# NABTEB Further Mathematics Questions 2022 Objective and Theory Update

Filed in Exam on February 17, 2022

– NABTEB Further Mathematics Questions –

NABTEB Further Mathematics NABTEB Expo Questions is out now on our website. In this article, I will show you past NABTEB Further Mathematics random repeated questions for free. All you need to do is to stay focused and follow this guide.

The National Business and Technical Examinations Board was founded in 1992 to domesticate craft level examinations previously conducted by City & Guilds, Pittman’s, and the Royal Society of Arts throughout the United Kingdom in conformity with the principles of the National Policy on Education.

Its formation was the culmination of a 15-year evolutionary process in which FOUR Government Panels were established at various periods to assess the place and structure of public examinations in our educational system.

#### Mission

Develop as a globally recognised assessment body for craftsmen and technicians.

#### Vision

NABTEB is a vision-led, mission-driven public institution with the following vision: To be a Globally Recognised Assessment and Certification Body Preparing Candidates for the Workplace and Academic/Professional Excellence.

The management of the National Business and Technical Examinations Board (NABTEB) is yet to release the 2021 November/December NBC/NTC and ANBC/ANTC examinations results.

Candidates who intend to take part in the 2021 NABTEB GCE can learn how to check their results online.

NABTEB is one of the examination bodies set up by the Federal Government in 1992 to reduce the burden of conducting examinations, which involves a lot of technical and trade-related practicals.

The board conducts the National Technical Examination (NTC), National Business Certificate (NBC) and their respective advance level examinations (ANBC and ANTC).

### NABTEB Further Mathematics Questions and Answers

FURTHER MATHEMATICS OBJ:

11-20: ACBABDBDAB
21-30: BDDBCBACCD
31-40: ACBDACDBAB

#### FURTHER MATHEMATICS THEORY:

1a)
8m + 2³m = 1/4
2²m + 2²m = 2-²
4³m = 2-²
Equate the base
6m = – 2
m = -2/4 = -1/3

1b)
log15 base 4 = x
x = log15 base/log4 base 10 = 1.1761/0.6020
= 1.9536
therefore x approximately 1.954

2a)
(-2) = m(-2)² + n(-2)+2=0
4m-2n = – – – – – – – – (1)
F(1) = m(1)²+n(1)+2=3
m+n+2=3
m+n=3-2
m+n=1 – – – – – – – – – – – -(2)
X Equate (2) by 4
4m+4n=4 – – – – – – – – – – (3)
4m-2n= -2
-4m+4n=4/-6n = – 6
n=1

4)
Given y= 5/x² + 3
Y = 5(x² +3)-1
dy/dx = anxn-1
y = 5(x² + 3)-1
dy = 5(-1) (x2 +3)-1^1
dy/dx = 10x (x² + 3)-²
dy/dx = – 10/(x² +3)²

5a)
^nP5 ÷ ^nC4 = 24
n!/(n-5)! ÷ n!/(n-4)!4! =24
n!/(n-5)! * (n-4)!4!/n! =24
(n-4)!4!/(n-5)!=24
(n-4)(n-5)!4!/(n-5)! = 24
n-4=1
n=4+1
n=5

b)PR= 5C3 (1/6)³ (5/6)^5-3
5!/(5-3)!3! (1/6)³ (5/6)²
5!/2!3! (1/6)³ (5/6)²
10 (1/216) (25/36)
=0.03215
Pr = 1-0.03215
=0.9678

6) make a table containing
MARKS,TALLY,F & FX
UNDER F – 2,9,4,2,2,1
UNDER FX – 2,18,12,8,10,6
€F= 20
€FX = 56

6b) Mean = €fx / €f
= 56 / 20
= 2.8

7)
Given
h=15.4t-4.9t
Velocity V=dh/dt =15.4-9.8t
At maximum height V=0
Therefore 15.4-9.8t=0
9.8t=15.4
t=15.4/9.8
t=1.6secs
Time to reach maximum height is 1.6secs

7b)
maximum h=15.4t-9.4t^2
15.4(1.6)-4.9(1.6)^2
22.64-12.544
=12.096
Max height = 12.1m

9a)X+6/(x+1)^2 = A/x+1 + B/(x+1)² + C/(x+1)³
X+6/ ~(x+1)³~ =A(x+1)² + B(x+1) + C / ~(x+1)³~
X+6 =A (x+1)² + B(x+1) + C
Let X+1=0 , X=-1
-1+6= A (-1+1)² + B(-1+1) +C
5= C
Therefore:- C=5

X+6= A(X2+2x+1) + Bx + B + C
X+6= Ax²+2Ax+A+Bx+B+C
comparing the coefficient of X²
A=0
Comparing the coefficient of X
1=2A+B
1=2(0)+B
B=1
x+6/(x+3)² = o/x+1 + 1/(x+1)² + 5(x+1)³
1/(x+1)² + 5/(x+1)³

10a)
3x^2+x-2 <= 0 3x^2+3x-2x-2 <= 0 3x(x+1) -2 (x+1) <= 0 (3x-2)(x+1) <= 0
3x-2 <= 0 or 8+1 <= 0

DISCLAIMER! These are not real NABTEB Further Mathematics Questions, but likely repeated questions over the years to help candidates understand the nature of their examinations. Ensure to note every question provided on this page.

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