Problems (F/M): Part 2 - Solutions (16 pages; 29/4/17)

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1 Problems (F/M): Part 2 - Solutions (16 pages; 29/4/17) (11) Show that n ( n r=0 r ) = 2n Solution Method 1: Consider (1 + 1) n Method 2: Pascal's triangle The sum of each row is twice the sum of the previous one. eg = ( )[alternate terms] + ( ) = 2( ) = 2(1 + [4 + 6] + [4 + 1]) & = ( ) + ( ) = (1 + [5 + 10] + [10 + 5] + 1) +([1 + 5] + [ ] + [5 + 1]) Method 3: Counting ways of selecting any number of items 1st counting method: n ( n r=0 r ) 2nd counting method: For each object, there are 2 choices: include or exclude; giving 2 n [Note: 1 way of choosing no objects is included in the total.] Method 4: Induction If true for n = k, so that k ( k r=0 r ) = 2k, 1

2 then k+1 ( k + 1 r=0 r ) = ( k + 1 k + 1 ) + { r=1 (k 0 r )} + ( k + 1 k + 1 ) = 1 + k {( k r=1 r 1 ) + (k r )} + 1 k 1 = 1 + { ( k r 1 )} + [{ k r=0 (k r )} ( k r 1=0 0 )] + 1 k 1 = 1 + { ( k R=0 R )} + [2k 1] + 1 = 1 + { k ( k R=0 R )} (k k ) + 2k = k k = 2 k+1 (12) (i) For f(x) = ax 3 + bx 2 + cx + d, what is the x-coordinate of the point of inflexion? (ii) Give examples of cubic functions for which the PoI is at the Origin, and the gradient at the Origin is (a) 1 (b) 1. How do the shapes of the two graphs differ? Solution (i) f (x) = 3ax 2 + 2bx + c ; f (x) = 6ax + 2b f (x) = 0 x = b 3a (ii) As the graph passes through the Origin, we can consider f(x) = ax 3 + bx 2 + cx From (2), PoI is at x = b f (x) = 3ax 2 + c f (0) = 1 c = 1 3a, so that b = 0 2

3 eg y = 2x 3 + x = x(2x 2 + 1) f (0) = 1 c = 1 eg y = 2x 3 x = x(2x 2 1) Fig. 2: y = 2x 3 + x Fig. 3: y = 2x 3 x (13) ABC is a triangle circumscribed by a circle of radius R, as shown in the diagram below. Show that (i) a sina = 2R (ii) the area of the triangle is abc 4R 3

4 Solution (i) Drawing radii from B and C to the centre of the circle, as in the diagram below, and noting that the angle at the centre is twice the angle at the circumference, sin A = (a 2 ) R, so that a sina = 2R, as required (ii) Area of ABC = 1 2 bcsina = 1 2 a bc ( ) = abc 2R 4R (14) Find the equation of the circle passing through the points A (2,8), B (7,3) and D (5,7) Solution The first step is to find the centre of the circle, using the fact that the perpendicular bisector of each chord passes through the centre. The chord AB has mid-point (9/2, 11/2) and gradient = 1 The perpendicular bisector of AB therefore has equation y 11/2 x 9/2 = 1 1 2y 11 = 2x 9 4

5 y = x + 1 The chord BD has mid-point (6, 5) and gradient = 2 The perpendicular bisector of BD therefore has equation y 5 = 1 x 6 2 y = 1 2 x + 2 The centre of the circle C is then found from the intersection of these lines: x + 1 = 1 2 x + 2 so that x = 2 and y = 3 The radius is then the distance CA (for example) = (2 2) 2 + (3 8) 2 = 5 Hence the equation of the circle is (x 2) 2 + (y 3) 2 = 25 (Check: B and D satisfy the equation.) (15) Angle Bisector Theorem Referring to the diagram below, the Angle Bisector theorem says that BD = AB DC AC Prove the Angle Bisector Theorem. 5

6 Solution By the Sine rule for triangle ABD, BD sinθ = AB sinadb (1) and, for triangle ADC, DC sinθ = AC sinadc = AC sinadb (2) Then (1) so that BD = DC AB AC and hence sinθ = BD sinadb AB BD = AB DC AC sinθ and (2) = DC sinadb AC (16) Show that e 3 > 4e 3 2 Solution An equivalent result to prove is e 3 2 > 4 (dividing both sides by e 3 2, which is positive) [you can never be sure what counts as being obvious] e 3 > 16 (as the function y = x 2 is increasing for x > 0) e 3 > ( ) 3 > (2 2 )(0.7) = > 16, so that the original result is also true 6

7 (17) Prove that 1 dx = ln x for all x 0, x assuming that 1 dx = ln x for x > 0 x Solution Method 1 If 1 x dx = ln x for x > 0, then d For the case where x < 0: dx (lnx) = 1 x for x > 0 Let y = x, so that d dy (lny) = 1 y, as y > 0 [To convert back to xs: ] Hence d dx (lny). dx dy = 1 ( x) giving d dx (ln[ x])( 1) = 1 ( x) and so d dx (ln x ) = 1 x for x < 0 (*) and therefore 1 dx = ln x for x < 0, as well as x > 0 x [Note that the function y = ln x for x < 0 is the reflection in the y-axis of y = ln x (for x > 0), and therefore has a negative gradient, which agrees with (*).] Method 2 Referring to the diagram below, where u = x > 0 & c > 0, 7

8 x 1 c dt t = u c 1 t dt = (positive) area between graph and t-axis on LHS = (positive) area between graph and t-axis on RHS c = 1 u dt = t u 1 c t dt = lnu lnc x c As 1 dx only differs from 1 dt by an arbitrary constant, it x t follows that, when x < 0, 1 dx = ln u + C = ln x + C, as x required. (18) Linear Interpolation By approximating the graph of y = log 2 x by a straight line between x = 2 and x = 4, find an approximate value for log 2 ( 5 2 ) 8

9 Solution Approach 1: weighted average log 2 ( 5 2 ) ( ) log 22 + ( ) log 24 = (0.75)(1) + (0.25)(2) = 1.25 Approach 2: similar triangles Referring to the diagram below (for the general function f(x)) x = f(b) f(a) c a b a 9

10 For our example, x = 2 1, so that x = (0.5)(0.5) = 0.25, and hence log 2 ( 5 2 ) = 1.25 Approach 3: Equation of line The gradient of the line is ( f(b) f(a) ) b a Then f(c) f(a) + ( f(b) f(a) ) (c a) b a In this case, log 2 ( 5 ) (2 1 ) (2.5 2) = 1.25 again. 4 2 Also f(a) + ( f(b) f(a) ) (c a) b a = ( 1 ) ((b a)f(a) + (c a)f(b) (c a)f(a)) b a = ( 1 ) ((b c)f(a) + (c a)f(b)), b a which is the weighted average approach 10

11 (19) Cubics (i) What possible shapes might a cubic have (ignoring its position relative to the axes)? A, B & C have 2, 1 & 0 stationary points respectively. These are for cases where the coefficient of x 3 is positive; so inverted shapes are also possible. (ii) How many stationary points does the cubic function, f(x) = x 3 + x 2 2x + 3 have? f (x) = 3x 2 + 2x 2 To find the number of solutions to f (x) = 0, consider the discriminant: 2 2 4(3)( 2) > 0 Thus there are 2 stationary points. 11

12 (iii) What is the condition for there to be 2 stationary points for the general cubic f(x) = ax 3 + bx 2 + cx + d? 2 sol ns of f (x) = 3ax 2 + 2bx + c = 0 (2b) 2 4(3a)c > 0 b 2 3ac > 0 (iv) For f(x) = ax 3 + bx 2 + cx + d, find the x-coordinate of any turning points of the gradient. For a stationary point of the gradient, we want d (f (x)) = 0; ie dx f (x) = 0: f (x) = 3ax 2 + 2bx + c f (x) = 6ax + 2b f (x) = 0 x = b 3a And d2 dx2 (f (x)) = f (x) = 6a > 0, so that the stationary point is a minimum (ie it is a turning point). A turning point of the gradient is the definition of a point of inflexion (or inflection). Thus, all cubics have one point of inflexion. They can be shown to have rotational symmetry about this point. If the cubic has turning points, how could they be used to find the point of inflexion? By symmetry, the coordinates of the point of inflexion will be halfway between those of the turning points. 12

13 (v) For f(x) = ax 3 + bx 2 + cx + d, find conditions for the shape of the curve to be each of the 3 possibilities shown in (i), by considering the gradient at the point of inflexion. f ( b ) = b2 2b2 b2 + c = c 3a 3a 3a 3a Diagram (A): Either (i) a > 0 & f ( b 3a ) < 0 or (ii) a < 0 & f ( b 3a ) > 0 (i): 3ac b 2 < 0 b 2 3ac > 0 [agreeing with part (iii)] (ii): 3ac b 2 < 0 also Diagram (B): Stationary point of inflexion f ( b 3a ) = 0 b 2 3ac = 0 Diagram (C): Either (i) a > 0 & f ( b 3a ) > 0 or (ii) a < 0 & f ( b 3a ) < 0, so that b 2 3ac < 0 (20) (i) Suppose that y = f(x) is reflected in the line x = a, to give y = f(u). Find u in terms of x. (ii) Find the equation of the line resulting from the reflection of y = 2x + 1 in the line x = 1. Solution (i) A particular point can be reflected in the line x = a by considering a translation of both line and point by an amount a to the left, then performing a reflection in the y-axis and translating everything back, by a to the right. 13

14 In mathematical terms, x is first of all replaced by x + a; then x is replaced by x, and finally x is replaced by x a. Thus f(x) f(x + a) f( x + a) f( [x a] + a) = f(2a x) [As an aid to memory, consider the reflection of y = sinx about x = π, which is y = sin (π x)] 2 (ii) The transformed line is y = 2(2 x) + 1 = 5 2x Check: The transformed line will pass through the point where y = 2x + 1 meets the line x = 1; ie at (1, 3), and will have a gradient of 2; hence its equation is y 3 = 2 etc x 1 (21) Show that degrees. d dφ π sinφ = cosφ, when φ is measured in

15 Solution If φ is the angle in degrees, and θ is the angle in radians, so that φ = ( 180 ) θ, then π d sin dφ degφ = d sin dφ radθ = [ d sin dθ radθ] dθ = (cos dφ radθ) ( π ) 180 = (cos deg φ) ( π 180 ) [See "Trigonometry - Part 2" for alternative derivations, and related discussion.] (22) Derive a formula for the area of a triangle with corners at (0,0), (a, b), (c, d), using matrix transformations. Solution The formula for the area of a triangle with corners (0,0), (a, b), (c, d) can be obtained by considering the matrix transformation ( a c ): (a, b) is the image of (1,0) and (c, d) is b d the image of (0,1); the area of the triangle with corners (0,0), (1,0), (0,1) is 1/2, and the area scale factor is ad bc, since ad bc is the determinant of the matrix (the modulus sign only being needed when the order of the corners becomes reversed in the course of the transformation). (23) Show that 1 standard deviation to either side of the mean of the Normal distribution occurs at the point of inflexion of the Normal curve (where the pdf of the Normal distribution is φ(x) = 1 σ 2π e 1 2 (x μ σ )2 ). 15

16 Solution Considering N(0, 1), φ(x) = 1 φ (x) = x 1 2π e 2 x2 and φ (x) = 1 1 2π {e 1 2π e 2 x2 2 x2 +x( x)e 1 2 x2 } A point of inflexion is a turning point of the gradient, for which a necessary condition is that the gradient is stationary; ie d dx φ (x) = 0 or φ (x) = 0 [Technically, to confirm that it is a turning point of the gradient, we should check that d2 dx 2 φ (x) 0; ie φ (x) 0 (this is a sufficient condition; a necessary condition is that the first nonzero derivative of φ (x) is an even derivative). However, we can see from the curve that there is a point of inflexion.] φ (x) = 0 1 x 2 = 0 x = ±1, as required. 16

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