SECTION 1Question 1Answer:
CPeriod =
Question 2Answer:
EAverage rate of change =
-f(1)}{3-1}=9)
Question 3Answer:
DMinimum at x = 1
=-9)
and
so the left endpoint is the global max, as the x-value is furthest from the minimum
=0)
, so the range is
Question 4Answer:
A)=\frac{\mathrm{d}}{\mathrm{d}x}(e^{kx})g'(e^{kx})=kg'(e^{kx})e^{kx})
Question 5Answer:
B^{2}>0)
when
, so the maximal domain of the composite is
.
The range is
since the inner function spans all positive reals.
Question 6Answer:
CDilation by a factor of
parallel to the x-axis, followed by a translation of
units in the positive direction of the x-axis is a possible transformation, which was option C.
Question 7Answer:
DAt 2 am, t = 2
At 2 pm, t = 14
So the average temperature is
\mathrm{d}t=22)
Question 8Answer:
AThe given facts about p and q don't actually tell you anything about the answer, since the intercepts of
)
do not affect the gradient.
But, we have
and
, which means it is a negative cubic with two distinct stationary points and
and
, and
. Doing a quick sketch of this will show that the gradient is negative for
\cup(n,\infty))
.
Question 9Answer:
D\cdot(\frac{1}{2})\cdot(\sqrt{b-x^{2}})^{-\frac{1}{2}}=-\frac{x}{\sqrt{b-x^{2}}})
If the normal has a gradient of 3 at x=1, the tangent has a gradient of
(negative reciprocal)
Question 10Answer:
CSolve
\mathrm{d}x=0.4)
with
for
.
Question 11Answer:
EInverse normal with p = 0.6, µ = 252 and σ = 12 (as
=1-\Pr(X>x) = 1-0.4 = 0.6)
)
Question 12Answer:
APossible combos are WL and LW
She has just won, so
 = 0.7 \cdot (1 - 0.7) = 0.21)
And
 = (1 - 0.7) \cdot (1 - 0.6) = 0.12)
Adding these gives 0.33
Question 13Answer:
B+\Pr(A\cap B')=\Pr(A)=\frac{2}{5}+\frac{3}{7}=\frac{29}{35})
=\frac{\Pr(A\cap B')}{\Pr(A)}=\frac{(\frac{3}{7})}{(\frac{29}{35})}=\frac{15}{29})
Question 14Answer:
D
All rectangles are of width 1, so the area is
Question 15Answer:
BThe antiderivative of a quadratic is always a cubic so this eliminates A, C and E.
=0)
has two distinct solutions so the cubic graph will have two stationary points, which eliminates D.
Question 16Answer:
DIf c = 3 the stationary point at (a,-3), which is 'higher up' than the other stationary point, will 'touch' the x-axis, giving another solution, but if c < 3 there will still be a single solution as this highest stationary point will still lie below the x-axis.
If c = 8, the stationary point at (b,-8), which is 'lower down', will touch the x-axis, so any c > 8 will shift this lowest stationary point above the x-axis. Hence, c < 3 or c > 8 for there to be a single solution.
Question 17Answer:
BThe determinant of the coefficient matrix is
, which is 0 when m = 1 or m = -3.
However if m = 1 we have
These two lines are clearly the same, so there are infinitely many solutions. Checking the other option:
Which are two parallel but distinct lines, so they have no intersection. Hence the system will have no solutions for m = -3.
Question 18Answer:
A
At x = a,
, so the equation of the tangent is
Using the point
))
,
=\frac{1}{a}a+c\Rightarrow c=\ln(a)-1=\ln(\frac{a}{e}))
So the tangent has equation
)
At the point
)
,
=0)
)
)
Now,
<0\Rightarrow \ln(\frac{e}{a})<0)
(as
)
, so option A is false.
Question 19Answer:
B=-\cos(\theta)=-\cos(-\theta))
Question 20Answer:
E = (1-p)^{0}p=p)
=(1-p)^{1}p=p-p^2)
=1-\Pr(X=0)-\Pr(X=1)=1-p-(p-p^{2})=1-2p+p^{2}=(1-p)^{2})
Question 21Answer:
A
and
At h = 5,
, so the rate of decrease is
.
Question 22Answer:
B(a,b) is in the second quadrant and (c,d) is in the fourth quadrant.
The function has been shifted 2 units in the positive x direction, the absolute value has been taken and then the whole graph has been reflected in the x-axis, in that order.
Both stationary points are translated: (a+2,b) and (c+2,d), retaining their min/max nature
(a+2,b) is already above the x-axis so the modulus has no effect. However, (c+2,d) is part of the section below the x-axis, which is now reflected in the x-axis and so the point is moved above the x-axis to (c+2,-d) and becomes a maximum, so now both points are maxima.
Now the whole graph is reflected in the x-axis, so we have (a+2,b) becoming (a+2,-b) and (c+2,-d) becoming (c+2,d). Both were maxima, so they are now minima.
SECTION 2Question 1 (9 marks)
part a. (2 marks)
, as required.
part b. (2 marks)
We have
>0)
and
, so
}{14}>0)
(as
)
part c. (3 marks)
}{14}=\frac{16200x}{7}-\frac{25x^{3}}{14})
part d. (2 marks)
when
, given
When
,
^{2}}{7\cdot12\sqrt{3}}=\frac{120\sqrt{3}}{7})
Hence the block has maximum volume when
and
Question 2 (14 marks)
part a. (3 marks)
Hyperbola graph, asymptotes at y = 3 and x = 2, intercepts at
)
and
)
, found by evaluating
)
and solving
=0)
, respectively.
part b. i. (1 mark)
=\frac{-2}{(2x-4)^{2}}=\frac{-1}{2(x-2)^{2}})
(either one)
part b. ii. (1 mark)
)
part b. iii. (1 mark)
The gradient of f is never equal to 0, so f has no stationary points.
part c. (2 marks)
=\frac{1}{2p-4}+3)
so the tangent passes through
)
The gradient of the tangent is
=\frac{-2}{(2p-4)^{2}})
Let c be the y-intercept of the tangent. Then the equation of the tangent is:
^{2}}\cdot x+c)
The tangent passes through
, so:
^{2}}\cdot p+c)
^{2}}+3+\frac{2p}{(2p-4)^{2}}=\frac{4p-4}{(2p-4)^{2}}+3)
Hence the equation of the tangent is:
^{2}}\cdot x+\frac{4p-4}{(2p-4)^{2}}+3)
^{2}}\cdot x+\frac{4p-4}{(2p-4)^{2}})
^{2}(y-3)=-2x+4p-4)
, as required.
(2marks for all this shit wtf?)
part d. (2 marks)
Subbing in
and
(to find the values of p for which the tangent can pass through this point):
^{2}(\frac{7}{2}-3)=2+4p-4)
(\frac{1}{2})=4p-2)
(p-5)=0)
or
When
,
 = \frac{5}{2})
And when
,
 = \frac{19}{6})
So the points are
)
and
)
part e. (2 marks)
Let
=\frac{1}{2x-4}+3)
, and let
=\begin{bmatrix}ax+c\\y+d\end{bmatrix}=\begin{bmatrix}x'\\y'\end{bmatrix})
, where
and
are the images of the ordinates of
(i.e. the ordinates of
).
Then (
) and (
)
The equation for the image of f is then
-4}+3\Leftrightarrow y'-d-3=\frac{1}{2(\frac{x'-c}{a})-4})
, which is equivalent to
.
So we have
and
-4=x')
and
and
and
and
and
and
and
Question 3 (15 marks)
part a. i. (1 mark)
)
If n = 3, then
=1\cdot(\frac{1}{4})^{3}\cdot(\frac{3}{4})^{0}=\frac{1}{64})
part a. ii. (2 marks)
If n = 20, then
=0.0139)
part a. iii. (1 mark)
=np(1-p))
, as required.
part b. i. (3 marks)
Let
stand for a correct answer for question k, and
stand for an incorrect answer.
=\Pr(C_{2}C_{3}C_{4}C_{5})+\Pr(C'_{2}C_{3}C_{4}C_{5}))
\cdot(\frac{3}{4})^3+\frac{2}{3}\cdot(1-\frac{2}{3})\cdot(\frac{3}{4})^2=\frac{17}{64})
part b. ii. (2 marks)
Let the transition matrix
&\Pr(C_{k}\mid C'_{k-1})\\\Pr(C'_{k}\mid C_{k-1})&\Pr(C'_{k}\mid C'_{k-1})\end{bmatrix}=\begin{bmatrix}\frac{3}{4}&\frac{1}{3}\\\frac{1}{4}&\frac{2}{3}\end{bmatrix})
, and let the initial state matrix
Then
\\ \Pr(C'_{k}) \end{bmatrix})
\\ \Pr(C'_{25}) \end{bmatrix}=\begin{bmatrix}0.5714\\ 0.4286 \end{bmatrix})
to four decimal places.
Hence the probability that she answers question 25 correctly is 0.5714, correct to four decimal places.
part c. (2 marks)
)
=\Pr(Y=24)+\Pr(Y=25)=25p^{24}(1-p)+p^{25}=p^{24}(25-24p))
=6p^{25})
Therefore if
=6\Pr(Y=25))
, then
=6p^{25})
(as p > 0)
part d. (4 marks)
=\Pr(Z\geq\frac{20-a}{b})=\Pr(Z\leq\frac{a-20}{b}))
Inverse normal with p = Pr(Y≥18), µ = 0 and σ = 1 gives
)
=\Pr(Z\geq\frac{25-a}{b})=\Pr(Z\leq\frac{a-25}{b}))
Inverse normal with p = Pr(Y≥22), µ = 0 and σ = 1 gives
)
Equating b from (1) and (2) gives
Substituting this value of a into (1), we have
-20}{1.698233...}=2.50049...)
Hence a = 24.246 and b = 2.500 (or 2.501, seems to vary depending on decimals retained, friggen VCAA)
Question 4 (12 marks)
part a. i. (1 mark)
(similar triangles)
part a. ii. (1 mark)
^{2}\cdot h=\frac{\pi}{3}\cdot\frac{h^{2}}{25}\cdot h=\frac{\pi h^{3}}{75})
, as required.
part b. (1 mark)
At t = 20,
=10-\frac{16000}{1600}=10-10=0)
. Since V = 0 when h = 0, the tank is empty when t = 20, as required.
part c. i. (1 mark)
When t = 5,
=\frac{405}{64})
part c. ii. (3 mark)
)
and
}{40000})
When t = 5,
, so
^{2}(5^{2}-400)}{40000}\approx -3.5)
Hence the volume is decreasing at a rate of 3.5 cubic metres per minute, correct to one decimal place.
part d. (2 marks)
When h = 2 and 0 ≤ t ≤ 20,
=2\Rightarrow t=12.2)
minutes, correct to one decimal place.
part e. (2 marks)
When h = 2,
^{3}}{75}=\frac{8\pi}{75})
, so
(since V = 0 when t = 0, where t is the time in minutes after the tank is first empty)
So when h = 2,
minutes
part f. (1 mark)
He enters the tank when t = 12.2, which is 20 - 12.2 = 7.8 minutes before the tank is first empty.
Hence he has
minutes, correct to one decimal place.
Question 5 (8 marks)
part a. i. (1 mark)
\mathrm{d}x=1-e^{-2})
part a. ii. (1 mark)
\mathrm{d}x=1-e^{-2})
part a. iii. (1 mark)
\mathrm{d}x+\int_{0}^{1}f(x)\mathrm{d}x=1-e^{-2}+e-1=e-e^{-2})
part b. i. (2 marks)
=-\ln(a-x))
=\ln(\frac{1}{a-x}))
part b. ii. (1 mark)
The points are distinct for
(as a > 0)
Hence the set of values is
)
part c. (2 marks)
The x-coordinate of the midpoint is the midpoint of the x-coordinates, so the x-coordinate of the midpoint of AB is:
=\frac{a}{2})
If
, then
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