law to determine the examinations required in the public interest in the English-speaking West African countries, to conduct the examinations and to award certificates comparable to those of equivalent examining authorities internationally

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### WAEC Further Mathematics Questions and Answers 2022

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#### WAEC Further Mathematics OBJECTIVES (OBJ) ANSWERS Loading

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WAEC Further Mathematics OBJ Answers:

Further Mathematics -OBJ

1-10 AADDBACBCC

11-20 BACCDCCADC

21-30 BABACDDBCC

31-40 CCCCBDABBC

41-50 CBBDABBDCB

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### WAEC Further Mathematics Answers

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WAEC Further Mathematics

1. Evaluate log927

Solution; letâ€™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 =1Â½

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 2Ã—2 + 9x =5.

Solving solution: 2Ã—2 + 9x =5

2Ã—2 + 9x -5 =0

(2x â€“ 1) (x + 5) = 0

Either 2x â€“ 1 = 0 or x + 5 = 0

Therefore 2x = 1 or x = -5 so, x = Â½ or x = -5 Ans.

4. Solve 3Ã—2 â€“xy = 0 and 2y â€“ 5x = 1 simultaneously

Solution solving

3Ã—2 â€“ xy = 0 â€”â€”â€”â€”â€”â€”â€”- equation 1

2y â€“ 5x = 1 â€”â€”â€”â€”â€”â€”â€”-equation 2

From equation 2, we have 2y = 1 + 5x

Y = Â½ (1 + 5x) â€”â€”â€”â€”â€”â€”-equation 3

Substitute Â½(1 + 5x) for y in equation 1

3Ã—2 â€“ x [Â½(1 + 5x)] = 0

3Ã—2 â€“ Â½x â€“ 2Â½x2 = 0

Â½x2 â€“ Â½x = 0

Â½x (x â€“ 1) = 0

Therefore, x =0 or x =1

Substituting for x in equation 3

When x =0, y = Â½(1 + 5 X 0) = Â½

And, when x =1, y = Â½(1 + 5 X 0) =3

Therefore, the solutions are x =0 and y =Â½

x =1 and y =3

Or, in ordered pairs (0, Â½) and (1, 3) Ans.

5. An item is marked â€˜cost N7000 VAT inclusiveâ€™. 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 â€“ 4x + 3 at the point (3, 1).

Solution: y = x2 â€“ 4x + 3

= dy/dx = 2x â€“ 4, the gradient function.

If x = 3, the gradient of the curve at this point is 2 X 3 â€“ 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

â†’ y â€“ 1 = 2(x-3)

â†’ y = 2x â€“ 5 Ans.

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(13ai)

Given: mass ,m =10kg

Force,F = 40N

Time, t = 0.5secs

Impulse, I = Ft = 40Ã—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)Ã—0.5

S= 0.5 metres

Distance = 1/2 metre or 50cm

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