Differentiate `(e^x + x)^5`. (2 marks)
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Differentiate `(e^x + x)^5`. (2 marks)
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`5(e^x + 1)(e^x + x)^4`
| `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` |
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
`D`
| `text{Pr(YYY)}` | `= 7/10 xx 6/9 xx 5/8` |
| `= 7/24` |
`=> D`
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)
`a = 1/e`
`b = e`
| `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` |
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)
`1/2 e^4 + e^(−2) – 3/2\ \ \ text(u²)`
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)` |
For the function `f:R→R,\ \ f(x)= 2e^x + 3x`, determine the coordinates of the point `P` at which the tangent to `f(x)` is parallel to the line `y = 5x - 3`. (3 marks)
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`(0, 2)`
`y = 5x-3\ \ =>\ m=5`
| `y` | `= 2e^x + 3x` |
| `(dy)/(dx)` | `= 2e^x + 3` |
`text(Find)\ \ x\ \ text(when)\ \ (dy)/(dx) = 5,`
| `5` | `= 2e^x + 3` |
| `2e^x` | `= 2` |
| `e^x` | `= 1` |
| `x` | `= 0` |
`text(When)\ \ x = 0,`
| `y` | `= 2e^0 + (3 xx 0)=2` |
`:.P\ \ text{has coordinates (0, 2)}`
Solve the following equation for `x`:
`2e^(2x) - e^x = 0`. (2 marks)
`x = ln\ 1/2`
`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` |
Write `log2 + log4 + log8 + log16 + … + log128` in the form `a logb` where `a` and `b` are integers greater than 1. (2 marks)
`28log2`
`log2 + log4 + log8 + … + log 128`
`= log2^1 + log2^2 + log2^3 + … + log2^7`
`= log2 + 2log2 + 3log2 + … + 7log2`
`= 28log2`
Solve `5^x = 4` for `x`. (2 marks)
`(log_e 4)/(log_e 5)`
| `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)` |
Let `a = e^x`
Which expression is equal to `log_e(a^2)`?
`C`
| `log_e(a^2)` | `= log_e(e^x)^2` |
| `= log_e(e^(2x))` | |
| `= 2xlog_e e` | |
| `= 2x` |
`=> C`
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. --- 5 WORK AREA LINES (style=lined) --- --- 4 WORK AREA LINES (style=lined) ---
Let `f : (0, ∞) → R`, where `f(x) = log_e(x)` and `g: R → R`, where `g (x) = x^2 + 1`.
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ii. State the domain and range of `h`. (2 marks)
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| a.i. | `h(x)` | `= f(x^2 + 1)` |
| `= log_e(x^2 + 1)` |
a.ii. `text(Domain)\ (h) =\ text(Domain)\ (g) = R`
| `text(For)\ x ∈ R` | `-> x^2 + 1 >= 1` |
| `-> log_e(x^2 + 1) >= 0` |
`:.\ text(Range)\ (h) = [0,∞)`
| 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^{prime}(x) = 0`
`text(Using Chain Rule:)`
| `h^{prime}(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)`
`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)`
b.ii. `text(Range)\ (k^(-1)) =\ text(Domain)\ (k) = (-∞,0]`
`text(Domain)\ (k^(-1)) =\ text(Range)\ (k) = [0,∞)`
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. --- 4 WORK AREA LINES (style=lined) --- --- 4 WORK AREA LINES (style=lined) --- --- 4 WORK AREA LINES (style=lined) ---
a. `text(Let)\ \ X =\ text(Number of tagged sheep,)` `X ~\ text(Bi)(4,1/3)` c. `text(Let)\ \ Y =\ text(Days that no tagged sheep selected,)` `Y ~\ text(Bi)(6,16/81)`
`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`
`text(Pr)(Y = 6)`
`= ((6),(6)) xx (16/81)^6 xx (65/81)^0`
`= (16/81)^6 = (2/3)^24`
Let `f: R text{\}{1} -> R` where `f(x) = 2 + 3/(x - 1)`.
Let `f: (-∞,1/2] -> R`, where `f(x) = sqrt(1-2x)`.
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| a. | `f(x)` | `= (1-2x)^(1/2)` |
| `f^{prime}(x)` | `= 1/2(1-2x)^(-1/2) (-2)qquadtext([Using Chain Rule])` | |
| `= (-1)/(sqrt(1-2x))` |
| b. | `tan theta` | `= f^{prime}(-1)` |
| `= (-1)/(sqrt(1-2(-1)))` | ||
| `= (-1)/(sqrt3)` |
`text(S)text(ince)\ theta ∈ [0,pi],`
`=> theta = (5pi)/6`
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.
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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.
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a.i. `1/200 (8a + 4b + c) = 0`
a.ii. `1/200 (12a + 4b) = (– 3)/50`
a.iii. `1/200 (48a + 8b) = 0`
`text(Proof)\ \ text{(See Worked Solutions)}`
a.i. `N (2, 0),`
`1/200 (8a + 4b + c) = 0\ \ text{… (1)}`
`(dy)/(dx) (x = 2) = (– 3)/50,`
`1/200 (12a + 4b) = (– 3)/50\ \ text{… (2)}`
`(dy)/(dx)(x = 4) = 0,`
`1/200 (48a + 8b) = 0\ \ text{… (3)}`
ii. `text(Using the above equations,)`
| `12a + 4b` | `=-12` | `\ \ \ …\ (2^{′})` |
| `24 a + 4b` | `=0` | `\ \ \ …\ (3^{′})` |
`text{Solve simultaneous equations (By CAS):}`
`a=1, \ b=-6, \ c=16`
`\text{Solving manually:}`
`(3^{′})-(2^{′}):`
`12a=12 \ \Rightarrow\ \ a=1`
`text(Substitute)\ \ a = 1\ \ text(into)\ \ (3^{′}):`
`124(1)+4b=0 \ \Rightarrow\ \ b=-1`
`text(Substitute)\ \ a = 1,\ \ b = – 6\ \ text(into)\ \ (1):`
`8(1) + 4 (– 6) + c=0\ \ \Rightarrow\ \ 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)`
| `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)`
`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)`
`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)`
| `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)`
| `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)`
Let `f: R^+ uu {0} -> R,\ f(x) = 6 sqrt x-x-5.`
The graph of `y = f (x)` is shown below.
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Find the length of `AD` such that the area of rectangle `ABCD` is equal to the area of the shaded region. (2 marks)

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a. `x in [9,oo)`
b. `8/3`
c.i. `m=-1/3`
c.ii. `a=81/4`
| a. | ![]() |
| `text(Stationary point when)\ \ f^{′}(x)` | `= 0` |
| `x` | `= 9` |
`:.\ text(Strictly decreasing for)\ \ x in [9, oo)`
| b. | ![]() |
| `AD` | `= y_text(average)` |
| `= 1/(25-1) int_1^25\ f(x)\ dx` | |
| `= 8/3\ \ text{[by CAS]}` |
| c.i. | `m_(PB)` | `= (3-0)/(16-25)` |
| `:. m_(PB)` | `=-1/3` |
| c.ii. | `text(Solve)\ \ f^{′}(a)` | `= -1/3\ \ text(for)\ \ a in [16, 25]` |
| `:. a` | `= 81/4` |
The graph of a function `f`, with domain `R`, is as shown.
The graph which best represents `1 - f (2x)` is
`E`
`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`
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
`B`
| `text(Solve:)\ \ 1/(k-0) int_0^k 1/(2x + 1)\ dx` | `= 1/6 log_e (7)` |
| `text(for)\ \ k` | `> 0` |
`:. k = 3`
`=> B`
The inverse of the function `f: R^+ -> R,\ f(x) = e^(2x + 3)` is
`C`
`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`
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)))`
`E`
`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`
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`
`C`
`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`
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
`D`
`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`
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
`D`
| `text(Solve)\ \ int_a^(1/2) f (x)\ dx` | `= 0.2,\ \ a in [0, 1/2]` |
| `:. a` | `~~ 0.35` |
`=> D`
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
`D`
| `text(Pr) (X > 17)` | `= text(Pr) (Z > (17 – 14)/2)` |
| `= text(Pr) (Z > 1.5)` | |
| `= text(Pr) (Z < – 1.5)` |
`=> D`
Let `f: R -> R,\ f (x) = x^2`
Which one of the following is not true?
`D`
`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`
The general solution to the equation `sin (2x) = -1` is
`A`
| `2x` | `= 2n pi – pi/2,\ \ n in Z` |
| `x` | `= n pi – pi/4,\ \ n in Z` |
`=> A`
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.
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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.
`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)
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The value of `k` varies from day to day depending on the weather conditions.
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a. `text(S)text(ince)\ \ (m,n)\ \ text(lies on)\ \ y=x^2-1,`
`=> 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`
`\Rightarrow m = sqrt6/2`
`text(Substitute into:)\ \ D(m, m^2-1),`
`:. text(Desalination plant at)\ \ (sqrt6/2, 1/2)`
| 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)`
`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`
`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`
`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:)`
`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`
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a.i. `f^{prime}(x) = 12x^2 + 5`
a.ii. `text(S)text(ince)\ \ x^2>=0\ \ text(for all)\ x,`
| ` 12x^2` | `>= 0` |
| `12x^2 + 5` | `>= 5` |
| `f^{prime}(x)` | `>= 5\ \ text(for all)\ x` |
b.i. `p(x) = text(is a cubic)`
`:. 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)}`
| d.i. | `g^{prime}(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`
`text(Stationary point when)\ \ g^{prime}(x)=0`
`g^{prime}(x) = 3x^2+4x+4/3`
`text(Solve:)\ \ g^{prime}(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` |
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):}` --- 3 WORK AREA LINES (style=lined) --- --- 3 WORK AREA LINES (style=lined) --- --- 5 WORK AREA LINES (style=lined) --- --- 5 WORK AREA LINES (style=lined) --- 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. --- 6 WORK AREA LINES (style=lined) ---
a.i. `0.4938` a.ii. `0.4155` b. `4.333` c. `0.1812` d. `0.4103`
a.i. `X ∼\ N(3,0.8^2)` `text(Pr)(3 <= X <= 5) = 0.4938\ \ text{(4 d.p.)}` c. `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)`
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.)}`
`text(Pr)(W = 4)`
`=((10),(4)) (9/32)^4 (23/32)^6`
`=0.1812`
MARKER’S COMMENT: Students who used tree diagrams were the most successful.
d.

`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.)}`
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`
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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`
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| 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.)`
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The point `P(c, d)` is on the graph of `f`.
Find the exact values of `c` and `d` such that the distance of this point to the origin is a minimum, and find this minimum distance. (3 marks)
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Let `g: (−k, oo) -> R, g(x) = (kx + 1)/(x + k)`, where `k > 1`.
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Find the values of `k` such that `s(k) >= 1`. (2 marks)
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Let `A(k)` be the rule of the function `A` that gives the area of this enclosed region. The domain of `A` is `(1, oo)`.
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a. `text(Solution 1)`
| `(2x + 1)/(x + 2)` | `= 2-3/(x + 2)` |
| `= 2 + {(-3)}/(x + 2) qquad [text(CAS: prop Frac) ((2x + 1)/(x + 2))]` |
`text(Solution 2)`
| `(2x + 1)/(x + 2)` | `=(2(x+2)-3)/(x+2)` |
| `=2+ (-3)/(x+2)` |
b.i. `text(Let)\ \ y = f(x)`
`text(For Inverse: swap)\ \ x ↔ y`
| `x` | `=2-3/(y + 2)` |
| `(x-2)(y+2)` | `=-3` |
| `y` | `=(-3)/(x-2)-2` |
`text(Range of)\ f(x):\ \ y in R text(\){2}`
`:. f^(-1) (x) = (-3)/(x-2)-2, \ x in R text(\){2}`
b.ii. `text(Find intersection points:)`
| `f(x)` | `= x` |
| `(2x+1)/(x+2)` | `=x` |
| `2x+1` | `=x^2+2x` |
| `:. x` | `= +- 1` |
| `:.\ text(Area)` | `= int_(-1)^1 (f(x)-x)\ dx` |
| `=int_(-1)^1 (2-3/(x + 2)-x)\ dx` | |
| `= 4-3 ln(3)\ text(u²)` |
| b.iii. | `text(Area)` | `= int_(-1)^1 (f(x)-f^(-1) (x)) dx` |
| `=2 xx (4-3 ln(3))\ \ \ text{(twice the area in (b)(ii))}` | ||
| `= 8-6 log_e (3)\ text(u²)` |
| c. | ![]() |
|
| `text(Let)\ \ z` | `= OP, qquad P(c, -3/(c + 2) + 2)` | |
| `z` | `= sqrt (c^2 + (2-3/(c + 2))^2), \ c > -2` | |
`text(Stationary point when:)`
`(dz)/(dc) = 0, c > -2`
`:.\ c = sqrt 3-2 overset and (->) d = 2-sqrt 3`
`:. text(Minimum distance) = (2 sqrt 2-sqrt 6)`
d. `text(Given:)\ \ -k < x_1 < x_2`
`text(Must prove:)\ \ g(x_2)-g(x_1) > 0`
`text(LHS:)`
`g(x_2)-g(x_1)`
`= (kx_2 +1)/(x_2+k)-(kx_1 +1)/(x_1+k)`
`=((kx_2 +1)(x_1+k)-(kx_1 +1)(x_2+k))/((x_2+k)(x_1+k))`
`=(k^2(x_2-x_1)-(x_2-x_1))/((x_2+k)(x_1+k))`
`=((k^2-1)(x_2-x_1))/((x_2+k)(x_1+k))`
`x_2-x_1 >0,\ and \ k^2-1>0`
`text(S)text(ince)\ \ x_2>x_1> -k,`
`=> -x_2<-x_1<k`
`=>k+x_1 >0, \ and \ k+x_2>0`
`:.g(x_2)-g(x_1) >0`
`:. g(x_2) > g(x_1)`
| e.i. | `g(x)` | `= -x` |
| `(kx +1)/(x+k)` | `=-x` | |
| `kx+1` | `=-x^2-xk` | |
| `x^2+2k+1` | `=0` | |
| `:. x` | `=(-2k +- sqrt(4k^2-4))/2` | |
| `= sqrt (k^2-1)-k\ \ text(for)\ \ x > -k` |
`:. X (-k + sqrt(k^2-1),\ \ k-sqrt(k^2-1))`
e.ii. `text(Equate)\ \ x text(-coordinates:)`
| `-k + sqrt(k^2-1)` | `= -1/2` |
| `sqrt(k^2-1)` | `=k-1/2` |
| `k^2-1` | `=k^2-k+1/4` |
| `:. k` | `= 5/4` |
| e.iii. | `s(k)` | `= (1/2 xx YZ xx XO)^2` |
| `= 1/4 xx (YZ)^2 xx (XO)^2` |
`ZO = sqrt(1^2+1^2) = sqrt2`
`YZ=2 xx ZO = 2sqrt2`
`(YZ)^2 = 8`
| `(XO)^2` | `=(-k + sqrt(k^2-1))^2-(k-sqrt (k^2-1))^2` |
| `=2(-k + sqrt(k^2-1))^2` | |
`text(Solve)\ \ s(k) >= 1\ \ text(for)\ \ k >= 1,`
| `1/4 xx 8 xx 2(-k + sqrt(k^2-1))^2` | `>=1` |
| `(-k + sqrt(k^2-1))^2` | `>=1/4` |
| `k-sqrt(k^2-1)` | `>=1/2` |
| `k-1/2` | `>= sqrt(k^2-1)` |
| `k^2-k+1/4` | `>=k^2-1` |
| `:.k` | `<= 5/4` |
`:. 1<k<= 5/4`
| f.i. | `A(k)` | `= int_(-1)^1 (g(x)-x)\ dx,\ \ k > 1` |
| `= int_(-1)^1 ((kx+1)/(x+k) -x)\ dx` | ||
| `= int_(-1)^1 (k+ (1-k^2)/(x+k) -x)` | ||
| `=[kx + (1-k^2) log_e(x+k)-x^2/2]_(-1)^1` | ||
| `=(k+(1-k^2)log_e(1+k)-1/2)-(-k+(1-k^2)log_e(k-1)-1/2)` | ||
| `=2k+(1-k^2)log_e ((1+k)/(k-1))` | ||
| `= (k^2-1) log_e ((k-1)/(k + 1)) + 2k` |
| f.ii. | ![]() |
|
| `0` | `< A(k) < text(Area of)\ Delta ABC` | |
| `0` | `< A(k) < 1/2 xx AC xx BO` | |
| `0` | `< A(k) < 1/2 sqrt(2^2 + 2^2) xx (sqrt(1^2 + 1^2))` | |
| `0` | `< A(k) < 1/2 xx 2 sqrt 2 xx sqrt 2` | |
| `:. 0` | `< A(k) < 2` | |
A school has a class set of 22 new laptops kept in a recharging trolley. Provided each laptop is correctly plugged into the trolley after use, its battery recharges. On a particular day, a class of 22 students uses the laptops. All laptop batteries are fully charged at the start of the lesson. Each student uses and returns exactly one laptop. The probability that a student does not correctly plug their laptop into the trolley at the end of the lesson is 10%. The correctness of any student’s plugging-in is independent of any other student’s correctness. --- 5 WORK AREA LINES (style=lined) --- --- 5 WORK AREA LINES (style=lined) --- The time for which a laptop will work without recharging (the battery life) is normally distributed, with a mean of three hours and 10 minutes and standard deviation of six minutes. Suppose that the laptops remain out of the recharging trolley for three hours. --- 5 WORK AREA LINES (style=lined) --- A supplier of laptops decides to take a sample of 100 new laptops from a number of different schools. For samples of size 100 from the population of laptops with a mean battery life of three hours and 10 minutes and standard deviation of six minutes, `hat P` is the random variable of the distribution of sample proportions of laptops with a battery life of less than three hours. --- 6 WORK AREA LINES (style=lined) --- It is known that when laptops have been used regularly in a school for six months, their battery life is still normally distributed but the mean battery life drops to three hours. It is also known that only 12% of such laptops work for more than three hours and 10 minutes. --- 5 WORK AREA LINES (style=lined) --- The laptop supplier collects a sample of 100 laptops that have been used for six months from a number of different schools and tests their battery life. The laptop supplier wishes to estimate the proportion of such laptops with a battery life of less than three hours. --- 2 WORK AREA LINES (style=lined) --- The laptop supplier finds that, in a particular sample of 100 laptops, six of them have a battery life of less than three hours. --- 2 WORK AREA LINES (style=lined) --- `qquad qquad f(x) = {(((210-x)e^((x-210)/20))/400, 0 <= x <= 210), (0, text{elsewhere}):}` --- 2 WORK AREA LINES (style=lined) ---
a. `text(Solution 1)` `text(Let)\ \ X = text(number not correctly plugged),` `X ~ text(Bi) (22, .1)` `text(Pr) (X >= 1) = 0.9015\ \ [text(CAS: binomCdf)\ (22, .1, 1, 22)]` `text(Solution 2)` b. `text(Pr) (X < 5 | X >= 1)` `= (text{Pr} (1 <= X <= 4))/(text{Pr} (X >= 1))` `= (0.83938…)/(0.9015…)\ \ [text(CAS: binomCdf)\ (22, .1, 1,4)]` `= 0.9311\ \ text{(4 d.p.)}` c. `text(Let)\ \ Y = text(battery life in minutes)` `Y ~ N (190, 6^2)` `text(Pr) (Y <= 180)= 0.0478\ \ text{(4 d.p.)}` `[text(CAS: normCdf)\ (−oo, 180, 190,6)]` d. `text(Let)\ \ W = text(number with battery life less than 3 hours)` `W ~ Bi (100, .04779…)` e. `text(Let)\ \ B = text(battery life), B ~ N (180, sigma^2)` g. `text(95% confidence int:) qquad quad [(text(CAS:) qquad qquad 1-text(Prop)\ \ z\ \ text(Interval)), (x = 6), (n = 100)]` `p in (0.01, 0.11)`
`text(Pr) (X>=1)`
`=1-text(Pr) (X=0)`
`=1-0.9^22`
`=0.9015\ \ text{(4 d.p.)}`
`text(Pr) (hat P >= .06 | hat P >= .05)`
`= text(Pr) (X_2 >= 6 | X_2 >= 5)`
`= (text{Pr} (X_2 >= 6))/(text{Pr} (X_2 >= 5))`
`= (0.3443…)/(0.5234…)`
`= 0.658\ \ text{(3 d.p.)}`
`text(Pr) (B > 190)`
`= .12`
`text(Pr) (Z < a)`
`= 0.88`
`a`
`dot = 1.17499…\ \ [text(CAS: invNorm)\ (0.88, 0, 1)]`
`-> 1.17499`
`= (190-180)/sigma\ \ [text(Using)\ Z = (X-u)/sigma]`
`:. sigma`
`dot = 8.5107`
f.
`text(Pr) (MML)`
`= 1/2 xx 1/2 xx 1/2`
`= 1/8`
h.
`mu`
`= int_0^210 (x* f(x)) dx`
`:. mu`
`dot = 170.01\ text(min)`
Consider the function `f(x) = -1/3 (x + 2) (x-1)^2.`
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The diagram below shows part of the graph of `y = g(x)`, the tangent to the graph at `x = 2` and a straight line drawn perpendicular to the tangent to the graph at `x = 2`. The equation of the tangent at the point `A` with coordinates `(2, g(2))` is `y = 3-(4x)/3`.
The tangent cuts the `y`-axis at `B`. The line perpendicular to the tangent cuts the `y`-axis at `C`.
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i. Find the coordinates of `D`. (2 marks)
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| a.i. | `g(x)` | `= int f(x)\ dx` |
| `=-1/3 int (x + 2) (x-1)^2\ dx` | ||
| `=-1/3int(x^3-3x+2)\ dx` | ||
| `:.g(x)` | `= -x^4/12 + x^2/2-(2x)/3 + c` |
`text(S)text(ince)\ \ g(0) = 1,`
| `1` | `= 0 + 0-0 + c` |
| `:. c` | `= 1` |
`:. g(x) = -x^4/12 + x^2/2-(2x)/3 + 1\ \ …\ text(as required)`
a.ii. `text(Stationary point when:)`
`g^{′}(x) = f(x) = 0`
`-1/3(x + 2) (x-1)^2=0`
`:. x = -2, 1`
| b.i. | ![]() |
`B\ text(is the)\ y text(-intercept of)\ \ y = 3-4/3 x`
`:. B (0, 3)`
b.ii. `m_text(norm) = 3/4, \ text(passes through)\ \ A(2, 1/3)`
`text(Equation of normal:)`
| `y-1/3` | `=3/4(x-2)` |
| `y` | `=3/4 x-7/6` |
`:. C (0, -7/6)`
| b.iii. | `text(Area)` | `= 1/2 xx text(base) xx text(height)` |
| `= 1/2 xx (3 + 7/6) xx 2` | ||
| `= 25/6\ text(u²)` |
| c.i. | `text(Solve)\ \ \ g^{′}(x)` | `= -4/3\ \ text(for)\ \ x < 0` |
| `=> x` | `= -1` | |
| `g(-1)` | `=-1/12+1/2+2/3+1` | |
| `=25/12` |
`:. D (−1, 25/12)`
c.ii. `text(T) text(angent line at)\ \ D:`
| `y-25/12` | `=-4/3(x+1)` |
| `y` | `=-4/3x + 3/4` |
`DE\ \ text(intersects)\ \ AE\ text(at)\ E:`
| `-4/3 x + 3/4` | `= 3/4 x-7/6` |
| `25/12 x` | `= 23/12` |
| `x` | `=23/25` |
`:. E (23/25, -143/300)`
| `:. AE` | `= sqrt((2-23/25)^2 + (1/3-(-143/300))^2)` |
| `= 27/20\ text(units)` |
Let `f: [0, 8 pi] -> R, \ f(x) = 2 cos (x/2) + pi`.
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Find the value of `a` and the value of `b`. (3 marks)
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a. `text(Period)= (2pi)/n = (2 pi)/(1/2) = 4pi`
`text(Range:)\ [pi-2, pi + 2]`
b. `f^{′} (x) = text(−sin) (x/2)`
c. `[text(CAS: tangentLine)\ (f(x), x, pi)]`
`y = -x + 2 pi`
d. `text(Solve)\ \ f^{′} (x) = 1\ \ text(for)\ x in [0, 8 pi]`
`-> x = 3 pi or 7 pi`
`:. y = x-2 pi and y = x-6 pi\ \ [text(CAS)]`
e. `text(Using the transition matrix,)`
| `x_T` | `=x-pi` |
| `x` | `=x_T+pi` |
| `y_T` | `=ay+b` |
| `y` | `=(y_T-b)/a` |
`f(x)= cos (x/2) + pi/2\ \ ->\ \ f{′}(x) = -sin(x/2)`
| `(y_T-b)/a` | `=2cos((x_T+pi)/2)+pi` |
| `y_T` | `=2a cos((x_T+pi)/2)+a pi +b` |
| `=-2a sin(x_T/2)+a pi + bqquad [text(Complementary Angles)]` | |
| `-2a` | `=-1` |
| `:. a` | `=1/2` |
| `1/2 pi +b` | `=0` |
| `:.b` | `=-pi/2` |
f. `text(Solve)\ \ f(x) = 2 f^{′} (x) + pi\ \ text(for)\ \ x in [0, 8 pi]`
| `2 cos (x/2) + pi` | `= -2 sin(x/2)+pi` |
| `tan(x/2)` | `=-1` |
| `x/2` | `=(3pi)/4, (7pi)/4, (11pi)/4, (15pi)/4` |
| `:.x` | `= (3 pi)/2, (7 pi)/2, (11 pi)/2, (15 pi)/2` |
The function `f` is differentiable for all `x in R` and satisfies the following conditions.
Which one of the following is true?
`B`
The continuous random variable `X` has a normal distribution with mean 20 and standard deviation 6. The continuous random variable `Z` has the standard normal distribution.
The probability that `Z` is between – 2 and 1 is equal to
`B`
The continuous random variable `X` has a probability density function given by
`f(x) = {(cos(2x), if (3 pi)/4 < x < (5 pi)/4), (qquad qquad quad 0,\ \ \ text(elsewhere)):}`
The value of `a` such that `text(Pr) (X < a) = 0.25` is closest to
`C`
`text(Solve)\ \ int_((3 pi)/4)^a cos (2x)\ dx = 1/4\ \ text(for)\ \ a in ((3 pi)/4, (5 pi)/4)`
| `:. a` | `= (11 pi)/12\ \ \ text([by CAS.])` |
| `~~ 2.88` |
`=> C`
The average value of the function `f(x) = e^(2x) cos (3x)` for `0 <= x <= pi` is closest to
A. `– 82.5`
B. `26.3`
C. `– 26.3`
D. `– 274.7`
E. `pi`
`C`
| `y_text(avg)` | `= 1/(pi – 0) int_0^pi e^(2x) cos (3x)\ dx` |
| `~~ – 26.3` |
`=> C`
The function `f:\ (–oo, a] -> R` with rule `f(x) = x^3 - 3x^2 + 3` will have an inverse function provided
`A`
For `f(x) = x^3 + 2x`, the average rate of change with respect to `x` for the interval `[1, 5]` is
A. `18`
B. `20.5`
C. `24`
D. `32.5`
E. `33`
`E`
| `text(Average ROC)` | `= (f(5) – f(1))/(5 – 1)` |
| `=(135-3)/4` | |
| `= 33` |
`=> E`
The function with rule `f(x) = 4 tan (x/3)` has period
`D`
| `text(Period) ` | `=pi/n` |
| `= pi/(1/3)` | |
| `= 3 pi` |
`=> D`
The continuous random variable, `X`, has a probability density function given by
`qquad f(x) = {(1/4 cos (x/2), 3 pi <= x <= 5 pi), (0, text{elsewhere}):}`
The value of `a` such that `text(Pr) (X < a) = (sqrt 3 + 2)/4` is
`B`
| `int_(3 pi)^a f(x)\ dx` | `= (sqrt 3 + 2)/4` |
| `[1/2 sin (x/2)]_(3pi)^a` | `= (sqrt 3 + 2)/4` |
| `1/2 sin (a/2) +1/2` | `= (sqrt 3 + 2)/4` |
| `sin (a/2)` | `=sqrt3/2` |
| `a/2` | `=pi/3, (2pi)/3, (4pi)/3, (5pi)/3, (7pi)/3, …` |
| `:. a` | `= (14 pi)/3\ \ text(for)\ a in (3 pi, 5 pi)` |
`=> B`
Inside a container there are one million coloured building blocks. It is known that 20% of the blocks are red.
A sample of 16 blocks is taken from the container. For samples of 16 blocks, `hat P` is the random variable of the distribution of sample proportions of red blocks. (Do not use a normal approximation.)
`text(Pr) (hat P >= 3/16)` is closest to
A. `0.6482`
B. `0.8593`
C. `0.7543`
D. `0.6542`
E. `0.3211`
`A`
`text(Let)\ \ X = text(Number of red blocks),`
`X ~\ text(Bi) (16, 0.2)`
`text(Pr) (hat P >= 3/16)`
`= text(Pr) (X >= 3) qquad [text(binomCdf) (16, .2, 3, 16)]`
`= 0.6482`
`=> A`
Consider the graphs of the functions `f` and `g` shown below.
The area of the shaded region could be represented by
A. `int_a^d (f(x) - g(x))\ dx`
B. `int_0^d (f(x) - g(x))\ dx`
C. `int_0^b (f(x) - g(x))\ dx + int_b^c (f(x) - g(x))\ dx`
D. `int_0^a f(x)\ dx + int_a^c (f(x) - g(x))\ dx + int_b^d f(x)\ dx`
E. `int_0^d f(x)\ dx - int_a^c g(x)\ dx`
`E`
`text(Shaded Area:)`
`text(Area under)\ f(x)\ text(between)\ [0, d]\ text(less)`
`text(Area under)\ g(x)\ text(between)\ [a, c]`
`=> E`
The graph of a function `f` is obtained from the graph of the function `g` with rule `g(x) = sqrt (2x - 5)` by a reflection in the `x`-axis followed by a dilation from the `y`-axis by a factor of `1/2`.
Which one of the following is the rule for the function `f`?
`D`
`text(Let)\ \ y=sqrt(2x-5)`
`text(1st transformation:)`
`y = – sqrt(2x-5)`
`text(2nd transformation:)`
| `y` | `=-sqrt(2(2x)-5)` |
| `=- sqrt(4x-5)` | |
| `:. f(x)` | `= −sqrt(4x – 5)` |
`=> D`
For the curve `y = x^2 - 5`, the tangent to the curve will be parallel to the line connecting the positive x-intercept and the y-intercept when `x` is equal to
A. `sqrt 5`
B. `5`
C. `−5`
D. `sqrt 5/2`
E. `1/sqrt 5`
`D`
`text{Intercepts: (0, −5) and}\ (sqrt 5, 0)`
`text(Gradient between intercepts:)`
`m = (0 – (−5))/(sqrt 5 – 0) = 5/sqrt 5`
`text(Solve:)`
| `f prime (x)` | `= 5/sqrt 5` |
| `2x` | `= 5/sqrt 5` |
| `:. x` | `=5/(2sqrt5) xx sqrt5/sqrt5` |
| `= sqrt 5/2` |
`=> D`
Consider the graph of the function defined by `f: [0, 2 pi] -> R,\ f(x) = sin (2x).`
The square of the length of the line segment joining the points on the graph for which `x = pi/4 and x = (3 pi)/4` is
`A`
`text(When)\ \ x=pi/4,\ \ f(x) = sin(pi/2)=1`
`text(When)\ \ x=(3pi)/4,\ \ f(x) = sin((3pi)/2)=-1`
| `text(Let)\ \ z` | `= text(distance between)\ (pi/4, 1) and ((3pi)/4, −1)` |
| `z^2` | `= ((3pi)/4 – pi/4)^2 + (−1 −1)^2` |
| `=pi^2/4 + 4` | |
| `= (pi^2 + 16)/4` |
`=> A`
A part of the graph of `g: R -> R, g(x) = x^2 - 4` is shown below.
The area of the region marked `A` is the same area of the region marked `B`.
The exact value of `a` is
A. `0`
B. `6`
C. `sqrt6`
D. `12`
E. `2sqrt3`
`=> E`
`text(S)text(ince Area)\ A = text(Area)\ B,`
| `int_0^a (x^2-4)\ dx` | `=0` |
| `[x^3/3 – 4x]_0^a` | `=0` |
| `a^3/3 – 4a` | `=0` |
| `a^2` | `=12` |
| `:.a` | `=2sqrt3,\ \ text(for)\ a > 0` |
`=> E`
The equation `x^3 - 9x^2 + 15x + w = 0` has only one solution for `x` when
A. `−7 < w < 25`
B. `w <= −7`
C. `w >=25`
D. `w < −7` or `w > 25`
E. `w > 1`
`=> D`
The graph shown could have equation
`=> B`
`text{Amplitude = 2 (range from – 1 to 3)}`
`text(Median) = 1`
`:.\ text(Solution is)\ A\ text(or)\ B.`
`text(From graph:)`
`text(Period) = (2pi)/3 – pi/6 = pi/2`
`text(Consider option)\ B,`
`text(Period)= (2pi)/n= (2pi)/4 = pi/2`
`=> B`
The continuous random variable `X` has a normal distribution with mean 30 and standard deviation 5. For a given number `a, text(Pr)(X > a) = 0.20`.
Correct to two decimal places, `a` is equal to
`=> D`
Consider the function `f: R -> R, \ f(x) = x(x - 4)` and the function
`g: [3/2,5) -> R, \ g(x) = x + 3`.
If the function `h = f + g`, then the domain of the inverse function of `h` is
`=> D`
For the continuous random variable `X` with probability density function
`f(x) = {{:(log_e(x),1 <= x <= e),(qquad0,text(elsewhere)):}`
the expected value of `X, text(E)(X)`, is closest to
`=> E`
| `text(E)(X)` | `= int_1^e x(log_e(x))dx` |
| `= 2.097` |
`=> E`
The derivative of `log_e(2f(x))` with respect to `x` is
`A`
`text(Chain Rule:)`
| `text(If)\ \ h(x)` | `= f(g(x))` |
| `h′(x)` | `= f′(g(x)) xx g′(x)` |
| `d/(dx)(log_e(2f(x)))` | `= 1/(2f(x)) xx 2f′(x)` |
| `= (f′(x))/(f(x))` |
`=> A`
If `x + a` is a factor of `4x^3 - 13x^2 - ax` where `a ∈ R text(\{0})`, then the value of `a` is
A. `−4`
B. `−3`
C. `−1`
D. `1`
E. `2`
`=> B`
`text(Let)\ \ p(x) = 4x^3 – 13x^2 – ax`
`text(If)\ \ (x + a)\ \ text(is a factor,)`
| `p(−a)` | `= 0` |
| `0` | `=4(-a)^3-13(-a)^2-a(-a)` |
| `=-4a^3-13a^2+a^2` | |
| `=-4a^2(a+3)` |
`:. a = −3,\ \ \ (a != 0)`
`=> B`
The gradient of a line perpendicular to the line which passes through `text{(−2, 0)}` and `text{(0, −4)}` is
`=> A`
| `m` | `= (−4 – 0)/(0 – (−2))` |
| `= −2` | |
| `:. m_⊥` | `= 1/2` |
`=> A`
For the function with rule `f(x) = x^3 - 4x`, the average rate of change of `f(x)` with respect to `x` on the interval `[1,3]` is
A. `1`
B. `3`
C. `5`
D. `6`
E. `9`
`=> E`
| `text(Average ROC)` | `= (f(3) – f(1))/(3 – 1)` |
| `=(15-(-3))/2` | |
| `= 9` |
`=> E`
The decreasing value of a depreciating asset is shown in the graph below.
Let `A_n` be the value of the asset after `n` years, in dollars.
What recurrence relation below models the value of `A_n`?
`D`
`text(The asset is decreasing at 12.5% per year)`
`text(on a decreasing balance basis.)`
| `A_1` | `= 120\ 000(1 – 0.125)^1` |
| `vdots` | |
| `A_n` | `= 120\ 000(1 – 0.125)^n` |
`=> D`