KYUEM Mathematics Competition

1. Expand and simplify (3+2√5 )4

2. Find 4√97+56√3 in the form of a+b√33. Find the real roots of the equation x8−194 x4+1=0

(20 Marks)

1. Show that dydx(cot x )=cosec 2 x

2. Hence, find ∫ 1sin 2x

dx

3. Find ∫2cot x ln sin x dx(10 Marks)

1. Show that ∫0

2π

−cosn x sin x dx= 2n+1

2. Show that ∫0

π

ncot x sinn x−n tan xcosn xdx=0∨1, and state the conditions for

each possible solution

(20 Marks)

Lines and circles stated in each part of this question do not affect one another. I.e. lines ‘drawn’ in part one can be ignored in part two.

A circle, radius r and centre origin (0,0), is drawn in the Cartesian plane. A line,l1: y=−mx+c, where m and c are positive, real numbers, and c>r is also drawn. The line, l1, does not intersect the circle. Find the shortest distance between the circle and line l1 in terms of m ,r∧c.

A new line, l2: y=−mx+√2, is drawn. Line l2 touches the circle, radius 1 and centre (0,0), at one point (i.e. is tangent to the curve). Find the value of m, given that m is positive.

A new circle, radius 1 and centre (2,1), is drawn. A line, l3: y=mx+5, with a variable gradient is plotted. It is given that the line l3 intercepts the circle at two distinct points. Find the range of values of m for which the given condition is met.

(20 Marks)

This question concerns the foci of an ellipse.

An ellipse is drawn. This ellipse has the

equation e : x2

52+ y

2

32=1. For two given points,

the foci, an ellipse is the locus of points such that the sum of the distance to each focus (from the perimeter) is constant.

Given the above information, find the foci of ellipse e.

(10 Marks)

Expand and simplify (3+2√5 )4

Answer: 1561+696√5This was a very easy part of the question, however the risk of arithmetical error is high. Two simple approaches would be binomial expansion and considering that (3+2√5 )4=((3+2√5 )2 )2, with the latter being slightly more complicated, but

useful for those who are not confident with or aware of the calculation of (nr ).

Find 4√97+56√3 in the form of a+b√3

Answer: 2+√3To answer this part, candidates needed to recognise that this part is similar to part (i) in the sense that it involves a power of 4. Raising both sides to the power 4, then comparing coefficients is the easiest method to find the answer. Candidates should then obtain these two equations,

1. a4+18a2b2+9b4=972. a3b+3ab3=14

As it is quite difficult to solve these equations by substitution, candidates should consider the following inequality

a4+9b4<97

From here, candidates can find suitable values by trial and error, where the possible combinations are

a = 1, b = 1; a = 2, b = 1; a = 3, b = 1

The combinations should then be checked with equation (2.), to reach the conclusion that

a = 2, b = 1

Find the real roots of the equation x8−194 x4+1=0

Answer: 2+√3 ,2−√3 ,−2−√3 ,−2+√3Candidates should first use the substitution y=x4 to get the equation y2−194 y+1=0

Then, using the quadratic formula −b±√b2−4 ac2a

to get 194±√1942−42

More able candidates can then recognise this as 194±√1942−22

2=194±√ (194+2 ) (194−2 )

2=194 ±√196×192

2

By prime factorisation, it can be obtained that

196=22×72

192=26×3=(2¿¿3)2×3¿

Using those results

194±√196×1922

=194±√22×72× (23 )2×3

2=194±112√3

2

194±112 √32

=97+56√3

Returning to the original substitution,

y=97+56√3

x4=97+56√3

x=± 4√97+56√3From part (ii), x=±(2±√3)

An alternative to the above method would be to write 194±√1942−42

in the form

1942± √1942−4

2=97± √37632

2

Candidates should realise that the value of a in a+b√3 matches the result from part (ii), which may lead some towards correctly, but unjustly assuming that the simplified form is the result obtained form part (ii).

From here, candidates should write 56√3 and √1942−42

in the form √c or √c2

From the results, candidates can justly conclude that y=97+56√3, then deduce the values of x as above.

Show that dydx(cot x )=−cosec2 x

Candidates simply need to write cot x= cos xsin x and differentiate using quotient rule

v dudx

−u dvdx

v2, where u=cos x∧v=sin x, to obtain the answer. The identity

sin2 x+cos2 x=1 is needed for this question.

Hence, find ∫ −2sin 2x

dx

Answer: ln (cot x )+c

Noting that the question began with ‘Hence’, candidates should be able to recognise that the first and second parts are connected. First, candidates should expand sin 2 x=2 sin x cos x. The 2 in the numerator and denominator can

cancel out. Next, to obtain a fraction in the form of a

sin2 x×b, where b is a real

number, −1

sin xcos x× sin xsin x

= −1sin 2x

× sin xcos x

Then, it can be converted into the form f' ( x )f (x )

.

−1sin2 x

× sin xcos x

=−cosec2 xcot x

As it is in the form f' ( x )f (x )

, and ∫ f ' ( x )f (x)

dx=ln ( f ( x ) )+c, it can be deduced that

∫ −2sin 2x

dx=ln ¿

Find ∫2cot x ln sin x dxAnswer: ln 2sin x+c

A big clue in this question is the presence of ln sin xin the integral. The easiest

method to solve this is to recognise that cot x=dydx¿¿, and therefore immediately

recognise the integral to be in the form of f ' ( x )=2 dydxy and realise that the

original function is f ( x )= y2.

An alternative, and also quite simple method, is to use integration by parts. Taking

u=ln sin x∧v '=2cot x, the answer can be realised in merely 5 lines of working.

Show that ∫π

2π

−cosn x sin x dx= 2n+1

It is important to recognise that −sin x=dydx¿¿. With that result in mind,

candidates can find the original function by reverse differentiation. A good start

would be to consider that dydx(cosnx )=−ncosn−1 xsin x. Seeing that we have a

coefficient n instead of 1, and a power n−1 instead of n, it can be deduced that the initial power of cos x is n+1. But to eliminate the n+1 which appears after

differentiation of the original function, an additional 1n+1 needs to be added.

Therefore, the original function can be stated as f ( x )= 1n+1

cosn+1 x.

Then, complete the question by definite integration, where I=∫π

2π

−cosnx sin xdx .

I=[ 1n+1 cosn+1 x ]2π0 = 1n+1

cosn+12π - 1n+1cosn+1π

I= 1n+1

−( −1n+1 )

I= 2n+1

Show that ∫0

π

ncot x sinn x−n tan xcosn xdx=0∨−2, and state the conditions for each

possible solution, where n is a positive integer.

Initially, to reduce clutter, the n can be factorised out of the function to be

integrated, as it is just a constant. That gives us I=n∫0

π

cot x sinn x dx−tan x cosn x dx.

What we are focused on now is what remains. So, let g' (x)=cot x sinn x−tan x cosn x.

Firstly, expand cot x∧ tan x. So, g' (x)=cos xsin xsinnx− sin x

cos xcosn x.

Now note that g '(x ) can be further simplified into g' (x)=cos x sin n−1x−sin x cosn−1 x.

Notice that the function is now in the form of g' (x )=a dydxy+b dt

dxt.

Similarly to the above part, it can now be deduced that g ( x )=1nsinn x+ 1

ncosn x

Bringing back the constant taken out from the start, we find that I=[sinn x+cosn x ] π

0I=[sinn π+cosnπ ]−[sin n0+cosn0 ]

As cos π=−1, it is important to note that for even powers of n, cosnπ=1. Therefore, the value of I may change depending on the value of n. I=(0 )n+ (−1 )n−(0 )n−(1 )n

I=(−1 )n− (1 )n

From here, it can be seen that for even values of n, I=0. And for odd values of n, I=−2.

An ellipse is drawn. This ellipse has the equation

e : x2

52+ y

2

32=1,where e is∈the form

x2

a2+ y

2

b2=1. For two given points, the

foci, an ellipse is the locus of points such that the sum of the distance to each focus is constant.

Given the above information, find the foci of ellipse e.

Answer: (−4,0 ) ,(4,0)

Although candidates are dealing with an unfamiliar type of curve, finding the answer should be quite straightforward. Having being given a diagram, this question becomes even easier. Given that if the sum of the distance between a point to the ellipse to another point is constant, the two points are the foci, candidates can imagine a string being attached to each point. This string will always have the same length and can be dragged along the perimeter of the ellipse perfectly!

Aside from the formula c2=a2−b2, the foci can be found by finding the distance between the two foci, (−x ,0 )∧(x ,0), and two known points of the ellipse. The easiest points would be (0,3 )∧(5,0 ), which can be found either graphically or by substituting x=0∨ y=0 into the equation e. Once two equations in x have been obtained, it is possible to equate them together to obtain x=±4.