STEP Solutions Mathematics STEP 9465, 9470, 9475

Transcription

1 STEP Solutions 9 Mathematics STEP 965, 97, 975

2 The Cambridge Assessment Group is Europe's largest assessment agency and plays a leading role in researching, developing and delivering assessment across the globe. Our ualifications are delivered in over 5 countries through our three major eam boards. Cambridge Assessment is the brand name of the University of Cambridge Local Eaminations Syndicate, a department of the University of Cambridge. Cambridge Assessment is a not-for-profit organisation. This mark scheme is published as an aid to teachers and students, to indicate the reuirements of the eamination. It shows the basis on which marks were awarded by the Eaminers. It does not indicate the details of the discussions which took place at an Eaminers meeting before marking commenced. All Eaminers are instructed that alternative correct answers and unepected approaches in candidates scripts must be given marks that fairly reflect the relevant knowledge and skills demonstrated. Mark schemes should be read in conjunction with the published uestion papers and the Report on the Eamination. Cambridge Assessment will not enter into any discussion or correspondence in connection with this mark scheme. UCLES 9 More information about STEP can be found at:

3 Contents Step Mathematics (965, 97, 975) Report Page STEP Mathematics I STEP Mathematics II STEP Mathematics III 5

4 STEP I, Solutions 9

5 Question A proper factor of an integer N is a positive integer, not or N, that divides N. (i) Show that 5 has eactly proper factors. It is not by accident that this uestion writes 5 and not 5 : it is aiming to suggest that it is much more straightforward to think about factors of a number if we are given its prime factorisation to begin with. Also, note that the uestion does not ask us to multiply out the factorisations at any point. In fact, there is no need to even give the factors eplicitly if you do not need to. Determining the proper factors of 5 is straightforward: any factor must be of the form r 5 s with r and s, giving the factors: 5 ( ) (this is not a proper factor) 5 ( 5) 5 ( 5) 5 ( 5) 5 ( ) 5 ( 5) 5 ( 75) 5 ( 75) 5 ( 9) 5 ( 5) 5 ( 5) 5 ( 5) (this is not a proper factor) Therefore there are proper factors in total. Alternatively, we could simply note that there are possible values for the power of (namely, and ) and for the power of 5 (namely,, and ), making factors. Of these, two are not proper ( and the number 5 itself), leaving proper factors. (i) (cont.) Determine how many other integers of the form m 5 n (where m and n are integers) have eactly proper factors. Now that we have done this and understood how to count the factors of 5, we can answer the second part: the number of proper factors of a 5 b is (a + )(b + ), as the power of in a factor can be,,..., a, and the power of 5 can be,,..., b. So we reuire (a + )(b + ), or (a + )(b + ). Here are the possibilities: 5

6 a + b + a b n a 5 b so there are 6 possibilities in total. This means that there are 5 other integers with the reuired properties. We use these same ideas in part (ii). (ii) Let N be the smallest positive integer that has eactly 6 proper factors. Determine N, giving your answer in terms of its prime factors. Following the same ideas as in part (i), let n a b 5 c 7 d be the prime factorisation of the positive integer n. (Note that we should use a letter other than N to distinguish our arbitrary integer from the special one that we seek.) Then the number of factors of n is (a + )(b + )(c + )(d + ), and we must subtract to get the number of proper factors. Assuming now that n has 6 proper factors, we must have (a + )(b + )(c + )(d + ) 6, so (a + )(b + )(c + )(d + ) 8. Now we can factorise 8 7, and 7 is prime. So the possible factorisations of 8 are 8 7 7, so there can be at most three prime factors in n. We are seeking the smallest such n, so we choose the smallest possible primes, giving the smaller ones higher powers and larger ones smaller powers. So the smallest values of n for each possible factorisation of 8 are as follows: With 8 8: n 7 With 8 : n With 8 7: n 6 With 8 7: n 6 5 Since we seek the smallest possible value, our answer is clearly 6 5, as 7 > 7 >

7 Question A curve has the euation y + a + b, where a and b are positive constants. Show that the tangent to the curve at the point ( a, b) is b y a a + b. Differentiating the euation of the curve with respect to gives y dy d. Substituting a and y b gives b dy d euation of a straight line gives a, so dy d a /b. Then the standard y b a ( + a), b which easily rearranges into the form b y a a + b, as reuired. In the case a and b, show that the -coordinates of the points where the tangent meets the curve satisfy In the case a, b, the curve has euation y + 9, and the tangent at (, ) has euation y 9. We therefore substitute y + 9 into y + 9 as follows (after multiplying by ): 6y ( + 9) , on dividing by 9, and this is the euation reuired. Hence find positive integers p,, r and s such that p + r + s. Now our euation looks hard to solve, but we know that there is a solution at, as the curve and line are tangent at this point. In fact, since they are tangent, must be a double root. So we can take out a factor of ( + ) to get ( + )(7 7) ( + ) (7 7). 7

8 Thus either, which we aleady know, or 7. Since this point lies on the line 7 y 9, the y-coordinate is ( 7 + 9) /. Thus, as this point also lies on the curve 7 7 y + a + b, we have ( ) ( 7 7 ) Now multiplying both sides by 7 gives us our reuired result: , so a solution is p, 7, r 7, s. 8

9 Question (i) By considering the euation + a, show that the euation (a ) has one real solution when a and no real solutions when a <. This looks somewhat confusing at first glance; why might (a ), which can be rearranged as the given uadratic, only have one solution whereas the uadratic can have two? But we must remember that this euation involves a suare root, and by convention, this is the positive suare root; therefore, real solutions must satisfy both and a, even if the uadratic has other solutions in addition. Consider the euation (a ). If this is true, then suaring gives a, or + a. The solutions of this uadratic are given by ± + a, and these will be real solutions of ( ) if and only if and a, that is a. But as a, we will always have a for real solutions of the uadratic, so we need only check that. For a solution to ( ), we therefore reuire the plus sign in the uadratic formula, and we also need + a, so a. Thus, for a <, there are no real solutions to ( ), and for a, ( + + a ) / is the uniue real solution. An alternative approach is graphical. After we have shown that ( ) leads to the uadratic + a, we see that the real solutions of ( ) correspond to those of the uadratic where. We can therefore sketch the graph of y + a and observe where the roots are. The uadratic y + has roots at and with a line of symmetry at. The euation y + a is a simple translation by a vertically downwards, like this: ( ) a a a It is therefore clear that, for a, there is one root with, and for a <, there is no such root. 9

10 (i) (cont.) Find the number of distinct real solutions of the euation ( ( + a) a ) in the cases that arise according to the value of a. Since cube-rooting is invertible, we have ( ( + a) a ) ( + a) a. We are thus trying to solve the cubic euation ( + a) + a. Inspection reveals one root,, so we can factorise the cubic as ( )( + a). Using the discriminant of the uadratic factor, + a, we find that + a has, or real roots according to whether + a <, + a or + a >, respectively. Hence the original euation has real root if a <, distinct real roots if a (being and ), and real roots if a >. In the latter case, there is the possibility that they are not all distinct, though, if is a root of + a. This only happens when a, and in this case, there are also only distinct real roots. (ii) Find the number of distinct real solutions of the euation (b + ) in the cases that arise according to the value of b. This is very similar to part (i), with the only difference being that this time we have b + rather than a. The argument should therefore be fairly similar to part (i). Starting with the euation we again suare this to get b +, or b. (b + ), ( ) From the first form, we see that to have any solutions, we must have and b +. From the second, we see that the discriminant + b and b + as long as is real. So if b <, there are no solutions. The solutions to the uadratic are ± + b. In the case b, the repeated solution is, so there is one solution in this case.

11 In the case b >, we still reuire for solutions. The smaller root is + b, which is negative if b > and non-negative if b. Thus the euation ( ) has no solutions if b <, one solution if b solutions if < b. or b >, and two An alternative approach to this part of the uestion is again to draw a graph of the functions involved. As in part (i), we draw the graph of y b and determine how many roots it has with. The graph y has roots at and, and the graph of y b is a translation of this by b in the y-direction, as shown in this sketch: b b b b b From these sketches it is clear that for b < b, there are no positive real solutions; at b b there is one (repeated) positive real solution, for b > b there are two non-negative real solutions, and for b > there is one positive real solution. Finally, determining b is easy: we want b to have a repeated root, and this means that b ( ), so that b.

12 Question The sides of a triangle have lengths p, p and p +, where p > >. The largest and smallest angles of the triangle are α and β, respectively. Show by means of the cosine rule that ( cos α)( cos β) cos α + cos β. In situations like this, it s often useful to draw a sketch to gain some clarity about what is happening. Note that the largest angle is always opposite the longest side, and the smallest angle is always opposite the shortest side. p C α p β A p + B Using the cosine rule with the angles at A and C gives, respectively: (p + ) p + (p ) p(p ) cos α () (p ) p + (p + ) p(p + ) cos β () Then we need to manipulate these two euations in order to reach the desired result. There are several ways to do this; we show two of them. Approach : Determining the cosines and substituting Euation () gives, upon rearranging: cos α p + (p ) (p + ) p(p ) p + (p p + ) (p + p + ) p(p ) p p p(p ) p (p ). ()

13 Likewise, euation () yields: cos β p + (p + ) (p ) p(p + ) p + (p + p + ) (p p + ) p(p + ) p + p p(p + ) p + (p + ). () Using euations () and (), we now evaluate ( cos α)( cos β) and cos α + cos β: ( ( cos α)( cos β) p ) ( p + ) (p ) (p + ). p + (p ). p (p + ) p p and cos α + cos β p (p ) + p + (p + ) (p )(p + ) + (p + )(p ) (p )(p + ) p p + p + p (p ) (p ) (p ) p p. Therefore we have the reuired euality ( cos α)( cos β) cos α + cos β. ( ) Approach : Determining /p and euating From euation () above, we can epand to get p + p + p + p p + p(p ) cos α, so that We now divide by p to get p p p(p ) cos α. /p ( /p) cos α.

14 We can now rearrange this to find /p: ( cos α )/p cos α, so p cos α cos α. Doing the same for euation () gives: p p + p + p + p + p(p + ) cos β, so that p + p p(p + ) cos β. Again, dividing by p brings us to + /p ( + /p) cos β, yielding so ( cos β)/p cos β, p cos β cos β. Euating these two epressions for /p now gives us cos α cos α cos β cos β, so that (cross-multiplying and dividing by two): Now we epand the brackets to get so that ( cos α )( cos β) (cos α )( cos β ). cos α cos α cos β + cos β cos α cos β cos β cos α +, cos α cos β + cos α cos β cos α + cos β, and the left hand side factorises to give us our desired result: ( cos α)( cos β) cos α + cos β. In the case α β, show that cos β sides of the triangle. and hence find the ratio of the lengths of the Substituting α β into ( ) gives: ( cos β)( cos β) cos β + cos β.

15 We use the double angle formula for cos β to write this epression in terms of cos β, giving: so ( cos β)( cos β) cos β + cos β, 8( + cos β)( cos β) ( cos β )(cos β + ). Since cos β, we can divide by cos β + to get so we can rearrange to get which factorises as 8( cos β) cos β, 8 cos β 8 cos β + 9, ( cos β )( cos β ). Since cos β is impossible, we must have cos β, as reuired. We now substitute this result into euation () to get Epanding this gives (p ) p + (p + ) p(p + ).. p p + p + p + p + p p, so p 5 p, which gives p 5. Hence the side lengths are p, p 5 and p + 6, which are in the ratio : 5 : 6. An alternative way to do this last part is as follows. We have α β, so cos α cos β ( ). It follows that sin α and sin β 7. We can now use the sine rule to get a sin A c sin C, or a/c sin A/ sin C. It follows that p p + sin β sin α 7 8 7, giving (p ) (p + ), or p 5. The rest of the result follows as above. A third way of doing this, and arguably the simplest, is to substitute into euation (), which gives: p + (p + ). Multiplying both sides by (p + ) to clear the fractions gives (p + ) (p + ), so that p 5. The rest of the argument again follows as above. 5

16 Question 5 A right circular cone has base radius r, height h and slant length l. Its volume V, and the area A of its curved surface, are given by V πr h, A πrl. (i) Given that A is fied and r is chosen so that V is at its stationary value, show that A π r and that l r. Since A is fied and r is allowed to vary, we rearrange A πrl as l A/πr. Also, we can draw a side view of the cone to determine the relationship between r, h and l: h l r So clearly, l h + r. Substituting this into l A/πr gives l h + r A π r, which we can rearrange to give h in terms of r and A: Now V πr h, so we have h A π r r. V 9 π r h ( ) π r A 9 π r r 9 (A r π r 6 ). (Working with V rather than just V allows us to avoid suare roots.) Differentiating with respect to r gives V dv dr 9 (A r 6π r 5 ). When V is at its stationary value, dv/dr, so we reuire A r 6π r 5. As r, we must have 6π r A, or A π r, as wanted. 6

17 Substituting this into our formula for l gives so l r, as we wanted to show. l A π r π r π r r, (ii) Given, instead, that V is fied and r is chosen so that A is at its stationary value, find h in terms of r. We have l h + r A /π r and V πr h. This time, V is fied, so h V/πr. Thus Differentiating as before gives A π r (h + r ) ( ) 9V π r π r + r 9V r + π r A da dr 8V r + π r, so da/dr when π r 6 8V, so πr V. Finally, substituting this into our formula h V/πr gives h πr / πr r. 7

18 Question 6 (i) Show that, for m >, m /m + d (m ) (m + ) + ln m. m We note that the numerator of the fraction ( ) has a higher degree than the denominator ( + ), so we divide them first, getting ( + )( ) +, so our integral becomes m /m m + d + /m + d [ + ln + ] m /m ( m m + ln m + ) ( m m + ln (/m) + ) m m + m m + + ln m (/m) + (m + )(m m + m ) + ln m (m + )(m ) + ln m. m m(m + ) + m (An alternative approach is to use the substitution u +, which leads to eactly the same result.) (ii) Show by means of a substitution that m m n ( + ) d u n /m u + du. /m Comparing the two integrals suggests that we should try the substitution u /. If we do this, we get /u and d/du /u. Also, the limits /m and m become u m and u /m respectively. So we have m /m /m n ( + ) d d (/u) n (/u + ) du du m /m m /m m m /m u n /u + u n u du u( + u) du u n u + du. 8

19 (iii) Evaluate: (a) / 5 + ( + ) d. This clearly relies on the earlier parts of the uestion, where m. We can break the integral into two parts and use the results of (i) and (ii) as follows: / 5 + ( + ) d / / 5 ( + ) d + + d + / / + d ( (m + )(m ) m ( + ln ) 8 + ln. / ( + ) d u u + du ) + ln m An alternative way to approach this uestion is to ignore what has gone before and to use partial fractions. We first divide to get 5 + ( + ) + + ( + ) and then epress the final term using partial fractions: Integrating then gives: / 5 + ( + ) d / + ( + ) d [ + ln + ] ln( + ) / ( + ln + ln ) ( + ln ln ) ln. (iii) Evaluate: (b) ( + ) d. 9

20 It is no longer so obvious how to proceed, as the limits are not of the form /m to m. So we break up the integral as in (a) and repeat the substitution of part (ii) once again, noting that the only change here is that the limits are different. We have: 5 ( + ) d + d, and there is not much we can do at this point, short of evaluating this integral as in part (i). Net, we have The third part gives us ( + ) d + d [ ln + ] ln ln. ( + ) d u / u + du, by using the substitution of part (ii), but noting the the limits transform into / and /, which are then reversed by the minus sign. Adding all three terms and using part (i) with m now gives: ( + ) d 8 8 / d + ln ln + + d + ln ln + + ln + ln ln + ln. / u u + du Again, this uestion can also be approached using partial fractions. Dividing gives ( + ) and then partial fractions epansion gives us: We now integrate to reach our final answer: 5 + ( + ) d + + ( + ) + ( + ) [ d + ln + + ln( + ) ( + ln + + ln ) ( ln ) 8 + ln. 8 ]

21 Question 7 Show that, for any integer m, π e cos m d ( e π ). m + This is a standard integral with standard techniues for solving it. The approach used here is not the only possible one, but is fairly general. We write I π e cos m d and apply integration by parts twice, taking great care of the signs, as follows: I π e cos m d [ e ] π m sin m π ( m eπ sin πm ) ( m e sin ) π [ e e sin m d m ] π m ( cos m) + ) ( m eπ cos πm m ( e π ) m I. Multiplying throughout by m gives e sin m d m π π e sin m d m e ( cos m) d m ( ) π m e cos e m I ( e π ) I, cos m d m and now adding I to both sides and dividing by m + gives the desired result. We performed these integrations by writing e cos m d in the form v du d, where d du cos m and v d e. We could eually well have chosen du d e and v cos m, and would have ended up with the same conclusion. (i) Epand cos(a + B) + cos(a B). Hence show that π e cos cos 6 d 9 65( e π ).

22 We have cos(a + B) + cos(a B) cos A cos B sin A sin B + cos A cos B + sin A sin B cos A cos B. Matching this to the integral we have been given, we set A 6 and B (so that A B is positive, though this is not critical), giving Thus our integral becomes as reuired. π cos cos 6 cos 7 + cos 5. π e cos cos 6 d e (cos 7 + cos 5) d ( ( e π ) + ( e π )) ( 5 + ) (e π ) ( e π ) 65( 9 e π ), (ii) Evaluate π e sin sin cos d. We are clearly asked to do the same type of trick again. Here is one way to proceed. We are looking for the product of two sines, and this appears in the compound angle formula for cosine. So we consider cos(a + B) cos(a B) cos A cos B sin A sin B cos A cos B sin A sin B sin A sin B. We can therefore write sin sin cos cos 6, using A and B. This gives the integral as π e sin sin cos d π π e (cos cos 6) cos d e cos cos d π e cos 6 cos d. Now the second integral is eactly the one we evaluated in (i), and the first integral can be approached in the same way, with A and B, so cos cos cos + cos.

23 Therefore π π e cos cos d e (cos + cos ) d ( ( e π ) + ( e π )) + + ( + ) (e π ) Finally, subtracting the two integrals gives π ( e π ). π π e sin sin cos d e cos cos d e cos 6 cos d ( e π ) ( 9 65 e π ) ( 95 )( e π ) ( e π ) 5( e π ).

24 Question 8 (i) The euation of the circle C is ( t) + (y t) t, where t is a positive number. Show that C touches the line y. This is a circle with centre (t, t) and radius t, therefore it touches the -ais, which is distance t from the centre. Alternatively, we are looking to solve the simultaneous euations ( t) + (y t) t y. Substituting the second euation into the first gives ( t) + t t, or ( t). Since this only has one solution, t, the line must be tangent to the circle. (i) (cont.) Let α be the acute angle between the -ais and the line joining the origin to the centre of C. Show that tan α and deduce that C touches the line y. We begin by drawing a sketch of the situation. C α α t t Clearly, therefore, tan α t/t, so tan α tan α tan α.

25 From the sketch, it is clear that by symmetry, the line through the origin which makes an angle of α with the -ais touches the circle C. Since the gradient of this line is tan α and it passes through the origin, it has euation y, or y. (ii) Find the euation of the incircle of the triangle formed by the lines y, y and y + 5. Note: The incircle of a triangle is the circle, lying totally inside the triangle, that touches all three sides. This circle has the properties described in part (i), in that it touches both the -ais (i.e., the line y ) and the line y, and it lies above the -ais, so it must have the form given. The only remaining condition is that it must touch the line y + 5. Two circles with centre (t, t) touch these three lines, but only one of them lies within the triangle, as illustrated on this sketch, so we must take the one with the smaller value of t: y y + 5 B I O A Approach : Algebraic substitution Substituting y + 5 into the euation of C will give us the intersections of the line and the circle. The line will be a tangent to C if and only if the discriminant of the resulting uadratic euation is. Doing this, from the euation of C: we get: ( t) + (y t) t ( t) + ((5 )/ t) t. Multiplying both sides by and epanding gives: so 6 6t + 6t + (5 ) 8(5 )t + 6t 6t, 6 6t + 6t t + t + 6t 6t. Collecting terms in gives: 5 (9 + t) + 5 t + 6t. 5

26 Since this has a repeated root, as the line is reuired to be tangent to the circle, the discriminant must be zero, so so or giving (9 + t) 5(5 t + 6t ), (9 + t) (5 t + 6t ), 6t + 7t + 8 (5 t + 6t ), 8t + 9t. Dividing all the terms by 8 gives t t +, which factorises as (t )(t ). Thus the two circles in the sketch are given by t and t, and the incircle is clearly the one with t. So the incircle has euation ( ) + (y ). Alternative : Distance of a point from a line In the formula book, we are given the result: The perpendicular distance of (α, β, γ) from n +n y+n z+d is n α + n β + n γ + d. n + n + n Therefore the distance of the centre of the circle, (t, t, ), from the line +y +z 5 is given by t + t + 5 t 5 t But the line is tangent to the circle, which has radius t, so we must have t t. This euation has two solutions: t t gives t, and t t gives t. As we reuire the smaller solution, we must have t, so that the incircle has euation ( ) + (y ). Alternative : Finding another angle bisector As in part (i), using the notation in the diagram above, the line from A to I bisects the angle OAB. So if we write OÂI β, then OÂB β. Now tan β, so we solve tan β tan β tan β to find tan β. We get the uadratic euation tan β 8 tan β, 6

27 which factorises as ( tan β )(tan β + ), so tan β or tan β. But β is acute, so tan β. Thus the line AI has euation which can be written as y + 5. y ( 5) The point I has coordinates (t, t), and substituting this in gives t + t 5, so t as before. Alternative : Using Euclidean geometry We find that the intersection of the line y + 5 with the -ais is at (5, ). We can also find the intersection of the two lines y + 5 and y by solving them simultaneously: y y 6, so 5 5, or 9, so y. Thus the side of the triangle from the origin to the point 5 5 of intersection has length, the side from (5, ) to the point of intersection has length and the base has length 5 (using Pythagoras), so we have a --5 triangle. (The triangle is right-angled as y + 5 and y are perpendicular.) Consider now this figure, where we have drawn radii from the incentre to the three sides of the triangle. Note that, since the triangle s sides are tangents to the circle, the radii meet at right angles, so ICBD is a suare with all sides eual to the radius, t. D t B t C t t I t t t O E A Now AC AE t since these are both tangents from A to the circle; likewise, OE OD t, so OA OE + EA ( t) + ( t) 7 t. But OA 5, so t and t, giving the radius and hence the euation of the incircle. Alternative 5: More Euclidean geometry We begin in the same way by finding the side lengths of the triangle. We can then uote the result that if r is the incircle radius and the side lengths of the triangle are a, b, c, then these are related by Area r(a + b + c). In this case, the area is 6 and (a + b + c) 6, so r as reuired. 7

28 Question 9 Two particles P and Q are projected simultaneously from points O and D, respectively, where D is a distance d directly above O. The initial speed of P is V and its angle of projection above the horizontal is α. The initial speed of Q is kv, where k >, and its angle of projection below the horizontal is β. The particles collide at time T after projection. Show that cos α k cos β and that T satisfies the euation (k )V T + dv T sin α d. We begin by drawing a sketch showing the initial situation. D β +ve d Q kv O P α V For components, we will write u P and u Q for the horizontal components of the velocities of P and Q respectively, and v P and v Q for the vertical components (measured upwards). For the displacements from O, we write P and Q for the horizontal components and y P and y Q for the vertical components. Resolving horizontally, using the suvat euations, we have u P V cos α, u Q kv cos β; P V t cos α, Q kv t cos β. Likewise, vertically we have v P V sin α gt, v Q kv sin β gt; y P V t sin α gt, y Q d kv t sin β gt. At time T, the particles collide, so P Q, giving V T cos α kv T cos β, 8

29 so that cos α k cos β. Also, y P y Q, so which gives V T sin α gt d kv T sin β gt, V T sin α d kv T sin β. Now k sin β k k cos β k cos α from the above, so we rearrange and then suare both sides of the previous euation to make use of this conclusion: kv T sin β d V T sin α, so which then gives us k V T sin β (d V T sin α), V T (k cos α) d dv T sin α + V T sin α. Finally, subtracting the right hand side from the left gives V T (k cos α sin α) d + dv T sin α, or as reuired. V T (k ) + dv T sin α d, ( ) Given that the particles collide when P reaches its maimum height, find an epression for sin α in terms of g, d, k and V, and deduce that gd ( + k)v. At the maimum height, we have v P, so V sin α gt. Substituting for T in ( ) gives us ( ) ( ) V sin α V sin α V (k ) + dv sin α d. g g Multiplying through by g and epanding brackets gives (k )V sin α + gdv sin α g d. Thus sin α g d (k )V + gdv. For the ineuality, the only thing we know for certain is that sin α, and this gives so g d, (k )V + gdv g d (k )V + gdv. 9

30 It is not immediately clear how to continue, so we try completing the suare for gd: (gd V ) V (k )V, so that (gd V ) k V. Now if a b, then a b, so gd V kv. But kv >, so we are almost there: gd V kv, which finally gives us the reuired gd ( + k)v.

31 Question A triangular wedge is fied to a horizontal surface. The base angles of the wedge are α and π α. Two particles, of masses M and m, lie on different faces of the wedge, and are connected by a light inetensible string which passes over a smooth pulley at the ape of the wedge, as shown in the diagram. The contacts between the particles and the wedge are smooth. M α π α m (i) Show that if tan α > m M the particle of mass M will slide down the face of the wedge. We start by drawing the forces on the picture, and labelling the masses as A (of mass M) and B (of mass m). We let T be the tension in the string. T N A A a α T mg a Mg π α B N B We now resolve along the faces of the wedge at A and B. normally to the face doesn t help at all for this problem.) (It turns out that resolving R A ( ) T Mg sin α Ma () R B ( ) T mg cos α ma () We are not interested in the tension in the string, so we subtract these euations (as () ()) to eliminate T, yielding Mg sin α mg cos α ma + Ma. Thus, dividing by M + m gives a Mg sin α mg cos α. () M + m The mass M will slide down the slope if and only if a >, that is if and only if Mg sin α mg cos α >. Rearranging and dividing by g cos α gives M tan α > m, or tan α > m M.

32 (ii) Given that tan α m, show that the magnitude of the acceleration of the particles M is g sin α tan α + and that this is maimised at m M. We simply substitute m/m tan α into our formula () for a to find the magnitude of the acceleration: Mg sin α mg cos α a M + m g sin α (m/m)g cos α + (m/m) g sin α g tan α cos α + tan α g sin α g sin α + tan α g sin α + tan α multiplying by /M To maimise this with respect to α, we differentiate with respect to α (using the uotient rule) and solve da/dα : so we reuire da dα ( + tan α)g cos α sec α.g sin α ( + tan α) g( cos α + sin α sec α sin α) ( + tan α) cos α + sin α sec α sin α. We can simplify this by dividing through by cos α, so that we are able to epress everything in terms of tan α: + tan α sec α tan α, so + tan α ( + tan α) tan α, so that tan α. Remembering that tan α m/m, we finally get which leads immediately to M m. 8m M, We finally need to ensure that this stationary point gives us a maimum; this is clear, since as α, a and as α π, a.

33 Question Two particles move on a smooth horizontal table and collide. The masses of the particles are m and M. Their velocities before the collision are ui and vi, respectively, where i is a unit vector and u > v. Their velocities after the collision are pi and i, respectively. The coefficient of restitution between the two particles is e, where e <. (i) Show that the loss of kinetic energy due to the collision is and deduce that u p. m(u p)(u v)( e), Before the collision: m ui M vi After the collision: m pi M i Conservation of momentum gives: Newton s Law of Restitution gives: Now the loss of kinetic energy due to the collision is mu + Mv mp + M. () E initial KE final KE p e(u v). () ( mu + Mv ) ( mp + M ) m(u p ) + M(v ) m(u p)(u + p) + M(v )(v + ). Now from (), we get M(v ) m(p u), so we get E m(u p)(u + p) + m(p u)(v + ) m(u p)( (u + p) (v + ) ) m(u p)( (u v) ( p) ) m(u p)( (u v) e(u v) ) m(u p)(u v)( e). Now, since the loss of energy cannot be negative, we have E. But we are given that e < and u > v, so we must have u p, or u p.

34 (ii) Given that each particle loses the same (non-zero) amount of kinetic energy in the collision, show that u + v + p +, and that, if m M, e (M + m)u + (M + m)v. (M m)(u v) The first particle loses an amount of kinetic energy eual to m(u p ); the second loses M(v ), so we are given so Again, we use M(v ) m(p u), so that m(u p ) M(v ), m(u p ) M(v ). m(u p ) M(v ) m(u + p)(u p) M(v + )(v ) m(u + p)(u p) m(v + )(p u) m(p + + u + v)(u p). Since the amount of kinetic energy lost is non-zero, we have E > (in the notation of part (i)), so that u > p. Thus we must have p + + u + v. We can also euate the loss of energy of each particle with E, so: m(u p ) m(u p)(u v)( e), giving u + p (u v)( e). () We now need to eliminate p, and rearrange to get an epression for e. Euation () gives M Mp Me(u v), and subtracting this from (), mp + M mu + Mv, gives We multiply () by m + M to get: (m + M)p (m Me)u + (M + Me)v. (m + M)p + (m + M)u (M + m)(u v)( e), and then substitute in our epression for (m + M)p to give us: (m Me)u + (M + Me)v + (m + M)u (M + m)(u v)( e). We epand and rearrange to collect terms which are multiples of e: (mu Mue + Mv + Mve + mu + Mu) (Mu Mv + mu mv)( e)

35 so (Mv Mu + Mu Mv + mu mv)e Mu Mv + mu mv mu Mv Mu, which leads to (M m)(v u)e Mu Mv mu mv. We can write the right hand side as (M + m)u (M + m)v, so that assuming M m, we can divide by (M m)(v u) to get e (M + m)u + (M + m)v, (M m)(v u) as we wanted. 5

36 Question Prove that, for any real numbers and y, + y y. We can rearrange the ineuality to get y + y. But the left hand side is just ( y), so this ineuality becomes ( y). This is clearly true, as any real number suared is non-negative, and since this is just a rearrangement of the desired ineuality, that must also be true. (i) Carol has two bags of sweets. The first bag contains a red sweets and b blue sweets, whereas the second bag contains b red sweets and a blue sweets, where a and b are positive integers. Carol shakes the bags and picks one sweet from each bag without looking. Prove that the probability that the sweets are of the same colour cannot eceed the probability that they are of different colours. We can draw a tree diagram to represent this situation: b a + b R a a + b b a + b R a a + b b a + b B R B a a + b B So P(same colour) P(different colours) a a + b ab (a + b) a a + b a + b (a + b). b a + b + a a + b + b a + b b a + b a a + b b a + b Since a + b ab, it follows that the probability that the sweets are of the same colour cannot eceed the probability that they are of different colours, as reuired. 6

37 (ii) Simon has three bags of sweets. The first bag contains a red sweets, b white sweets and c yellow sweets, where a, b and c are positive integers. The second bag contains b red sweets, c white sweets and a yellow sweets. The third bag contains c red sweets, a white sweets and b yellow sweets. Simon shakes the bags and picks one sweet from each bag without looking. Show that the probability that eactly two of the sweets are of the same colour is (a b + b c + c a + ab + bc + ca ) (a + b + c), and find the probability that the sweets are all of the same colour. Deduce that the probability that eactly two of the sweets are of the same colour is at least 6 times the probability that the sweets are all of the same colour. We argue in the same way: P(eactly same) P(RRW) + P(RRY) + P(RWR)+ P(RYR) + P(WRR) + P(YRR)+ P(WWR) + P(WWY) + P(WRW)+ P(WYW) + P(RWW) + P(YWW)+ P(YYR) + P(YYW) + P(YRY)+ P(YWY) + P(RYY) + P(WYY) aba (a + b + c) + abb (a + b + c) + acc (a + b + c) + aac (a + b + c) + bbc (a + b + c) + cbc (a + b + c) + bcc (a + b + c) + bcb (a + b + c) + bba (a + b + c) + baa (a + b + c) + aca (a + b + c) + cca (a + b + c) + cac (a + b + c) + caa (a + b + c) + cbb (a + b + c) + ccb (a + b + c) + aab (a + b + c) + bab (a + b + c) (a b + b c + c a + ab + bc + ca ) (a + b + c). More simply, the probability that all three are the same colour is given by P(all same) P(RRR) + P(WWW) + P(YYY) abc (a + b + c) + bca (a + b + c) + cab (a + b + c) abc (a + b + c). 7

38 Now to find the ineuality we want, we apply the initial ineuality again: we have a + b ab, so a c + b c abc. Similarly, b + c bc, so b a + c a abc, and finally, c b + a b abc. Thus (a b + b c + c a + ab + bc + ca ) (a c + b c + b a + c a + c b + a b) (a + b + c) (a + b + c) (abc + abc + abc) (a + b + c) 6(abc) (a + b + c), showing that the probability that eactly two of the sweets are of the same colour is at least 6 times the probability that the sweets are all of the same colour. 8

39 Question I seat n boys and girls in a line at random, so that each order of the n + children is as likely to occur as any other. Let K be the maimum number of consecutive girls in the line so, for eample, K if there is at least one boy between each pair of girls. (i) Find P(K ). There are two euivalent ways to approach this uestion: either to regard the boys and girls as all distinct, so that there are (n + )! possible orders, or to regard all boys as indistinguishable and all girls as indistinguishable, so that there are ( ) n+ possible orders. We use the latter approach here. We note that, regarding the boys as indistinguishable and the girls as indistinguishable, there are ( ) n + (n + )(n + )(n + ) 6 possible arrangements of the students. If K, this means that all three girls are adjacent. So the situation must be that there are r boys, followed by girls, followed by n r boys, where r,,..., n, so there are n + possibilities. Thus P(K ) n + (n + )(n + )(n + ) 6 6 (n + )(n + ). (ii) Show that P(K ) n(n ) (n + )(n + ). Approach : Counting eplicitly To have K, we must have each pair of girls separated by at least one boy, like this: B. }{{.. B} G B. }{{.. B} G } B. {{.. B} G } B. {{.. B} r r r r where r, r >, r >, r and r + r + r + r n. If r and r are fied, then we must have r + r n (r + r ), so we can have r,,..., n (r + r ), giving n (r + r ) possibilities for r and r. 9

40 Thus, if r is fied, r could be,,..., n r (but not n r, as we must have r > ). Thus the number of possibilities for a fied value of r is given by n r r (n r r ) n r r (n r ) n r r (n r )(n r ) (n r )(n r ) (n r )(n r ) r (n nr + r n + r ). Now, r can take the values,,,..., n (as we need r > and r > ), giving the total number of possibilities as n r (n nr + r n + r ) n r (n n) (n ) n r r + n (n )(n n) (n ) (n )(n ) + (n )(n )(n ) (n )( 6(n n) (n )(n ) + (n )(n ) ) (n )(6n 6n 6n + 5n 6 + n 7n + 6) (n )(n + n) n(n )(n + ). 6 r r Thus we can finally deduce P(K ) 6 n(n )(n + ) 6 (n + )(n + )(n + ) n(n ) (n + )(n + ). Approach : A combinatorial argument We have to place each of the three girls either between two boys or at the end of the line, and we cannot have two girls adjacent. We can think of the line as n boys with gaps between them and at the ends, like this: B B... B B Note that there are n + gaps (one to the right of each boy, and one at the left of the line). Three of the gaps are to be filled with girls, giving ( ) n+ (n+)n(n ) ways of choosing 6 them. Therefore P(K ) 6 (n + )n(n ) 6 (n + )(n + )(n + ) n(n ) (n + )(n + ).

41 Approach : Another combinatorial argument We add one more boy at the right end of the line. In this way, we have a boy to the right of each girl, as follows: B. }{{.. B} GB B. }{{.. B} GB B. }{{.. B} GB B. }{{.. B} r r r r where this time, we have r, r, r and r. Also, as there are now n + boys, we have r + r + r + r n + n. So if we think of GB as one person, there are n boys and GBs, giving n + people in total. There are ( ) n+ (n + )n(n ) ways of arranging them, giving 6 P(K ) 6 (n + )n(n ) 6 (n + )(n + )(n + ) n(n ) (n + )(n + ). (iii) Find E(K). We could attempt to determine P(K ) directly, but it is far easier to note that K can only take the values, or. Thus P(K ) P(K ) P(K ) (n + )(n + ) n(n ) 6 (n + )(n + ) n + 5n + 6 n + n 6 (n + )(n + ) 6n (n + )(n + ). Since E(K) k k.p(k k), we can now calculate E(K): E(K).P(K ) +.P(K ) +.P(K ) n(n ) + 6n + 6 (n + )(n + ) n n + n + 8 (n + )(n + ) n + n + 8 (n + )(n + ) (n + )(n + 9) (n + )(n + ) n + 9 n +.

42 STEP II, Solutions 9

43 Both graphs are symmetric in the lines y, and + y u is also symmetric in the - and y-aes. These facts immediately enable us to write down the coordinates of B(, ), C(, ) and D(, ). Remember to keep the cyclic order A, B, C, D correct, else this could lead to silly calculational errors later on. The easiest way to show that ABCD is a rectangle is to work out the gradients of the four sides (which turn out to be either or ) and then note that each pair of adjacent sides is perpendicular using the product of gradients result. Working with distances is also a possible solution-approach but, on its own, only establishes that the uadrilateral is a parallelogram. However, the net part reuires you to calculate distances anyhow, and we find that CB, DA have length ( + ) while BA, DC are of length ( ). Multiplying these then give the area of ABCD as ( ). All of this is very straightforward, and the only tricky bit of work comes net. It is important to think of and as particular values of and y satisfying each of the two original euations. It is then clear that ( ) + ( ) u v, so that Area ABCD u v. Substituting u 8, v into this formula then gives Area 8 6, which is intended principally as a means of checking that your answer is correct. (i) (ii) It is perfectly possible to differentiate a^(sin[ e ]) by using the Chain Rule (on a function of a function of a function) but simplest to take logs. and use implicit differentiation. Then, setting d y and noting that e and ln a are non-zero, we are left solving the en. cos( e ) for d the turning points. This gives e (n + ) ln n, y a or a, depending upon whether n is even or odd. Although not actually reuired at this point, it may be helpful to note at this stage that the evens give maima while the odds give minima. There is, however, a much more obvious approach to finding the TPs that doesn t reuire differentiation at all, and that is to use what should be well-known properties of the sine function: namely, that sin(.ep ) y a has maima when sin( e ), i.e. e (n + ), and ln(n + ) for n,,, with y ma a. Similarly, minima occur when sin( e ), i.e. e (n ), and ln(n ) for n,,, with y min a. Using the addition formula for sin(a + B), and the approimations given, we have sin( e ) sin( + ) sin( ) for small, ln a leading to y a e. ln a. (iii) Firstly, we can note that, for <, the curve has an asymptote y (as, y +). Net, for >, the curve oscillates between a and a, with the peaks and troughs getting ever closer together. The work in (i) helps us identify the TPs: the first ma. occurs when n at a negative value of [N.B. ln < ] at y a; while the result in (ii) tells us that the curve is approimately negative linear as it crosses the y-ais. (iv) The final part provides the only really tricky part to the uestion, and a uick diagram might be immensely useful here. Noting the relevant -coordinates lnk, lnk, and lnk, the area is the sum of two trapezia (or rectangle triangle), and manipulating k k ln ln ln k leads to the final, given answer. k k a a

44 Using the addition formula for tan(a B), tan cos sin cos LHS tan tan cos sin cos sin cos cos sin sin sin cos cos sin sin (since c + s sin ) sec tan RHS. cos Alternatively, one could use the t tan( -angle) formulae to show that t RHS t t t ( t) t ( t)( t) t tan tan LHS. (i) Setting in (*) tan. Then, using the addition formula for tan(a + B) with 8 A and B 8, we have tan tan 8, as reuired. 6 (ii) Now, in the spirit of maths, one might reasonably epect that one should take the given epression, rationalise the denominator (twice) and derive the given answer, along the lines However, with a given answer, it is perfectly legitimate merely to multiply across and verify that 6 6. (iii) Having got this far, the end is really very clearly signposted. Setting in (*) gives tan 8 sec tan t t (i) Writing p() ().( ) 5, where () is a uartic polynomial, immediately gives p(). (ii) Diff g. using the product and chain rules leads to p() ().5( ) + ().( ) 5 ( ).{5 () + ( ) ()}, so that p() is divisible by ( ). (iii) Similarly, we have that p() is divisible by ( + ) and p( ). Thus p() is divisible by ( + ).( ) ( ). However, p() is a polynomial of degree eight, hence p() k( ) 8 6 for some constant k. That is, p() k Integrating term by term then gives p() k C, and use of both 5 p() and p( ) help to find k and C; namely, k and C. 8

45 5 The very first bit is not just a giveaway mark, but rather a helpful indicator of the kind of result or techniue that may be used in this uestion: happens here. Most particularly, the fact that ; but pay attention to what does NOT necessarily mean that since positive numbers have two suare-roots! Recall that and not just. Notice that, during the course of this uestion, the range of values under consideration switches from (5, ) to 5,, and one doesn t need to be particularly suspicious to wonder why this is so. A modicum of investigation at the outset seems warranted here, as to when things change sign. (i) So while seems a perfectly acceptable thing to write, since is a necessary condition in order to be able to take suare-roots at all here (for real numbers), simply writing down that may cause a problem. A tiny amount of eploration shows that changes from negative to positive around. Hence, in part (i), we can ignore any negative considerations and plough ahead: I 5 d 5. (ii) Here in (ii), however, you should realise that the area reuested is the sum of two portions, one of which lies below the -ais, and would thus contribute negatively to the total if you failed to take this into account. Thus, Area d + d ( ) d + ( ) d.5 ¼..5.5 (iii) Now so we have no cause for concern here. Then I.5 d ( ) ( ).5 d. 5 6 If you don t know about the Fibonacci Numbers by now, then shame on you! Nevertheless, the first couple of marks for writing down the net few terms must count as among the easiest on the paper. (F, F ), F, F, F 5 5, F 6 8, F 7, F 8, F 9 and F 55. (i) If you re careful, the net section isn t particularly difficult either. Using the recurrence relation gives since F i < F i for i. Splitting off the first few terms then Fi Fi Fi Fi n leads to S >... or..., where the i Fi F F F F F long bracket at the end is the sum-to-infinity of a GP. These give, respectively, S > + or + +. A simpler approach could involve nothing more complicated than adding the terms until a sum greater than is reached, which happens when you reach F 5. 5

46 A similar approach yields F i F i for i and splitting off the first few terms, this time separating the odd- and even-numbered terms, gives S n i i F F F F F F F < (ii) To show that S >., we simply apply the same approaches as before, but taking more terms initially before summing our GP (or stopping at F 7 in the simpler approach mentioned previously). Something like S >... 5 F F F F F 7 5 >. 6 does the job pretty readily. Then, to show that S <, a similar argument to those you have been directed towards by the uestion, works well with little etra thought reuired: S < Returning to the initial argument, F i < F i or F i F i for i, we can etend this to F i > Fi or F i F i for i 5, F i < 5 Fi or 5 F i F i for i 6, etc., simply by using the defining recurrence relation for the Fibonacci Numbers, leading to the general results F n > k k F F F n or n k k n F F F F for n k + and F n < k k F F F n or n k k n F F F F for n k +. Since the terms n n F F 5, the golden ratio, (being the positive root of the uadratic euation +, we can deduce the approimation S n n i i F F since the geometric progression.... Taking n 9, (i.e. just using the first Fibonacci Numbers which you were led to write down at the start), S 9 F F i i , which is correct to 5 d.p. For further information on this number, try looking up the Reciprocal Fibonacci constant on Wikipedia, for instance. 6

47 7 It is easy to saunter into this uestion s opening without pausing momentarily to wonder if one is going about it in the best way. Whilst many can cope with differentiating a triple -product with ease, many others can t. However, even for interests sake, one might stop to consider a general approach to such matters. Differentiating y pr (all implicitly functions of ) as, initially, p(r) and applying the product-rule twice, one obtains y p r + p r + p r, and this can be used here with p ( a) n, e b and r without the need for a lot of the mess (and subseuent mistakes) that was (were) made by so many candidates. Here, y ( a) n e b gives d y ( a) n e b + ( a) n b e b + n( a) n e b d n b ( a) e Factorising out the given terms ( a) b( a)( ) n( ), and we are only reuired to note that the term in the brackets is, indeed, a cubic; though it may prove helpful later on to simplify it by multiplying out and collecting up terms, to get () b + (n + ab) + (b a) + (n ab). ( ) e (i) The first integral, I d, might reasonably be epected to be a very straightforward application of the general result, and so it proves to be. With n 5, and taking a b, so that () (which really should be checked eplicitly), we find I ( ) 5 e (+ C). ( ) e (ii) This second integral, I d, is clearly not so straightforward, since the bracketed term is now uartic. Of the many things one might try, however, surely the simplest is to try to factor out a linear term, the obvious candidate being ( ). Finding that ( )( ), we now try n, a, b to obtain () and I ( ) e (+ C). 6 ( ) e (iii) The final integral, I d, is clearly intended to be even less simple than its predecessor. However, you might now suspect that the net case up is in there somewhere. So, if you try n 8, a, b, which gives d 7 y 8 6 ( ) e ( ) e 7, d as well as the obvious target n 7, a, b, which yields d 6 y ( ) e 7, d It may now be clear that both are involved. Indeed, I d y8 dy 7 d y 8 + y 7 ( ) 7 e (+ C). d d 7

48 8 For the diagram, you are simply reuired to show P on AB, strictly between A and B; and Q on AC on other side of A to C. The two given parameters indicate that CQ AC and BP AB. Substituting these into the given epression, CQ BP AB AC AC. AB AB. AC. [Notice that CQ, BP, etc., are scalar uantities, and hence the cannot be the vector product!] Writing the euation of line PQ in the form r t p + ( t) for some scalar parameter t and substituting the given forms for p and gives r t a + t( )b + ( t) a + ( t)( )c. t Eliminating r t a t( ) b ( t) c. Comparing this to the given answer, we note that when t from the b-component, t, etc., so that we do indeed get r a b c, as reuired. Since d c b a, one pair of sides of opposite sides of ABDC are eual and parallel, so we can conclude that ABDC is a parallelogram 9 (i) If you break the lamina up into a rectangle and a triangle (shapes whose geometric centres should be well-known to you), with relative masses and, and impose (mentally, at least) a coordinate system onto the diagram, then the -coordinate of the centre of mass is given by mi 9 i 7. m i (ii) A more detailed approach, but still along similar lines, might be constructed in the following, tabular way: Shape Mass Dist. c.o.m. from OZ LH end 5 7 Note that each mass has been RH end 5 7 calculated as Front d 7 area density () Back d Base 9d 9 (5) 7 d 7 9dp Then E 8 9d d ( d ). ( d) 5( d) 9, which (after much cancelling) simplifies to A similar approach for the full tank gives Object Mass Dist. c.o.m. from OZ Tank 88 7 Water 8k 7 and F kp k 7 5k. 5k 8

49 The standard approach in collision uestions is to write down the euations gained when applying the principles of Conservation of Linear Momentum (CLM) and Newton s Eperimental Law of Restitution (NEL or NLR), and then what can be deduced from these. For P, : CLM m u m v + m v and NEL eu v v. Solving to determine the final speeds of P and P then yields ( m em ) m ( e) v u and v u. m m m m Similarly, for P, : CLM m u m v + m v and NEL eu v v, leading to m ( e) ( m em) v u and v u. m m m m If we now write X OP and Y OP initially, and euate the times to the following collisions at O, we have ( st collision): m m X m m Y m ( e) u m ( e) u and ( nd collision): m m X m m Y. m em u m em u Cancelling u s and ( + e) s m m X m m Y and m m X m m Y. (*) m m m em m em Dividing these two (or euating for X / Y) m m m m m em m m em, which simplifies to m. Finally substituting back into one of the euations (*) then gives m m X Y X Y. m m Rather surprisingly, however, the momentum euations turn out to be totally unnecessary here. Consider Collision P, : NEL eu v v Collision P, : NEL eu v v so that v v v v (*). X Y X Y Net, the two euated sets of times are and X v Y v and X v Y v. v v v v X v v Y v X Y from (*). Subtracting: v 9