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Mathematics Grade 11 P1 Free State June 2025 Question Paper

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Grade 11 Mathematics Paper 1 question paper from the Free State June Exam 2025, 7 pages. Read online or get the free PDF download, memo linked when available.

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Free State June Exam 2025

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Testpapers.co.za Downloaded from Stanmorephysics.com education Department of Education FREE STATE PROVINCE MARKS: 100 HOURS: 2 HOURS This question paper consists of 7 pages and a formula sheet Copyright reserved
Testpapers.co.za Mabewnoaded fram iStanmorephysics.com FSDoE/June 2025 INSTRUCTIONS AND INFORMATION Read the following instructions carefully before answering the questions. I. 2: This question paper consists of 6 questions. Answer all the questions. Answer ALL the questions in the SPECIL ANSWER BOOK provided. Clearly show ALL calculations, diagrams, graphs, etc. which you have used in determining your answers. yarded full marks. Answers only will NOT )grammable and non- tated. An approved scientii ji graphical) may be If necessary, answers | ff to TWO decimal places, unless stated otherwise. Diagrams are not necess ‘awn to scale. An information sheet with formulae is included at the end of the question paper. Write neatly and legibly. Copyright reserved
Testpapers.co.za Mabewmnoaded fram 1Stanmorephysics.com QUESTION 1 113 2. Art grt gros +e 2m gions 1.2 Prove that ¥2..32-4 QUESTION 2 2.1 Solve for x: (answer correct to two decimal places) 215. [eseavJe 2.2 Solve for n if dn =8 2.3 Solve for x and y: x +4y? =2xy+7 and 3+x=2y Copyright reserved FSDoE/June 2025 (2) (3) (3) (4) Q) [15] (2) (3) (3) (4) (5) (3) (6) [26]
Testpapers.co.za Mabewmnoaded fram 1Stanmorephysics.com FSDoE/June 2025 QUESTION3 3.1. Determine the nature of the roots of the following equation, without solving the equation: x° +5x-9=0. : (3) 3.2. The solutions of a quadratic equation are given by: x = (2) For which value(s) of k will the equation have two real and equé 3.3. Given: S (x) =ax’ +bx-te. has equal roots, name = (5) and it is further giver - ; termine the [10] QUESTION 4 imorephysics.com Given; g(x)=——+2 x— 4.1. Write down the equation of the asymptotes of g. (2) 4.2 Calculate the intercepts with the axes of g. (3) 4.3. Draw a graph of g, indicating any intercepts with the axes and the asymptotes. (3) 4.4 Determine the value(s) of x where g(x) >0. (2) 4.5 Determine the equation of the axis of symmetry of g which has a positive gradient. (2) 4.6 Write down the domain of h if h(x)=g(x+l). 2) [14] Copyright reserved
Testpapers.co.za Mahowneoaded fram iStanmorephysics.com FSDoE/June 2025 QUESTIONS The graph of h(x)=—x* +x+2and k(x)=4x—2 are drawn below. A and B are the x-intercepts of h. D and C are the y-intercepts of / and k respectively. P and Q are the points of intersection of and k. 5.1 Calculate the coordinates of A and B. (4) 5.2. Determine the length of CD. (3) 5.3 Determine the range of h(x). (4) 5.4 Determine the maximum value of airs (2) 5.5. Write down the coordinates of the turning point of f, if f is the reflection of / in the x axis. (2) Copyright reserved
Testpapers.co.za Mahowneoaded fram iStanmorephysics.com FSDoE/June 2025 5.6 If the coordinates of P and Q are (-4;—18) and (1; 2) respectively, determine for which values(s) of x is: 5.6.1 h(x)-k(x)>0? (2) 5.6.2 xh(x)<0. (2) 5.7 Write down the value(s) of n for which h(x) =n will have equal roots? (4) [23] QUESTION 6 The graph of g(x v. T (1;2) is the point on the graph. 6.1 Determine the equation of (3) 6.2 Write down the equation of the asymptote of g. (1) 6.3. Write down the equation of f if f is the graph of g shifted 2 units to the right. (1) 6.4 Calculate the average gradient of g between point T and the origin. (2) 6.5 If the domain of g(x) is restricted to be x €[-2;3], write down the range of g. (3) [10] TOTAL: 100 MARKS Copyright reserved
Testpapers.co.za Mabewnoaded fram iStanmorephysics.com FSDoE/June 2025 INFORMATION SHEET/INLIGTINGSBLAD _ =b£Nb? -4ac 2a A=P(l+ni) A=P(I-ni) A=P(1-i)" A= P(1+i)" T, =a+(n-l)d S, =F Ra+ (nna m=tan0 ab c In AABC: = a’ =b* +c? —2be.cos A sind sinB sinC area AABC = Sabin Cc cos @—sin* a cos2a ={1—2sin> @ sin 2a = 2sin a.cosa 2cos* a-l P YG, -37 xa oa os n n n(A) P(A) = n(S) P(A or B) = P(A) + P(B) — P(A and B) n peathx p= 2-0-9) DY@-»x) 6 Copyright reserved

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