STEP Paper I Questions

1)1998/1 How many integers between 10 000 and 100 000 (inclusive) contain exactly two different digits?(23 332 contains exactly two different digits but neither of 33 333 and 12 331 does)_____________________________________________________________________________________1) There are 36 different pairs of non-zero digits. Denote them by a and bFor any given pair a and b there are 5 ways of arranging one a and four bs or one b and four asand 10 ways of arranging two as and three bs or two bs and three ashence there are 30 36 1080 different 5 digit numbers using exactly two non-zero digits.With 0 as one of the digits, the other may be any of the remaining 9since the first (left most) digit cannot be 0 there is just one way of having one non-zero digit,4 ways of having two non-zero digits, 6 ways of having three and 4 ways of having four.hence number of 5 digit numbers using just two digits, one of which is 0 is 9 15 135And total number of 2-digit numbers, including 100 000 is 1080 + 135 + 1 = 1216_____________________________________________________________________________________2)1999/1 How many integers greater than or equal to zero and less than a million are not divisible by 2 or5? What is the average value of these integers?

How many integers greater than or equal to zero and less than 4179 are not divisible by 3 or 7? What is theaverage value of these integers?____________________________________________________________________________________2) number of integers z such that 0 z 1 000 000 divisible by 2 is 500 000Number divisible by 5 is 200 000 and number divisible by 10 is 100 000Hence, number not divisible by 2 or 5 is 1 000 000 – 600 000 = 400 000We may group these integers in pairs thus where each pair has a sum of 1, 999 999, 3, 999 9971 000 000 and there will be 200 000 such pairs, hence average value of these integers is200 0001 000 000

400 000 500 000Consider now the integers z such that 0 z 4179Working as before, the number of integers such that z is divisible by 3 is 1393Number divisible by 7 is 597 and by 21, is 199 so number divisible by 3 or 7 is 2388Grouping in pairs 1 & 4178, (2 & 4177), 4 & 4175) etc we have 1194 pairs of numberse ach with asum of 4179 so average value is 41791194

2388 41792 2089.5

_____________________________________________________________________________________

STEP Paper I Questions

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3)1999/4 Sketch the following subsets of the plane.x y(i) |x| |y| 1 :(ii) |x 1| |y 1| 1 :(iii) |x 1| |y 1| 1 :(iv) |x||y 2| 1._____________________________________________________________________________________3)

explanation

(i) area between x y 1 and the axes reflected in both axes.

(ii) area in (i) translated by 1 unit upwards and 1 unit to the right.

(iii) area between x y 1 and x -axis, reflected in both axes and translated 1 unit right and 1 unit down

(iv) The area between the rectangular hyperbola xy 1 and the axes translated 2 units upwards_____________________________________________________________________________________


1

1–1

–1

–11

1

2

0 2

0

–2

1(0,0–1) (2,–1)

(i) (ii)

(iii)

2

0

(iv)




4)2000/6 Show that and hence, or otherwise, indicate by meansx2 y2 x 3y 2 x y 2x y 1of a sketch the region of the x-y plane for which

x2 y2 x3y 2.Sketch also the region of the x-y plane for which x2 4y2 3x 2y 2.Give the co-ordinates of a point for which both inequalities are satisfied or explain why no such pointexists._____________________________________________________________________________________4) x y 2x y 1 x yx y 1 2x y 1 (x y)(x y) x y 2x 2y 2

as required x2 y2 x 3y 2x2 y2 x 3y 2 x y 2x y 1 0Drawing first x y 2 0 and x y 1 0 and noting that x y 2 and x y 1 must have

the same sign, we obtain the shaded region in the first graph on the next page.

Point of intersection is at 12 , 3

2By inspectionx2 4y2 3x 2y 2 (x 2y 1)x 2y 2so we draw x 2y 1 0 and x 2y 2 0which intersect at 3

2 , 14

Region satisfying inequality is shown shadedin diagram below

Clearly (2, 2) satisfies both inequalities.

_____________________________________________________________________________________


1

2

1

–1




5)2004/1 (i) Express (3 2 5 )3 in the form a b 5 where a and b are integers.

(ii) Find the positive integers c and d such that 3 99 70 2 c d 2.

(iii) Find the two real solutions of x6 198x3 1 0_____________________________________________________________________________________5). (i) 3 2 5 3 33 3(3)22 5 3.3(2 5 )2 (2 5 )3

27 54 5 180 40 5 207 94 5

(ii) 3 99 70 2 c d 2 99 70 2 (c d 2 )3

i.e. 99 70 2 c3 3c2d 2 6cd2 2d3 2 c3 2cd2 (3c2d 2d3 ) 2so we require c3 6cd2 99 and 3c2d 2d3 70d must be a factor of 70 and 2d3 70 d 1 or 2d 1 gives 3c2 68 so c is non integral. d 2 gives 6c2 54 c 3

Hence 3 99 70 2 3 2 2

(iii) x6 198x3 1 0 x3 12 198 1982 4

1982 4 1982 22 196 200 140 2 so x3 99 70 2hence, solutions are x 3 2 2 and 3 2 2 .__________________________________________________________________________________





6)2003/1 It is given that can be written in the form where p,q,r and s arer1

nr2 pn3 n2 rn s,

numbers. By setting obtain four equations that must be satisfied byn 1, 0, 1 and 2,p,q.r and s and hence show that

r0

nr2 1

6 nn 12n 1.

Given that can be written in the form show similarly thatr2

nr3 an4 bn3 cn2 dn e,

r0

nr3 1

4 n2n 12.

_____________________________________________________________________________________

6).Assume that r1

nr2 pn3 qn2 rn s

Taking n 1, 0, 1 and 2 we have

(1)1 p q r s (2) 1 s (3) 2 p q r s (4) 6 8p 4q 2r s

Adding (1) and (3) gives (5) 3 2q 2s so since s 1, q 12

So from (3) p r 12 and from (4) 8p 2r 3 so 6p 2 p 1

3 and r 16

Hence, r1

nr2 1

3 n3 12 n2 1

6 n 1 r0

nr2 1

3 n3 12 n2 1

6 n 16 n(2n2 3n 1)

i.e. r0

nr2 1

6 nn 12n 1

Assuming r2

nr3 an4 bn3 cn2 dn e then taking n 2,1, 0, 1 and 2 gives

(1) 8 16a 8b 4c 2d e (2) 9 a b c d e (3) 9 e (4) 8 a b c d e and

(5) 0 16a 8b 4c 2d e

(1) + (5) gives 32a 8c 2e 8, and (2) + (4) gives 2a 2c 2e 17

also e 9 so 32a 8c 10 and 2a 2c 1 24a 6 so a 14 and c 1

4

(2) now gives b d 12 and (1) gives 8b 2d 4 so 6b 3 b 1

2 and d 0

So r2

nr3 1

4 n4 12 n3 1

4 n2 9 r0

nr3 1

4 n4 12 n3 1

4 n2

n2

4 (n2 2n 1) 14 n2n 12

__________________________________________________________________________________





7)2002/1 Show that the equation of any circle passing through the points of intersection of the ellipse and the ellipse can be written in the formx 22 2y2 18 9x 12 16y2 25

x2 2ax y2 5 4a._____________________________________________________________________________________7). We solve the equations of the ellipses simultaneously to find the points of intersectionThe equations are so eliminating yx2 4x 2y2 14 0 and 9x2 18x 16y2 16 0x2 50x 96 0 x 2x 48 0 x 2 or 48x 2 2y2 2 0 so y 1, and x 48 2y2 0 so no solutionshence, ellipses intersect at 2, 1 and 2,1Any circle passing through these points must clearly have its centre on the x-axis, at a, 0 sayhence, radius of such a circle is (2 a)2 1equation of such a circle is thus x a2 y2 2 a2 1 i.e. x2 2ax y2 5 4a as required._____________________________________________________________________________________8)2000/1 To nine decimal places, log102 0.301029996 and log103 0.477121255

(i) Calculate By taking logs, or otherwise, log105 and log106 to three decimal places.show that 5 1047 3100 6 1047.Hence write down the first digit of 3 100

(ii) Find the first digit of each of the following numbers: 21000 , 210 000 , and 2100 000

_____________________________________________________________________________________8) (i) We are given the values of log102 and log103 and of course we know that log1010 1So we simply have to use the fact that 5 10 2 and 6 2 3Hence by the laws of logarithms, log 5 log 10 log 2 and log 6 log 2 log 3i.e. andlog105 log1010 log102 1 0.3010 0.699 (3 dp)log106 log102 log103 0.3010 0.4771 0.778 (3dp)Taking logs as suggested, we have log10(5 1047 ) log105 47 log1010 47 log105 47.699and similarly log10(6 1047 ) 47 log106 47.778Hence 47.699 log10(3100 ) 47.778 so, since it is the decimal part that determines the significantfigures in a number it follows that the first digit of 3100 must be 5

(ii) similarly it follows thatlog10(21000 ) 1000 log102 301.030 (3dp) so since log101 0.030 log102the first digit of 21000 is 1log10(210 000 ) 3010.300 so first digit of 210 000 is again 1

the first digit of log10(2100 000 ) 30102.9996 and since log109 2 log103 0.95 (2dp) 2100 000

must be 9._____________________________________________________________________________________





9)2000/2 Show that the coefficient of in the expansion of x12 x4 1x2

5x 1

x6. is 15,

and calculate the coefficient of x2.Hence, or otherwise, calculate the coefficients of in the expansion ofx4 and x38

x2 111(x4 x2 1)5 ._____________________________________________________________________________________9) (x4 x2 )5 x20 5x14 10x8 10x2 5x4 x10

and (x x1 )6 x6 6x4 15x2 20 15x2 6x4 x6

clearly the only product that produces x12 is (x10 )(15x2 )so coefficient of x12 in (x4 x2 )5(x x1 )6 is 1 15 15three products produce x2 i.e. (10x8 )(x6 ), (10x2 )(20) and (5x4 )(x6 )so coefficient of x2 is 10 200 5 215Noting that (x2 1)(x4 x2 1) x6 1 it follows thatalso x2 1 x(x x1 ) and x6 1 x2(x4 x2 )Hence, (x2 1)11(x4 x2 1)5 (x6 1)5(x2 1)6 x16(x4 x2 )5(x x1 )6

so coefficient of x4 in (x2 1)11(x4 x2 1)5 is the coeeficient of x12 in (x4 x2 )5(x x1 )6

i.e. 15, and similarly, coefficient of x38 is that of x22 in former expansion, i.e. 15_____________________________________________________________________________________10)2003/5 (i) In the binomial expansion of show that the term which is independent(2x 1/x2 )6 for x 0,of x is 240.

Find the term which is independent of x in the binomial expansion of (ax3b/x2)5n.

(ii) Let By considering the expansion of show that the term which isf(x) (x6 3x5 )1/2 . (1 3/x)1/2

independent of x in the expansion of in powers of f(x) 1/x, for |x| 3, is 27/16.

Show that there is no term independent of x in the expansion of in powers of x, for f(x) |x| 3_____________________________________________________________________________________10). (i) 2x 1

x26

has general term 6Cr(2x)6r 1x2

rwith power of x, 6 r 2r 6 3r

So term is independent of x if r 2 i.e. term is 6C2(2x)4 1x2

2 15 16 240

ax3 bx2

5nhas general term 5nCr(ax3 )5nr b

x2r

with power of x, 33n r 2r

So term independent of so term is x if 15n 5r 0 r 3n 5n3n a2nb3n

(ii) (x6 3x5 )1/2 x3 1 3x

1/2 x3 1 1

2 3x

12 1

22!

9x2

12 1

2 32

3!27x3 ...

so term independent of x is 2716

If x 3 then we have (x6 3x5 )1/2 3 x5/2 1 x3

1/2and since the expansion of 1 x

31/2

consists entirely of positive powers of x then there is no term independent of x._____________________________________________________________________________________





11)2002/6 A pyramid stands on horizontal ground. Its base is an equilateral triangle with sides of length a,the other three sides of the pyramid are of length b and its volume is V. Given that the formula for thevolume of a pyramid is show that 1

3 area of base height,V 1

12 a2(3b2 a2 )12 .

The pyramid is then placed so that a non-equilateral face lies on the ground. Show that the new height, h,of the pyramid is given by

h2 a2(3b2a2)4b2a2 .

Find, in terms of a and b, the angle between the equilateral triangle and the horizontal._____________________________________________________________________________________11). Area of base is 1

2 a2 sin60o a2 3

4height of pyramid is PV AV2 AP2

and AP 23 AD 2

3 a cos 30o 3 a3

so height is b2 a2

3 13

(3b2 a2 )12

and so volume of pyramid is 13

a2 34 1

3(3b2 a2 )

12

i.e. 112 a2(3b2 a2 )

12 as required.

base is now BCV with area 12 a b2 a2

4 14 a 4b2 a2

hence, since volume must be the same13

14 a(4b2 a2 )

12 h 1

12 a2(3b2 a2 )12

so h2 a4(3b2a2)144 144

a2(4b2a2) a2(3b2a2)

4b2a2 as required.angle between equilateral triangle and the horizontal is AD̂Vi.e. sin1 h

AD sin1 a(3b2a2)1/2

(4b2a2 )1/2 23 a

sin1 2(3b2a2)1/2

3 (4b2a2 )1/2

_____________________________________________________________________________________


A

B

C

V

Pa

bb

b

aD

A

B

C

V

D




12)2001/2 Solve the inequalities(i) 1 2x x2 2/x x 0(ii) 3x 10 2 x 4 x 10/3_____________________________________________________________________________________12). (i) 1 2x x2 2

x x 2x2 x3 2 if x 0 or x 2x2 x3 2 if x 0i.e. if x 0 then x3 2x2 x 2 0 x 1(x2 x 2) 0 x 1x 1x 2 0and by inspection x 1x 1x 2 0 for 1 x 2 or x 1 i.e. 1 x 2 since x 0similarly, if x 0 then we have x 1x 1x 2 0 1 x 1 or x 2but x 0 so we can only have 1 x 0 and putting results together we have1 2x x2 2

x for 1 x 0 or 1 x 2Alternative method.1 2x x2 2

x x2x2x32x 0 x32x2x2

x 0 x1x1x2x 0

and again, by inspection, considering possible ranges of values of x we have result as before.(ii) 3x 10 2 x 4 3x 10 x 4 2 3x 10 2 (3x 10)(x 4) x 4 4 4x 10 2 3x2 22x 40 16x2 80x 100 12x2 88x 160 4x2 8x 60 0 x2 2x 15 0 x 5x 3 0 x 5 < 3 or x 5checking these, x 3 gives 3x 10 1 contradicting the condition x 10/3he3nce, solution is x 5_____________________________________________________________________________________13) 2006/2 A small goat is tethered by a rope to a point at ground level on a side of a square barn whichstands in a large horizontal field of grass. The sides of the barn are of length 2a and the rope is of length4a. Let A be the area of grass that the goat can graze. Prove that

and determine the minimum value of A.A 14a2

_____________________________________________________________________________________13).

If the goat is tethered to a point distance x m from one corner of the barn then the area that can be grazedis that enclosed by the curves in the diagram.i.e. A 2 1

4 (4a)2 14 (2a x)2 1

4 (4a x)2 14 (2a x)2 1

4 x2

4 (32a2 4a2 4ax x2 16a2 8ax x2 4a2 4ax x2 x2 )

4 (56a2 8ax 4x2 ) (14a2 2ax x2 ) (x a)2 13a2

clearly this has a minimum value of 13a2 when x a and since 0 x a, the maximum value is14a2 hence, A 14a2 as required and minimum value of A is 13a2____________________________________________________________________________________


x 4a–x

2a–x

2a+x

x

2a

2a




14)1998/4 Prove that the rectangle of greatest perimeter which can be inscribed in a given circle is asquare.

The result changes if, instead of maximising the sum of the lengths of the sides of the rectangle, we seek tomaximise the sum of the powers of the lengths of those sides forn th

What happens if Justify your answers.n 2. n 2?. What happens if n 3?_____________________________________________________________________________________14) Let one side of the rectangle be x and the diameter d then other side is d2 x2

Perimeter is thus 2 x d2 x2 so we require to maximise x d2 x2

Let f(x) x d2 x2 then fx 1 xd2x2 0 if d2 x2 x 2x2 d2 x d

2

fx d2x2 x2d2x21/2

d2x2 d 2

d2x23/2 0 so f d2

is a maximum.

x d2 x2 1

2 d2 d2 x2 12 d2 and so the rectangle is a square.

Let g(x) x2 d2 x2 then g(x) is constant.Now let h(x) x3 d2 x23/2 hx 3x2 3xd2 x21/2 0 if x d

2

But hx 6x 3d2 x21/2 3x2

d2x21/2 6x 3d 2x23x2

d2x21/2 6d2 0 0

So the value is now a minimum._____________________________________________________________________________________





15)2008/1 What does it mean to say that a number is irrational?Prove by contradiction statements A and B below, where p and q are real numbers.A: If pq is irrational, then at least one of p and q is irrational.B: If p + q is irrational, then at least one of p and q is irrationalDisprove by means of a counterexample statement C below, where p and q are real numbers.C: If p and q are irrational, then p + q is irrational. If the numbers are irrational, prove that at most one of the numberse, , 2, e2, and e

Is rational. e, e, 2 e2, 2 e2

_____________________________________________________________________________________15). An irrational number is one which cannot be expressed in the form x

y where x and y are integers.A. Suppose pq is irrational but p and q are both rational, i.e. p a

b and q cd where a, b, c and d are

integers then pq acbd and ac and bd are integers so pq is rational, a contradiction, hence at least one of

p and q must be irrational.B. Suppose p q is irrational with p and q both rational as before, then p q adbc

bd which is rational,again a contradiction so at least one of p and q must be irrational.C. For a counterexample consider p 1 2 and q 1 2 then p and q are both irrational but p q 2which is rational.We are given that , e, 2, e2 and e are all irrational.Suppose two of e, e, 2 e2 and 2 e2 are rational. We consider the six possible cases:(a) e and e rational ( e) ( e) 2 which is not rational, a contradiction.(b) e and 2 e2 rational 2e2

e e is rational and we have case (a) again.(c) e and 2 e2 rational 2e2

e e is rational and we have case (a) again.(d) 2 e2 and 2 e2 both rational (2 e2 ) (2 e2 ) 22 which is not rational, contradiction(e) e and 2 e2 rational ( e)2 rational ( e)2 (2 e2 ) is rational, i.e. 2e is rational but at least one of and e is irrational, so again a contradiction.(f) ) e and 2 e2 rational e (2 e2 ) ( e)2 2e so a contradiction as in (e)This covers all cases of two rational numbers so no two of the numbers e, e, 2 e2 & 2 e2

can be rational and hence at most one of them is rational._____________________________________________________________________________________





16)2007/3 Prove the identities cos4 sin4 cos2 and cos4 sin4 1 12 sin22.Hence, or otherwise evaluate

0

2 cos4 d and 0

2 sin4 d

Evaluate also 0

2 cos6 d and 0

2 sin6 d

_____________________________________________________________________________________16). cos4 sin4 (cos2 sin2)(cos2 sin2) cos 2cos4 sin4 (cos2 sin2)2 2 sin2 cos2 1 1

2 sin22so cos4 1

2 1 12 sin22 cos 2 1

2 18 (1 cos 4) 1

2 cos 2

so 0

/2cos4d 3

8 1

32 sin4 14 sin2 0

/2 3

16

0

/2(cos4 sin4)d 1

2 sin2 0/2

0 so 0

/2sin4d

0

/2cos4d 3

16

Similarlycos6 sin6 (cos2 sin2)(cos4 cos2 sin2 sin4) cos 2 1 1

2 sin22 14 sin22 cos 2 1 1

4 sin22cos6 sin6 (cos2 sin2)(cos4 cos2 sin2 sin4) 1 1

2 sin22 14 sin22 1 3

4 sin22cos6 1

2 cos 2 1 14 sin22 1 3

4 sin22

so 0

/2cos6d 1

2 0

/2cos 2 1 1

4 sin22 1 34 sin22 d

12

0

/2cos 2 1

4 sin22 cos2 1 38 (1 cos4) d

12

12 sin2 1

24 sin32 58

332 sin 4 0

/2 5

32

0

/2(cos6 sin6)d

0

/2cos 2 1 1

4 sin22 d 12 sin2 1

24 sin32 0/2

0

so 0

/2sin6d

0

/2cos6d 5

32

_____________________________________________________________________________________





17)2005/4 (a) Given that cos 35 and that 3

2 2, show that sin 2 2425 , and evaluate cos 3.

(b) Prove the identity tan3 3 tantan3

13 tan2 .Hence evaluate tan, given that tan3 11

2 and that 4 2 .

_____________________________________________________________________________________17). (a) cos 3

5 and 32 2 sin 4

5 so sin2 2 45

35 24

25cos2 2cos2 1 18

25 1 725

so cos 3 cos 2 cos sin2 sin 21125 96

125 117125

(b) tan3 tan(2 ) tan 2tan 1tan 2 tan

2 tan 1tan2

tan

1 2 tan21tan2

2 tan tantan31tan22 tan2 3 tan tan3

13 tan2

so tan3 112 3 tantan3

13 tan2 112 i.e. 6 tan 2 tan3 11 33 tan2

or 2 tan3 33 tan2 6 tan 11 0By inspection tan 1

2 satisfies this equation so (2 tan 1) is a factor4

2 tan 1 so 2 tan 1 0 so removing this factor we have

tan2 16 tan 11 0 tan 12 16 256 44 1

2 (16 300 ) 8 5 3so since tan 1 we must have tan 8 5 3_____________________________________________________________________________________18)2004/5 The positive integers can be split into five distinct arithmetic progressions, as shown:

A: 1, 6, 11, 16, ...B: 2, 7, 12, 17, ...C: 3, 8, 13, 18, ...D 4, 9, 14, 19, ...E: 5, 10, 15, 20, ...

Write down an expression for the value of the general term in each of the five progressions. Hence provethat the sum of any term in B and any term in C is a term in E.Prove also that the square of every term in B is a term in D. State and prove a similar claim about thesquare of every term in C.(i) Prove that there are no positive integers x and y such that .x2 5y 243 723.(ii) Prove also that there are no positive integers x and y such that x4 2y4 26 081 974._____________________________________________________________________________________18). A: un 5n 4, B: un 5n 3, C: un 5n 2. D: un 5n 1, E: un 5nWriting a, b, c, d, e for arbitrary members of sets A,B,C,D,Eb c 5x 3 5y 2 for some integers x and y 5x y 5 5x y 1 i.e. a term of E.b2 5x 3 25x2 30x 9 55x2 6x 2 1 for some x i.e a member of Dc2 (5x 2)2 25x2 20x 4 55x2 4x 1 1 i.e. a member of DSimilarly a2 5x 42 25x2 40x 16 55x2 8x 4 4 i.e. a member of Aand d2 5x 12 25x2 10x 1 55x2 2x 1 4 i.e also type Aand obviously e2 is of type E(i) x2 must be of type A,D or E so x2 5y is of type A,D or EBut 243723 is of type C and so cannot be x2 5y(ii) Since x2 is type A,D or E, x4 must be type A or ESimilarly y4 is type A or E so 2y4 is type B or Ehence, x4 2y4 is of type A,B or E but 26081974 is type D so cannot be x4 2y4

_____________________________________________________________________________________





19)2001/5 Show that (i) (ii) (for t 0). 01 11tx2 dx 1

1t , 01 2x1tx3 dx 1

1t2 ,Noting that the right hand side of (ii) is the derivative of the right hand side of (i) , conjecture the value of

01 6x2

1x4 dx(You need not verify your conjecture)_____________________________________________________________________________________19). (i) 0

1 1(1tx)2 dx 1

t1tx 0

1 1

t(1t) 1t

tt1t

11t

(ii) 01 2x

(1tx)3 dx taking 1 tx y we have dydx t, x 0 y 1 and x 1 y 1 t

so 01 2x

(1tx)3 dx 11t 2y1

ty31t dy 1

1t 2

t2y2 2

t2y3 dy 2t2y

1t2y2 1

1t

2t21t

1t21t2 2

t2 1t2

21t11t2

t21t2 t2

t21t2 1

1t2We note that the right hand side of (ii) is the derivative of the right hand side of (i) w.r.to tand the integrand in (ii) is the derivative of the integrand in (i) w.r.to t so since 6x2

1tx4 is the derivative

w.r.to t of 2x1tx3 we conjecture that 0

1 6x2

1tx4 dx is ddt

11t2 2

1t3

and hence that 01 6x2

1x4 dx is 223 1

4 (by putting t 1_____________________________________________________________________________________20)2002/2 where m , n are both integers greater than 1. Show that the curve Let f(x) xmx 1n,y f(x)has a stationary point with By considering show that this stationary point is a maximum if0 x 1. fx,n is even and a minimum if n is odd.Sketch the graphs of in the four cases that arise according to the values of m and n.f (x)_____________________________________________________________________________________20). f(x) xmx 1n fx mxm1(x 1)n nxm(x 1)n1 xm1x 1n1[mx 1 nx]so fx 0 when x 0, 1 or m

mn and since m, n are both positive, 0 mmn 1

so there is a stationary point between 0 and 1 as required.fx m 1xm2x 1n1[m nx m] n 1xm1x 1n2[m nx m]

m nxm1x 1n1 but at stationary point [m nx m] 0 sof(x) m nxm1x 1n1 also x m

nn x 1 nmn and so

f(x) m n( mmn )m1( n

mn )n1 which is positive if n 1 is even, i.e. if n is odd, and negativeif n is even, i.e. f(x) is a maximum if n is even and a minimum if n is odd.

x 0 or x 1 then f x 0 so there is either a turning point or a point of inflexion at 0, 0 & 1, 0m and n both even f(x) 0 for x 0 and x 1 also for 0 x 1 so we must have minimumturning points at 0, 0 and 1, 0Similar reasoning shows that we have a maximum at 0, 0 and an inflexion at 1, 0 if m is even

.and n oddInflexion at 0, 0 and minimum at 1, 0 if m is odd and n even andinflexions at 0, 0 and 1, 0 if m and n are both odd.(i) (ii) m and n both even m even, n odd


0 1

0 1




(iii) (iv) m odd, n even m and n both odd

_____________________________________________________________________________________21)1998/2 Show, by means of a suitable change of variable, or otherwise, that

0

f (x2 1)1/2 x dx 12 1

(1 t2 )f(t)dt.

Hence, or otherwise, show that0 (x2 1)1/2 x 3

dx 38 .

_____________________________________________________________________________________21) To obtain f(t) in the transformed integral we must obviously take t (x2 1)1/2 xi.e. t x (x2 1)1/2 t x2 x2 1 t2 2tx 1 so x t21

2tx 0 t 1 and x t so we also get the correct limits.

Differentiating our substitution, dxdt

4t2(t21)4t2 t21

2t2 12 (1 t2 )

thus the given integral becomes 1

12 (1 t2 ) f(t)dt 1

2 1

1 t2f(t)dt

Hence, 0

(x2 1)1/2 x 3

dx 12

1

t31 t2dt 12 1

2t2 1

4t4 1

3

8

_____________________________________________________________________________________22)1998/3 Which of the following statements are true and which are false? Justify your answers.

(i) a ln b b ln a for all a, b 0.(ii) cossin sincos for all real .(iii) There exists a polynomial P such that |P() cos| 106 for all real .(iv) x4 3 x4 5 for all x 0.

_____________________________________________________________________________________22) (i) True, since a ln b (e ln a ) ln b (e ln b ) ln a b ln a

(ii) False, e.g. cossin cos 0 1 but sincos sin1 1(iii) False, P() will tend to as and cos 1(iv) True since x4 3 x4 5 x4 2 x4 (x2 x2 )2 0 x4 3 x4 5_____________________________________________________________________________________


0 1

0 1




23)1999/2 A point moves in the x-y plane so that the sum of the squares of its distances from the threefixed points Find the equation of the locus of the point and(x1, y1 ), (x2, y2 ) and (x3, y3 ) is always a2 .interpret it geometrically. Explain why cannot be less than the sum of the squares of the distances of thea2

three points from their centroid.

[ The centroid has co-ordinates (x, y) where 3x x1 x2 x3, 3y y1 y2 y3,_____________________________________________________________________________________

23) Let point be x, y then i1

3[x xi2 y yi2 ] a2

i.e. 3x2 2xx1 x2 x3 x12 x2

2 x32 3y2 2yy1 y2 y3 y1

2 y22 y3

2 a2

x2 y2 23 x1 x2 x3x 2

3 y1 y2 y3y 13 x1

2 x22 x3

2 y12 y2

2 y32 a2 0

Which is the equation of a circle, centre and radius 13 x1 x2 x3, 1

3 y1 y2 y3 i.e. (x, y)

x2 y2 13 x1

2 x22 x3

2 y12 y2

2 y32 a2

12

i.e. 3r2 3x2 3y2 x12 x2

2 x32 y1

2 y22 y3

2 a2 which is positive ifa2 x1

2 x22 x3

2 y12 y2

2 y32 3x2 y2

Sum of squares of the distances of the three points from the centroid is (x1 x)2 (y1 y)2 (x2 x)2 (y2 y)2 (x3 x)2 (y3 y)2

x12 x2

2 x32 y1

2 y22 y3

2 3x2 3y2 2xx1 x2 x3 2yy1 y2 y3i.e. x1

2 x22 x3

2 y12 y2

2 y32 3x2 3y2 6x2 6y2 a2 by previous work

So cannot be less than the sum of the squares of the distances of the three points from the centroid.a2

_____________________________________________________________________________________24)1999/3 The n positive numbers x1, x2, ..., xn, where n 3, satisfy

x1 1 1x2 , x2 1 1

x3 , ... xn1 1 1xn , and also xn 1 1

x1 .Show that

(i) x1, x2, ..., xn 1,(ii) x1 x2

x2x3x2x3 ,

(iii) x1 x2 , , , xn._____________________________________________________________________________________24) (i) xk 0 so xk1 1 1

xk 1 for all k

(ii) x1 x2 1 1x2 1 1

x3 1x2

1x3 x3x2

x2x3 x2x3x2x3

(iii) similarly, x2 x3 x3x4x3x4 x1 x2

x3x4

x2x32x4

, x3 x4 x4x5x4x5 x1 x2

x4x5

x2x32x4

2x5etc and finally xn1 xn

xnx1xnx1 and xn x1

x1x2x1x2 so x1 x2 1n x1x2

x12x2

2x32...xn2

And hence, since x12x2

2x32...xn

2 1 by (i) we must have x1 x2 and it follows immediately that x2 x3,x3 x4 etc and in fact x1 x2 x3 ... xn

(taking positive root since thus x1 1 1x1 x1

2 x1 1 0 x1 1 5

2 x1 1_____________________________________________________________________________________





25)1999/6 (i) Find the greatest and least values of distinguishing carefully betweenbx a for 10 x 10,the cases b 0, b 0 and b 0. (ii) Find the greatest and least values of cx2 bx a, where c 0, for 10 x 10,distinguishing carefully between the cases that can arise for different values of b and c._____________________________________________________________________________________25) (i) If b 0, bx a is constant and equal to a so greatest and least values are both aIf b 0, bx a is decreasing so least value is at x 10, i.e. a 10b and greatest is at x 10,i.e. a 10bIf b 0, bx a is an increasing function so greatest value is at x 10, i.e. 10b a and least is at x 10i.e. a 10b(ii) Let if f(x) cx2 bx a c 0 we have the same as in (i)if c 0, then the apex of the parabola is at b

2c , 4acb2

4cif b

2c 10, least value is at x 10 i.e. 100c 10b a and greatest is 100c 10b a10 b

2c 0 least value at x b2c , i.e. b2

4c b2

2c a a b2

4c and greatest 100c 10b a0 b

2c 10 least value a b2

4c and greatest 100c 10b a b

2c 10 least value 100c 10b a and greatest 100c 10b a_____________________________________________________________________________________26)1999/7 Show that sin(k sin1x), where k is a constant, satisfies the differential equation

(*)(1x2) d2ydx2 x dy

dx k2y 0In the particular case when find the solution of equation (*) of the formk 3,

y Ax3 Bx2 CxD,that satisfies y 0 and dy

dx 3 at x 0.Use this result to express sin3 in terms of powers of sin._____________________________________________________________________________________26) y sink sin1x dy

dx 11x2

cos(k sin1x) i.e. 1 x2 dydx k cosk sin1x 0

hence, differentiating again, 1 x2 d2ydx2

x1x2

dydx

k2

1x2 sink sin1x 0

so, multiplying through by 1 x2 and noting that y sinksin1x we have1 x2

d2ydx2 x dy

dx k2y 0 as required

k 3 1 x2d2ydx2 x dy

dx 9y 0, we try y Ax3 Bx2 Cx Dsubstituting in d.e we have 1 x26Ax 2B x3Ax2 2Bx C 9Ax3 Bx2 Cx D 0equating coefficients of powers of x, for x3 , 6A 3A 9A 0for x2, 2B 2B 9B 0 B 0, for x, 6A C 9C 0 C 3

4 Aand from constant terms, 2B 9D 0 D 0 so general solution is y A

4 4x3 3xclearly x 0 at y 0, and dy

dx A4 12x2 3 so dy

dx 3 at x 0 12 3A so A 4hence, solution is y 3x 4x3

now take sin1x so y sin3 is a solution of the equation by the first part of the question.so since x sin we have sin 3 3 sin 4 sin3_____________________________________________________________________________________





27)2000/3 For any number x, the largest integer less than or equal to x is denoted by [x]. For example 3.7 3 and 4 4.

Sketch the graph of y x for 0 x 5 and evaluate 05xdx.

Sketch the graph of y [ex ] for 0 x lnn, where n is an integer, and show that0

ln n[ex ]dx n ln n ln(n!)_____________________________________________________________________________________27) graph is a step graph as shown

0

5[x]dx is the area under the graph

i.e. integral 1 2 3 4 10

y [ex ] will also be a step graph with the stepsoccurring at x ln2, x ln3 etc.finishing at lnn

hence 0

ln n[ex ]dx sum of rectangles

i.e. ln2 ln1 2ln 3 ln 2 ... n 1lnn lnn 1 n 1 lnn (ln2 ln3 .... lnn 1) n 1 lnn lnn 1! n ln n ln n lnn 1! lnn lnn!_____________________________________________________________________________________


0 1 2 3 4 5

1

234

0 ln1 ln2 ln3 ln4

12

3

ln n

n–1




28)2000/5 Arthur and Bertha stand at a point O on an inclined plane. The steepest line in the plane throughO makes an angle with the horizontal. Arthur walks uphill at a steady pace in a straight line which makesan angle with the steepest line. Bertha walks uphill at the same speed in a straight line which makes anangle with the steepest line (and is on the same side of the steepest line as Arthur). Show that, whenArthur has walked a distance d, the distance between Arthur and Bertha is Show also2d sin 1

2 ( ) .that, if the line joining Arthur and Bertha makes an angle with the vertical, where ,

cos sin sin 12 ( )

_____________________________________________________________________________________28) The secret here is a good clear diagram Let Arthur and Bertha be at points A and Brespectively, then, by elementary trigonometry

the modulus symbol beingAB 2d sin 12

Required since we do not know which of and is the larger.Height of A above horizontal plane through

and of Bstarting point is AN d cos sinBM d cos sin

hence, angle of AB to vertical is given by cos ANBMAB

d cos sin d cos sin 2d sin 1

2 sin cos cos

2 sin 12

i.e. cos sin 2 sin 1

2 () sin 12

2 sin 12 b

sin sin 12 ( ) as required

_____________________________________________________________________________________29)2000/7 Show that

11 xex dx 1

0 xexdx 01 xexdx

and hence evaluate the integral.Evaluate the following integrals:

(i) 04 x3 2x2 x 2 dx;

(ii)

sinx cos x dx._____________________________________________________________________________________29) f(x) ax x3

1x2 fx a 3x21x22x4

1x22 a x43x2

1x22

Now consider 98

x43x2

1x22 918x29x48x424x2

81x22 x46x2981x22

x232

81x22 0 x43x2

1x22 98

(Note that it is usually easier to show that a b 0 rather than a b

i.e. fx a 98 if a 9

8 then fx 0 for all x_____________________________________________________________________________________


A

Bd

d N M




30)2001/8 Given that y x and y 1 x2 satisfy the differential equation(*)d2y

dx2 p(x) dydx q(x)y 0

show that p(x) 2x(1 x2 )1 and q(x) 2(1 x2 )1.Show also that ax b(1 x2 ) satisfies the differential equation for any constants a and b.

Given instead that y cos2(x2/2) and y sin2(x2/2) satisfy the equation (*), find p(x) and q(x)._____________________________________________________________________________________30). If y x and y 1 x2 satisfy d2y

dx2 p(x) dydx q(x)y 0 then

0 p(x) xq(x) 0 and 2 2xp(x) (1 x2 )q(x) 0so 2xp(x) 2x2q(x) 0 and 2xp(x) (1 x2 )q(x) 2 (1 x2 )q(x) 2 q(x) 2

1x2

and p(x) xq(x) 2x1x2 as required..

y ax b1 x2 dydx a 2bx and d2y

dx2 2b so substituting in d2ydx2 p(x) dy

dx q(x)y we have

2b 2xa2bx1x2 2(axb1x2)

1x2 2b1x22xa2bx2axbbx2

1x2

2b2bx22ax4bx22ax2b2bx2

1x2 0 so ax b1 x2 satisfies diff. eq.n for any a, bIf y cos2(x2/2) and y sin2(x2/2) are solutions then so also arecos2(x2/2) sin2(x2/2) 1 andcos2(x2/2) sin2(x2/2) cos(x2 )y 1 gives q(x) 0 and then y cos(x2 ) gives y 2x sin(x2 ), y 2 sin(x2 ) 4x2 cos(x2 ) so2 sin(x2 ) 4x2 cos(x2 ) p(x)2x sin(x2 ) 0 p(x) sin(x2)2x2 cos(x2)

x sin(x2) 1x 2x cot(x2 )

____________________________________________________________________________________





31)2001/6 A spherical loaf of bread is cut into parallel slices of equal thickness. Show that, after anynumber of slices have been eaten, the area of crust remaining is proportional to the number of slicesremaining.

A European ruling decrees that a parallel-sliced spherical loaf can only be referred to as ‘crusty’ if the ratioof volume V (in cubic metres) of bread remaining to area A (in square metres) of crust remaining after anynumber of slices have been eaten satisfies V/A <1. Show that the radius of a parallel-sliced spherical crustyloaf must be less than 2 2

3 metres.

[The area A and volume V formed by rotating a curve in the x - y plane round the x - axis from to x a are given by x a t

]A 2 aat y 1 dy

dx2

12

dx, V aat

y2dy_____________________________________________________________________________________31).. Consider the loaf formed by rotating the curve x2 y2 a2

about the x axis with a section cut off by the plane x t a (see diagram)

Surface area of crust remaining is then 2 ata y 1 dy

dx2

dx

i.e. A 2 ata y 1 x2

y2 dx since dydx

xy

so A 2 ata y2 x2 dx 2 a

ta adx 2[ax]ata

2ata a2 2atSo area remaining is proportional to length of loaf remaining and hence to number of slices remainingsince slices are of equal thickness.Volume remaining a

ta y2dx ata(a2 x2 )dx a2x 1

3 x3ata

i.e. V a2t a 13 t a3 a3 1

3 a3 a2t a3 13 t3 at2 a2t 1

3 a3 23 a3

at2 13 t3

t23at3

so VA

t3at6a 1 if 3at t2 6a t2 3at 6a 0

which is true for any t if 9a2 24a 0 (discriminant of quadratic)i.e. 0 a 24

9 so a 83 as required.

_____________________________________________________________________________________


–a –a+t




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