Unit 1 Revision Questions

Unit 1 Outcome 1

A line passes through the points (1 , –6) and (–2 , 0) Find the equation of this line.

A line makes an angle of 68o with the positive direction of the x -axis, as shown in diagram. The scales on the axes are equal. Find the gradient of the line giving your answer correct to 3 significant figures. y From m = tan θ m = tan 68o 68o x m = 2∙48

A line L has equation y = 3x – 2 Write down the gradient of a line which is: (a) parallel to L (b) perpendicular to L. From y = mx + c m = 3 a) Parallel  m1 = m2 So m = 3 b) Perpendicular  m1 × m2 = – 1 So m = – 1/3

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A line passes through the points (–3 , 4) and (7 , –1) Find the equation of this line.

A line makes an angle of 52o with the positive direction of the x -axis, as shown in Diagram. The scales on the axes are equal. Find the gradient of the line giving your answer correct to 3 significant figures. y From m = tan θ m = tan 52o 52o x m = 1∙28

A line L has equation y = –2x + 5 Write down the gradient of a line which is: (a) parallel to L (b) perpendicular to L. From y = mx + c m = –2 a) Parallel  m1 = m2 So m = –2 b) Perpendicular  m1 × m2 = – 1 So m = 1/2

Unit 1 Outcome 2

The graph of a cubic y = f(x) is shown in diagram. On separate diagrams sketch the graphs of: (a) y = – f(x) (b) y = f( x + 1) y (–3 , 4) x –4 y y (–4 , 4) –4 x (–3 , –4) x –5 –1

The graphs with equations y = 2x and y = ax are shown in diagram. If the graph with equation y = ax passes through the point (1 , 6), find the value of a y y = ax y = 2x (1 , 6) (2 , 4) 1 x If graph passes through (1 , 6), y = 6 when x = 1 So 6 = a1 a = 6

The graphs of y = 7x and its inverse function are shown in the diagram. Write down the equation of the inverse function. y = 7x y 1 1 x Remember 23 = 8 → log28 = 3 7x → y = log7 x

Obtain an expression for f(g(x)). Functions f and g are defined on suitable domains by f(x) = x3 and g(x) = 3x – 2. Obtain an expression for f(g(x)). f(g(x)) = f(3x – 2) = (3x – 2)3

Now try this set of examples

The graph of a cubic, y = f(x) is shown in diagram. (0 , 2) 3 –2 The graph of a cubic, y = f(x) is shown in diagram. On separate diagrams sketch the graphs of: (a) y = – f(x) (b) y = f( x + 2) x y 3 –2 x y (-2 , 2) (0 , -2) –4 1

The graphs with equations y = 2x and y = ax are shown in diagram. If the graph with equation y = ax passes through the point (1,8), find the value of a y y = ax y = 2x (1 , 8) (3 , 8) 1 x If graph passes through (1 , 8), y = 8 when x = 1 So 8 = a1 a = 8

The graphs of y = 4x and its inverse function are shown in the diagram. Write down the equation of the inverse function. y = 4x y 1 1 x Remember 23 = 8 → log28 = 3 4x → y = log4x

Obtain an expression for f(g(x)). Functions f and g are defined on suitable domains by f(x) = x2 and g(x) = 2x – 3. Obtain an expression for f(g(x)). f(g(x)) = f(2x – 3) = (2x – 3)2

Unit 1 Outcome 3

When x = 2, mtan = – 3

A curve has equation y = 1/3 x3 – 3x2 – 7x + 1 A curve has equation y = 1/3 x3 – 3x2 – 7x + 1. Find the coordinates of the turning points and determine their nature. + – + x = –1  y = 102/3 ( –1 , 102/3) Max ( 7 , –802/3) Min x = 7  y = –802/3

Now try this set examples

When x = 3, mtan = 11

A curve has equation y = 1/3 x3 – 4x2 + 12x – 5 A curve has equation y = 1/3 x3 – 4x2 + 12x – 5. Find the coordinates of the turning points and determine their nature. + – + x = 2  y = 52/3 ( 2 , 52/3) Max ( 6 , –5) Min x = 6  y = –5

Unit 1 Outcome 4

In a pond, 30% of the existing tadpoles are eaten by predators each day but during the night 900 tadpoles are hatched. There are un tadpoles at the start of a particular day. (a) Write down a recurrence relation for un+1 the number of tadpoles at the start of the next day. (b) It is known that the pond cannot sustain more than 2500 tadpoles at any one time. (i) Find the limit of the sequence generated by this recurrence relation as n → ∞. (ii) In the long term, can the pond sustain the number of tadpoles? 30% die  70% left un+1= 0∙7un + 900 In the long term there will be 3000 tadpoles, too many to be sustained in the pond

Now try this example

(b) If the amount of plant food in the container exceeds 8g, the A gardener is using a new plant food on a particular shrub grown in a container. He applies 4g at the first feed and then repeats this every week. Each week, the amount of food left in the container drops by 52%. Let un be the amount of plant food in the container immediately after a feed. (a) Write down a recurrence relation for un+1 the amount of plant food in the container immediately after the next feed. (b) If the amount of plant food in the container exceeds 8g, the shrub will be damaged. (i) Find the limit of the sequence generated by this recurrence relation as n → ∞. (ii) Can the plant food be applied at this rate in the long term without damaging the shrub?

52% drop  48% left un+1= 0∙48un + 4 In the long term there will be 7∙69 g plant food in the container, so it is safe to use in the long term

If lines are concurrent then ( 1 , – 1) also lies on 4x – 5y = 9 Are the lines with equations 2x – y = 3,  3x + y = 2 and 4x – 5y = 9 concurrent? (1) If lines are concurrent then ( 1 , – 1) also lies on 4x – 5y = 9 2x – y = 3 3x + y = 2 (2) Add (1&2) 4 × 1 – 5 × (– 1) 5x = 5 = 4 + 5 x = 1 Put x = 1 in (1) = 9 2 – y = 3 So lines are concurrent y = – 1 Lines intersect at ( 1 , – 1)

Hence lines Perpendicular    a) Find the equation of the line : i)    with gradient ¾ passing through (-4 , 1) ii)   passing through (7 , 3) and (10 , -1)  b) Prove the two lines in a) are perpendicular, and find their point of intersection. ¾ Perpendicular lines m1× m2 = –1 y – b = m(x – a) 1 –4 -4/3 y – 1 = ¾(x + 4) y – b = m(x – a) 3/4 × –4/3 = –1 3 7 4y – 4 = 3(x + 4) y – 3 = -4/3(x – 7) 4y – 4 = 3x + 12 3y – 9 = –4x + 28 Hence lines Perpendicular 3y + 4x = 37 4y – 3x = 16

Solving (1) & (2) gives intersection as (5 , 3) In triangle STU, S is (2 , 6), T(4 , –2) and U(13 , 7).  Find the coordinates of the point of intersection of the altitude SF and median UE. y U(13 , 7) S(2 , 6) E F x T(4 , –2) Perpendicular m1× m2 = –1 (2) Solving (1) & (2) gives intersection as (5 , 3) (1)

Use y = mx + c to find gradient of 3x – 2y = 1 Find the equation of the line through the point (3, 2) which is perpendicular to the line with equation 3x – 2y = 1 Use y = mx + c to find gradient of 3x – 2y = 1 Perpendicular m1× m2 = –1 m2 = –2/3 –2/3 – 2y = –3x + 1 y – b = m(x – a) – 2y = –3x + 1 2 3 y – 2 = – 2/3(x + 4) y = 3/2x – 1/2 3y – 6 = – 2x – 8 m1 = 3/2 2x + 3y = – 2

Find the gradient of the line OB in the diagram given the equation of OA is x – 2y = 0. x – 2y = 0  2y = x y y = 1/2 x A m = 1/2 30o ao x m = tan θ O tan ao = 1/2 OB meets x-axis at 56∙6o a = tan-1 1/2 m = tan 56∙6o a = 26∙6o m = 1∙52

Find the equation of the line which is parallel to the line 2x + 5y = 7 and passing through the point (2 , –1) Use y = mx + c to find m 5y = –2x + 7 y = – 2/5x + 7/5  m = – 2/5 For parallel lines m1 = m2 y – b = m(x – a) y – 2 = – 2/5(x + 1) 2 – 2/5 – 1 5y – 10 = – 2(x + 1) 5y – 10 = – 2x – 2 2x + 5y = 8

Given f(x) = x2 + 2x – 8, express f(x) in the form (x + a)2 – b. [ ] = [ (x + 1)2 ] – 8 –1 = (x + 1)2 – 9

Find the values of a and b and sketch the graph of y = log2(x – 1) – 3 b = log28  b = 3 ( 8 , b) y = log2 x 23 = 8 a = 1 (a , 0) ( 9 , 0) (2 , –3) Find the values of a and b and sketch the graph of y = log2(x – 1) – 3

A sequence is defined by the recurrence relation Functions f and g are given by f(x) = 2x – 3 and g(x) = x2. Find an expression for g(f(x)). g(f(x)) = g(2x – 3) = (2x – 3)2 = 4x2 – 12x + 9 A sequence is defined by the recurrence relation un+1 = 3un – 4, u0 = –1. What is the value of u2? u1 = 3u0 – 4 u2 = 3u1 – 4 u1 = –3 – 4 = –7 u2 = 3×(–7) – 4 u2 = –25

Part of the graph of y = f(x) is shown in the diagram. Sketch the graph of y = –3 + f(–x) y 3 f(–x) flips graph horizontally. – 3 moves graph 3 down. (0 , 0) x –2 1 Applying changes to points x → –x, y → y – 3 (–1 , –3) (–2 , 0) → (2 , –3) (2 , –3) (0 , 3) → (0 , 0) (1 , 0) → (–1 , –3)

A minimum at (4, 5) B maximum at (4, 5) Which of the following describes the stationary point on the curve with equation y = 3(x – 4)2 – 5? A minimum at (4, 5) B maximum at (4, 5) C minimum at (4, –5) D maximum at (4, –5) y = 3(x – 4)2 – 5  x2 term positive so ‘HAPPY’ C If 2x2 – 12x + 11 is expressed in the form 2(x – b)2 + c, what is the value of c? 2x2 – 12x + 11 = 2[x2 – 6x] + 11 = 2[(x – 3)2 – 9 ] + 11 = 2(x – 3)2 – 18 + 11 = 2(x – 3)2 – 7 c = –7

π/4 π/2 One complete cycle between 0 and π/2 (90o) Centre line is 1 The graph shown in the diagram has equation of the form y = sin(px) + q. What are the values of p and q? y 2 1 x π/4 π/2 One complete cycle between 0 and π/2 (90o) Centre line is 1 q = 1 So four cycles between 0 and 2π (360o) p = 4

(x + 1)2 can never be negative so g is never decreasing. Which one of the following is true for the function g where g′(x) = x2 + 2x + 1? A g is never increasing. B g is decreasing then increasing. C g is increasing then decreasing. D g is never decreasing. g′(x) = x2 + 2x + 1 g′(x) = (x + 1)(x + 1) (x + 1)2 can never be negative so g is never decreasing. g′(x) = (x + 1)2

The curve passes through the point (–5, 0). Sketch the curve. Find the stationary points on the curve with equation y = x3 + 3x2 – 9x + 5 and justify their nature. The curve passes through the point (–5, 0). Sketch the curve.

Find the stationary points on the curve with equation y = x3 + 3x2 – 9x + 5 and justify their nature. + – + ( –3 , –22) Max ( 1 , 0) Min