# Mathematics Grade 12 KZN March 2015 Memo

> Grade 12 Mathematics memorandum from the KZN March 2015, 12 pages. Read online or get the free PDF download, question paper linked when available.

- Page: https://www.hlayiso.com/library/maths-march-qp-memo-2015-gr12/5e8bf37f-7ab1-46b8-b694-05bb773f3532
- Updated: 2026-09-27
- Subject: Mathematics
- Grade: Grade 12
- Document type: Memorandum
- Year: 2015
- Exam period: KZN March
- Curriculum: CAPS
- Pages: 12

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## About this document

Mathematics · Grade 12 · KZN March · 2015. Memorandum, 12 pages. Read online or download the PDF.

Read it free online on the page above; download the PDF from there.

## Document text

Text extracted from the PDF. Layout, diagrams and tables may not survive extraction.

### Page 1

Basic Education

KwaZulu-Natal Department of Basic Education
REPUBLIC OF SOUTH AFRICA

QP RR RS RSS ES ER GRR RSS SSSR SR RR RR SRE R RRR RR ERE SEE E SS
| foes _. MATHEMATICS ~ oe
Cae COMMON TEST | ‘

()

NATIONAL
SENIOR CERTIFICATE

Marks: 100

Time: 2 hours

N.B. This question paper consists of 6 pages and an information sheet.

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Mathematics arch 2015 Common Test
NSC

INSTRUCTIONS AND INFORMATION

Read the following instructions carefully before answering the questions.

lL. This question paper consists of 6 questions.

2. Answer ALL the questions.

3. Clearly show ALL calculations, diagrams, graphs, et cetera that you have used in
determining your answers.

4. Answers only will not necessarily be awarded full marks.

5. An approved scientific calculator (non-programmable and non-graphical) may be

used, unless stated otherwise.

6. If necessary, answers should be rounded off to TWO decimal places, unless stated ¢ )
otherwise.

7. Diagrams are NOT necessarily drawn to scale.

8. Number the answers correctly according to the numbering system used in this

question paper. Write neatly and legibly.

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Mathematics March 2015 Common Test
NSC
QUESTION 1
Given: g(x)=-x* + 4x? +11x—30
=~(x—2)(x—5)(x +3)
1.1 Write down the y ~ intercept of g. qd)
1,2 Write down the x ~ intercepts of g . (2)
1.3. Determine g’(x). QB)
14 Hence, or otherwise, calculate the coordinates of the turning points of g. (5)
1.5 Sketch a neat graph of g . Clearly indicating the intercepts with the axes and the
coordinates of the turning points. (3)
1.6 Determine the value(s) of x for which g is concave up. (3)
[17]
QUESTION 2
21 Given: f(x)=—-x? +x, determine /’(x) from first principles, (5)
2.2 Determine:-
2.2.1 Sit y Aer -3, (4)
2.2.2 f'(a) if £0) (4)
[13]
QUESTION 3
Given the equation of the curve p(x) =x° +3x? +4x-8
3.1 Determine the gradient of the tangent to p at x =—1. (2)
3.2. Determine the equation of the tangent to p at x =—1. (4)
3.3. Show that the function p is increasing for all real value(s) of x. “)
f11]

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NSC

Mathematics

QUESTION 4

A container shaped in the form of a cylinder with no top has a volume of 340 ml.
It has a radius of x cm and a height of A.cm.

Note: 1 ml =1 cm’.

4.1 Write down the height (#) in terms of x. (2)

4.2 Show that the surface area (S) of the cylinder with no top is given by S = TH? + 680 . (2)
x

4.3 Calculate the value of x for which the surface area of the cylinder will be a minimum. (5)

19]

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NSC

Mathematics

QUESTION 5

In the diagram below, tangent RP, having equation y — x — 2 = 0, touches a circle with
centre M(4 ;.4) at P and cuts the x-axis at R. PT is a diameter of the circle.

V is a point on another tangent to the circle that passes through T such that MV = 10.

Determine:

5.1 The equation of PT, the diameter of the circle.

5.2 The coordinates of P, the point of contact of tangent PR and citcle M.
5.3. The equation of the circle.

5.4 The coordinates of T .

5.5 Calculate the length of TV.

Copyright Reserved

March 2015 Common Test

(4)
(3)
(3)
(3)
(3)
[16]

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Mathematics rch 2015 Common Test
NSC
QUESTION 6
' 6.1 Make use of sin (x+y) = sin xcos y + cos x sin y to derive an expansion of
cos (x — y) in terms of sines and cosines of x and y. (4)
62  Ifcos 62° = m, express cos 2° in terms of m. (5)

6.3 Prove that:

sin2(360° — A) + cos(90° + 2A) +cos*(180°—A) _ cos A sin A

= 9
sin(90° — 2A) cos A + sin A )
' 64 Given: 4sinx+6sin x cos x—3 cos x = 2.
6.4.1 Show that 4sin x + 6sin xcos x —3cos x =2
can be written as (2 sin x — 1) (2 +3 cos x) = 0 (2)
6.4.2. Hence or otherwise find the general solution of
4sin x +6sin x cos x —-3.cos x = 2. (5)
6.5 Inthe figure below, D, B and C are three points in the same horizontal plane.
AD isa vertical tower, The angle of elevation from B to A is x.
In addition, AB = AC, ACB = y and BD=q.
A
B
6.5.1 Express AB in terms of q and a trigonometric ratio of x. Q)
6.5.2. Determine, with reasons, the size of BACin terms of ¥y. (3)
6.5.3 Hence, show that BC. = 2g 608 ¥ (4)
cos x
[34]
TOTAL MARKS: [100]

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NSC

Mathematics March 2015 Common Test

INFORMATION SHEET: MATHEMATICS
INLIGTINGSBLAD: WISKUNDE

xe -b4Vb? -4ac

2a
A=P(ltni) A= P(l=ni) A= P(l-iy’ A= P(L+i)"
Yen yi T, =a+(n-Nd 8, =" @a+(n—Da)
i=l fl
T,, =ar""' S _ a” =I ; r#] S, =—-;-1le<r<l
. r-1l l-r

poe 1+)’ -1 purl d+]

a t

P= tim LEAD-L@)
h->0 h
xX, +x. +
d=y(%)—¥)" + -94) my “2 1 Ye
2 2
y=mete yry, =m(x—-%x,) m=22 FL m=tand
2 7X]
(x-a) +(y—b) =r?
inaasce; =? _© a? =b? 467 —2be.cosA area ABC =Lab.sinC
sind sinB sinc 2
sin(a: + B) = sina@.cos # + cosa.sin 8 sin(a -p) = sina.cos 6 —cosa.sin B
cos(a+ f)= cos a.cos 8 —sin asin B cos(a— #)= cosa.cos # +sin a.sin B

cos’ a-sin’ a
cos 2a =41—2sin’ a sin 2a = 2sin a.cos @

2cos* a —1

(x; y) > (xcos 6 + ysin@; ycos 0 —xsind) (x3 y) > (xcosO— ysind ; ycosO +xsin 0)

x= o
n n
P(A) = WS) P(A or B) = P(A) + P(B)— P(A and B)
nh
peatbx b= LE-HO-H)
Vi@-x)

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