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__Waec Further Mathematics Past Questions and Answers PDF__

__Waec Further Mathematics Past Questions and Answers PDF__
Hello to all my

It is known by all that further Mathematics is one of the toughest subject ever available, though some people can scatter this subject within a twinkle of an eye, but majority of the students cannot. Therefore, this is the reason I have decided to share with you this

*candidates, you are welcome. Today, I want to share with you some of the***2018 Waec Further Mathematics***2018 Waec Further Mathematics Past questions and Answers*and also, share with you the Free downloads of this past questions.It is known by all that further Mathematics is one of the toughest subject ever available, though some people can scatter this subject within a twinkle of an eye, but majority of the students cannot. Therefore, this is the reason I have decided to share with you this

*2018 Waec Further Mathematics Past questions and Answers*.
As you all know,

*Waec further Mathematics Syllabus*is set by the West African Examination Council (WAEC). This past questions are meant for you to understand and know how to read for the*Waec Further Mathematics 2018*. It will help you to know the type and the manner of questions you should expect for the*2018 Waec Further Mathematics questions*. This WASSCE Further Mathematics past questions will be given to you for free with no cost. All you need for this free download is you MB.Read also on: Waec Christian Religious studies questions and answers

**Note:**We are not giving you any

*Waec Futher Math Expo*. We are only giving you tips on how you should read on this particular subject. So, you should take note and use this questions with care. You are not advised to take this questions into the Exams hall, study them and know them.

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### 2018 Waec Further Mathematics Past questions and Answers

Table of Contents

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**Question 1**

If f(y) = log2 (y + 1) – log2 (y + 3),

find the domain of f(y);

solve f(y) = 1

Observation

Majority of the candidates were reportedly able to express the right hand side of the equality sign as a single log i.e. f(y) = log2 (y+ 1) – log2 (y + 3) = log 2(). However, candidates should know that f(y) would not exist when () is less than or equal to zero and this would happen when -3 £ y £ -1. Therefore, the domain of f(y) was Doman(f) = y : y R, y < -3 and y > -1 A good number of the candidates were reported not to get this right.

In part (b), candidates were reportedly able to show that if f(y) = 1, then log2 () =

log22 i.e. = 2. By cross multiplying and bringing like terms together, we obtain

y = – 5.

**Question 2**

If 2×2 + 3x + 3 = kx – k has real roots, find the range of values of k.

Observation

The Chief Examiner reported that this question was popular among the candidates and they performed well in it. However, a few candidates were reported not to recall that a quadratic equation ax2 + bx + c = 0 would have real roots if b2 ³ 4ac. In the given equation b = (3 – k), a = 2 and c = 3 + k. Therefore, the equation would have real roots if (3 – k)2 ³ 4(2)(3+k) i.e. k2 – 14k – 15 ³ 0. Solving this inequality gave the required values of k which were k £ -1 or k ³ = 15.

**Question 3**

The volume of a sphere is increased by 6%. Calculate the corresponding percentage increase in its area.

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Observation

This question was reported to be quite unpopular among the candidates. Majority of those who attempted it were also reported to have performed poorly. Candidates were expected to recall that

. If volume of sphere = V = r3, = 4pr2. Hence, dV = 4pr2dr. = = .

But = 6%. This implies that = 6 i.e = = 2%.

Surface area of a sphere = A = 4pr2. = 8pr. Therefore, dA = 8prdr. = = . Hence, = = 2 ´ 2 = 4%. i.e. The area is increasing by 4%.

**Question 4**

Find the equation of the tangent to the curve y = , when x = 1.

Observation

The Chief Examiner described the performance of candidates in this question as fair. Majority of them could not apply the quotient rule of differentiation correctly. Candidates were expected to show that if y = = , then = . Here, v = x + 1 and u = 1. = 1 and = 0. Hence, = = = . At x= 1, = = and y =

Therefore, the required equation of the tangent was y – = (x – 1) which simplified to x + 4y – 3 = 0.

**Question 5**

Events A and B are such that P(A) = and P(B) = .

Find P(A È B) if events A and B are:

(a) mutually exclusive;

(b) independent.

Observation

This question was reported to be quite popular among the candidates and they performed well in it. Majority of the candidates were reported to correctly recall the relation P(AÈB) = P(A) + P(B) – P(AÇB). However, it was also observed that in part (a) some candidates multiplied the respective probabilities instead of adding them.

In part (a), since the events were mutually exclusive, P(A Ç B) = 0. Therefore, P(A È B) = P(A) + P(B) = + = .

In part (b), since the events were independent, then P(A Ç B) = P(A) × P(B) = × = . Therefore, P = P(A) + P(B) – P(A – .

### 2018 Waec Further Mathematics Past questions and Answers / Free download

**Question 6**

The deviations from the mean of a set of numbers are

-2, x, (x + 7), (x + 2)2 and (x + 3)2, where x is a constant.

Find the value of x.

**Observation**

The Chief Examiner reported that this was also a very popular question among the candidates and they performed well in it. Majority of the candidates were reported to have recalled that for a given set of numbers, the sum of all the deviations from the mean = 0. i.e. -2 + x+7 (x+2)2 + (x+3)2 = 0. Expanding the squares and simplifying the equation gave x2 + 6x + 9 = 0 i.e. (x+3)2 = 0 Hence, x = -3.

**Question 7**

If the position vectors of A and B relative to the origin are and , find the angle between and .

**Observation**

The Chief Examiner reported that majority of the candidates attempted this question and their performance was described as fair. Majority of them were reported not to have applied the dot product correctly. Candidates were expected to show that if = and = , then êê = = 5 and êê = = . Furthermore, ٠ = ٠ = 20 – 6 = 14. If θ was the angle between and , then cos θ = = . Therefore, θ = cos-1() = 58.67o.

**Question 8**

(a) Resolve x2 + 1 into partial fractions.

3

(b) The gradient of a curve at point (x, y) is (2x – 3). If the minimum value on the curve

is 3, find the equation of the curve.

**Observation**

In part (a), candidates were expected to express as + + in partial fractions. This was not done by majority of those who attempted this question. If = + + , then x2 + 1 = A(x + 2)2 + B(x + 2) + C. i.e x2 + 1 = Ax2 + 4Ax + 4A + Bx + 2B + C. By comparing coefficients we have, A = 1 …eqn.(1); 4A + B = 0…. eqn.(2) and 4A+2B + C = 1….eqn3.

Substituting eqn(1) in to eqn.(2) gave 4(1) + B = 0. Which implied that B = -4. Substituting 1 for A and -4 for B in eqn(3) gave 4(1) + 2(-4) + C = 1 which implied that C=5. Therefore, the partial fractions of = – + .

In part (b), candidates were expected to integrate (2x – 3) with respect to x to have

y = x2 – 3x + c. The minimum value of y would occur at the point where the gradient function (2x – 3) = 0 i.e. when x = . Substituting this into the equation y = x2 – 3 x + c gave 3 = – 3() + c. Simplifying this gave c = . Therefore the required curve was 4y = 4×2 – 12x + 21.

**Question 9**

The sum to infinity of an exponential sequence (G.P.) with a positive common

ratio is 25 and the sum of the first two terms is 16. Find the:

(a) fifth term;

(b) sum of the first four terms.

Observation

The Chief Examiner reported that this question was very popular among the candidates and they performed very well in it. Candidates were reportedly able to show that if the sum to infinity was 25, then a = 25 – 25r, where a = first term and r = constant ratio. Also, since the sum of the first two terms was 16, then, a + ar = 16 i.e. a(1+ r) = 16. Solving these two equations simultaneously gave the quadratic equation 25(1 – r2) = 16. Solving this equation gave r = . Substituting for r in the equation a = 25 – 25r gave a = 10. The fifth term was obtained using the formula T5 = ar4 = 10. The sum of the first four terms = 10() = 10() = 25 ´ = = 21.

**Question 10**

A fair coin is tossed four times. Calculate the probability of obtaining:

at least one head;

equal number of heads and tails.

The probabilities that Sani, Kalu and Tato will hit a target are respectively. If all the three men shoot once, what is the probability that the target will be hit only once?

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Observation

This question was also reported to be attempted by majority of the candidates and

they performed well in it. Their performance in part (a) was also reported to be better than what it was in part (b).

In part (a) majority of the candidates were reported to show that the probability followed a binomial distribution nCrprqn-r, where n = 4, r = number of favourable outcome, p = probability of obtaining a head = and q = probability of not obtaining a head = .

Probability of getting at least one head = 1 – probability of getting no head

i.e. 1 – 4Co ()o ()4 = . Probability of equal number of heads and tails =

4C2 () 2()2 = x ()4 = .

In part (b), it was reported that majority of the candidates did not show that the probability that the target will be hit only once implied probability that (only Sani will hit or only Kalu will hit or only Talo will hit). The probability that Sani will not hit the target = 1 – probability that he will hit = 1 – = . In a similar way, Probability that Kalu will not hit = 1 – = while probability that Tato will not hit = 1 – = . Therefore probability that only one of them would hit =

( ´ ´ ´ ´ + ( ´ ´ ) = + + = .

**Question 11**

The position vectors of two points P and R are p = 4i – 2j and r = 2i + 3j respectively. Find:

(a) ê3p – 2rê;

(b) the values of the scalars m and n such that 8i + 8j = mp + nr;

(c) the cosine of the acute angle between p and r leaving the answer in surd form.

Observation

This question was reported to be quite unpopular with the candidates as majority of them did not attempt it. However, majority of those who attempted it performed well.

In part (a), candidates were expected to obtain 3p as 3(4i – 2j) = 12i – 6j.

2r = 2(2i + 3j) = 4i + 6j. Therefore, ê3p – 2rê = ê12i – 6j) – (4i + 6j) ê = ê8i – 12jê = 2 = = 4.

In part (b) candidates were expected to respond as follows:

mp = m(4i – 2j) = 4mi – 2mj; nr = n(2i + 3j) = 2ni + 3nj. Therefore mp + nr = (4m + 2n)i + (3n – 2m)j. Since mp + nr = 8i + 8j, it implied that 4m + 2n = 8 and 3n – 2m = 8. Solving these equations simultaneously gave n = 3 and m = .

In part (c), candidates were expected to apply the dot product thus:

If θ was the acute angle between p and r, then = . êpê = = ; êrê= = and p ٠r = (4i – 2j).(2i+3j) = 2 Therefore, cos θ = = .

### 2018 Waec Further Mathematics Past questions and Answers / Free download

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