C2 24th May 2012 REVISION THREAD

All of these I consider to be relatively difficult questions and some I got stuck on, please answer these questions and show working.

Thank you everyone!

All of these I consider to be relatively difficult questions and some I got stuck on, please answer these questions and show working.

Thank you everyone!

For the third question, question 8 about logs:

Part a:

You have to know the log rule:

log_an = x is the same as a^x = n

So if we apply this rule to log₂y = -3 then this becomes:

2^-3=y

Work out 2^-3 and we get the value for y and hence the answer to this question.

The value of y = 0.125


Part b:

Firstly, I would multiply both sides by log₂x to get rid of that fraction so the equation becomes:

log₂32 + log₂16 = (log₂x)²

For this question you will have to know the log multiplication rule: log_ax + log_ay = log_axy

So the equation becomes:

log₂(32*16) = (log₂x)²

Simplifying it:

log₂(512) = (log₂x)²

We then square root both sides. You can find the square root of log₂(512) on your calculator, which equals 3 so the equation becomes:

3 = log₂x

We can then apply the rule from part a: log_an = x is the same as a^x = n

So the equation then becomes:

2³ = x

And the final answer:

x = 8


Note that you can check if your answers are correct by substituting your answer back into the equation.

There are two answers to the 2nd part, $x =\dfrac18 \ and \ x = 8$

I will do it in this way,
$\displaystyle \frac{log_2 32 + log_2 16}{log_2 x} = log_2 x$

$log_2 32 + log_2 16 = log_2 2^5 + log_2 2^4 = 5log_2 2 +4 log_22 = 5+4=9$

$\displaystyle \frac{log_2 32 + log_2 16}{log_2 x} = log_2 x \implies \frac{9}{log_2 x} = log_2 x \implies 9 = (log_2 x)^2 \\ \implies log_2 x = \pm \sqrt9 \implies log_2 x = \pm 3$

$log_2 x =3$ gives $x = 2^3 = 8$
$log_2 x = -3$ gives $x = 2^{-3} = \dfrac18$

So $\displaystyle \boxed{ x = 8 \ and \ \frac18 }$

All of these I consider to be relatively difficult questions and some I got stuck on, please answer these questions and show working.

Thank you everyone!

okay, q.6:

a) complete the square:
(x-3)^2 -9 + (y+2)^2 -4 =12
(x-3)^2 + (y+2)^2 =25
centre at (3,-2)
radius is root25 = 5 (+ve value because it's a length)

b) P at (-1,1) and Q at (7,-5)
midpoint PQ should be (3,-2) if a diameter
midpoint= (7-1/2, -5-1/2) = (3,-2)

c) I'm not really sure how you'd do this. There's a right-angled triangle between RPQ and you can draw the whole thing onto a set of axes but although R is (0,y), I'm not sure how you'd find the y coordinate

For the second question, the (ii):

4sinx=3tanx

For this question you have to know the trigonometric identity that tanx is equal to sinx/cosx. So we substitute tanx for sinx/cosx, which transforms the equation into:

4sinx = 3(sinx/cosx)

We then multiply both sides by cosx:

4sinx(cosx) = 3 sinx

There is sinx on both sides so we can cancel them out so the equation becomes:

4cosx = 3

Then we use our calculator to find the value of x so this becomes:

x = 41.4 (1 decimal place)

Then I think there are various ways to do this next step, but I use a CAST graph. I could draw it out if you want me to, but I'm assuming you know what this is/know what to do as this stage.

So the answer is 41.4 and 318.6 degrees.

If you have any queries, quote me.

For the third question, question 8 about logs:

Part a:

You have to know the log rule:

log_an = x is the same as a^x = n

So if we apply this rule to log₂y = -3 then this becomes:

2^-3=y

Work out 2^-3 and we get the value for y and hence the answer to this question.

The value of y = 0.125


Part b:

Firstly, I would multiply both sides by log₂x to get rid of that fraction so the equation becomes:

log₂32 + log₂16 = (log₂x)²

For this question you will have to know the log multiplication rule: log_ax + log_ay = log_axy

So the equation becomes:

log₂(32*16) = (log₂x)²

Simplifying it:

log₂(512) = (log₂x)²

We then square root both sides. You can find the square root of log₂(512) on your calculator, which equals 3 and -3 so the equation becomes:

3 = log₂x and -3 = log₂x

We can then apply the rule from part a: log_an = x is the same as a^x = n

So the equation then becomes:

2³ = x and 2^-3 = x

And the final answer:

x = 8 and x = 1/8


Note that you can check if your answers are correct by substituting your answer back into the equation.

For the last part, draw a diagram.

Here is the diagram:




$|PR|^2 = (0+1)^2 + (y-1)^2 = y^2 -2y+2 \\ |RQ|^2 = (7-0)^2 + (-5-y)^2 = y^2 +10y +74$

As it is a right angled triangle, so applying Pythagoras theorem gives, $|PQ|^2 = |PR|^2 + |RQ|^2 \\ 10^2 = y^2-2y+2 + y^2 +10y+74 \implies y^2+4y-12=0 \\ \implies (y-2)(y+6)=0 \implies y= 2 \ or -6 \implies \boxed{y=2}$

Also please help

$y =sin(ax-b)$

We know the coordinates of P, Q and R.

Sub in P,
$\displaystyle 0 = sin \left( \frac{\pi a }{10} - b \right) \implies \frac{\pi a }{10} - b = 0 \implies \frac{\pi a }{10} = b \longrightarrow (1)$

Sub in Q,
$\displaystyle 0 = sin \left( \frac{3 \pi a }{5} - b \right) \implies \frac{3 \pi a }{5} - b = \pi \longrightarrow (2)$

Sub in R,
$\displaystyle 0 = sin \left( \frac{11 \pi a }{10} - b \right) \implies \frac{11 \pi a }{10} - b = 2 \pi \longrightarrow (3)$

(3) - (2) gives,
$a=2$
Sub 'a=2' in (1), you will get, $b = \dfrac{\pi}5$

All makes sense now! Thank you, hopefully i'll remember to apply this on Thursday

For the second question, the (ii):

4sinx=3tanx

For this question you have to know the trigonometric identity that tanx is equal to sinx/cosx. So we substitute tanx for sinx/cosx, which transforms the equation into:

4sinx = 3(sinx/cosx)

We then multiply both sides by cosx:

4sinx(cosx) = 3 sinx

There is sinx on both sides so we can cancel them out so the equation becomes:

4cosx = 3

Then we use our calculator to find the value of x so this becomes:

x = 41.4 (1 decimal place)

Then I think there are various ways to do this next step, but I use a CAST graph. I could draw it out if you want me to, but I'm assuming you know what this is/know what to do as this stage.

So the answer is 41.4 and 318.6 degrees.

If you have any queries, quote me.

Guys you can't discuss the exam yet!!!!!!!


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