2010 Paper 2 Q5 - coord geometry

The curve y = 5 – e^2x intersects the coordinate axes at the points A and B.

(i) Given that the line AB passes through the point with the coordinates (ln 5, k), find the value of k.

(ii) In order to solve the equation x = ln √(9-x), a graph of a suitable straight line in drawn on the same set of axes as the graph of y = 5 – e^2x.  Find the equation of the straight line.

At first thought, we might get stumped by the second question. That might paralyse your thinking for the first one. But, in all problems, we start from where we are and see how to get to where we want to go. So we don’t have to really worry about it now. But we need to look at solving the first one first. So let’s get that one started…

(i)

Since A and B are points on the x- and y-axes, we will find the coordinates of A and B.

At x-axis, y = 0

So,  0 = 5 – e^2x

e^2x = 5

2x = ln 5

x = ½ln 5

Hence, A(½ln 5, 0)

At the y-axis, x = 0

So, y = 5

Hence, B(0, 5)

Gradient of AB is

Therefore, the equation of AB is  y = (10/ln 5)x + c

When x = ½ln 5, y = 0.  Therefore, c = 5

So, the equation of AB is  ln5y = 10x + 5ln5

Hence, when x = ln 5,   ln5y = 10ln5 + 5ln5

Dividing throughout by ln5, we get   y = 15

Therefore k = 15.

(ii)  Now, let’s see what we can do about the next section.

The thing is, when you want to solve the equation of x = ln √(9-x) by drawing a straight line through the curve y = 5 – e^2x, what that means is we have to find a straight line y = mx + c, such that when

                mx + c = 5 – e^2x,

the resulting equation can be simplified to give x = ln √(9-x).

So, let us work with the above equation then.

                mx + c = 5 – e^2x

                e^2x = (5-c) – mx

                e^x = √[(5-c) – mx]

                x = ln√[(5 – c) – mx]

Equating this with x = ln √(9-x), we can see that

5 – c = 9

c = -4

m = 1

Hence, the equation of the straight line to be drawn on the same graph of y = 5 – e^2x is

                y = x – 4

2010 Paper 2 Q4 - Binomial theorem

(i) Given that the constant term in the binomial expansion of 

is 7, find the value of the positive constant k.


 

(ii)  Using the value of k found in part (i), show that there is no constant term in the expansion of

(i) For the first expansion, we recall the general term of a binomial expansion, which is

and this simplifies to

Hence, the term independent of x (constant term) is when 8 – 4r = 0

or when  r = 2

therefore, the constant term is  ⁸C₂ (-k)² = 7

                                                                28k² = 7

                                                                k = ½

(ii)  This is a tricky one. At first glance, one might think that the question is wrong. Afterall, if the constant term of the binomial expansion is 7 and if it is multiplied by a 1 from the expression (1 + x⁴), that will surely give a constant of 7. So why would there not be a constant term? Well, that is because we also need to look at what the x⁴ term multiplies with. If the corresponding x^-4 term is -7, then when the two constants are added together, it is zero!

So, let us now find out what the coefficient of x^-4 is in the binomial expansion.

For the term in x^-4, then 8 – 4r = -4

                                                4r = 12

                                                r = 3

Substitute r = 3 into the expression, we get the coefficient of x^-4 as  ⁸C₃(-½)³ = 56(-1/8) = -7!

Hence, we have shown that there would not be a constant term in the expansion.

Maths is cool, isn't it??

2010 Paper 2 Q3 - simultaneous exponential

Without using a calculator, solve, for x and y, the simultaneous equations

For questions like these, whenever you seem stumped, start with the most fundamental idea. In this case, I would look at the common base. For the first equation, the common base is 2; for the second equation, the common base is 3. Let us convert the above questions to their common base, then…

For the first equation, this converts to

Or,          5x + y = 0 ---- (1)

The next equation converts to

And this becomes

Multiplying by x, we get

x² – 12x – 3xy = 4 ----- (2)

From (1),             y = -5x   ---- (3)

Substituting (3) into (2), we get

x² – 12x – 3x(-5x) = 4

x² – 12x + 15x² – 4 = 0

16x² – 12x – 4 = 0

Dividing throughout by 4, we get

4x² – 3x – 1 = 0

(4x + 1)(x – 1) = 0

x = - ¼ or x = 1

When x = -¼, y = 5/4

When x = 5, y = 5.

 

2010 Paper 2 Q2 - min / max areas

The diagram shows a triangular piece of land PQR in which angle PQR = 90°, PQ = 8m and QR = 12m. A rectangle QUVW is to be used as the base of a greenhouse, where U, V and W lie on QR, RP and PQ respectively, QU = x m and QW = y m.

(i) Show that 

 

(ii) Express the area A m², of the base of the greenhouse in terms of x.

(iii) Given that x can vary, find the maximum value of A.

(i) Whenever you see a right angled triangle, or any other triangle for that matter, then the use of similar triangles must come to your mind quickly. In this case, we can indeed use similar triangles.

Since PQ = 8m, then PW = 8 – y m

So, using similar triangles,

(ii)  A = xy

A = x (8 – 2x/3)

A = 8x – 2x²/3  m²   (don’t forget your units!!)

(iii) Although we could use differentiation to find the maximum value of A, we note that this is a quadratic function, so we can just use completing the square to get it.

Since this is a negative x² function, it means that there is a max area.

Using completing the square method:

A = ⅔(12x – x²)

   = -⅔[(x – 6)² – 36]

   = -⅔(x – 6)² + 24

Hence, the max area is 24 m² and occurs when x = 6m

2010 Paper 2 Q1 - trignometric equations

We are now on our way to the second paper. Let’s look at the first question…

Solve the equation  3cot²θ + 10cosecθ = 5  for  0° ≤ θ ≤ 360°

You will realize at the very outset, that there are two trigonometric functions,  cot and cosec. We need to put the whole equation into one function only.  To do this, we need to refer to our identities. Luckily there is the identities formula given in the formula sheet (not that you did not know this at this time).  We will use the following identity:

cot²θ = cosec²θ – 1

So, substituting that into the above equation, we get

3(cosec²θ – 1) + 10cosecθ = 5 

3 cosec²θ – 3 + 10cosecθ – 5 = 0

3 cosec²θ + 10cosecθ – 8 = 0

Now, we have a quadratic equation in cosecθ, and so we can solve it using normal quadratic functions.

3 cosec²θ + 10cosecθ – 8 = 0

(3cosecθ – 2)(cosecθ + 4) = 0

cosecθ = 2/3  or cosecθ = -4

Remembering that cosecθ = 1/sinθ, then finding the reciprocal of each of the answers to cosecθ will give us the answer for sinθ.

Therefore,

sinθ = 3/2 (inadmissible)  or  sinθ = -¼

So we only solve for sinθ = -¼

The basic angle is arc sin (0.25) = 14.5°

since sinθ is negative, then θ must lie in the 3^rd and 4^th quadrants.

Therefore,

θ = 180° + 14.5°  ,   θ = 360° - 14.5° 

   =  194.5°                     = 345.5°

θ = 194.5° , 345.5°