Cameroon GCE A level June 2024 pure mathematics with statistics 1

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Subject Code	0770
Subject Title	Pure Math with Statistics
Session	June 2024 Paper 1

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Cameroon GCE A Level Pure Math Statistics Paper 1 Solution - June 2024

1. Two functions f, g:ℝ → ℝ.     Given that f(x) = x² - 1   and   g(x) = 2x + 1,     g ∘f(2) =

A. 11

B. 7

C. 24

D. 13

The correct answer is B.

First we compute f(2) = 2² - 1 = 3
Now compute g∘f(2) = 2(3) + 1 = 7

2. If log_xY = 2 and XY = 125, then the values of X and Y are respectively

A. 3 and 9

B. 9 and 3

C. 5 and 25

D. 25 and 5

The correct answer is A.
log_xY = 2 ⇒ Y = X²
thus substituting give X(X²) = 125
X³ = 125 ⇒ X = 5

3. The vertical assymptote of the curve y = 1(x - 2)(x + 3) are:

A. x = 3 , x = -2

B. x = 2, x = -3

C. x = -3, x = -2

D. x = 2, x = 3

The denominator equeal zero when (x - 2)(x + 3) = 0
thus, wen x = 2 or when x = -3, the denomnator is zero.
therefore, the VAs are x = 2, -3

4. When the polynomial P(x) = ax³ + 5x² + 2x - 8 is divided by (x - 1), the remainder is -3. a =

A. -4

B. -2

C. 2

D. 3

• Firstly, x - 1 = 0 ⇒ x = 1.
• To get the remainder, we know P(1) = -3 , and implies
• a(1) + 5(1) + 2(1) - 8 = -3
• a - 1 = -3 thus a = -2

5. The set of values of x which (x + 1)² + 3 (x + 3)(x - 2) < 0

A. {x : x < -3 or x > 2}

B. {x : x ≤ -3 or x > 2}

C. {x : -3 ≤ x < 2}

D. {x : -3 < x < 2}

You can ckeck directly by testing the regions and you will see that the key is D. You can use any numbers between -3 and 2 to test. C is not correct because of the equality sign. e.g. texting with x = -3 will make the function undefined.

ALternatively, we can use tables. we can leave out te numerator since (x + 1)² + 3 is always positive, meaning the expression will only be negative when denointor is negative. The boundary solutions are -3 and 2

x	x < -3	-3 < x < 2	x > 2
(x + 3)	-	+	+
(x - 2)	-	-	+
f(x)	+	-	+

6. A function f:ℝ → ℝ is defined by

If f(x) has a limit at x = 1, the value of k is

A. 2

B. 1

C. 3

D. -2

equating limits from both sides at x = 1, we get:
4k + 4(1) = 4(1) - k² - 4
4k + 4 = - k²
solving for k in k² + 4k + 4 = 0 gives k = -2

7. Let p and q be statements:

P	Q	⊕
T	T	T
T	F	T
F	T	T
F	F	F

The compound statement on the table could represent

A. P ⇒ Q

B. P ⟺ Q

C. P ∨ Q

D. P ∧ Q

Analyzing the truth table:,
This is an OR statement as the results is only false when both are false

8. Given the lines:
L₁: r = 3i - 2j + 4k + d(4i + 2j - 3k)
L₂: r = i + 3j + k + d(8i + 4j - 6k)
L₃: r = 2i + 6j + 2k + b(i - 4j - 4k)
L₄: r = 4i + 12j + 4k - u(i + 3j + k)

which pair of lines are parallel?

A. L1 and L2

B. L1 and L4

C. L2 an L4

D. L2 and L3

Compare direction vectors and find parallel ones. If you multiply the directon vector of L1 by 2, you get the direction vector of L2. When a direction vector is a scalar multiple of another, the two lines are parallel
Answer: A (L1 and L2 are parallel).

9. A relation R defined on a non-empty set S is called a partial order if R is

A. Reflexive, symmetric and anti-symmetric,

B. Reflexive, anti-symmetric and transitive

C. Reflexive, symmetric and transitive

D. symmetric, anti-symmetric and transitive

Solution: Partial order relations must be reflexive, antisymmetric, and transitive. Answer: B (Reflexive, antisymmetric and transitive)

10. Disequilibrium unemployment is said to exist when:

A. 10

B. 5

C. 15

D. 7

The given summation is a geometric series
The general formula for the sum of an infinite geometric series is: S = a/(1 - r), where a is the first term and r is the common ratio.
THis gives m1 - 1/2 = 10
⟹ m = 5

11. The first order differential equation obtained from the function y = Ax + ln x is

A.

x dy dx = y + lnx + 1

B.

dy dx = y - lnx + 1

C.

x dy dx = y - ln(x + 1)

D.

x dy dx = y - lnx + 1

Step 1: Differentiate both sides with respect to x
dy/dx = a + 1/x
Step 2: Express the result as a first-order differential equation

x dy dx = ax + 1

but ax from original equation equals y - lnx thus, we replace it to have

x dy dx = y - lnx + 1

12. The shaded region represents:

A. A ∩ B

B. A ∩ B'

C. A' ∩ B

D. (A U B)'

The shaded region is that outside A that is A' and its intersection with B

13. A function f: ℝ --> ℝ is given by

f(x) = x + 6 x² - 2x - 15

. The domain of f is

A. ℝ - {3, 5}

B. ℝ - {-3, 5}

C. ℝ - {-5, 3}

D. ℝ - {-5, -3}

The function will be undeined when the denominator equals zero, and tis occurs wen x = -3 or when x = 5, the the domain is the set of all real numbers excluding -3 and 5

14. A curve is defined parametrically by: x = 2t, y = 2/t. The cartesian equation of the curve is

A. xy = 4

B. xy = 2

C. xy = 1

D. xy = 3

from x = 2t we make t the subject and obtain t = x/2. substituting t = x/2 into y = 2/t gives y = 2/(x/2) = 4/x thus xy = 4

15. If r = i + 2j + 2k and S = -2i + 3j + k then r x s is

A. -4i - 5j - 7k

B. -4i + 5j - 7k

C. -4i - 5j + 7k

D. -4i + 5j + 7k

USe the cross product 3 x 3 determinant formular or any other method and obtain: r×s = −4i−5j+7k.

16. Given that f(θ) = sinθ - √3cosθ
Expressing f(θ) in the form Rsin(θ - α) where R > 0 and α is and acute ngle, we obtain f(θ) =

A. 2sin(θ - π/2)

B. √10sin(θ-π/3)

C. 2sin(θ - π/3)

D. √10sin(θ - π/4)

firstly, Rsin(θ - α) = Rsinθcosα - Rcosθsinα
equating with sinθ - √3cosθ
we obtain Rsinθcosα = sinθ ---> Rcosα = 1 ......[1]
and - Rcosθsinα = - √3cosθ ---> Rsinα = √3 ..... [2]
from [1] and [2] R = 2 and tanα = √3 thus α = π/3

17. If f(x) = x² + kx - 3 as a root in te interval [1, 2] then the range of k is

A. k < -1/2

B. k > 2

C. -1/2 < k < 2

D. k < -1/2 or k > 2

By applying the intermediate theorem we obtain :
[1² + k(1) - 3] . [2² + k(2) - 3] < 0
(k - 2)(2k + 1) < 0
the boundary solutions are -1/2 and +2 and testing with k = 0, the region inbetween is satisfied. thus [-1/2, 2] key C

18. Given that the sum and difference of the roots of the equation 2x² - 2x + q = 0 are equal, the value of q is

A. -2

B. 1/2

C. 2

D. -1/2

None

None: Let the roots be A and B. First rewrite the equation as x² - x + 0.5q = 0
x² - (sum of roots)x + product of roots = 0
From both forms, sum of roots A + B = 1 ....[1] ; product of roots AB = 0.5q ....[2]
also A - B = 1 ....[3] thus, adding [1] and [3] ---> 2A = 2 ---> A = 1, B = 0
thus q = 0

19. The sum of the first n terms of an arithmetic progression is n(2n + 3). The nth term of the progression is

A. 6n + 1

B. 2n + 1

C. 4n - 1

D. 4n + 1

T1 = S1 = 1[2(1) + 3] = 5
T2 = S2 - S1 = 2[2(2) + 3] - 5 = 9
T3 = S3 - S2 = 3[2(3) + 3] - 2[2(2) + 3] = 13
Thus the first term is 5, the common difference is +4
Tn = a + (n-1)d = 5 + 4(n - 1) = 4n + 1

20.

A. 5

B. √5

C. √125

D. 6

Just by looking at the question, the magnitude of the numberator is five times that of the denominator |-5 + 10i| = 5|-1 + 2i|
Equally |1 + 2i| = |-1 + 2i| thus, the result of numerator divided by denominator = 5

21. The line L has equation r = 5i - 8j + k + δ(i + 3j + 2k) and the plane π has equation 2x + 2y - z = 5. The sine of the acute angle between the line and the plane is

A. 3√147

B. √147

C. 2√147

D. 4√147

E. None

The acute angle is the angle between the direction vetcto of the line i + 3j + 2k and the normal to the plane 2i + 2j - k
sin(θ) = |(A x B)| / (|A| * |B|) where the A x B = -7i + 5j - 4k
|A| = 3; |B| = √14; |A x B| = √90 = 3√10;
sin(θ) = 3√10 / 3 √14 = √10/√14

22.

dy dx (3x + 2)⁴

A. 12x(3x + 2)³

B. 12(3x + 2)⁴

C. 12(3x + 2)³

D. 43 (3x + 2)³

differentiating the core, 3x + 2 gives 3
Final derivative = 4(3) (3x + 2)³

23. Find the equation of the circle passing through the origin and points (2,0) and (0,-4).

A. x² + y² - 2x + 4y = 0

B. x² + y² - 2x - 4y = 0

C. x² + y² + 2x - 4y = 0

D. x² + y² + 2x + 4y = 0

The general equation of a circle is x² + y² + Ax + By + C = 0
if circle passes through center, (0, 0), then, C = 0
Substitute the point (2, 0) into the equation of a circle, we get A = -2
Substitute the point (0,-4) into the equation of a circle, we obtain B = 4
The equation of the circle is x² + y² - 2x + 4y = 0

24. The area of the finite region bounded by the curve y = 2x - 1/x and the lines x = 1 and x = 3 is

A. 8 - ln3

B. 8 + ln3

C. 8 - ln2

D. 8 + ln2

25. If a root of the equation x² + 2x + 1 = 0 is also a root of the equation x³ - 2x² + 6x + k = 0, then the value of the constant k is

A. 9

B. -9

C. 5

D. -5

the equation x² + 2x + 1 = 0 or the equation (x + 1)² = 0 has single root x = -1.
substituting in the cubic equation gives: (-1)³ - 2(-1)² + 6(-1) + k = 0
-1 - 2 - 6 + k = 0
k = 9

26.

A. 1

B. 3

C. 11

D. 81

The period of the function is 4 - 0 = 4
f(9) = f(9 - 4)
= f(5)
= f(5 - 4) = f(1)
= 1² = 1

27. If two functions f,g:R → R are such that f(x) = 2x - 1 and (f ∘ g)(x) = x - 2, then g(x) =

A. x - 22

B. x - 12

C. 2x - 12

D. x + 12

(f ∘ g)(x) = x - 2 ..... [1]
(f ∘ g)(x) = 2g(x) - 1 ... [2]
equating both gives: 2g(x) - 1 = x - 2
g(x) = (x - 1)/2

28. Find the equation of the tangent to the curve y = 3x(2 - x²) at the point (1, 3) is

A. 3x - y - 6 = 0

B. 3x + y - 6 = 0

C. y - 3x - 6 = 0

D. 3x - y + 6 = 0

y' = 6 - 9x² and at x = 1, y'(1) = 6 - 9(1) = -3
equation of tangent is: y = -3(x - 1) + 3
giving y + 3x - 6 = 0 or 3x + y - 6 = 0

29. Simplifying:

sin5x - sinx sin4x - sin2x     gives

A. 2cosx

B. 2cos2x

C. 2secx

D. 2sinx

Simplify the numerator: sin5x − sinx = 2cos(3x)sin(2x).
Simplify the denominator: sin4x − sin2x = 2cos(x)sin(x).

2 cos3x sin2x 2 cos3x sinx

= 2cos3x . 2sinx cosx 2 cos3x sinx = 2cosx

30. The number of arrangements of the letters of the word "ARREARS" is

A. 210

B. 420

C. 640

D. 840

The word "ARREARS" consists of 7 letters, with some of them repeating. The distinct letters and their frequencies are:
A appears 2 times.
R appears 3 times.
E appears 1 time.
S appears 1 time.
The total number of arrangements of the letters in the word is calculated using the formula for permutations of multiset elements:

= 7! 2! x 3! x 1! x 1! = 420

31. In the expansion of √3 + x, the range of values for which the complete expansion is valid.

A. -3 ≤ x ≤ 3

B. -3 < x < 3

C. -1 < x < 1

D. -1/3 < x < 1/3

32. The value 1 + 1/3 + 1/9 + 1/27 + . . . is

A. 1/3

B. 3/2

C. 40/27

D. 3

a = 1; r = 1/3;
Sum to infinity is a/(1 - r) = 1/(1 - 1/3) = 3/2

33.

A. 0

B. 1

C. 2

D. 3

As x tends to infinity, consider just the most important term 4x²- 1 is appoximately 4x². And for denominator is just 2x² as adding 1 has negligile effect. now dividing both results in 2.
the liit of 2 as x tends to infinity is 2

34.

A. 3 + 4ln3

B. 3 + 3ln4

C. 4 + 4ln3

D. 4 + 3ln4

= x + 3 x = 1 + 3/x

the integral of 1 from 1 to 4 is 4 - 1 = 3
the integram of 3/x from 1 to 4 is 3ln4 - 3ln1 = 3ln4
finally, the integral is 3 + 3ln4

35. If the matrix M, then the determinant of M is

A. -18

B. -8

C. 0

D. 18

---

SECTION B: STATISTICS

36. The mean of the scores of 10 students in a physics evaluation is 54, and the mean of the scores of 40 different students in a mathematics evaluation is 60. The combined mean of the scores of the two subjects is:

A. 55.2

B. 58.8

C. 57.5

D. 57.0

Solution: Total scores for physics = 10 × 54 = 540
Total scores for mathematics = 40 × 60 = 2400
Combined mean = (540 + 2400) / (10 + 40) = 58.8
Answer: B. 58.8

37. If two events A and B are independent, then:

A. P(A ∩ B) = 0

B. P(A) = P(B)

C. P(A ∪ B) = 1

D. P(A ∩ B) = P(A)P(B)

If two events are independent, then the AND of their probabilities is the product of their indepenent probabilities

38. Consider the data below: 24, 19, 20, 12, 3, 14, 8, 9, 6, 5, 3
The upper quartile for the data is:

A. 24

B. 19

C. 9

D. 5

Step 1: Arrange the data in ascending order
3, 3, 5, 6, 8, 9, 12, 14, 19, 20, 24
There are 11 values in the ordered data set.

Step 2: Find the position of the upper quartile (Q3)
The formula for finding the position of Q3 (75th percentile) is:
Q3 = 3(n+1)/4 and substituting n = 11 gives Q3 = 9
The value at the 9th position is 19.

39. A random variable X is such that X ~ B(33, 0.4). Var(2X - 1) =

A. 30.68

B. 15.84

C. 31.68

D. 14.84

For a binomial distribution 𝑋 ∼ 𝐵(𝑛,𝑝)
X∼B(n,p), the variance is given by: Var(X)=np(1−p)
Var(X) = 33 x 0.4 x 0.6 = 7.92

Secondly. Var(aX + b) = a² Var(X) = 2² x 7.92 = 31.68

40. The number of calls received by a certain police station on a certain day is a random variable with a Poisson distribution, having a mean of 5 calls per hour. The probability that the police station will receive 10 calls per hour is:

A.

B.

C.

D.

X ∼ Poisson(λ = 5). The probability mass function (PMF) for a Poisson distribution is given by:
P(X = 10) = e^-λ λ^kk!

= e^-5 5¹⁰10!

41. A continuous random variable X has probability density function , defined by

The value of the constant k is:

A. 1/18

B. 2/27

C. 1/9

D. 1/27

42. A random variable X is normally distributed with mean 23 and variance 9. P(X > 20) =

A. Φ(-1)

B. Φ(1)

C. Φ(-1/3)

D. Φ(1/3)

43. A random sample of size 25 is drawn from a population with mean μ\muμ and variance 100. If the sample mean is 765, then the 95% confidence interval for the population mean is:

A. 765 ± 1.96

B. 765 ± 3.92

C. 765 ± 1.645

D. 765 ± 3.29

44. The data for 10 pairs of values of x and y are as follows: ∑x = 30, ∑y = 20, and ∑xy = 90. The covariance of x and y, Cov(x,y) =

A. 15

B. 4.5

C. 3

D. 6

45. The geometric mean of 70, 75, 80, 85, 90 is:

A. 79.69

B. 7969

C. 56.683

D. 56683

70 × 75 × 80 × 85 × 90 = 3213000000
geometric mean of the five values is the 5th root of 3213000000 = 79.69

46. Two events A and B are such that: P(A) = 1/4, P(B)=2/5 and P(A∩B)=1/5. Find P(B∣A′)

A. 4/15

B. 4/5

C. 1/12

D. 1/8

47. A true null hypothesis is rejected when it should have been accepted at a 2% level of significance. This is a:

A. type I error

B. type II error

C. 2% level of error

D. rando error

Rejecting a true null hypothesis is a Type I error.
A type I error (false-positive) occurs if an investigator rejects a null hypothesis that is actually true in the population; a type II error (false-negative) occurs if the investigator fails to reject a null hypothesis that is actually false in the population.

48. A discrete random variable X has the following probability distribution:

x	0	1	2
P(X = x)	1/8	6/8	1/8

E(2X + 5) =

A. 10

B. 2

C. 6

D. 7

E(X) = 0(1/8) + 1(6/8) + 2(1/8) = 1
E(2X + 5) = 2E(X) + 5 = 2 x 1 + 5 = 7

49. Two independent variables X and Y are such that Var(X) = 15/2 and Var(Y) = 4. Find Var(2X − Y) =

A. 11

B. 26

C. 34

D. 19

50. For four pairs of rankings, ∑d² = 2. The Spearman's coefficient of rank correlation is:

A. 4/5

B. 19/30

C. 1/2

D. 1/5

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