Mathematics Grade 11 P2 Free State June 2025 Question Paper

Testpapers.co.za education Department of Education FREE STATE PROVINCE MARKS: 100 TIME: 2 HOURS This question paper consists of 10 pages and | information sheet. Copyright reserved Please turn over

Testpapers.co.za FSDoE/June 2025 INSTRUCTIONS AND INFORMATION Read the following instructions carefully before answering the questions. i= This question paper consists of 7 questions. RP: Answer ALL the questions in the SPECIAL ANSWER BOOK provided. 3 Clearly show ALL calculations, diagrams, graphs, etc. which you have used in determining your answers. 4, Answers only will NOT necessarily be awarded full marks. 5. You may use an approved scientifi calculator (non-programmable and non- graphical), unless 6. If necessary, ro i sorrect to TWO decimal places, unless stated otherwise. Te Diagrams are NO drawn to scale. 8. An information s rmulae is included at the end of the question paper.

Testpapers.co.za FSDoE/June 2025 QUESTION 1 (s ;— 1) are vertices of ACAB. E is the py =—3x+7. > x \" 4 \aP’norephysics.cum 1.1 Determine the length of AC. (2) 1.2. Determine the values of s and tif M (—3 ; 4) is the midpoint of BC. (3) 1.3. Prove that ACAB is a right-angled triangle. (4) 1.4 Determine the equation of a line passing through C and parallel to AB. (4) 1.5 Calculate the size of BAE . (6) 1.6 Determine p if CA is extended to D (p ; 1) such that C, A and D are (3) collinear. [22]

Testpapers.co.za FSDoE/June 2025 u \drilateral having AD || BC. BA produced has an x-intercept at E. BD and \ tersect at M. N(t; 0) isa point on AD. F.afimorephisies.com D(x; y) 2.1 Calculate the gradient of BC. (2) 2.2 Determine the equation of AD. (3) 2.3. Determine the value of ¢. (2) 2.4 Determine the coordinates of D if ABCD is a parallelogram. (3) 2.5. Determine the equation of the perpendicular bisector of AD (4) [14]

Testpapers.co.za FSDoE/June 2025 QUESTION 3 3. 4£7 sin A—3 =0 and 90° < A < 270°, determine the following without using a calculator and with the aid of a sketch: aah 1 cos A (4) sinA B42 tan A (2) 3.2 If cos26° =k , express the following in terms of k. 3.2.1 sin 26° (2) 3.2.2 (2) [10] QUESTION 4 4.1 Simply without usin tan 330° -sin 120° (5) 4.2 Simplify the following to-ofte trigonometric ratio : sin(180° — @) cos(—@) cos 25° sin(90° + 6)sin 425° (5) sin 0 — tan 0.cos’ 0 Prove the following identity: —————————- = tan0 43 E 7 cosO—1+sin’ 0 (4) 4.4 Determine the general solution of the following equation: 6cos’ x-7cosx-3=0 (6) [20]

Testpapers.co.za FSDoE/June 2025 QUESTION 5 5.1 Complete the theorem: The perpendicular drawn from the centre of the circle to the ford .......... the chord (1) 5.2.— In the diagram below, O is the centre of the circle. AB is a chord, and OM is drawn such that OM | AB. Radius OA = 4 units and OM = 3 units. Calculate the length of AB, giving reasons. (4) [5]

Testpapers.co.za FSDoE/June 2025 QUESTION 6 6. iagram below ‘ t states that: entre of the circle is »f the circle (5)

ce) Testpapers.co.za FSDoE/June 2025 [ | D Nii umference of the circle and TQ is a diameter of the circle. Chord PT and radius OS are drawn. B, = 70° Lon) Determine, with reasons, the size of the following angles: 6.2.1 R (2) 6.2.2 B (2) 6.2.3 6, (2) [11]

Testpapers.co.za FSDoE/June 2025 animorephysics.com Determine, with reasons, the size of: 711. ic (2) 7.1.2 Bz (2) 713 A (2)

2 Testpapers.co.za FSDoE/June 2025 In the diagram, LK and LM are tangents to the circle at K and M respectively. Q is a point on the circle and P is a point on chord QK such that PL || QM. KM and PM are drawn. PL intersects KM at N. Let Ry =m : am > merepl YSIES.COr1 7.2.1 Write down with reasons, THREE other angles EACH equal to x. (6) 7.2.2 Prove, giving reasons, that: (a) PKLM is a cyclic quadrilateral (2) (b) AQMP is isosceles (4) [18] TOTAL 100 10

Testpapers.co.za FSDoE/June 2025 INFORMATION SHEET ganz bst vb" -4ac 2a A= P(1+ni) A= P(l1—ni) A=P(1-i)" A=P(1+i)" T, =a+(n-l)d S, =F a+ (n—nd] fsar™* Ss _ ar’ -1) sr#l §, =; -I<r<l " r-1 l=? m=tan 0 (x-a) +(y—b) =r? In AdBC: 4_=_' _=_© sind sinB sinC a =b? +c? —2be.cos A area AABC = Sabin sin(a+ £)=sina.cos B+cosa.sin B sin(a—)=sina.cos B—cosa.sin B cos(@+f)=cosa.cos #—sina.sin B cos(a@—£)=cosa.cos f+ sin@.sin 2 cos’ asin’? a cos 2a =41—2sin? a sin 2a, = 2sina.cosa 2cos*a@-1 pa? @-3? ¥=+t— a n n n(A) P(A) =7(S) P(A or B) = P(A) + P(B) — P(A and B) n dX -*y-35) S(x—xP pH=atbx b= 11