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:
(1)Possible methods:
where
where
(Same as above but u = g(x))
b) Differentiate using the quotient rule:
(1) (1)Using the quotient rule on the formulae sheet:
2. (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
and the
(1) (1)b) The probability that Daniel receives only one call on each of the days is
(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) Correct graph shape
(1)Correct and labelled intercepts
(1)Correct and labeled endpoints
(1)b) i. Transformations as follows:
Therefore the point
will become
(1)ii. As shown before:
Sub into equation:
Alternatively:
(1) (1)6. a) (1) is a solution which occurs when
(1)b) is the other x-coordinate for intersection.
(1)7. ,
Using logarithmic laws:
(1)Taking the exponential of each side:
(1)However
(1)8.a) (1) (1)b) (1) (1)However
(1)9. a) Use product rule given on formula sheet:
(or the other way around)
(1)b) (1) (1) (1)10. a) i. Differentiate and solve to equal zero for stationary point:
(1)Taking logarithm of each side:
(1)ii. We want to find the values of
for which the value of
for which
is greather than zero i.e.
We have
And
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:
One point we know the tangent goes through is
, another can be found by substituting
into the original equation
(1)We can now use the equation:
(where m is the gradient)
We want it to go through the point (x,y) = (0,0) so
(1) has no solutions, so solve for
(1)