Exam Questions Second Term SS 2 Further Mathematics
Further Mathematics SS 2 Second Term Examination Questions
Section A: Objective Questions (Fill in the Blanks)
-
Given that x = -3 and y = -7, evaluate (x² – y) / (y² – x)
a) -1/11
b) 1/23
c) 4/13
d) 12/17 -
Solve the equation x² – 2x – 8 = 0
a) x = 0 or 12
b) x = -2 or 4
c) x = 2
d) x = 2 or 4 -
The conjugate of 2 + √3 is:
a) 2
b) √3
c) √3 + 2
d) 2 – √3 -
If f(x) = 10x – 3, find f(2)
a) 7
b) 20
c) 3
d) 17 -
Find the 4th term of an arithmetic progression (AP) where the first term is 2 and the common difference is 0.5
a) 0.5
b) 2.5
c) 3.5
d) 4 -
Cot θ can be expressed as:
a) 1/sin θ
b) 1/cos θ
c) 1/tan θ
d) tan θ -
Solve the inequality y + 9 > 5y – 7
a) y > 4
b) y > -4
c) y < 4
d) y < -4 -
Find the distance between the points P (1,6) and Q (5,3)
a) 6
b) 5
c) 4
d) 3 -
Find the sum of the first 12 terms of the arithmetic series 5 + 9 + 13 + …
a) 612
b) 702
c) 119
d) 324 -
Solve the simultaneous equations 3x – y = 23 and 2x + 5y = 4 for y
a) -20
b) -2
c) 2
d) 7 -
If 2ˣ = 4, find the value of x
a) 5
b) 10
c) 12
d) 9 -
Find the square root of 400
a) 20
b) 200
c) 40
d) 2000 -
If P = √(4 × 9 × 16), find the value of P
a) 24
b) 6
c) 48
d) 12 -
Express 27^(-2/3)
a) 9
b) 1/9
c) 0.9
d) 1 -
Simplify a⁶ × a⁻³
a) a³
b) a⁴
c) a⁵
d) a⁶ -
Find the 40th term of the arithmetic sequence 6, 11, 16, 21, …
a) 201
b) 200
c) 75
d) 150 -
Solve the equation (1/2)ˣ = 8 for x
a) 3
b) 6
c) -3
d) -6 -
Simplify log₁₀(√8) / log₁₀(108)
a) 1/3
b) 1/2
c) log₁₀(√2)
d) 1/3 log₁₀(√8) -
Solve (x + 2)(x + 7) = 0
a) x = 1 or 8
b) x = -2 or 7
c) x = -4 or 5
d) x = -3 or 6 -
The nth term of a sequence is given by (-1)ⁿ⁻² × 2ⁿ⁻¹. Find the sum of the second term
a) 3
b) 4
c) 2
d) 10 -
Simplify log₅(8) / log₅(√8)
a) -2
b) -1/2
c) 1/2
d) 2 -
Simplify (3√2 – 1)(3√2 + 1)
a) 17
b) 18
c) 1
d) 6 -
If E = MN / (S + N) and E = 75, M = 120, N = 5000, find S
a) 1000
b) 200
c) 3000
d) 4000 -
Simplify log₃(7) + log₃(15) – log₃(5)
a) log₃(19)
b) log₃(5)
c) 3
d) 1 -
Find the common ratio of the geometric sequence 6, 12, 24, 48, …
a) 6
b) 12
c) 2
d) 48 -
Divide 4x³ – 6x² + 2x + 7 by 2x – 3, find the quotient
a) 6x + 8
b) 2x² + 6x + 8
c) 2x² + 6x
d) 2x + 8 -
Find the sum of the roots of the equation 2x² + 3x – 9 = 0
a) -18
b) -6
c) 9/2
d) -3/7 -
Find the distance between A(3,2) and B(4,6)
a) 5
b) 4
c) 3
d) 2 -
Find the common ratio of the geometric sequence 36, 12, 4, 4/3, …
a) 3
b) 1/3
c) -3
d) -1/3 -
Find the equation of a straight line that passes through the points P(2, -3) and Q(-4, 2)
a) y = 5x + 3
b) y = -5/6x – 3
c) y = -5/6x + 2
d) y = 5/6x – 3
Section B: Theory Questions
-
Determine if AB is parallel or perpendicular to PQ in the following cases:
a) A(3,1), B(4,3), P(4,6), Q(5,8)
b) A(5,1), B(3,2), P(2,4), Q(5,6)
c) A(4,7), B(6,8), P(3,5), Q(5,6) -
Find the distance between the following pairs of points:
a) A(3,2) and B(4,6)
b) C(-1,3) and D(2,-7)
c) P(5,-8) and Q(-4,-2)
d) S(-3,-2) and T(1,-4) -
Find the equation of the straight line that passes through P(2,-3) and Q(-4,2).
-
Find the common ratio in the following exponential sequences:
a) 6, 12, 24, 48, …
b) 36, 12, 4, 4/3, …
c) 4, -8, 16, -32, …
d) 54, -18, 6, -2, …
Further Mathematics SS 2 Second Term Examination – Answers and Explanations
Below are the correct answers to the questions, along with step-by-step explanations in simple language.
Section A: Objective Questions
1. Given that x = -3 and y = -7, evaluate (x² – y) / (y² – x)
First, we calculate the values inside the expression:
(-3)² – (-7) = 9 + 7 = 16
(-7)² – (-3) = 49 + 3 = 52
Now, we divide:
16 / 52 = 4 / 13
✅ Answer: (C) 4/13
2. Solve the equation x² – 2x – 8 = 0
We factorize the equation:
(x – 4)(x + 2) = 0
Setting each bracket to zero:
x – 4 = 0 → x = 4
x + 2 = 0 → x = -2
✅ Answer: (B) x = -2 or 4
3. The conjugate of 2 + √3 is:
In mathematics, the conjugate of a + b√c is a – b√c.
So, the conjugate of 2 + √3 is 2 – √3.
✅ Answer: (D) 2 – √3
4. If f(x) = 10x – 3, find f(2).
Substituting x = 2 into the function:
f(2) = 10(2) – 3 = 20 – 3 = 17
✅ Answer: (D) 17
5. Find the 4th term of an arithmetic sequence where the first term is 2 and the common difference is 0.5
Using the formula for the nth term of an arithmetic sequence:
Tn = a + (n-1)d
T4 = 2 + (4-1) × 0.5
= 2 + 1.5 = 3.5
✅ Answer: (C) 3.5
6. Cot θ can be expressed as:
cotθ = cosθ / sinθ = 1 / tanθ
✅ Answer: (C) 1/tanθ
7. Solve the inequality y + 9 > 5y – 7
Rearrange the equation:
y + 9 – 5y > -7
-4y > -16
Dividing by -4 (remember to reverse the inequality sign):
y < 4
✅ Answer: (C) y < 4
8. Find the distance between P(1,6) and Q(5,3).
Using the distance formula:
d = √[(x₂ – x₁)² + (y₂ – y₁)²]
d = √[(5-1)² + (3-6)²]
d = √(16 + 9) = √25 = 5
✅ Answer: (B) 5
9. Find the sum of the first 12 terms of the arithmetic sequence 5, 9, 13, …
Using the sum formula:
Sn = (n/2) [2a + (n-1)d]
S12 = (12/2) [2(5) + (12-1) × 4]
= 6 (10 + 44) = 6 × 54 = 324
✅ Answer: (D) 324
10. Solve 3x – y = 23 and 2x + 5y = 4 for y.
Express y in terms of x:
y = 3x – 23
Substituting into the second equation:
2x + 5(3x – 23) = 4
2x + 15x – 115 = 4
17x = 119
x = 7
Substituting x into y = 3x – 23:
y = 3(7) – 23 = 21 – 23 = -2
✅ Answer: (B) -2
11. If 2^x = 4, find x.
Since 4 = 2², we rewrite:
2^x = 2²
x = 2
✅ Answer: (D) 2
12. Find the square root of 400.
√400 = 20
✅ Answer: (A) 20
13. If P = √(4 × 9 × 16), find P.
P = √576 = 24
✅ Answer: (A) 24
14. Express 27^(-2/3).
27 = 3³
(3³)^(-2/3) = 3^(-2) = 1/9
✅ Answer: (B) 1/9
15. Simplify a⁶ × a⁻³.
a⁶ × a⁻³ = a^(6-3) = a³
✅ Answer: (A) a³
16. Find the 40th term of the arithmetic sequence 6, 11, 16, 21, …
Tn = a + (n-1)d
T40 = 6 + (40-1) × 5
= 6 + 195 = 201
✅ Answer: (A) 201
17. Solve (1/2)^x = 8 for x.
(2^(-1))^x = 2³
2^(-x) = 2³
- x = 3
x = -3
✅ Answer: (C) -3
18. Simplify log₁₀(√8) / log₁₀(108).
= (1/2) log₁₀ 8 / log₁₀ 108
= (1/3) / 1 = 1/3
✅ Answer: (A) 1/3
19. Solve (x + 2)(x + 7) = 0.
x + 2 = 0 → x = -2
x + 7 = 0 → x = -7
✅ Answer: (B) x = -2 or -7
20. The nth term of a sequence is given by (-1)ⁿ⁻² × 2ⁿ⁻¹. Find the second term.
T₂ = (-1)^(2-2) × 2^(2-1) = (1) × (2) = 2
✅ Answer: (C) 2
This explanation provides clear, step-by-step solutions in simple language, making it easy for students and teachers to understand and apply these concepts in future problems.