2001 Higher Maths Non-Calculator PAPER 1 ( Non-Calc. )

Transcription

1 001 PAPER 1 ( Non-Calc. ) 1

2 1) Find the equation of the straight line which is parallel to the line with equation x + 3y = 5 and which passes through the point (, 1). Parallel lines have the same gradient. x + 3y = 5 3y = x + 5 y = x The parallel line will have gradient and passes through the point (, 1). 3 y b = m (x a) y + 1 = (x ) 3 ( multiply by 3 both sides ) 3y + 3 = (x ) 3y + 3 = x + 4 3y = x y = x + 1 3y + x = 1 ) For what value of k does the equation x 5x + (k + 6) = 0 have equal roots? For equal real roots b 4ac = 0 a = 1 ( 5) (4)(1)(k + 6) = 0 b = 5 5 4(k + 6) = 0 c = (k + 6) 5 4k + 4 = 0 1 4k = 0 1 = 4k 4k = 1 k = 1 4

3 3) (a) Roadmakers look along the tops of a set of T-rods to ensure that straight sections of road are being created. Relative to suitable axes he top left corners of the T-rods are the points A( 6, 10, ), B(, 1, 1) and C(6, 11, 5). A B C Determine whether or not the section of road ABC has been built in a straight line. (b) A further T-rod is placed such that D has co-ordinates (1, 4, 4). Show that DB is perpendicular to AB. A B C D (a) AB = b a BC = c b = 8 = = + 8 = = 6 = = 3 = AB = BC 3 AB and BC are parallel, in same direction and have common point B. A, B and C are collinear. The section of road ABC has been built in a straight line. 3

4 (b) Required to show that DB is perpendicular to AB. B A D cos ABD = BA. BD BA BD BA = a b BD = d b = 8 = = 8 + = = 6 = BA. BD = ( 6 3) + ( 9 3) + ( 3 3) = ( 18) + (7) + ( 9) = ( 7) + (7) = 0 cos ABD = BA. BD = 0 = 0 BA BD BA BD cos ABD = 0 cos 90 0 = 0 ABD = 90 0 DB is perpendicular to AB. A B 90 0 D 4

5 Additional BA = 6 BD = BA = ( 6) + ( 9) + ( 3) BA = (3) + ( 3) + (3) = = = 16 = 7 = 9 14 = 9 3 = 3 14 = 3 3 5

6 4) Given f(x) = x + x 8, express f(x) in the form (x + a) b. f(x) = x + x 8 Divide by = 1 f(x) = (x + 1) 9 = x + x = x + x 8 f(x) = (x + 1) 9 6

7 5) (a) Solve the equation sin x o cos x o = 0 in the interval 0 x 180. (b) The diagram shows parts of two trigonometric graphs, y = sin x o and y = cos x o. Use your solutions in (a) to write down the co-ordinates of the point P. y y = sin x o 90 P 180 x y = cos x o (a) sin x o cos x o = 0 sin x o = sin(x + x) o sin x o cos x o cos x o = 0 = sinx o cos x o + cos x o sin x o cos x o ( sin x o 1) = 0 = sin x o cos x o cos x o = 0 or sin x o 1 = 0 x o = 90 0, x o = 70 0 or sinx o = 1 sinx o = 1 x o = 30 0, x o = (180 30) 0 x o = sine is positive sin 30 0 = 1 sin (180 30) 0 = 1 in the 1 st and nd quadrants sin = 1 S T A C x Solutions x o = 30 0, 90 0,

8 (b) y y = sin x o x P y = cos x o P is a point of intersection of y = sin x o and y = cos x o sin x o = cos x o sin x o cos x o = 0 From (a) x o = 30 0, 90 0, From the above diagram, P has x co-ordinate Using y = cos x o y = cos y = S A T C cos 30 0 = 3 cosine is negative in the nd quadrant P is the point (150 0, 3 ) cos = cos (180 30) 0 = 3 8

9 P 6) A company spends x thousand pounds a year on advertising and this results in a profit of P thousand pounds. A mathematical model, illustrated in the diagram, suggests that P and x are related by P = 1x 3 x 4 for 0 x 1. Find the value of x which gives the maximum profit. o (1, 0 ) x P (x) = 36x 4x 3 = 4x (9 x) Set P (x) = 0 4x (9 x) = 0 4x = 0 or (9 x) = 0 x = 0 or 9 = x x = 9 Nature x p (x) 4(9 + 1) 0 4(8) 0 400( 1) = 4x (9 x) = 40 = 3 = 400 The point (0, 0) is a point of inflection (increasing). At x = 9 there is a maximum turning point. x = 9 gives the maximum profit. 9

10 7) Functions f(x) = sin x, g(x) = cos x and h(x) = x + π are defined on a suitable 4 set of real numbers. (a) Find expressions for: (i) f(h(x)); (ii) g(h(x)). (b) (i) Show that f(h(x)) = 1 sin x + 1 cos x. (ii) Find a similar expression for g(h(x)) and hence solve the equation f(h(x)) g(h(x)) = 1 for 0 x π. (a)(i) f(h(x)) = sin x + π (ii) g(h(x)) = cos x + π 4 4 (b)(i) f(h(x)) = sin x cos π + cos x sin π (ii) g(h(x)) = cos x cos π sin x sin π cos π = 1 sin π = π 4 a h 1 1 o f(h(x)) = 1 sin x + 1 cos x g(h(x)) = 1 cos x 1 sin x 10

11 Solve f(h(x)) g(h(x)) = 1 for 0 x π 1 sin x + 1 cos x 1 cos x 1 sin x = 1 1 sin x + 1 cos x 1 cos x + 1 sin x = 1 sin x = 1 (multiply by both sides) sin x = (divide by both sides) sin x = sin x = sin x = 1 π 4 a h 1 1 o π 0 sin π = 1 4 sin π π = sin 4π π = sin 3π = S T A C sin is positive in both the 1 st and nd quadrants x = π or x = 3π

12 8) Find x if 4 log x 6 log x 4 = 1 4 log x 6 log x 4 = 1 log x 6 4 log x 4 = 1 log x 6 4 = 1 4 log x (6 )(6 ) = 1 (4)(4) log x (36)(36) = 1 (4)(4) log x (9)(9) = 1 log x 81 = 1 x 1 = = 81 x = 81 1

13 9) The diagram shows the graphs of two quadratic functions y = f(x) and y = g(x). Both graphs have a minimum turning point at (3, ). Sketch the graph of y = f (x) and on the same diagram sketch the graph of y = g (x). y O (3, ) y = f(x) y = g(x) x The graph of y = f(x) has negative gradient as it approaches x = 3 from the left. At x = 3 the gradient of y = f(x) is 0. The graph of y = f(x) has positive gradient after x = 3. The same is true for y = g(x). However y = g(x) has shallower negative gradient as it approaches x = 3 from the left and shallower positive gradient after x = 3. At x = 3 the gradient of y = g(x) is also 0. y y = f (x) y = g (x) O (3, 0) x 13

14 10) The diagram shows a sketch of part of the graph of y = log (x). y y = log (x) (a) State the values of a and b. (b) Sketch the graph of y = log (x + 1) 3 O (a, 0) (8, b) x (a) y = log (x) y = log (x) (a, 0) (8, b) y = 0 x = a y = b x = 8 0 = log (a) b = log (8) 0 = a b = 8 a = 1 ln b = ln 8 (a, 0) is the point (1, 0) b ln = ln 8 b = ln 8 ln b = 3 (8, b) is the point (8, 3) 14

15 (b) y = log (x + 1) 3 is the graph of y = log (x) moved 1 unit left and 3 units down. y = log (x) y = log (x + 1) y = log (x + 1) 3 [subtract 1 from the [subtract 3 from the x coordinates of y = log (x)] y coordinates of y = log (x + 1)] (1/4, ) ( 3/4, ) ( 3/4, 5) (1/, 1) ( 1/, 1) ( 1/, 4) (1, 0) (0, 0) (0, 3) (, 1) (1, 1) (1, ) (4, ) (3, ) (3, 1) (8, 3) (7, 3) (7, 0) y (0, 3) O (3, 1) (7, 0) x y = log (x + 1) 3 15

16 11) Circle P has equation x + y 8x 10y + 9 = 0. Circle Q has centre (, 1) and radius. (a) (i) Show that the radius of circle P is 4. (ii) Hence show that circles P and Q touch. (b) Find the equation of the tangent to circle Q at the point ( 4, 1). (c) The tangent in (b) intersects circle P in two points. Find the x coordinates of the points of intersection, expressing your answers in the form a ± b 3. (a)(i) Equ n of Circle P x + y 8x 10y + 9 = 0 Equ n of a circle x + y + gx + fy + c = 0 g = 4 C P = ( g, f ) = (4, 5) f = 5 c = 9 r P = g + f c = ( 4) + ( 5) 9 = = 3 = 16 r P = 4 (ii) Circle P has centre (4, 5) C P (4, 5) Circle Q has centre (, 1) C Q (, 1) If circles P and Q touch then the distance between the centres of the two circles will be the sum of the two radii C P C Q = r P + r Q = 4 + = 6 C P C Q = (x x 1 ) + (y y 1 ) C P (4, 5) C Q (, 1) = ( 4) + ( 1 5) = ( 6) + ( 6) = = 7 = 36 C P C Q = 6 16

17 C P C Q = r P + r Q = 6 circles P and Q touch (b) T ( 4, 1) radius tangent C Q (, 1) m rad m tan = 1 m rad = y y 1 C Q (, 1) x x 1 T ( 4, 1) = 1 ( 1) ( 4) ( ) = ( 4) + = ( ) m rad = 1 } m tan = 1 Equation of tangent T ( 4, 1) m tan = 1 y b = m (x a ) y 1 = 1 (x + 4 ) y 1 = x + 4 ( add 1 both sides ) y = x

18 (c) Required to find the x-coordinates of the points of intersection of x + y 8x 10y + 9 = 0 and y = x + 5 Substitute the line into the circle x + (x + 5) 8x 10(x + 5) + 9 = 0 x + x + 10x + 5 8x 10x = 0 x 8x 16 = 0 (x 4x 8) = 0 x 4x 8 = 0 Quadratic Formula x = b ± b 4ac a = 4 ± 16 (4)(1)( 8) ()(1) = 4 ± = 4 ± 48 = 4 ± ( 16)( 3) = 4 ± 4 3 = ± 3 x = + 3 or x = 3 18

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