2010 Paper 1 Q5 - curve sketching of absolute functions

(i) Sketch the curve of y = |9 – x²| for  -5 ≤ x ≤ 5.

(ii)  Find the x-coordinates of the points of intersection of the curve y = |9 – x²| and the line y = 27.

This is a fundamental Sec 3 curve sketching question. It is usually paired up with quadratic functions so that we will have to do some completing the square. That’s right folks…completing the square so that we can find the turning point. 

Fortunately, in this case, we do not need to complete the square because there are roots for the quadratic function and we can simply calculate the turning point.

So, let’s get on with it…

(i)  Let y = 9 – x²

Therefore,  y = (3 + x)(3 – x). Hence, the roots are x = 3 and -3.

The turning point will occur midpoint between -3 and 3, and that would be at x = 0.

Hence, when x = 0, y = 9.

When x = 5, then y = 9 – 25 = - 16

But for the absolute value function of y, then when x = 5 or -5, y = 16.

With all these information, we will now be able to sketch the curve of y = |9 – x²| for  -5 ≤ x ≤ 5.

(ii)  To find the x – coordinates of the intersection of the curve with y = 27, we simply let y = - 27 for the quadratic function, solve for x. Why y= -27? Because, remember the function is an absolute value function. The portion of the curve below the x-axis got reflected up. Since for y = 27, it would actually have been in the y = -27 if it had not been reflected up. Hope you understood that. If not, just drop me a line again.

So,  to solve:

                                9 – x² = -27

                                x² = 36

                                x = ± 6

2010 Paper 1 Q4 - differentiation of trig functions

Today we will do Q4, 5 and 8. We had already done Q6 and for Q7, it would require plotting on the graph paper, which I shall skip for technical reasons. We can discuss this at tuition class if we want to.

OK, Q4 first. It says:

A curve has the equation y = sin 2x – 3 cos x.

(i) Find the gradient of the curve when x = π/6

(ii) When x is increasing at a constant rate of 0.06 units per second, find the rate of change of y when x = π/6.

(i)When we see the question talking about gradients, we know we need to have the gradient function. The gradient function is the differential. Hence, we will need to find dy/dx.

Hence,  if  y = sin 2x – 3 cos x, then

When x = π/6

(ii)  This second part is talking about rates of increase.  This means that we have to go into rate of change which is always stated as:

Given that dy/dx = 2.5, and that dx/dt = 0.06 units per second, then

therefore,

Therefore, rate of change of y = 2.5 x 0.06 = 0.15 units per second.

2010 Paper 1 Q3 - Inequalities

OK…let’s do Q3 and call it a day in terms of A maths blogging….

3.  Using a separate diagram for each part, represent on the number line the solution set of

(i) 3(2 – x) < x + 18,

(ii)  3(x² – 5) > x – 1

State the set of values of x which satisfy both these inequalities

For each of these, we need to simplify each of the inequalities. Let’s do the first one first:

(i)            3(2 – x) < x + 18

                6 – 3x < x + 18

                -3x – x < 18 – 6

                -4x < 12

                4x > -12

                x > -3

Hence, the number line looks like this:

(ii)  for the second inequality, we must remember that this is a quadratic inequality. So we must be careful with the signs.

3(x² – 5) > x – 1

3x² – 15 > x – 1

3x² – x – 14 > 0

(3x – 7)(x + 2) > 0

Hence the number line is:

 Which becomes

Hence, for the set of values of x which satisfies both inequalities, we simply look for the values of x which has an overlap. In this case, when x > 3, both inequalities are satisfied.

Hence, the solution set is {x : x Є R,  x > 3 }

Do remember that you have to put it in set notation, as they asked for the solution set. If not, you would lose marks. And since this is a one mark question, you might even lose half. Or worse, all of it!

See you tomorrow when we talk about another 3 questions….

2010 Paper 1 Q2 - Trig functions & intregration

2.  (i) Show that (sin x + cos x)² = 1 + sin 2x.

     (ii) Hence find, in terms of π, the value of

(i) We need to remember our identities for this one. Whenever we are asked to “show”, we need to remember that we have to use either the left-hand side (LHS) and work our way to the right-hand side (RHS) of the equation, or vice versa.

In this case, we will start with the LHS.

Why LHS? Well, it is because the LHS will be able to provide us with more information to work with. Hence, whenever you need to prove, always choose the one that will be able to provide more information.

So here goes…

Taking the LHS,

(sin x + cos x)² = sin² x + 2sinxcosx + cos² x

                                = sin² x + cos² x + 2sinxcosx

                                = 1 + 2sinxcosx

                                = 1 + sin 2x

                                = RHS  (shown)

2010 Paper 1 Q1 - Remainder & Factor Theorem

Let’s look at the first question of the 2010 paper 1…

1. The function f is defined by  f(x) = x⁴ – x³ + kx -4, where k is a constant.

(i) given that x – 2 is a factor of f(x), find the value of k.

(ii) Using the value of k found in part (i), find the remainder when f(x) is divided by x + 2.

We must remember that, if x – 2 is a factor, then f(2) = 0. And this is the way we solve for k.

f(2) = 2⁴ – 2³ + 2k – 4 = 0

Therefore,  16 – 8 + 2k – 4 = 0

                                4 + 2k = 0

                                2k = -4

                                k = -2

Once we have found the value of k, it would be useful to end this part of the question by putting it into the f(x) expression.

Hence,

                f(x) = x⁴ – x³ – 2x – 4

(ii)  To find the remainder when f(x) is divided by  x + 2, we find the value of f(-2).  Why f(-2) and not f(2)? Well, if we equate the divisor, x + 2, to zero, we will then be able to solve for x which is -2.  Hence, this is the value which we substitute for x in the expression for f(x).

So,

                f(-2) = (-2)⁴ – (-2)³ -2(-2) – 4

                        = 16 –(-8) + 4 – 4

                        = 24

Hence, the remainder when f(x) is divided by x + 2 is 24.

This is really a slam dunk question which, hopefully, everyone got correct for the O levels.

Check out the answers to the rest of the questions…..