Sorry about the delay, I see some solutions are up already. Too bad I had an exam this morning, not to worry
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!
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:
^3(2x-5))
(1)Possible methods:
where
^3(2x-5))
 = y'(g(x))g'(x))
where
) = g(x)^4, g(x) = x^2 - 5x)
(Same as above but u = g(x))
b) Differentiate using the quotient rule:
 = \frac{\sin(x)\frac{d}{dx}(x) - x \frac{d}{dx}(\sin(x))}{(\sin(x))^2} = \frac{\sin(x) - x\cos(x)}{\sin^2(x)})
(1) = \frac{\sin\left(\frac{\pi}{2} \right) - \left(\frac{\pi}{2} \right)\cos\left(\frac{\pi}{2} \right)}{\sin^2(\left(\frac{\pi}{2} \right))} = \frac{1 - 0}{1} = 1)
(1)Using the quotient rule on the formulae sheet:
)
2. ^3} \ dx)
(2x-1)^2})
(1)^2})
(1)Using
^n \ dx = \frac{(ax+b)^{n+1}}{a(n+1)}, n \neq -1)
3. For inverse swap x and y
(1) = h^{-1}(x))
 = \left(\frac{1}{2}\left(x-1\right)\right)^{\frac{1}{3}})
(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
and the
)
)
(1) + 1 (0.2) + 2(0.5) + 3(0.1) = 1.5)
(1)b) The probability that Daniel receives only one call on each of the days is
 \times \Pr(x=1) \times \Pr(x=1))
)^3 = (0.2)^3 = \frac{1}{125})
(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:
 \times \Pr(x = 3) + \Pr(x = 3) \times \Pr(x = 1) + \Pr(x = 2) \times \Pr(x = 2) = 0.2 \times 0.1 + 0.1 \times 0.2 + 0.5 \times 0.5 = 0.29)
(1)The probability that he receives one call on each days is:
 \times \Pr(x \geq 1) = 0.8 \times 0.8 = 0.64)
(1)Therefore the overall probability is:
(1)5. a) 
Correct graph shape
(1)Correct and labelled intercepts
(1)Correct and labeled endpoints
(1)b) i. Transformations as follows:
 \rightarrow (x,-y) \rightarrow (x + 5,-y))
Therefore the point
)
will become
+5,-(2)) = (8,-2))
(1)ii. As shown before:
 \rightarrow (x + 5,-y))
 = (x+5),-y))
Sub into equation:
-3| + 2)
Alternatively:
 = -f(x) = |x - 3| - 2)
(1) = g(x-5) = |(x-5) - 3| - 2 =|x-8| - 2)
(1)6. a)  = a\sin(x))
}{\cos(x)} = a\frac{\sin(x)}{\cos(x)})
 = 1)
 = \frac{1}{a})
(1)
is a solution which occurs when
 = \sqrt{3})
(1)b) 
is the other x-coordinate for intersection.
(1)7.  - \log_e(x) = \log_e(2x+1))
,
Using logarithmic laws:
^2) - \log_e(x) = \log_e(2x+1))
^2}{x}\right) = \log_e(2x+1))
(1)Taking the exponential of each side:
^2}{x} = 2x+1)
(1)^2 = 2x^2 + x)
(x-4))
However
(1)8.a)  = q)
 = \Pr(100<X<106))
 = \frac{1}{2})
(1) = q - \frac{1}{2})
 = q - \frac{1}{2} = \frac{1}{2}\left(2q - 1\right))
(1)b)  \ dx = \int_{0}^b \frac{x+1}{12} \ dx = \frac{1}{12} \int_0^b (x+1) dx = \frac{5}{8})
(1) dx = \frac{1}{12}\left[\frac{x^2}{2}+x\right]^b_0 = \frac{1}{12}\left[\left[\frac{(b)^2}{2}+(b)\right] - \left[\frac{(0)^2}{2}+(0)\right]\right] = \frac{1}{12}\left(\frac{b^2}{2} +b\right))
 = \frac{5}{8})
(1)
(b+5))
However
(1)9. a) Use product rule given on formula sheet:
)
(or the other way around)
 = \sin(x) + x\cos(x))
(1)b) ) = \sin(x) + x\cos(x))
 = \frac{d}{dx}(x\sin(x)) - \sin(x))
 \dx = x\sin(x) - \int \sin(x) \ dx)
 \dx = x\sin(x) + \cos(x) \ dx)
(1) +\cos(x)]^{\frac{\pi}{2}}_{\frac{\pi}{6}} = <br /><br />\left[\left(\frac{\pi}{2}\right)\sin\left(\frac{\pi}{2}\right) + c\os\left(\frac{\pi}{2}\right)\right] - \left[\left(\frac{\pi}{6}\right)\sin\left(\frac{\pi}{6}\right) + \cos\left(\frac{\pi}{6}\right)\right])
(1) + 0\right] - \left[\left(\frac{\pi}{6}\right)\left(\frac{1}{2}\right) + \left(\frac{\sqrt{3}}{2}\right)\right])
(1)
(1)10. a) i.  = e^{-mx} + 3x)
Differentiate and solve to equal zero for stationary point:
 = -me^{-mx} + 3 = 0)
(1)
Taking logarithm of each side:
 = \log_e\left(\frac{3}{m}\right))
)
)
^{-1})
)
(1)ii. We want to find the values of
for which the value of
for which
 = 0)
is greather than zero i.e.
>0)
We have
And
 \ \text{for all} \ m>3)
Therefore
(1)b. The interpretation of this question is that the line of the tangent to the graph
)
at the point
will go through the point
)
Firstly find the gradient of the tangent:
 = -me^{-m(-6)} + 3 = -me^{6m}+3)
One point we know the tangent goes through is
, another can be found by substituting
into the original equation
 = e^{-m(-6)} + 3(-6) = e^{6m} - 18)
(1)We can now use the equation:
)
(where m is the gradient)
 = (-me^{6m}+3)(x-(-6)))
We want it to go through the point (x,y) = (0,0) so
 = (-me^{6m}+3)(0+6))
(1)
e^{6m} = 0)
has no solutions, so solve for
(1)
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