Band 4 – Page 59

Graphs, MET2 2007 VCAA 10 MC

The graph of `y = kx-3` intersects the graph of `y = x^2 + 8x` at two distinct points for

`k = 11`
`k > 8 + 2 sqrt 3 or k < 8-2 sqrt 3`
`5 <= k <= 6`
`8-2 sqrt 3 <= k <= 8 + 2 sqrt 3`
`k = 5`

Show Answers Only

`B`

Show Worked Solution

`text(Intersection occurs when:)`

`kx-3`	`= x^2 + 8x`
`x^2 + (8-k)x + 3`	`= 0`

`text(For 2 points of intersection:)`

`Delta`	`> 0`
`(8-k)^2-4 (3)`	`> 0`

`:. k < 8-2 sqrt 3\ uu\ k > 8 + 2 sqrt 3`

`=> B`

Algebra, MET1 SM-Bank 24

The polynomial `p(x) = x^3-ax + b` has a remainder of 2 when divided by `(x-1)` and a remainder of 5 when divided by `(x + 2)`.

Find the values of `a` and `b`. (3 marks)

--- 6 WORK AREA LINES (style=lined) ---

Show Answers Only

`a`	`= 4`
`b`	`= 5`

Show Worked Solution

`p(x)`	`= x^3-ax + b`
`P(1)`	`= 2`
`1-a + b`	`= 2`
`b`	`= a+1\ \ \ …\ text{(1)}`
`P (-2)`	`= 5`
`-8 + 2a + b`	`= 5`
`2a + b`	`= 13\ \ \ …\ text{(2)}`

`text(Substitute)\ \ b = a+1\ \ text(into)\ \ text{(2)}`

`2a + a+1`	`= 13`
`3a`	`= 12`
`:. a`	`= 4`
`:. b`	`= 5`

Graphs, MET2 2008 VCAA 22 MC

The graph of the function `f` with domain `[0, 6]` is shown below.

Which one of the following is not true?

The function is not continuous at `x = 2` and `x = 4.`
The function exists for all values of `x` between `0` and `6.`
`f(x) = 0` for `x = 2` and `x = 5.`
The function is positive for `x ∈ [0, 5).`
The gradient of the function is not defined at `x = 4.`

Show Answers Only

`C`

Show Worked Solution

`f(x) > 0\ \ text(for)\ \ x = 2`

`:.\ text(Option)\ \ C\ \ text(is not true)`

`=> C`

Probability, MET2 2008 VCAA 14 MC

The minimum number of times that a fair coin can be tossed so that the probability of obtaining a head on each trial is less than 0.0005 is

A. `8`

B. `9`

C. `10`

D. `11`

E. `12`

Show Answers Only

`D`

Show Worked Solution

`text(Let)\ \ X = text(Number of heads,)`

`X ∼ text(Bi) (n, 1/2)`

`text(Pr) (X = n)`	`< 0.0005`
`((n), (n)) (1/2)^n (1/2)^0`	`< 0.0005`
`n`	`> 10.97`

`:. n_min = 11`

`=> D`

Probability, MET2 2008 VCAA 13 MC

According to a survey, 30% of employed women have never been married.

If 10 employed women are selected at random, the probability (correct to four decimal places) that at least 7 have never been married is

A. `0.0016`

B. `0.0090`

C. `0.0106`

D. `0.9894`

E. `0.9984`

Show Answers Only

`C`

Show Worked Solution

`text(Let)\ \ X =\ text(Number who have not been married)`

`X∼\ text(Bi) (10, 0.3)`

`text(Pr) (X >= 7)\ \ \ [text(CAS: binomCdf) (10, 0.3, 7, 10)]`

`= 0.0106`

`=> C`

Probability, MET2 2008 VCAA 5 MC

Let `X` be a discrete random variable with a binomial distribution. The mean of `X` is 1.2 and the variance of `X` is 0.72

The values of `n` (the number of independent trials) and `p` (the probability of success in each trial) are

A. `n = 4,\ \ \ \ \ p = 0.3`

B. `n = 3,\ \ \ \ \ p = 0.6`

C. `n = 2,\ \ \ \ \ p = 0.6`

D. `n = 2,\ \ \ \ \ p = 0.4`

E. `n = 3,\ \ \ \ \ p = 0.4`

Show Answers Only

`E`

Show Worked Solution

`X∼\ text(Bi) (n, p)`

`text(E) (X)`	`= 1.2`
`np`	`= 1.2\ …\ (1)`
`text(Var) (X)`	`= 0.72`
`np (1 – p)`	`= 0.72\ …\ (2)`

`text(Solve simultaneous equations:)`

`:. n = 3,\ \ p = 0.4`

`=> E`

Algebra, 2UG AM3 SM-Bank 06

Solve these simultaneous equations to find the values of `x` and `y`.

`4x - 2y = 18`

`3x + 2y = 10` (3 marks)

Show Answers Only

`x = 4,\ \ y = -1`

Show Worked Solution

`text(Solution 1 – Eliminations)`

`4x – 2y = 18\ …\ text{(i)}`

`3x +2y = 10\ …\ text{(ii)}`

`text{(i)} + text{(ii)}`

`7x`	`= 28`
`x`	`= 4`

`text(Substitute)\ \ x = 4\ \ text(into)\ text{(i)}`

`4 xx 4 – 2y`	`= 18`
`- 2y`	`= 2`
`y`	`= -1`

`:.\ text(Solution is)\ \ x = 4\ \ text(and)\ y = -1`

`text(Alternative Solution – Substitution)`

`4x – 2y = 18\ …\ text{(i)}`

`3x + 2y = 10\ …\ text{(ii)}`

`text{Divide (i) by 2}`

`2x – y`	`= 9`
`y`	`= 2x – 9`

`text(Substitute)\ \ y = 2x – 9\ \ text(into)\ text{(ii)}`

`3x + 2(2x – 9)`	`= 10`
`3x + 4x – 18`	`= 10`
`7x`	`= 28`
`x`	`= 4`

`text(Substitute)\ \ x= 4\ \ text{into (i)}`

`4 xx 4 – 2y`	`= 18`
`- 2y`	`= 2`
`y`	`= -1`

Algebra, MET1 SM-Bank 25

Solve these simultaneous equations to find the values of `x` and `y`. (3 marks)

`y = 2x + 1`

`x − 2y − 4 = 0`

Show Answers Only

`x = -2,\ y = -3`

Show Worked Solution

`text(Solution 1 – Substitution)`

`y = 2x + 1\ \ \ \ \ …\ text{(i)}`

`x\ – 2y\ – 4 = 0\ \ \ \ \ …\ text{(ii)}`

`text(Substitute)\ \ y = 2x + 1\ \ text(into)\ text{(ii)}`

`x\ – 2(2x + 1)\ – 4`	`= 0`
`x\ – 4x\ – 2\ – 4`	`= 0`
`-3x\ – 6`	`= 0`
`3x`	`= -6`
`x`	`= -2`

`text(Substitute)\ \ x = –2\ \ text(into)\ text{(i)}`

`y = 2(–2) + 1 = -3`

`:.\ text(Solution is)\ x = -2,\ y = -3`

`text(Alternative Solution – Elimination)`

`y = 2x + 1\ \ \ \ \ …\ text{(i)}`

`x\ – 2y\ – 4 = 0\ \ \ \ \ …\ text{(ii)}`

`text(Multiply)\ text{(i)} xx 2`

`2y`	`= 4x + 2`
`-4x + 2y\ – 2`	`= 0\ \ \ \ \ …\ text{(iii)}`

`text{Add (ii) + (iii)}`

`-3x\ – 6`	`= 0`
`x`	`= -2`
`y`	`= -3\ \ text{(see Solution 1)}`

Calculus, MET1 SM-Bank 28

The function `f` has the rule `f(x) = 1 + 2 cos x`.

Show that the graph of `y = f(x)` cuts the `x`-axis at `x = (2 pi)/3`. (1 mark)
Sketch the graph `y = f(x)` for `x in [-pi,pi]` showing where the graph cuts each of the axes. (2 marks)
Find the area under the curve `y = f(x)` between `x = -pi/2` and `x = (2 pi)/3`. (3 marks)

Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`
`((7 pi)/6 + sqrt 3 + 1)\ text(u²)`

Show Worked Solution

a. `f(x) = 1 + 2 cos x`

`f(x)\ text(cuts the)\ x text(-axis when)\ f(x) = 0`

`1 + 2 cos x`	`= 0`
`2 cos x`	`=-1`
`cos x`	`= -1/2`

`:. x = (2 pi)/3\ …\ text(as required)`

c. `text(Area)`	`= int_(-pi/2)^((2 pi)/3) 1 + 2 cos x\ \ dx`
	`= [x + 2 sin x]_(-pi/2)^((2 pi)/3)`
	`= [((2 pi)/3 + 2 sin (2 pi)/3) – ((-pi)/2 + 2 sin (-pi)/2)]`
	`= ((2 pi)/3 + 2 xx sqrt 3/2) – ((-pi)/2 +2(- 1))`
	`= (2 pi)/3 + sqrt(3) + pi/2 + 2`
	`= ((7 pi)/6 + sqrt(3) + 2)\ text(u²)`

Graphs, MET1 SM-Bank 27

The graph shown is `y = A sin bx`.

Write down the value of `A`. (1 mark)
Find the value of `b`. (1 mark)
Copy or trace the graph into your writing booklet.

On the same set of axes, draw the graph `y = 3 sin x + 1` for `0 <= x <= pi`. (2 marks)

Show Answers Only

`A = 4`
`b = 2`
`text(See Worked Solutions for sketch)`

Show Worked Solution

a. `A = 4`

b. `text(S)text(ince the graph passes through)\ \ (pi/4, 4)`

`text(Substituting into)\ \ y = 4 sin bx`

`4 sin (b xx pi/4)`	`=4`
`sin (b xx pi/4)`	`= 1`
`b xx pi/4`	`= pi/2`
`:. b`	`= 2`

MARKER’S COMMENT: Graphs are consistently drawn too small by many students. Aim to make your diagrams 1/3 to 1/2 of a page.

Calculus, MET1 SM-Bank 25

Evaluate `int_0^(pi/4) cos 2x\ dx`. (2 marks)

Show Answers Only

`1/2`

Show Worked Solution

`int_0^(pi/4) cos 2x`

`= [1/2 sin\ 2x]_0^(pi/4)`

`= [1/2 sin\ pi/2 – 1/2 sin\ 0]`

`= 1/2 – 0`

`= 1/2`

Calculus, MET1 SM-Bank 21

Find the equation of the tangent to the curve `y = cos 2x` at the point whose `x`-coordinate is `pi/6`. (3 marks)

Show Answers Only

`y = – sqrt 3 x + ((sqrt 3 pi)/6 + 1/2)`

Show Worked Solution

`y = cos 2x`

`dy/dx = -2 sin 2x`

`text(When)\ \ x = pi/6,`

`y`	`= cos (2 xx pi/6)`
	`= cos (pi/3)`
	`= 1/2`

`dy/dx`	`= -2 sin (pi/3)`
	`= -2 xx sqrt 3 / 2`
	`= – sqrt 3`

`:. text(Equation of tangent,)\ \ m = – sqrt 3, text(through)\ \ (pi/6, 1/2) :`

`y – y_1`	`= m(x – x_1)`
`y – 1/2`	`= – sqrt 3 ( x – pi/6)`
`y – 1/2`	`= – sqrt 3 x + (sqrt 3 pi)/6`
`y`	`= – sqrt 3 x + ((sqrt 3 pi)/6 + 1/2)`

Calculus, MET1 SM-Bank 20

If `f(x)= 2 sin 3x - 3 tan x`, find `f′(0)`. (2 marks)

Show Answers Only

`3`

Show Worked Solution

`y`	`= 2 sin 3x – 3 tan x`
`(dy)/(dx)`	`= 6 cos 3x – 3 sec^2 x`

`text(When)\ \ x = 0,`

`(dy)/(dx)`	`= 6 cos 0 – 3 sec^2 0`
	`= 6 (1) – 3/(cos^2 0)`
	`= 6 – 3`
	`= 3`

Algebra, MET1 SM-Bank 24

The rule for `f` is `f(x) = e^x − e^(-x)`.

Show that the inverse function is given by

`f^(-1)(x) = log_e((x + sqrt(x^2 + 4))/2)` (3 marks)

Show Answers Only

`text{Proof (See Worked Solutions.)}`

Show Worked Solution

`y = e^x − e^(-x)`

`text(Inverse: swap)\ x harr y`

`x`	`= e^y − 1/(e^y)`
`xe^y`	`= e^(2y) − 1`
`e^(2y) − xe^y − 1`	`= 0`

`text(Let)\ \ A = e^y`

`:.A^2 − xA − 1 = 0`

`text(Using the quadratic formula)`

`A`	`=(x ± sqrt((-x)^2 − 4 · 1 · (-1)))/(2 · 1)`
	`=(x ± sqrt(x^2 + 4))/2`

`text(S)text(ince)\ \ (x – sqrt(x^2 + 4))/2<0\ \ text(and)\ \ e^y>0,`

`:.e^y`	`= (x + sqrt(x^2 + 4))/2`
`log_e e^y`	` = log_e((x + sqrt(x^2 + 4))/2)`
`y`	`= log_e((x + sqrt(x^2 + 4))/2)`
`:.f^(-1)(x)`	`= log_e((x + sqrt(x^2 + 4))/2)\ \ …\ text(as required)`

Algebra, MET1 SM-Bank 23

The function `f: [0,oo) → R` with rule `f(x) = 1/(1 + x^2)` is drawn below.

Copy or trace this diagram into your writing booklet.
On the same set of axes, sketch `y=f^(-1)(x)` where `f^(-1)` is the inverse function of `f(x)`. (1 mark)
Find the domain of the inverse `f^(-1)`. (1 mark)
Find an expression for `y = f^(-1)(x)` in terms of `x`. (2 marks)
The graphs of `y = f(x)` and `y = f^(-1)(x)` meet at exactly one point `P`.Let `α` be the `x`-coordinate of `P`. Explain why `α` is a root of the equation
`x^3 + x-1 = 0`. (1 mark)

Show Answers Only

`text(See Worked Solutions)`
`0 < x ≤ 1`
`y = sqrt((1-x)/x), y > 0`
`text(See Worked Solutions)`

Show Worked Solution

b. `text(Range of)\ \ f(x):\ (0,1]`

`:.\ text(Domain of)\ \ f^(−1)(x): (0,1]`

c. `f(x) = 1/(1 + x^2)`

`text(Inverse: swap)\ x harr y`

`x`	`= 1/(1 + y^2)`
`x(1 + y^2)`	`= 1`
`1 + y^2`	`= 1/x`
`y^2`	`= 1/x-1`
	`= (1-x)/x`
`y`	`= ± sqrt((1-x)/x)`

`:.y = sqrt((1-x)/x), \ \ y >= 0`

d. `P\ \ text(occurs when)\ \ f(x)\ \ text(cuts)\ \ y = x`

`text(i.e. where)`

`1/(1 + x^2)`	`= x`
`1`	`= x(1 + x^2)`
`1`	`= x + x^3`
`x^3 + x-1`	`= 0`

`=>\ text(S)text(ince)\ α\ text(is the)\ x\ text(-coordinate of)\ P,`

`text(it is a root of)\ \ \ x^3 + x-1 = 0`

Calculus, MET1 SM-Bank 21

The rule for function `f` is `f(x) = e^(-x^2)`. The diagram shows the graph `y = f(x)`.

The graph has two points of inflection.

Find the `x` coordinates of these points. (2 marks)
Explain why the domain of `f(x)` must be restricted if `f(x)` is to have an inverse function. (1 mark)
Find the rule for the inverse function `f^(-1)` if the domain of `f(x)` is restricted to `x ≥ 0`. (2 marks)
Find the domain for `f^(-1)`. (1 mark)
Sketch the curve `y = f^(-1) (x)`. (1 mark)

Show Answers Only

`x = +- 1/sqrt2`
`text(There can only be 1 value of)\ y\ text(for each value of)\ x.`
`f^(-1)x = sqrt(ln(1/x))`
`0 <= x <= 1`

Show Worked Solution

a.	`y`	`= e^(-x^2)`
	`dy/dx`	`= -2x * e^(-x^2)`
	`(d^2y)/(dx^2)`	`= -2x (-2x * e^(-x^2)) + e ^(-x^2) (-2)`
		`= 4x^2 e^(-x^2)-2e^(-x^2)`
		`= 2e^(-x^2) (2x^2-1)`

`text(P.I. when)\ \ (d^2y)/(dx^2) = 0`

`2e^(-x^2) (2x^2-1)`	`= 0`
`2x^2-1`	`= 0`
`x^2`	`= 1/2`
`x`	`= +- 1/sqrt2`

COMMENT: It is also valid to show that `f(x)` is an even function and if a P.I. exists at `x=a`, there must be another P.I. at `x=–a`.

`text(When)\ \ `	`x < 1/sqrt2,`	`\ (d^2y)/(dx^2) < 0`
	`x > 1/sqrt2,`	`\ (d^2y)/(dx^2) > 0`

`=>\ text(Change of concavity)`

`:.\ text(P.I. at)\ \ x = 1/sqrt2`

`text(When)\ \ `	`x <-1/sqrt2,`	`\ (d^2y)/(dx^2) > 0`
	`x >-1/sqrt2,`	`\ (d^2y)/(dx^2) < 0`

`=>\ text(Change of concavity)`

`:.\ text(P.I. at)\ \ x =-1/sqrt2`

b.	`text(In)\ f(x), text(there are 2 values of)\ y\ text(for)`
	`text(each value of)\ x.`
	`:.\ text(The domain of)\ f(x)\ text(must be restricted)`
	`text(for)\ \ f^(-1) (x)\ text(to exist).`

c. `y = e^(-x^2)`

`text(Inverse: swap)\ x harr y`

`x`	`= e^(-y^2),\ \ \ x >= 0`
`lnx`	`= ln e^(-y^2)`
`-y^2`	`= lnx`
`y^2`	`= -lnx`
	`=ln(1/x)`
`y`	`= +- sqrt(ln (1/x))`

`text(Restricting)\ \ x>=0,\ \ =>y>=0`

`:. f^(-1) (x)=sqrt(ln (1/x))`

d.	`f(0) = e^0 = 1`

`:.\ text(Range of)\ \ f(x)\ \ text(is)\ \ 0 < y <= 1`

`:.\ text(Domain of)\ \ f^(-1) (x)\ \ text(is)\ \ 0 < x <= 1`

Graphs, MET1 SM-Bank 20

The rule for `f` is `f(x) = x-1/2 x^2` for `x <= 1`. This function has an inverse, `f^(-1) (x)`.

Sketch the graphs of `y = f(x)` and `y = f^(-1) (x)` on the same set of axes. (Use the same scale on both axes.) (2 marks
Find the rule for the inverse function `f^(-1) (x)`. (2 marks)
Evaluate `f^(-1) (3/8)`. (1 mark)

Show Answers Only

`y = 1-sqrt(1-2x)`
`1/2`

Show Worked Solution

b.	`y = x-1/2 x^2,\ \ \ x <= 1`

`text(For the inverse function, swap)\ \ x↔y,`

`x`	`= y-1/2 y^2,\ \ \ y <= 1`
`2x`	`= 2y-y^2`
`y^2-2y + 2x`	`= 0`

`text(Using quadratic formula,)`

`y`	`= (2 +- sqrt( (-2)^2-4 * 1 * 2x) )/2`
	`= (2 +- sqrt(4-8x))/2`
	`= (2 +- 2 sqrt(1-2x))/2`
	`= 1 +- sqrt (1-2x)`

`:. y = 1-sqrt(1-2x), \ \ (y <= 1)`

c.	`f^(-1) (3/8)`	`= 1-sqrt(1 -2(3/8))`
		`= 1-sqrt(1-6/8)`
		`= 1-sqrt(1/4)`
		`= 1-1/2`
		`= 1/2`

Algebra, MET1 SM-Bank 11

Let `f: R→R` where `f(x)= x^3 - 2`.

Evaluate `f^(-1)(25),` where `f^(-1)` is the inverse function of `f`. (2 marks)

Show Answers Only

`3`

Show Worked Solution

`text(Let)\ \ y = x^3 – 2`

`text(For inverse),\ \ x harr y`

`x`	`= y^3 – 2`
`y^3`	`= x + 2`
`y`	`= (x + 2)^(1/3)`

`:. f^(-1)(25)`	`=(25+2)^(1/3)`
	`=3`

Calculus, MET1 2006 ADV 2bi

If `f′(x)= (x^2)/(x^3 + 1)` and `f(1)= log_e 2,` find `f(x)`. (3 marks)

Show Answers Only

`f(x) = 1/3 log_e(x^3 + 1) + 2/3 log_e 2`

Show Worked Solution

`f(x)`	`=int f′(x)\ dx`
	`=int (x^2)/(x^3 + 1)\ dx`
	`= 1/3 int (3x^2)/(x^3 + 1)\ dx`
	`=1/3 log_e \|(x^3 + 1)\|+c`

`text(S)text(ince)\ \ f(1)= log_e 2,`

`1/3 log_e 2+c`	`= log_e 2`
`c`	`=2/3 log_e 2`

`:.\ f(x) = 1/3 log_e(x^3 + 1) + 2/3 log_e 2`

Calculus, MET1 2011 ADV 4b

Find the value of `int_e^(e^3) 5/x` with respect to `x` (2 marks)

Show Answers Only

`10`

Show Worked Solution

`int_e^(e^3) 5/x\ dx`

`= 5 int_e^(e^3) 1/x\ dx`

`= 5[lnx]_e^(e^3)`

`= 5(ln e^3 – ln e)`

`= 5(3 – 1)`

`= 10`

Calculus, MET1 2008 ADV 3b

Differentiate `log_e(cos x)` with respect to `x`. (2 marks)
Hence, or otherwise, evaluate `int_0^(pi/4) tan x\ dx`. (2 marks)

Show Answers Only

`−tan x`
`−log_e(1/sqrt2)\ text{or 0.35 (2 d.p.)}`

Show Worked Solution

a.	`y`	`= log_e(cos x)`
	`(dy)/(dx)`	`= (−sin x)/(cos x)`
		`= −tan x`

b. `int_0^(pi/4) tan x\ dx`

`= −[log_e(cos x)]_0^(pi/4)`

`= −[log_e(cos(pi/4)) – log_e(cos 0)]`

`= −[log_e(1/sqrt2) – log_e 1]`

`= −[log_e(1/sqrt2) – 0]`

`= −log_e(1/sqrt2)`

`= 0.346…`

`= 0.35\ \ (text(2 d.p.))`

Calculus, MET1 2016 ADV 12d

Differentiate `y = xe^(3x)`. (1 mark)
Hence find the exact value of `int_0^2 e^(3x) (3 + 9x)\ dx`. (2 marks)

Show Answers Only

`e^(3x) (1 + 3x)`
`6e^6`

Show Worked Solution

a. `y = xe^(3x)`

`text(Using product rule,)`

`(dy)/(dx)`	`= x · 3e^(3x) + 1 · e^(3x)`
	`= e^(3x) (1 + 3x)`

b. `int_0^2 e^(3x) (3 + 9x)\ dx`

`= 3 int_0^2 e^(3x) (1 + 3x)\ dx`

`= 3 [x e^(3x)]_0^2`

`= 3 (2e^6 – 0)`

`= 6e^6`

Calculus, MET1 2015 ADV 11e

Differentiate `(e^x + x)^5`. (2 marks)

Show Answers Only

`5(e^x + 1)(e^x + x)^4`

Show Worked Solution

`y`	`= (e^5 + x)^5`
`(dy)/(dx)`	`= 5(e^x + x)^4 xx d/(dx)(e^x + x)`
	`= 5(e^x + x)^4 xx (e^x + 1)`
	`= 5(e^x + 1)(e^x + x)^4`

Probability, MET2 2006 VCAA 5 MC

A bag contains three white balls and seven yellow balls. Three balls are drawn, one at a time, from the bag, without replacement.

The probability that they are all yellow is

`3/500`
`27/1000`
`21/100`
`7/24`
`243/1000`

Show Answers Only

`D`

Show Worked Solution

`text{Pr(YYY)}`	`= 7/10 xx 6/9 xx 5/8`
	`= 7/24`

`=> D`

Calculus, MET1 SM-Bank 5

The diagram shows the graph of the function `f: (0,oo) →R,` where `f(x) = 1/x`.

The area under `f(x)` between `x = a` and `x = 1` is `A_1`. The area under the curve between `x = 1` and `x = b` is `A_2`.

The areas `A_1` and `A_2` are each equal to 1 square unit.

Find the values of `a` and `b`. (3 marks)

Show Answers Only

`a = 1/e`

`b = e`

Show Worked Solution

`int_a^1 1/x dx`	`= 1`
`[ln x]_a^1`	`= 1`
`ln 1 – ln a`	`= 1`
`ln a`	`= −1`
`:. a`	`= e^(−1) = 1/e`

`int_1^b 1/x dx`	`= 1`
`[ln x]_1^b`	`= 1`
`ln b – ln 1`	`= 1`
`ln b`	`= 1`
`:. b`	`= e`

Calculus, MET1 SM-Bank 4

The curves `y = e^(2x)` and `y = e^(−x)` intersect at the point `(0,1)` as shown in the diagram.

Find the exact area enclosed by the curves and the line `x = 2`. (3 marks)

Show Answers Only

`1/2 e^4 + e^(−2) – 3/2\ \ \ text(u²)`

Show Worked Solution

Note: there’s marker’s comment in this part

`text(Area)`	`= int_0^2 e^(2x)\ \ dx – int_0^2 e^(−x)\ \ dx`
	`= int_0^2 (e^(2x) – e^(−x))dx`
	`= [1/2 e^(2x) + e^(−x)]_0^2`
	`= [(1/2 e^4 – e^(−2)) – (1/2 e^0 + e^0)]`
	`= 1/2 e^4 + e^(−2) – 3/2\ \ \ text(u²)`

Algebra, MET1 SM-Bank 9

Solve the following equation for `x`:

`2e^(2x) - e^x = 0`. (2 marks)

Show Answers Only

`x = ln\ 1/2`

Show Worked Solution

`text(Solution 1)`

`2e^(2x) – e^x = 0`

`text(Let)\ \ X = e^x`

`2X^2 – X`	`= 0`
`X (2X – 1)`	`= 0`

`X = 0 or 1/2`

`text(When)\ \ e^x = 0\ =>\ text(no solution)`

`text(When)\ \ e^x = 1/2`

`ln e^x`	`= ln\ 1/2`
`:. x`	`= ln\ 1/2`

`text(Solution 2)`

`2e^(2x)-e^x`	`=0`
`2e^(2x)`	`=e^x`
`ln 2e^(2x)`	`=ln e^x`
`ln 2 +ln e^(2x)`	`=x`
`ln 2 + 2x`	`=x`
`x`	`=-ln2`
	`=ln\ 1/2`

Algebra, MET1 SM-Bank 8

Write `log2 + log4 + log8 + log16 + … + log128` in the form `a logb` where `a` and `b` are integers greater than 1. (2 marks)

Show Answers Only

`28log2`

Show Worked Solution

`log2 + log4 + log8 + … + log 128`

`= log2^1 + log2^2 + log2^3 + … + log2^7`

`= log2 + 2log2 + 3log2 + … + 7log2`

`= 28log2`

Algebra, MET1 SM-Bank 2

Solve `5^x = 4` for `x`. (2 marks)

Show Answers Only

`(log_e 4)/(log_e 5)`

Show Worked Solution

	`text(Given)\ \ 5^x`	`= 4`
	`log_e 5^x`	`= log_e 4`
	`x xx log_e 5`	`= log_e 4`
	`:. x`	`= (log_e 4)/(log_e 5)`

A company produces motors for refrigerators. There are two assembly lines, Line A and Line B. 5% of the motors assembled on Line A are faulty and 8% of the motors assembled on Line B are faulty. In one hour, 40 motors are produced from Line A and 50 motors are produced from Line B. At the end of an hour, one motor is selected at random from all the motors that have been produced during that hour.

What is the probability that the selected motor is faulty? Express your answer in the form `1/b`, where `b` is a positive integer. (2 marks)
The selected motor is found to be faulty.

What is the probability that it was assembled on Line A? Express your answer in the form `1/c`, where `c` is a positive integer. (1 mark)

Show Answers Only

`1/15`
`1/3`

Show Worked Solution

a. `text(Construct tree diagram)`

`text(Pr)(AF) + text(Pr)(BF)`

`= 4/9 xx 1/20 + 5/9 xx 2/25`

`= 1/45 + 2/45`

`= 1/15`

`:. text(Pr)(F) = 1/15`

♦♦ Mean mark 32%.

b.	`text(Pr)(A\|F)`	`= (text(Pr)(A ∩ F))/(text(Pr)(F))`
		`= (4/9 xx 1/20)/(1/15)`
		`= 1/45 xx 15`
		`= 1/3`

Graphs, MET1 2016 VCAA 5

Let `f : (0, ∞) → R`, where `f(x) = log_e(x)` and `g: R → R`, where `g (x) = x^2 + 1`.

i. Find the rule for `h`, where `h(x) = f (g(x))`. (1 mark)
ii. State the domain and range of `h`. (2 marks)
iii. Show that `h(x) + h(−x) = f ((g(x))^2 )`. (2 marks)
iv. Find the coordinates of the stationary point of `h` and state its nature. (2 marks)
Let `k: (– ∞, 0] → R` where `k (x) = log_e(x^2 + 1)`.
i. Find the rule for `k^(−1)`. (2 marks)
ii. State the domain and range of `k^(– 1)`. (2 marks)

Show Answers Only

i. `log_e(x^2 + 1)`
ii. `[0,∞)`
iii. `text(See Worked Solutions)`
iv. `(0, 0)`
i. `−sqrt(e^x – 1)`
ii. `text(Domain)\ (k) = (−∞,0]`
`text(Range)\ (k) = [0,∞)`

Show Worked Solution

a.i.	`h(x)`	`= f(x^2 + 1)`
		`= log_e(x^2 + 1)`

a.ii. `text(Domain)\ (h) =\ text(Domain)\ (g) = R`

♦♦ Mean mark part (a)(ii) 30%.

`text(For)\ x ∈ R`	`-> x^2 + 1 >= 1`
	`-> log_e(x^2 + 1) >= 0`

`:.\ text(Range)\ (h) = [0,∞)`

MARKER’S COMMENT: Many students were unsure of how to present their working in this question. Note the layout in the solution.

a.iii.	`text(LHS)`	`= h(x) + h(−x)`
		`= log_e(x^2 _ 1) + log_e((−x)^2 + 1)`
		`= log_e(x^2 + 1) + log_e(x^2 + 1)`
		`= 2log_e(x^2 + 1)`

`text(RHS)`	`= f((x^2 + 1)^2)`
	`= 2log_e(x^2 + 1)`

`:. h(x) + h(−x) = f((g(x))^2)\ \ text(… as required)`

a.iv. `text(Stationary points when)\ \ h′(x) = 0`

♦ Mean mark part (a)(iv) 41%.
MARKER’S COMMENT: Solving a fraction is zero

`text(Using Chain Rule:)`

`h′(x)`

`= (2x)/(x^2 + 1)`

`:.\ text(S.P. when)\ \ x=0`

`text(Find nature using 1st derivative test:)`

`:.\ text{Minimum stationary point at (0, 0)}.`

b.i. `text(Let)\ \ y = k(x)`

♦ Mean mark (b)(i) 49%.
MARKER’s COMMENT: Many students failed to consider the restrictions on the domain in `k(x)` and only select the negative root.

`text(Inverse: swap)\ x ↔ y`

`x`	`= log_e(y^2 + 1)`
`e^x`	`= y^2 + 1`
`y^2`	`= e^x – 1`
`y`	`= ±sqrt(e^x – 1)`

`text(But range)\ \ (k^(−1)) =\ text(domain)\ (k)`

`:.k^(−1)(x) = −sqrt(e^x – 1)`

♦ Mean mark part (b)(ii) 44%.

b.ii. `text(Range)\ (k^(−1)) =\ text(Domain)\ (k) = (−∞,0]`

`text(Domain)\ (k^(−1)) =\ text(Range)\ (k) = [0,∞)`

Probability, MET1 2016 VCAA 4

A paddock contains 10 tagged sheep and 20 untagged sheep. Four times each day, one sheep is selected at random from the paddock, placed in an observation area and studied, and then returned to the paddock.

What is the probability that the number of tagged sheep selected on a given day is zero? (1 mark)
What is the probability that at least one tagged sheep is selected on a given day? (1 mark)
What is the probability that no tagged sheep are selected on each of six consecutive days?

Express your answer in the form `(a/b)^c`, where `a`, `b` and `c` are positive integers. (1 mark)

Show Answers Only

`16/81`
`65/81`
`(2/3)^24`

Show Worked Solution

a. `text(Let)\ \ X =\ text(Number of tagged sheep,)`

`X ~\ text(Bi)(4,1/3)`

`text(Pr)(X = 0)`	`= ((4),(0)) xx (1/3)^0 xx (2/3)^4`
	`= 16/81`

b.	`text(Pr)(X >= 1)`	`= 1 – text(Pr)(X = 0)`
		`= 1 – 16/81`
		`= 65/81`

c. `text(Let)\ \ Y =\ text(Days that no tagged sheep selected,)`

`Y ~\ text(Bi)(6,16/81)`

`text(Pr)(Y = 6)`	`= ((6),(6)) xx (16/81)^6 xx (65/81)^0`
	`= (16/81)^6 = (2/3)^24`

Calculus, MET1 2016 VCAA 3

Let `f: R text{\}{1} -> R` where `f(x) = 2 + 3/(x - 1)`.

Sketch the graph of `f`. Label the axis intercepts with their coordinates and label any asymptotes with the appropriate equation. (3 marks)
Find the area enclosed by the graph of `f`, the lines `x = 2` and `x = 4`, and the `x`-axis. (2 marks)

Show Answers Only

`4 + 3log_e(3)\ text(units)`

Show Worked Solution

b.	`text(Area)`	`= int_2^4 2 + 3(x – 1)^(−1)\ dx`
		`= [2x + 3 log_e(x – 1)]_2^4`
		`= (8 + 3log_e(3)) – (4 + 3log_e(1))`
		`= 4 + 3log_e(3)\ \ text(u²)`

Calculus, MET1 2016 VCAA 2

Let `f: (−∞,1/2] -> R`, where `f(x) = sqrt(1 - 2x)`.

Find `f′(x)`. (1 mark)
Find the angle `theta` from the positive direction of the `x`-axis to the tangent to the graph of `f` at `x = – 1`, measured in the anticlockwise direction. (2 marks)

Show Answers Only

`(−1)/(sqrt(1 – 2x))`
`(5pi)/6`

Show Worked Solution

a.	`f(x)`	`= (1 – 2x)^(1/2)`
	`f′(x)`	`= 1/2(1 – 2x)^(−1/2) (−2)qquadtext([Using Chain Rule])`
		`= (−1)/(sqrt(1 – 2x))`

♦ Mean mark 40%.

b.	`tan theta`	`= f′(−1)`
		`= (−1)/(sqrt(1 – 2(−1)))`
		`= (−1)/(sqrt3)`

`text(S)text(ince)\ theta ∈ [0,pi],`

`=> theta = (5pi)/6`

Calculus, MET2 2009 VCAA 2

A train is travelling at a constant speed of `w` km/h along a straight level track from `M` towards `Q.`

The train will travel along a section of track `MNPQ.`

Section `MN` passes along a bridge over a valley.

Section `NP` passes through a tunnel in a mountain.

Section `PQ` is 6.2 km long.

From `M` to `P`, the curve of the valley and the mountain, directly below and above the train track, is modelled by the graph of

`y = 1/200 (ax^3 + bx^2 + c)` where `a, b` and `c` are real numbers.

All measurements are in kilometres.

The curve defined from `M` to `P` passes through `N (2, 0)`. The gradient of the curve at `N` is – 0.06 and the curve has a turning point at `x = 4`.
i. From this information write down three simultaneous equations in `a`, `b` and `c`. (3 marks)
ii. Hence show that `a = 1`, `b = – 6` and `c = 16`. (2 marks)
Find, giving exact values
i. the coordinates of `M and P` (2 marks)
ii. the length of the tunnel (1 mark)
iii. the maximum depth of the valley below the train track. (1 mark)

The driver sees a large rock on the track at a point `Q`, 6.2 km from `P`. The driver puts on the brakes at the instant that the front of the train comes out of the tunnel at `P`.

From its initial speed of `w` km/h, the train slows down from point `P` so that its speed `v` km/h is given by

`v = k log_e ({(d + 1)}/7)`,

where `d` km is the distance of the front of the train from `P` and `k` is a real constant.

Find the value of `k` in terms of `w`. (1 mark)
If `v = (120 log_e (2))/(log_e (7))` when `d = 2.5`, find the value of `w`. (2 marks)
Find the exact distance from the front of the train to the large rock when the train finally stops. (2 marks)

Show Answers Only

1. `1/200 (8a + 4b + c) = 0`
  
  `1/200 (12a + 4b) = (– 3)/50`
  
  `1/200 (48a + 8b) = 0`
2. `text(Proof)\ \ text{(See Worked Solutions)}`
1. `M (2 + 2 sqrt 3, 0),\ \ P (2 – 2 sqrt 3, 0)`
2. `2 sqrt 3\ text(km)`
3. `2/25\ text(km)`
`(– w)/(log_e (7))`
`120`
`0.2\ text(km)`

Show Worked Solution

a.i. `N (2, 0),`

`1/200 (8a + 4b + c) = 0\ \ text{… (1)}`

`{:frac (dy) (dx)|:}_(x = 2) = (– 3)/50,`

`1/200 (12a + 4b) = (– 3)/50\ \ text{… (2)}`

`{:frac (dy) (dx)|:}_(x = 4) = 0,`

`1/200 (48a + 8b) = 0\ \ text{… (3)}`

ii. `text(Using the above equations,)`

♦ Mean mark part (a)(ii) 41%.

`12a + 4b`	`=-12`	`\ \ \ …\ (2′)`
`24 a + 4b`	`=0`	`\ \ \ …\ (3′)`

`text(Solve simultaneous equations:)`

`(3′) – (2′):`

`12a`	`= 12`
`:. a`	`= 1`

`text(Substitute)\ \ a = 1\ \ text(into)\ \ (3′):`

`24(1) + 4b`	`= 0`
`:. b`	`= – 6`

`text(Substitute)\ \ a = 1,\ \ b = – 6\ \ text(into)\ \ (1),`

`8(1) + 4 (– 6) + c`	`= 0`
`:. c`	`= 16`

b.i. `text(For)\ \ x text(-intercepts),`

`text(Solve:)\ \ x^3 + -6x^2 + 16`	`= 0\ \ text(for)\ x,`
`x= 2 – 2 sqrt 3, 2, 2 + 2 sqrt 3`

`:. M (2 + 2 sqrt 3, 0),\ \ P (2 – 2 sqrt 3, 0)`

ii. `N (2, 0)`

♦ Mean mark part (a) 41%.

`bar (NP)`	`=\ text(Tunnel length)`
	`= 2 – (2 – 2 sqrt 3)`
	`= 2 sqrt 3\ text(km)`

iii. `text(Solve)\ \ frac (dy) (dx) = 0\ \ text(for)\ \ x in (2, 2 + 2 sqrt 3)`

♦ Mean mark part (b) 50%.

`x = 4`

`text(When)\ \ x=4,\ \ y = – 2/25\ text(km) = 80\ text{m (below track)}`

`:.\ text(Max depth below is)\ \ 80\ text(m.)`

c. `text(Solution 1)`

♦ Mean mark part (c) 35%.

`text(Let)\ \ v(d) = k log_e ({(d + 1)}/7)`

`text(S)text(ince)\ \ v=w\ \ text(when)\ \ d=0\ \ text{(given),}`

`k log_e ({(0 + 1)}/7)`	`=w`
`k`	`=w/log_e (1/7)`
	`=(-w)/log_e7`

`text(Solution 2)`

`text(Solve:)\ \ v (0)`	`= w\ \ text(for)\ \ k`
`:. k`	`= (– w)/(log_e (7))`

d. `v (2.5) = k log_e(1/2)`

♦ Mean mark part (d) 40%.

`text(Solve)\ \ v(2.5)`	`= (120 log_e (2))/(log_e (7))\ \ text(for)\ \ k,`
`:. k`	`= (– 120)/(log_e (7))`
`(– w)/(log_e (7))`	`= (– 120)/(log_e (7))`
`:. w`	`= 120`

e. `text(Define)\ \ v (d) = (– 120)/(log_e (7)) log_e ((d + 1)/7)`

♦♦ Mean mark part (e) 32%.

`text(Solve:)\ \ v(d)`	`= 0\ \ text(for)\ d,`
`:. d`	`= 6\ text(km from)\ \ P`

`:.\ text(Distance between train and)\ \ Q`

`= 6.2 – 6`

`= 0.2\ text(km)`

Calculus, MET2 2009 VCAA 1

Let `f: R^+ uu {0} -> R,\ f(x) = 6 sqrt x - x - 5.`

The graph of `y = f (x)` is shown below.

State the interval for which the graph of `f` is strictly decreasing. (2 marks)
No longer in syllabus.
Points `A` and `B` are the points of intersection of `y = f (x)` with the `x`-axis. Point `A` has coordinates `(1, 0)` and point `B` has coordinates `(25, 0)`.

Find the length of `AD` such that the area of rectangle `ABCD` is equal to the area of the shaded region. (2 marks)
The points `P (16, 3)` and `B (25, 0)` are labelled on the diagram.
1. Find `m`, the gradient of the chord `PB`. (Exact value to be given.) (1 mark)
2. Find `a in [16, 25]` such that `f prime (a) = m`. (Exact value to be given.) (2 marks)

Show Answers Only

a. `x in [9,oo)`

b. `text(Not in syllabus.)`

c. `8/3`

d.i. `m=-1/3`

d.ii. `a=81/4`

Show Worked Solution

`text(Stationary point when)\ \ f prime (x)`	`= 0`
`x`	`= 9`

`:.\ text(Strictly decreasing for)\ \ x in [9, oo)`

b. `text(No longer in syllabus.)`

`AD`	`= y_text(average)`
	`= 1/(25 – 1) int_1^25\ f(x)\ dx`
	`= 8/3\ \ text{[by CAS]}`

d.i.	`m_(PB)`	`= (3 – 0)/(16 – 25)`
	`:. m_(PB)`	`= – 1/3`

d.ii.	`text(Solve)\ \ f prime (a)`	`= -1/3\ \ text(for)\ \ a in [16, 25]`
	`:. a`	`= 81/4`

Graphs, MET2 2009 VCAA 19 MC

The graph of a function `f`, with domain `R`, is as shown.

The graph which best represents `1 - f (2x)` is

Show Answers Only

`E`

Show Worked Solution

`text(Determine transformations)`

`text(from)\ \ f -> – f (2x) + 1`

`text(- Dilate by factor)\ 1/2\ text(from)\ y text(-axis.)`

`text(- Reflect in)\ x text(-axis.)`

`text(- Shift up 1 unit.)`

`=> E`

Calculus, MET2 2009 VCAA 18 MC

The average value of the function `f: R\ text(\){text(−)1/2} -> R,\ f(x) = 1/(2x + 1)` over the interval `[0, k]` is `1/6 log_e (7).`

The value of `k` is

`(-6)/(log_e(7)) - 1/2`
`3`
`e^3`
`(-log_e(7))/(2(log_e(7) + 6))`
`171`

Show Answers Only

`B`

Show Worked Solution

`text(Solve:)\ \ 1/(k-0) int_0^k 1/(2x + 1)\ dx`	`= 1/6 log_e (7)`
`text(for)\ \ k`	`> 0`

`:. k = 3`

`=> B`

Algebra, MET2 2009 VCAA 16 MC

The inverse of the function `f: R^+ -> R,\ f(x) = e^(2x + 3)` is

`{:(f^-1:\ R^+ -> R, qquad qquad qquad qquad f^-1 (x) = e^(-2x - 3)):}`
`{:(f^-1:\ R^+ -> R, qquad qquad qquad qquad f^-1 (x) = e^((x - 3)/2)):}`
`{:(f^-1:\ (e^3, oo) -> R, qquad qquad f^-1 (x) = log_e (sqrt x) - 3/2):}`
`{:(f^-1:\ (e^3, oo) -> R, qquad qquad f^-1 (x) = e^((x - 3)/2)):}`
`{:(f^-1:\ (e^3, oo) -> R, qquad qquad f^-1 (x) = -log_e (2x - 3)):}`

Show Answers Only

`C`

Show Worked Solution

`text(Let)\ \ y = f(x)`

`text(Inverse: swap)\ \ x harr y`

`x`	`=e^(2y + 3)`
`log_e x`	`=2y+3`
`2y`	`=log_e x -3`
`y`	`=1/2 log_e (x) – 3/2`

`:. f^-1 (x) = log_e (sqrt x) – 3/2`

`=> C`

Calculus, MET2 2009 VCAA 15 MC

For `y = sqrt (1 - f(x)),\ \ (dy)/(dx)` is equal to

A. `(2 f prime (x))/(sqrt(1 - f(x))`

B. `(-1)/(2 sqrt (1 - f prime (x)))`

C. `1/2 sqrt (1 - f prime (x))`

D. `3/(2(1 - f prime(x)))`

E. `(-f prime (x))/(2 sqrt (1 - f (x)))`

Show Answers Only

`E`

Show Worked Solution

`text(Using Chain Rule,)`

`dy/dx`	`= – f′(x) xx 1/2 xx (1-f(x))^(- 1/2)`
	`=(- f′(x))/(2 sqrt (1 – f (x)))`

`=> E`

Probability, MET2 2009 VCAA 13 MC

A fair coin is tossed twelve times.

The probability (correct to four decimal places) that at most 4 heads are obtained is

A. `0.0730`

B. `0.1209`

C. `0.1938`

D. `0.8062`

E. `0.9270`

Show Answers Only

`C`

Show Worked Solution

`text(Let)\ \ X = text(Number of heads),`

`X∼\ text(Bi) (12, 1/2)`

`text(Pr) (X <= 4) ~~ 0.1938\ \ [text(CAS: binomCdf) (12, 1/2, 0, 4)]`

`=> C`

Graphs, MET2 2009 VCAA 12 MC

A transformation `T: R^2 -> R^2` that maps the curve with equation `y = sin (x)` onto the curve with equation `y = 1 - 3 sin(2x + pi)` is given by

`T [(x), (y)] = [(2, 0), (0, -3)] [(x), (y)] + [(pi), (1)]`
`T [(x), (y)] = [(– 1/2, 0), (0, 3)] [(x), (y)] + [(pi/2), (1)]`
`T [(x), (y)] = [(0, – 3), (2, 0)] [(x), (y)] - [(pi), (1)]`
`T [(x), (y)] = [(1/2, 0), (0, – 3)] [(x), (y)] + [(– pi/2), (1)]`
`T [(x), (y)] = [(1/2, 0), (0, 3)] [(x), (y)] + [(– pi/2), (– 1)]`

Show Answers Only

`D`

Show Worked Solution

`text(Let)\ \ f(x) = sin (x)`

`text(Let)\ \ g(x) = – 3 sin (2 (x + pi/2)) + 1`

`text(Find transformations taking)\ \ f -> g`

`3 f (2x) = 3 sin (2x) = h(x)`

`– h(x) = – 3 sin (2x) = k(x)`

`k (x + pi/2) + 1 = – 3 sin (2 (x + pi/2)) + 1 = g(x)`

`text(Dilate by factor 3 from)\ \ x text(-axis)`

`text(Dilate by factor)\ \ 1/2\ \ text(from)\ \ y text(-axis)`

`text(Reflect in)\ \ x text(-axis)`

`text(Translate left)\ \ pi/2\ \ text(up 1)`

`=> D`

Probability, MET2 2009 VCAA 11 MC

The continuous random variable `X` has a probability density function given by

`f(x) = {(pi sin (2 pi x), text(if)\ \ 0 <= x <= 1/2), (0, text(elsewhere)):}`

The value of `a` such that `text(Pr) (X > a) = 0.2` is closest to

`0.26`
`0.30`
`0.32`
`0.35`
`0.40`

Show Answers Only

`D`

Show Worked Solution

`text(Solve)\ \ int_a^(1/2) f (x)\ dx`	`= 0.2,\ \ a in [0, 1/2]`
`:. a`	`~~ 0.35`

`=> D`

Probability, MET2 2009 VCAA 10 MC

The discrete random variable `X` has a probability distribution as shown.

The median of `X` is

`0`
`1`
`1.1`
`1.2`
`2`

Show Answers Only

`B`

Show Worked Solution

`text(Pr) (X <= 0)`	`= 0.4`
`text(Pr) (X <= 1)`	`= 0.6`

`:.\ text(Median) = 1`

`=> B`

Probability, MET2 2009 VCAA 6 MC

The continuous random variable `X` has a normal distribution with mean 14 and standard deviation 2.

If the random variable `Z` has the standard normal distribution, then the probability that `X` is greater than 17 is equal to

`text(Pr) (Z > 3)`
`text(Pr) (Z < 2)`
`text(Pr) (Z < 1.5)`
`text(Pr) (Z < – 1.5)`
`text(Pr) (Z > 2)`

Show Answers Only

`D`

Show Worked Solution

`text(Pr) (X > 17)`	`= text(Pr) (Z > (17 – 14)/2)`
	`= text(Pr) (Z > 1.5)`
	`= text(Pr) (Z < – 1.5)`

`=> D`

Algebra, MET2 2009 VCAA 5 MC

Let `f: R -> R,\ f (x) = x^2`

Which one of the following is not true?

`f(xy) = f (x) f (y)`
`f(x) - f(-x) = 0`
`f (2x) = 4 f (x)`
`f (x - y) = f(x) - f(y)`
`f (x + y) + f (x - y) = 2 (f (x) + f(y))`

Show Answers Only

`D`

Show Worked Solution

`text(Solution 1)`

`text(Consider option)\ D:`

`f(x-y)`	`=(x-y)^2`
	`=x^2 -2xy+y^2`
`f(x)-f(y)`	`= x^2-y^2`
`:.f(x-y)`	`!=f(x)-f(y)`

`=>D`

`text(Solution 2)`

`text(Define)\ \ f(x) = x^2`

`text(Enter each functional equation on CAS)`

`text(until output does NOT read “true”.)`

`=> D`

Algebra, MET2 2009 VCAA 4 MC

The general solution to the equation `sin (2x) = -1` is

`x = n pi - pi/4,\ n in Z`
`x = 2n pi + pi/4 or x = 2n pi - pi/4,\ n in Z`
`x = (n pi)/2 + (-1)^n pi/2,\ n in Z`
`x = (n pi)/2 + (-1)^n pi/4,\ n in Z`
`x = n pi + pi/4 or x = 2n pi + pi/4,\ n in Z`

Show Answers Only

`A`

Show Worked Solution

`2x`	`= 2n pi – pi/2,\ \ n in Z`
`x`	`= n pi – pi/4,\ \ n in Z`

`=> A`

Calculus, MET2 2011 VCAA 4

Deep in the South American jungle, Tasmania Jones has been working to help the Quetzacotl tribe to get drinking water from the very salty water of the Parabolic River. The river follows the curve with equation `y = x^2 - 1`, `x >= 0` as shown below. All lengths are measured in kilometres.

Tasmania has his camp site at `(0, 0)` and the Quetzacotl tribe’s village is at `(0, 1)`. Tasmania builds a desalination plant, which is connected to the village by a straight pipeline.

If the desalination plant is at the point `(m, n)` show that the length, `L` kilometres, of the straight pipeline that carries the water from the desalination plant to the village is given by
`L = sqrt(m^4 - 3m^2 + 4)`. (3 marks)
If the desalination plant is built at the point on the river that is closest to the village
i. find `(dL)/(dm)` and hence find the coordinates of the desalination plant (3 marks)
ii. find the length, in kilometres, of the pipeline from the desalination plant to the village. (2 marks)

The desalination plant is actually built at `(sqrt7/2, 3/4)`.

If the desalination plant stops working, Tasmania needs to get to the plant in the minimum time.

Tasmania runs in a straight line from his camp to a point `(x,y)` on the river bank where `x <= sqrt7/2`. He then swims up the river to the desalination plant.

Tasmania runs from his camp to the river at 2 km per hour. The time that he takes to swim to the desalination plant is proportional to the difference between the `y`-coordinates of the desalination plant and the point where he enters the river.

Show that the total time taken to get to the desalination plant is given by

`qquadT = 1/2 sqrt(x^4 - x^2 + 1) + 1/4k(7 - 4x^2)` hours where `k` is a positive constant of proportionality. (3 marks)

The value of `k` varies from day to day depending on the weather conditions.

If `k = 1/(2sqrt13)`
i. find `(dT)/(dx)` (1 mark)
ii. hence find the coordinates of the point where Tasmania should reach the river if he is to get to the desalination plant in the minimum time. (2 marks)
On one particular day, the value of `k` is such that Tasmania should run directly from his camp to the point `(1,0)` on the river to get to the desalination plant in the minimum time. Find the value of `k` on that particular day. (2 marks)
Find the values of `k` for which Tasmania should run directly from his camp towards the desalination plant to reach it in the minimum time. (2 marks)

Show Answers Only

`text(See Worked Solutions)`
i. `(sqrt6/2, 1/2)`
ii. `sqrt7/2\ text(km)`
`text(See Worked Solutions)`
i. `(dT)/(dx) = (x(2x^2 – 1))/(2sqrt(x^4 – x^2 + 1)) – (sqrt13 x)/13`
ii. `(sqrt3/2, −1/4)`
`1/4`
`(5sqrt37)/74`

Show Worked Solution

a. `text(S)text(ince)\ \ (m,n)\ \ text(lies on)\ \ y=x^2-1,`

♦ Mean mark 47%.

`=> n=m^2-1`

`V(0,1), D(m,m^2 – 1)`

`L`	`= sqrt((m – 0)^2 + ((m^2 – 1) – 1)^2)`
	`= sqrt(m^2 + m^4 – 4m^2 + 4)`
	`= sqrt(m^4 – 3m^2 + 4)\ \ text(… as required)`

b.i. `(dL)/(dm) = (2m^2 – 3m)/(sqrt(m^4 – 3m^2 + 4))`

`text(Solve:)\ \ (dL)/(dm) = 0quadtext(for)quadm >= 0`

`:. m = sqrt6/2`

`text(Substitute into:)\ \ D(m, m^2 – 1),`

`:. text(Desalination plant at)\ \ (sqrt6/2, 1/2)`

♦ Mean mark part (b)(ii) 41%.

b.ii.	`L(sqrt6/2)`	`= sqrt(m^4 – 3m^2 + 4)`
		`=sqrt(36/16-3xx6/4+4`
		`=sqrt7/2`

c. `text(Let)\ \ P(x,x^2 – 1)\ text(be run point on bank)`

♦♦♦ Mean mark part (c) 16%.

`text(Let)\ \ D(sqrt7/2, 3/4)\ text(be desalination location)`

`T`	`=\ text(run time + swim time)`
	`= (sqrt((x – 0)^2 + ((x^2 – 1) – 0)^2))/2 + k(3/4 – (x^2 – 1))`
	`= (sqrt(x^2 + x^4 – 2x^2 + 1))/2 + k/4(3 – 4(x^2 – 1))`
`:. T`	`= (sqrt(x^4 – x^2 + 1))/2 + 1/4k(7 – 4x^2)`

d.i. `(dT)/(dx) = (x(2x^2 – 1))/(2sqrt(x^4 – x^2 + 1)) – (sqrt13 x)/13`

d.ii. `text(Solve:)\ \ (dT)/(dx) = 0`

♦♦ Mean mark part (d)(ii) 33%.

`x = sqrt3/2`

`y=x^2-1=-1/4`

`:. T_(text(min)) \ text(when point is)\ \ (sqrt3/2, −1/4)`

e. `(dT)/(dx) = (x(2x^2 – 1))/(2sqrt(x^4 – x^2 + 1)) – 2kx`

♦♦ Mean mark part (e) 39%.

`text(When)\ \ x=1,`

`text(Solve:)\ \ (dT)/(dx)`	`=0\ \ text(for)\ k,`
`1/2 -2k`	`=0`
`:.k`	`=1/4`

f. `text(Require)\ T_text(min)\ text(to occur at right-hand endpoint)\ \ x = sqrt7/2.`

`text(This can occur in 2 situations:)`

♦♦♦ Mean mark 13%.

`text(Firstly,)\ \ T\ text(has a local min at)\ \ x = sqrt7/2,`

`text(Solve:)\ \ (x(2x^2 – 1))/(2sqrt(x^4 – x^2 + 1)) – 2kx=0| x = sqrt7/2,\ \ text(for)\ k,`

`:.k = (5sqrt37)/74`

`text(S)text(econdly,)\ \ T\ text(is decreasing function over)\ x ∈ (0, sqrt7/2),`

`text(Solve:)\ \ (dT)/(dx) <= 0 | x = sqrt7/2,\ text(for)\ k,`

`:. k > (5sqrt37)/74`

`:. k >= (5sqrt37)/74`

Calculus, MET2 2011 VCAA 3

Consider the function `f: R -> R, f(x) = 4x^3 + 5x-9`.
1. Find `f^{′}(x)` (1 mark)
2. Explain why `f^{′}(x) >= 5` for all `x`. (1 mark)
The cubic function `p` is defined by `p: R -> R, p(x) = ax^3 + bx^2 + cx + k`, where `a`, `b`, `c` and `k` are real numbers.
1. If `p` has `m` stationary points, what possible values can `m` have? (1 mark)
2. If `p` has an inverse function, what possible values can `m` have? (1 mark)
The cubic function `q` is defined by `q:R -> R, q(x) = 3-2x^3`.
1. Write down a expression for `q^(−1)(x)`. (2 marks)
2. Determine the coordinates of the point(s) of intersection of the graphs of `y = q(x)` and `y = q^(−1)(x)`. (2 marks)
The cubic function `g` is defined by `g: R -> R, g(x) = x^3 + 2x^2 + cx + k`, where `c` and `k` are real numbers.
1. If `g` has exactly one stationary point, find the value of `c`. (3 marks)
2. If this stationary point occurs at a point of intersection of `y = g(x)` and `g^(−1)(x)`, find the value of `k`. (3 marks)

Show Answers Only

1. `f^{′}(x) = 12x^2 + 5`
2. `text(See Worked Solutions)`
1. `m = 0, 1, 2`
2. `m = 0, 1`
1. `q^(−1)(x) = root(3)((3-x)/2), x ∈ R`
2. `(1, 1)`
1. `4/3`
2. `−10/27`

Show Worked Solution

a.i. `f^{′}(x) = 12x^2 + 5`

a.ii. `text(S)text(ince)\ \ x^2>=0\ \ text(for all)\ x,`

♦ Mean mark 47%.

` 12x^2`	`>= 0`
`12x^2 + 5`	`>= 5`
`f^{′}(x)`	`>= 5\ \ text(for all)\ x`

b.i. `p(x) = text(is a cubic)`

♦♦♦ Mean mark part (b)(i) 9%, and part (b)(ii) 20%.
MARKER’S COMMENT: Good exam strategy should point students to investigate earlier parts for direction. Here, part (a) clearly sheds light on a solution!

`:. m = 0, 1, 2`

`text{(Note: part a.ii shows that a cubic may have no SP’s.)}`

b.ii. `text(For)\ p^(−1)(x)\ text(to exist)`

`:. m = 0, 1`

c.i. `text(Let)\ y = q(x)`

`text(Inverse: swap)\ x ↔ y`

`x`	`= 3-2y^3`
`y^3`	`= (3-x)/2`

`:. q^(−1)(x) = root(3)((3-x)/2), \ x ∈ R`

c.ii. `text(Any function and its inverse intersect on)`

`text(the line)\ \ y=x.`

`text(Solve:)\ \ 3-2x^3`	`= xqquadtext(for)\ x,`
`x`	`= 1`

`:.\ text{Intersection at (1, 1)}`

♦ Mean mark part (d)(i) 44%.

d.i.	`g^{′}(x)`	`= 0`
	`3x^2 + 4x + c`	`= 0`
	`Delta`	`= 0`
	`16-4(3c)`	`= 0`
	`:. c`	`= 4/3`

d.ii. `text(Define)\ \ g(x) = x^3 + 2x^2 + 4/3x + k`

♦♦♦ Mean mark part (d)(ii) 14%.

`text(Stationary point when)\ \ g^{′}(x)=0`

`g^{′}(x) = 3x^2+4x+4/3`

`text(Solve:)\ \ g^{′}(x)=0\ \ text(for)\ x,`

`x = −2/3`

`text(Intersection of)\ g(x)\ text(and)\ g^(−1)(x)\ text(occurs on)\ \ y = x`

`text(Point of intersection is)\ (−2/3, −2/3)`

`text(Find)\ k:`

`g(−2/3)`	`= −2/3\ text(for)\ k`
`:. k`	` = −10/27`

Probability, MET2 2011 VCAA 2

In a chocolate factory the material for making each chocolate is sent to one of two machines, machine A or machine B.

The time, `X` seconds, taken to produce a chocolate by machine A, is normally distributed with mean 3 and standard deviation 0.8.

The time, `Y` seconds, taken to produce a chocolate by machine B, has the following probability density function.

`f(y) = {{:(0,y < 0),(y/16,0 <= y <= 4),(0.25e^(−0.5(y - 4)),y > 4):}`

Find correct to four decimal places
1. `text(Pr)(3 <= X <= 5)` (1 mark)
2. `text(Pr)(3 <= Y <= 5)` (3 marks)
Find the mean of `Y`, correct to three decimal places. (3 marks)

The content in Part c(i) and c(ii) is no longer in this course.

It can be shown that `text(Pr)(Y <= 3) = 9/32`. A random sample of 10 chocolates produced by machine B is chosen. Find the probability, correct to four decimal places, that exactly 4 of these 10 chocolate took 3 or less seconds to produce. (2 marks)

All of the chocolates produced by machine A and machine B are stored in a large bin. There is an equal number of chocolates from each machine in the bin.

It is found that if a chocolate, produced by either machine, takes longer than 3 seconds to produce then it can easily be identified by its darker colour.

A chocolate is selected at random from the bin. It is found to have taken longer than 3 seconds to produce.
Find, correct to four decimal places, the probability that it was produced by machine A. (3 marks)

Show Answers Only

a.i. `0.4938`

a.ii. `0.4155`

b. `4.333`

d. `0.1812`

e. `0.4103`

Show Worked Solution

a.i. `X ∼\ N(3,0.8^2)`

`text(Pr)(3 <= X <= 5) = 0.4938\ \ text{(4 d.p.)}`

a.ii.	`text(Pr)(3 <= Y <= 5)`	`= text(Pr)(3 <= Y <= 4) + (4 < Y <= 5)`
		`= int_3^4 (y/16)dy + int_4^5(1/4 e^(−1/2(y – 4)))dy`
		`= 0.4155\ \ text{(4 d.p.)}`

b.	`text(E)(Y)`	`= int_0^4 y(y/16)dy + int_4^∞ 0.25ye^(−1/2(y – 4))dy`
		`= 4.333\ \ text{(3 d.p.)}`

The content in Part c(i) and c(ii) is no longer in this course.

d. `text(Solution 1)`

`text(Let)\ \ W = #\ text(chocolates from)\ B\ text(that take less)`

`text(than 3 seconds)`

`W ∼\ text(Bi)(10, 9/32)`

`text(Using CAS: binomPdf)(10, 9/32,4)`

`text(Pr)(W = 4) = 0.1812\ \ text{(4 d.p.)}`

`text(Solution 2)`

`text(Pr)(W = 4)`	`=((10),(4)) (9/32)^4 (23/32)^6`
	`=0.1812`

♦♦♦ Mean mark part (e) 19%.
MARKER’S COMMENT: Students who used tree diagrams were the most successful.

`text(Pr)(A \| L)`	`= (text(Pr)(AL))/(text(Pr)(L))`
	`= (0.5 xx 0.5)/(0.5 xx 0.5 + 0.5 xx 23/32)`
	`= 0.4103\ \ text{(4 d.p.)}`

CORE, FUR2 SM-Bank VCE 4

Damon runs a swim school.

The value of his pool pump is depreciated over time using flat rate depreciation.

Damon purchased the pool pump for $28 000 and its value in dollars after `n` years, `P_n`, is modelled by the recursion equation below:

`P_0 = 28\ 000,qquad P_(n + 1) = P_n - 3500`

Write down calculations, using the recurrence relation, to find the pool pump's value after 3 years. (1 mark)
After how many years will the pump's depreciated value reduce to $7000? (1 mark)

The reducing balance depreciation method can also be used by Damon.

Using this method, the value of the pump is depreciated by 15% each year.

A recursion relation that models its value in dollars after `n` years, `P_n`, is:

`P_0 = 28\ 000, qquad P_(n + 1) = 0.85P_n`

After how many years does the reducing balance method first give the pump a higher valuation than the flat rate method in part (a)? (2 marks)

Show Answers Only

`$17\ 500`
`6\ text(years)`
`4\ text(years)`

Show Worked Solution

a.	`P_1`	`= 28\ 000 – 3500 = 24\ 500`
	`P_2`	`= 24\ 500 – 3500 = 21\ 000`
	`P_3`	`= 21\ 000 – 3500 = 17\ 500`

`:.\ text(After 3 years, the pump’s value is $17 500.)`

b. `text(Find)\ n\ text(such that:)`

`7000`	`= 28\ 000 – 3500n`
`3500n`	`= 21\ 000`
`n`	`= (21\ 000)/3500`
	`= 6\ text(years)`

c. `text(Using the reducing balance method)`

`P_1`	`= 0.85 xx 28\ 000 = 23\ 800`
`P_2`	`= 0.85 xx 23\ 800 = 20\ 230`
`P_3`	`= 0.85 xx 20\ 230 = 17\ 195`
`P_4`	`= 0.85 xx 17\ 195 = 14\ 615.75`

`text{Using the flat rate method (see part (a))}`

`P_4 = 17\ 500 – 3500 = 14\ 000`

`14\ 615.75 > 14\ 000`

`:.\ text(After 4 years, the reducing balance method)`

`text(first values the pump higher.)`

`:.e^(2x)`	`=4`	`e^(2x)`	`=1`
`2x`	`=log_e 4`	`x`	`=0`
`x`	`=(2log_e 2)/2`
	`=log_e 2`

`k`	`= – [log_e x]_1^2`
	`=-(log_e(2)-log_e(1))`
	`= – log_e (2)`
	`= log_e (1/2)`
`:. e^k`	`= e^(log_e(1/2))`
	`= 1/2`

`P(2)`	`=0`
`2^2 + 2a + b`	`= 0`
`2a + b`	`= −4`	`…\ (1)`

`(−1)^2 − a + b`	`= 18`
`−a + b`	`= 17`	`…\ (2)`

`text(Area)`	`~~ pi/6 [sin (pi/6) + sin (pi/3)]`
	`~~pi/6(sqrt3/2 + 1/2)`
	`~~ (pi (sqrt 3 + 1))/12`

`X^2-5X+4`	`=0`
`(X-4)(X-1)`	`=0`
`X`	`=4 or 1`

`text(Average ROC)`	`= (f(3) – f(0))/(3 – 0)`
	`= 26/3`

`3a`	`= −21`
`a`	`= −7`