2010 Paper 2 Q10 - partial fractions and integrals

Given that

where A, B and C are constants, find the value of A and B and show that C = 0.

(ii) Differentiate ln(x² + 4) with respect to x.

(iii) Using the results from parts (i) and (ii), find

(i)  Since

Equating coefficients of x², we have

A + 2B = 3   --- (1)

Equating coefficients of x, we get

B + 2C = 4   --- (2)

Equating constants, we have

4A + C = -20   ----(3)

From (1), A = 3 – 2B --- (4)

Sub (4) into (3), we get     4(3 – 2B) + C = - 20

                                                12 – 8B + C = -20

                                                -8B + C = -32 ---(5)

                                                B + 2C = 4  --- (2)

(5) x 2                                    -16B + 2C = -64 --- (6)

(2) – (6)                                                17B = 68

                                                B = 4

Sub B = 4 into (2), we get C = 0 (shown)

From (3)   4A = -20

                A = -5

Hence, A = -5, B = 4 and C = 0

(ii)

(iii) So,

From (ii),

so finally,

2010 Paper 2 Q9 - coordinate geometry

The diagram shows a triangle ABC with vertices at A(0, 5), B(8, 14) and C(k, 15). Given that AB = BC,

(i) find the value of k,

A line is drawn from B to meet the x-axis at D such that AD = CD.

(ii) Find the equation of BD and the coordinates of D.

(iii) Show that the area of the triangle ABC is 2/7 of the area of the quadrilateral ABCD.

(i) Using Pythagoras’ theorem,

AB² = BC²

So,

8² + 9² = (k – 8)² + 1²

64 + 81 – 1 = (k – 8)²

144 = (k – 8)²

k – 8 = ± 12

Since k is positive, therefore,

k = 8 + 12 = 20

(ii) For AD = CD, the line BD must be perpendicular to AC.

Hence, the gradient of BD = - 20/10 = -2

Equation of BD is  y = -2x + c

At (8, 14),   14 = -16 + c

                c = 30

Hence, the equation of BD is  y = -2x + 30

For the coordinates of D, which lies on the x-axis, y = 0

Therefore  x = 15

Hence, D(15, 0)

(iii) There are several ways to go about this. One way, is to find the area of the triangle using the coordinates (remember the anti-clockwise method?) and then the area of the quadrilateral and then find their ratio. Can, but too long, and not a clever option.

The next way is to find the height of the triangle and find the area using ½ base x height. And then do the same for the quadrilateral. Again, not a clever option.

The last way is a variant of the second one. Since the base of the triangle is the same for the smaller one and the larger one (the quad is made up of 2 triangles), then, the way to calculate the ratio of the areas is simply find the ratio of BM/BD, where M is the midpoint of AC.

So, M = (10, 10)

BM = √(2² + 4²) = √(4 + 16) = √20 = 2√5

BD = √(7² + 14²) = √(49 + 196) = √245 = 7√5

Hence, BM/BD = 2√5/7√5 = 2/7  (shown)

Notice something here. I did not go about finding the actual answer for each length. The reason was because the answer, 2/7, is a nice rounded number. With the radical (√) numbers, we know we will not get a nice round number. Hence, it gave me a hint that I should keep the radical sign and I expected that they would cancel each other.

2010 Paper 2 Q8 - kinematics

Two particles, P and Q, leave a point O at the same time and travel in the same direction along the same straight line. Particle P starts with a velocity of 9 m/s and moves with a constant acceleration of 1.5m/s². Particle Q starts from rest and moves with an acceleration of a m/s², where a = 1 + t/2 and t seconds is the time since leaving O.  Find

(i) the velocity of each particle in terms of t,

(ii) the distance travelled by each particle in terms of t,

Hence find

(iii) the distance from O at which Q collides with P,

(iv) the speed of each particle at the point of collision.

Ahh…they rather naughty…gave the acceleration function. Hence, to get the velocity and displacement functions, we will need to integrate and find the constants of integration. OK, so let’s start…

(i)  The velocity function for P is quite straight forward…  v_P = 9 + 1.5t

and for Q it is:

When t = 0, v = 0 so c = 0

Hence,  v_Q = t + t²/₄.

(ii) The distance functions, denoted by s_P and s_Q , are obtained by integrating the velocity functions.

                s_P = 9t + 0.75t² + c

                s_Q = t²/₂ + t³/₁₂ + c

When t = 0, s = 0, so c = 0

Hence, s_P = 9t + 0.75t²   and         s_Q = t²/₂ + t³/₁₂

(iii) When both particles collide, it means that they will both have travelled the same distance. Hence, we will find the value of t when this happens and then find the distance travelled itself. So, we equate both the distance functions together and solve for t.

                9t + 0.75t² = t²/₂ + t³/₁₂

Multiplying throughout by 12 gives

                108t + 9t² = 6t² + t³

                t³ – 3t² – 108t = 0

                t² – 3t – 108 = 0

                (t – 12)(t + 9) = 0

                t = 12 or t = -9

Hence, the particles will collide at t = 12s

The distance travelled is  9(12) + 0.75(12)² = 216m

Hence, they are 216m from O when they collide.

(iv) to find the speed of each particle at each point of collision, we substitute t = 12s into the velocity functions.

Hence, P will be travelling at   9 + 1.5(12) = 27m/s  and

Q will be travelling at    12 + (12)²/4 = 48m/s.

2010 Paper 2 Q7 - area under the curve

The diagram shows part of the curve y = √(2x + 5)  passing through the point P and meeting the x-axis at the point Q. The line  x = 2 passes through P and intersects the x-axis at the point S. Lines from Q meet x = 2 at the points R and T such that QR is parallel to the tangent to the curve at P, and RS = ST. Find

(i) the equation of QR,

(ii) the area of the shaded region

(i) To find the equation QR, we will need the gradient of the tangent of the curve at the point where x = 2. To get the gradient, we will have to differentiate the equation of the curve.

When x = 2, gradient is 1/√(4+5) = 1/3

Therefore, the equation of QR is  y = (1/3)x + c

At the point Q, y = 0. Substituting this into the equation of the curve, we get

0 =  √(2x + 5)

2x = -5

x = -5/2

Substituting the coordinates of the point Q(-5/2 , 0) into the equation of QR, we get

0 = -5/6 + c

c = 5/6

Hence, the equation of QR is y = (¹/₃)x + ⁵/₆

(ii) For the area of the shaded region, we will have to find the area under the curve as well as the area of triangle QST. Let’s find the length of ST first. This is the same as SR.  Subsituting x = 2 into the equation of QR, we get  y = ²/₃ + ⁵/₆ = ⁹/₆ = ³/₂

Hence, the area of QST = ½(⁹/₂)(³/₂) = ²⁷/₈ units²

For area under the curve,

hence, the area of the shaded region is 30⅜ units².

2010 Paper 2 Q6 - circle properties

The diagram shows a point X on a circle and XY is a tangent to the circle. Points A, B and C lie on the circle such that XA bisects angle YXB and YAC is a straight line. The lines YC and XB intersect at D.

(i) Prove that AX = AB

(ii) Prove the CD bisects angle XCB.

(iii) Prove that the triangles CDX and CBA are similar.

Word to the wise: In questions like these, I normally find it useful to redraw the diagram in the answer sheet. That way, I can annotate on the diagram as part of the answer. But more importantly, when I have completed drawing it out as per the description, I have a better understanding of the diagram. Hence, I would strongly recommend that you do draw it out, even if it seems easy. Your understanding deepens a lot more when you do that.

Okay, now on to the answers…

(i) To show that AX = AB, we will need to prove that triangle XAB is isosceles.

Let angle YXA = x°

Angle YXA = Angle XBA   (alternate segment theorem)

Hence, angle AXB = angle XBA

Therefore, triangle XAB is isosceles and AX = AB.

(ii)  Angle YXB = Angle XCB (alt seg)

Therefore, angle XCB = 2x°

Since we have shown that angle AXB = angle XBA = x°

And that angle XCA = angle XBA (angles in the same segment) = x°

Therefore, angle ACB = angle XCA = x°

Hence, the line CD bisects the angle XCB.

(iii)  Since angle CXD = angle CAB  (A)

angle XCD = angle ACB (A)

Therefore, angle CDX = angle CBA (A)

Hence, triangle CDX is similar to CBA  (AAA)