2025 NECO GCE Further Mathematics Past Questions and Answers |Practice Guide

Introduction

Many students preparing for the exam are searching for the 2025 NECO GCE Further Mathematics past questions and answers, including NECO GCE Further Mathematics questions and answers 2025, as well as reliable NECO GCE Further Mathematics practice questions to boost their exam preparation.Studying 2025 NECO GCE Further Mathematics Past Questions and Answers is an essential way to prepare. It not only helps students understand the exam structure but also builds problem-solving speed and confidence.

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This guide provides practice questions, worked solutions, and exam tips tailored for the 2025 NECO GCE Further Mathematics examination.

Importance of Practicing NECO GCE Further Mathematics Questions

1. Identify Common Topics – You’ll discover areas like calculus, vectors, algebra, and mechanics that frequently appear.

2. Boost Confidence – Regular practice improves accuracy and reduces exam fear.

3. Time Management – The exam is time-sensitive, and practice helps improve speed.

4. Exam Readiness – Understanding the marking style prepares you for answering in a structured way.

Structure of NECO GCE Further Mathematics Paper

• Paper I (Objective/Multiple Choice): Tests wide knowledge across various topics.

• Paper II (Essay/Problem-Solving): Requires detailed solutions with correct steps.

NECO GCE Further Mathematics 2025 Practice Questions and Answers

Section A: Objective Questions (Sample)

1. If f(x)=2×2+3x−5f(x) = 2x^2 + 3x – 5f(x)=2x2+3x−5, find $f(-2)$.
A. –7
B. –3
C. 1
D. 3

Answer:
f(−2)=2(−2)2+3(−2)−5=8−6−5=−3f(-2) = 2(-2)^2 + 3(-2) – 5 = 8 – 6 – 5 = -3f(−2)=2(−2)2+3(−2)−5=8−6−5=−3
Correct Option: B

2. Evaluate ∫02(3×2+2x)dx\int_0^2 (3x^2 + 2x) dx∫02​(3x2+2x)dx.
A. 12
B. 16
C. 20
D. 24

Answer:
∫(3×2+2x)dx=x3+x2\int (3x^2 + 2x) dx = x^3 + x^2∫(3x2+2x)dx=x3+x2
Substitute limits: (23+22)−(03+02)=(8+4)−0=12(2^3 + 2^2) – (0^3 + 0^2) = (8 + 4) – 0 = 12(23+22)−(03+02)=(8+4)−0=12
Correct Option: A

3. Solve for $x$ if log⁡10(x)+log⁡10(2)=3\log_{10}(x) + \log_{10}(2) = 3log10​(x)+log10​(2)=3.
A. 50
B. 100
C. 200
D. 500

Answer:
log⁡10(2x)=3\log_{10}(2x) = 3log10​(2x)=3 → 2x=103=10002x = 10^3 = 10002x=103=1000 → $x=500$
Correct Option: D

4. The vector a⃗=2i^+3j^\vec{a} = 2\hat{i} + 3\hat{j}a=2i^+3j^​, b⃗=i^−j^\vec{b} = \hat{i} – \hat{j}b=i^−j^​. Find a⃗⋅b⃗\vec{a} \cdot \vec{b}a⋅b.
A. –1
B. 2
C. –3
D. 5

Answer:
a⃗⋅b⃗=(2)(1)+(3)(−1)=2−3=−1\vec{a} \cdot \vec{b} = (2)(1) + (3)(-1) = 2 – 3 = -1a⋅b=(2)(1)+(3)(−1)=2−3=−1
Correct Option: A

5. Differentiate y=x3−4x+1y = x^3 – 4x + 1y=x3−4x+1.
A. $2x-4$
B. 3×2−43x^2 – 43x2−4
C. 3×2+43x^2 + 43x2+4
D. 2×2−42x^2 – 42x2−4

Answer:
dydx=3×2−4\frac{dy}{dx} = 3x^2 – 4dxdy​=3x2−4
Correct Option: B

Section B: Essay/Problem-Solving Questions (Sample)

Question 1:
If a particle moves in a straight line with displacement s=2t2+3t+1s = 2t^2 + 3t + 1s=2t2+3t+1, find its velocity and acceleration at $t=2$.

Solution:

• Velocity v=dsdt=4t+3v = \frac{ds}{dt} = 4t + 3v=dtds​=4t+3.
  At $t=2$: $v=4(2)+3=11\text{m/s}$.

• Acceleration a=dvdt=4a = \frac{dv}{dt} = 4a=dtdv​=4.
  At $t=2$: a=4 m/s2a = 4 \, \text{m/s}^2a=4m/s2.

Final Answer: Velocity = 11 m/s, Acceleration = 4 m/s².

Question 2:
Solve the quadratic equation 2×2−3x−5=02x^2 – 3x – 5 = 02x2−3x−5=0.

Solution:
Using the quadratic formula:
x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}x=2a−b±b2−4ac​​
Here, $a=2,b=-3,c=-5$.

x=−(−3)±(−3)2−4(2)(−5)2(2)x = \frac{-(-3) \pm \sqrt{(-3)^2 – 4(2)(-5)}}{2(2)}x=2(2)−(−3)±(−3)2−4(2)(−5)​​
x=3±9+404x = \frac{3 \pm \sqrt{9 + 40}}{4}x=43±9+40​​
x=3±494x = \frac{3 \pm \sqrt{49}}{4}x=43±49​​
x=3±74x = \frac{3 \pm 7}{4}x=43±7​

So, x=104=2.5x = \frac{10}{4} = 2.5x=410​=2.5 or x=−44=−1x = \frac{-4}{4} = -1x=4−4​=−1.

Final Answer: $x=2.5$ or $x=-1$.

Question 3:
Find the derivative of $y=\sin x\cdot \cos x$.

Solution:
Using product rule:
dydx=(cos⁡x)(cos⁡x)+(sin⁡x)(−sin⁡x)\frac{dy}{dx} = (\cos x)(\cos x) + (\sin x)(- \sin x)dxdy​=(cosx)(cosx)+(sinx)(−sinx)
dydx=cos⁡2x−sin⁡2x\frac{dy}{dx} = \cos^2x – \sin^2xdxdy​=cos2x−sin2x

Final Answer: dydx=cos⁡2x−sin⁡2x\frac{dy}{dx} = \cos^2x – \sin^2xdxdy​=cos2x−sin2x.

Question 4:
Evaluate $\int (2x+5)dx$.

Solution:
∫(2x+5)dx=x2+5x+C\int (2x + 5) dx = x^2 + 5x + C∫(2x+5)dx=x2+5x+C, where $C$ is the constant of integration.

Final Answer: x2+5x+Cx^2 + 5x + Cx2+5x+C.

Tips to Excel in NECO GCE Further Mathematics 2025

1. Master the Basics: Algebra, trigonometry, and calculus form the foundation.

2. Practice Daily: Solve at least 5–10 questions per topic daily.

3. Use NECO Past Questions: This helps identify frequently tested areas.

4. Show Working Steps: In Paper II, marks are awarded for correct methods, not just the final answer.

5. Manage Time Wisely: Attempt easier questions first before tackling tougher ones.

Final Thoughts

The 2025 NECO GCE Further Mathematics practice questions and answers provided here are designed to help students prepare with confidence. With constant practice, mastery of core concepts, and proper exam techniques, candidates can perform excellently in the exam.

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Disclaimer

This material is for educational and revision purposes only. It is not intended to encourage examination malpractice. Students are advised to use it responsibly in preparation for the NECO GCE 2025 Further Mathematics exam.