Sorry about the delay, I see some solutions are up already. Too bad I had an exam this morning, not to worry
![Tongue :P](https://www.atarnotes.com/forum/Smileys/default/tongue.gif)
Here are some solutions I typed up for the first methods exam, hope you all did well, if not you've always got exam 2 to make up for it!
![Cheesy :D](https://www.atarnotes.com/forum/Smileys/default/cheesy.gif)
note: I've added a potential marking scheme that VCAA may use. However, this is not necessarily accurate.
1. a) Differentiate the function using the chain rule:
(1)Possible methods:
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx})
where
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?y = u^4, u = x^2 - 5x)
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?\frac{dy}{du} = 4u^3, \frac{du}{dx} = 3u^3, hence: \frac{dy}{dx} = 4(x^2-5x)^3(2x-5))
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?y'(x) = y'(g(x))g'(x))
where
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?y(g(x)) = g(x)^4, g(x) = x^2 - 5x)
(Same as above but u = g(x))
b) Differentiate using the quotient rule:
(1)
(1)Using the quotient rule on the formulae sheet:
2. ![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?\int \frac{1}{(2x-1)^3} \ dx)
(1)
(1)Using
3. For inverse swap x and y
(1)
(1)(Note: You will probably be deducted one mark for not stating "for inverse swap x and y" or similar)
4. a) The mean is given by the sum of the product of
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?x)
and the
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?\Pr(x))
(1)
(1)b) The probability that Daniel receives only one call on each of the days is
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?\Pr(x=1) \times \Pr(x=1) \times \Pr(x=1))
(1)c) Note: This question is quite ambiguous and I'm sure will cause some controversy. However, this is my interpretation of the question, due to it being worth 3 marks.
We want to find the combinations such that Daniel receives four calls over two days. We are told, however ambiguously, that Daniel has already received at least 1 phone call on each of the days.
The combinations possible are:
- 1 call on Monday, 3 calls on Wednesday
- 3 calls on Monday, 1 call on Wednesday
- 2 calls on Monday, 2 calls on Wednesday
The probability of the three combinations is the sum of the 3:
(1)The probability that he receives one call on each days is:
(1)Therefore the overall probability is:
(1)5. a) ![](http://content.screencast.com/users/phy124/folders/Jing/media/760f9290-3730-40be-8e5e-cae83ad9b704/2012-11-07_1525.png)
Correct graph shape
(1)Correct and labelled intercepts
(1)Correct and labeled endpoints
(1)b) i. Transformations as follows:
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?(x,y) \rightarrow (x,-y) \rightarrow (x + 5,-y))
Therefore the point
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?(3,2))
will become
(1)ii. As shown before:
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?(x,y) \rightarrow (x + 5,-y))
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?\therefore (x',y') = (x+5),-y))
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?x' =x+5)
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?x = x' - 5)
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?y' = -y)
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?y = -y')
Sub into equation:
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?-y' = -|(x' - 5)-3| + 2)
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?y' = |x' -8| - 3)
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?y = |x -8| - 3)
Alternatively:
(1)
(1)6. a) ![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?\cos(x) = a\sin(x))
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?\frac{\cos(x)}{\cos(x)} = a\frac{\sin(x)}{\cos(x)})
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?a\tan(x) = 1)
(1)![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?x = \frac{\pi}{3})
is a solution which occurs when
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?\tan(x) = \sqrt{3})
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?\therefore \frac{1}{a}= \sqrt{3})
(1)b) ![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?\tan(x) = \sqrt{3})
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?x = \frac{\pi}{3}, \frac{4\pi}{3})
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?\therefore x = \frac{4\pi}{3})
is the other x-coordinate for intersection.
(1)7. ![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?2\log_e(x+2) - \log_e(x) = \log_e(2x+1))
,
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?x>0)
Using logarithmic laws:
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?\log_e((x+2)^2) - \log_e(x) = \log_e(2x+1))
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?\log_e\left(\frac{(x+2)^2}{x}\right) = \log_e(2x+1))
(1)Taking the exponential of each side:
(1)![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?(x+2)^2 = 2x^2 + x)
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?x^2 + 4x + 4 = 2x^2 + x)
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?x^2 - 3x - 4 = 0)
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?(x+1)(x-4))
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?x=-1,4)
However
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?x>0)
(1)8.a) ![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?\Pr(X<106) = q)
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?\Pr(94<X<100) = \Pr(100<X<106))
(1)![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?\therefore \Pr(100<X<106) = q - \frac{1}{2})
(1)b)
(1)
(1)![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?\frac{b^2}{24}+\frac{b}{12}-\frac{5}{8}=0)
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?b^2 + 2b - 15 = 0)
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?(b-3)(b+5))
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?b=-5,3)
However
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?0 \leq b \leq 4)
(1)9. a) Use product rule given on formula sheet:
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?u =x, v = \sin(x))
(or the other way around)
(1)b) ![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?\frac{d}{dx}(x\sin(x)) = \sin(x) + x\cos(x))
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?\therefore x\cos(x) = \frac{d}{dx}(x\sin(x)) - \sin(x))
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?\therefore \int x\cos(x) \dx = x\sin(x) - \int \sin(x) \ dx)
(1)
(1)
(1)10. a) i. ![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?f(x) = e^{-mx} + 3x)
Differentiate and solve to equal zero for stationary point:
(1)![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?-me^{-mx} = -3)
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?e^{-mx} = \frac{3}{m})
Taking logarithm of each side:
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?\log_e(e^{-mx}) = \log_e\left(\frac{3}{m}\right))
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?-mx = \log_e\left(\frac{3}{m}\right))
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?x = -\frac{1}{m}\log_e\left(\frac{3}{m}\right))
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?x = \frac{1}{m}\log_e\left(\frac{3}{m}\right)^{-1})
(1)ii. We want to find the values of
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?m)
for which the value of
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?x)
for which
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?f'(x) = 0)
is greather than zero i.e.
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?\frac{1}{m}\log_e\left(\frac{m}{3}\right)>0)
We have
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?\frac{1}{m}>0 \ \text{for all} \ m>0)
And
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?\log_e\left(\frac{m}{3}\right) \ \text{for all} \ m>3)
Therefore
(1)b. The interpretation of this question is that the line of the tangent to the graph
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?f(x))
at the point
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?x = -6)
will go through the point
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?(0,0))
Firstly find the gradient of the tangent:
One point we know the tangent goes through is
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?(0,0))
, another can be found by substituting
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?x = -6)
into the original equation
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?f(x))
(1)We can now use the equation:
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?y-y_1 = m(x-x_1))
(where m is the gradient)
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?y - (e^{6m} - 18) = (-me^{6m}+3)(x-(-6)))
We want it to go through the point (x,y) = (0,0) so
(1)![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?-e^{6m} + 18 = -6me^{6m}+18)
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?(6m-1)e^{6m} = 0)
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?e^{6m} =0 )
has no solutions, so solve for
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?6m-1=0)
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?6m-1 = 0)
![](https://archive.atarnotes.com/cgi-bin/mathtex.cgi?6m = 1)
(1)