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Scholar Higher Mathematics Revision Session Thursday 18 th May 7:30pm You will need a pencil, paper and a calculator for some of the activities.

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Presentation on theme: "Scholar Higher Mathematics Revision Session Thursday 18 th May 7:30pm You will need a pencil, paper and a calculator for some of the activities."— Presentation transcript:

1 Scholar Higher Mathematics Revision Session Thursday 18 th May 7:30pm You will need a pencil, paper and a calculator for some of the activities

2 SCHOLAR online tutor for Maths and Author of the SCHOLAR National 5 & Higher courses Margaret Ferguson

3  Hints & Tips on the Exam  Optimisation  Logs & Exponentials  Vectors  Functions  Trigonometry  The Straight Line  Integration  The circle Tonight’s Session will cover You will need paper, pencil and a calculator for some of the activities.

4 The New Higher Maths Exam Paper I  Paper I is the non-calculator paper and lasts for 1 hour 10 minutes  There should be about 12 questions and the paper is worth 60 marks Paper II  Paper II is the calculator paper and lasts for 1 hour 30 minutes  There should be about 10 questions and the paper is worth 70 marks The Old Higher Maths Exam Paper I  Paper I is the non-calculator paper and lasts for 1 hour 30 minutes  Allow 50 mins (+5 mins) for part A, 20 objective questions (40 marks)  This leaves 35 minutes for the 3 questions in part B (30 marks) Paper II  Paper II is the calculator paper and lasts for 1 hour 30 minutes  There should be about 10 questions and the paper is worth 70 marks

5 PQRS is a rectangle formed according to the following conditions: it is bounded by the lines x = 6 and y = 12 P lies on the curve with equation y = between (1,8) and (4,2) R is the point (6,12) (a) Express the lengths of PS and RS in terms of x, the x-coordinate of P. PS =RS = 6 - x Hence show than the area, A units 2, of PQRS is given by A = PS x RS = (6 – x) = 72 = - 12x+ 8 ✓ ✓ ✓

6 PQRS is a rectangle formed according to the following conditions: it is bounded by the lines x = 6 and y = 12 P lies on the curve with equation y = between (1,8) and (4,2) R is the point (6,12) (b) Find the greatest and least values of A and the corresponding values of x for which they occur. What do you notice about the stationary points? x = -2 is outside our closed interval. ✓ ✓ ✓ ✓ How do we do this?

7 ✓ ✓ + 0 - When x = 1, A = 20 When x = 2, A = 32 When x = 4, A = 20 Hence the greatest area is 32 units 2 when x = 2 and the least area is 20 units 2 when x = 1 or 4. ✓ ✓ How do we determine the nature of the stationary point? shape x2 PQRS is a rectangle formed according to the following conditions: it is bounded by the lines x = 6 and y = 12 P lies on the curve with equation y = between (1,8) and (4,2) R is the point (6,12) Find the greatest and least values of A and the corresponding values of x for which they occur.

8 A0 B14 C26 D28 Vote for the correct answer now Vectors p and q are such that |p| = 3, |q| = 4 and p.q = 10. What is the value of q.(p + q)? q.(p + q) = q.p + q.q = 10 + 4 2 = 26 ✓ = p.q + q2q2

9 It is claimed that a wheel is made from wood which is over 1000 years old. To test the claim, carbon dating is used. The formula A(t) = A 0 e -0.000124t is used to determine the age of wood where : A 0 is the amount of carbon in any living tree A(t) is the amount of carbon in the wood being dated t is the age of the wood in years For the wheel it was found that A(t) was 88% of the amount of carbon in a living tree. Is the claim true? What is A(t) expressed in terms of A 0 ? A(t) = 0.88A 0 What is the formula expressed in terms of A 0 ? 0.88A 0 = A 0 e -0.000124t Simplify 0.88 = e -0.000124t How do we determine t? log e 0.88 = -0.000124t t = 1030.9 Hence the claim is true since 1030.9 > 1000. ✓ ✓ ✓ ✓

10 ABCDABCD What is the solution of the equation ? Vote for the correct answer now ✓ A S T C In which quadrant is our solution? x = x = 120° 60°

11 Two variables, x and y, are connected by the law y = a x. The graph of log 4 y against x is a straight line passing through the origin and the point A(6,3). Find the value of a. What is the gradient of the straight line? m = ½ What is the equation of the straight line? Y = ½x log 4 y = ½x Express y = a x in terms of log 4. log 4 y = log 4 a x log 4 y = x log 4 a What is log 4 a equal to?log 4 a = ½ 4 ½ = a So a = 2 ✓ ✓ ✓ ✓

12 The diagram shows a wire framework in the shape of a cuboid with edges parallel to the axes. Relative to those axes A, B, C and H have coordinates (1,3,4), (2,3,4), (2,7,4) and (1,7,9) respectively. State the lengths of AB, AD and AE. AB = 1 AD =4AE = 5 Write down the components of HB and HC and hence or otherwise calculate the size of angle BHC. ✓ HB = b - h HC = c - h |HB| = |HC| = HB. HC = 1 2 + (-4) x 0 + (-5) 2 = 26 = 38.1° ✓ ✓ ✓ ✓ ✓ ✓ ✓

13 ABCDABCD The diagram shows the line L. The angle between L and the positive direction of the x-axis is 135° as shown. What is the gradient of L? Vote for the correct answer now ✓ What is the associated acute angle? m L = tan 135° = - tan 45° = -1

14 Functions f, g and h are defined on suitable domains by f(x) = x 2 – x + 10, g(x) = 5 – x and h(x) = log 2 x. Find expressions for h(f(x)) and h(g(x)). h(f(x)) =h(x 2 – x + 10) = log 2 (x 2 – x + 10) h(g(x)) =h(5 – x) = log 2 (5 – x) Hence solve h(f(x)) - h(g(x)) = 3. h(f(x)) – h(g(x)) =log 2 (x 2 – x + 10) – log 2 (5 – x) So log 2 (x 2 – x + 10) – log 2 (5 – x) = 3 => 40 – 8x = x 2 – x + 10 0 = x 2 + 7x - 30 (x – 3)(x + 10) = 0 x = 3 or x = -10 ✓ ✓ ✓ ✓ ✓ ✓ ✓✓

15 The graphs of y = f(x) and y = g(x) are shown in the diagram. f(x) = -4cos(2x) + 3 and g(x) if of the form g(x) = mcos(nx). What are the values of m and n? m = n = 32 ✓ Find correct to 1 decimal place, the coordinates of the points of intersection of the two graphs in the interval 0 ≤ x ≤ π. -4cos(2x) + 3 = 3cos(2x) -7cos(2x) = -3 How do we proceed? 2x = 64.6° A S T C 2x = 295.4° x = 32.3° = 0.6 radians x = 147.7° = 2.6 radians ✓ g(32.3) = 1.3g(147.7) = 1.3 Hence the points of intersection are (0.6,1.3) and (2.6,1.3) ✓ ✓ ✓ ✓

16 The graphs of y = f(x) and y = g(x) are shown in the diagram. f(x) = -4cos(2x) + 3 and g(x) = 3cos(2x). Calculate the shaded area. How do we proceed? = 10.9 – (-1.5) = 12.4 ✓ ✓ ✓ ✓ ✓ ✓ This is why the angles must be in radians

17 Circle P has equation x 2 + y 2 – 8x – 10y + 9 = 0 Circle Q has centre (-2,-1) and radius. Do circles P and Q touch, giving a reason for your answer? What are the radius and the coordinates of the centre of circle P? Centre of circle P = (4,5) Radius of circle P = What is the distance between the centres of circles P and Q? What is the sum of the radii? Since the sum of the radii is equal to the distance between the centres, circles P and Q touch externally.

18 Question Time Work through the exam questions sequentially Use the paper provided like a book If you can’t do a question write down the question number and leave a good amount of space This will make it easier to come back to it when you have finished the paper Always answer all the questions that you can first If you have any questions about tonight’s session or the exam please ask Carol will provide a link for you to give us feedback


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