Which equation represents the line perpendicular to `2x-3y = 8`, passing through the point `(2, 0)`?
- `3x + 2y = 4`
- `3x + 2y = 6`
- `3x-2y = –4`
- `3x-2y = 6`
Aussie Maths & Science Teachers: Save your time with SmarterEd
Which equation represents the line perpendicular to `2x-3y = 8`, passing through the point `(2, 0)`?
`B`
`2x-3y` | `= 8` |
`3y` | `= 2x-8` |
`y` | `= 2/3x-8/3` |
`m` | `= 2/3` |
`:.\ m_text(perp)` | `= -3/2\ \ \ (m_1 m_2=text(-1 for)_|_text{lines)}` |
`text(Equation of line)\ \ m = -3/2\ \ text(through)\ \ (2,0):`
`y-y_1` | `= m (x-x_1)` |
`y-0` | `= -3/2 (x-2)` |
`y` | `= -3/2x + 3` |
`2y` | `= -3x + 6` |
`3x + 2y` | `= 6` |
`=> B`
What is the solution to the equation `log_2(x-1) = 8`?
`D`
`log_2 (x-1)` | `= 8` |
`x-1` | `= 2^8` |
`x` | `= 257` |
`=> D`
The gradient function of a curve `y = f(x)` is given by `f^{′}(x) = 4x-5`. The curve passes through the point `(2, 3)`.
Find the equation of the curve. (2 marks)
--- 6 WORK AREA LINES (style=lined) ---
`f(x) = 2x^2-5x + 5`
`f^{′}(x)` | `= 4x-5` |
`f(x)` | `= int 4x-5\ dx` |
`= 2x^2-5x + C` |
`text(Given)\ \ f(x)\ text(passes through)\ (2,3):`
`3` | `= 2(2^2)-5(2) + C` |
`3` | `= 8-10 + C` |
`C` | `= 5` |
`:.\ f(x) = 2x^2-5x + 5`
Evaluate `int_0^(pi/2) sin (x/2)\ dx`. (3 marks)
`2\ – sqrt2`
`int_0^(pi/2) sin (x/2)\ dx`
`= [-2cos (x/2)]_0^(pi/2)`
`= -2 [ cos (pi/4)\ – cos 0]`
`= -2 [ 1/sqrt2\ – 1]`
`= -2/sqrt2 + 2`
`= 2\ – sqrt2`
Find `int 1/((x + 3)^2)\ dx`. (2 marks)
`(-1)/((x + 3)) + C`
`int 1/((x + 3)^2)\ dx`
`= int (x + 3)^(-2)\ dx`
`= 1/(-1)*(x + 3)^(-1) + C`
`= (-1)/((x + 3)) + C`
The scatterplot shows the relationship between expenditure per primary school student, as a percentage of a country’s Gross Domestic Product (GDP), and the life expectancy in years for 15 countries.
--- 1 WORK AREA LINES (style=lined) ---
What is the interquartile range? (1 mark)
--- 1 WORK AREA LINES (style=lined) ---
Would this country be an outlier for this set of data? Justify your answer with calculations. (2 marks)
--- 4 WORK AREA LINES (style=lined) ---
--- 4 WORK AREA LINES (style=lined) ---
--- 1 WORK AREA LINES (style=lined) ---
--- 1 WORK AREA LINES (style=lined) ---
`text(correlation between the two variables.)`
`text(of 127.3. This line of best fit is only accurate)`
`text(in a lower range of GDP expediture.)`
i. | `text(It indicates there is a strong positive)` |
`text(correlation between the two variables)` |
ii. | `text(IQR)` | `= Q_U\ – Q_L` |
`= 22.5\ – 8.4` | ||
`= 14.1` |
iii. `text(An outlier on the upper side must be more than)`
`Q_u\ +1.5xxIQR`
`=22.5+(1.5xx14.1)`
`=\ text(43.65%)`
`:.\ text(A country with an expenditure of 47.6% is an outlier).`
iv. |
v. `text(Life expectancy) ~~ 73.1\ text{years (see dotted line)}`
`text(Alternative Solution)`
`text(When)\ x=18`
`y=1.29(18)+49.9=73.12\ \ text(years)`
vi. | `text(At 60% GDP, the line predicts a life)` |
`text(expectancy of 127.3. This line of best)` | |
`text(fit is only predictive in a lower range)` | |
`text(of GDP expenditure.)` |
Blood alcohol content of males can be calculated using the following formula
`BAC_text(Male) = (10N - 7.5H)/(6.8M)`
where `N` is the number of standard drinks consumed
`H` is the number of hours drinking
`M` is the person's mass in kilograms
What is the maximum number of standard drinks that a male weighing 84 kg can consume over 4 hours in order to maintain a blood alcohol content (BAC) of less than 0.05? (3 marks)
--- 6 WORK AREA LINES (style=lined) ---
`5`
`text(BAC)_text(male) = (10N\ – 7.5H)/(6.8M)`
`text(Find)\ \ N\ \ text(for BAC)<0.05,\ \ text(given)\ \ H = 4\ \ text(and)\ \ M = 84`
` (10N – 7.5(4))/(6.8(84))` | `< 0.05` |
`10N – 30` | `< 0.05 (571.2)` |
`10N` | `< 28.56 + 30` |
`< 58.56` | |
`N` | `< 5.856` |
`:.\ text(Max number of drinks is 5.)`
The cost of hiring an open space for a music festival is $120 000. The cost will be shared equally by the people attending the festival, so that `C` (in dollars) is the cost per person when `n` people attend the festival.
--- 1 WORK AREA LINES (style=lined) ---
--- 2 WORK AREA LINES (style=lined) ---
--- 2 WORK AREA LINES (style=lined) ---
i.
\begin{array} {|l|c|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex}\text{Number of people} (n) \rule[-1ex]{0pt}{0pt} & \ 500\ & \ 1000 \ & 1500 \ & 2000 \ & 2500\ & 3000 \ \\
\hline
\rule{0pt}{2.5ex}\text{Cost per person} (C)\rule[-1ex]{0pt}{0pt} & 240 & 120 & 80 & 60 & 48\ & 40 \ \\
\hline
\end{array}
ii. |
iii. `C = (120\ 000)/n`
`n\ text(must be a whole number)`
iv. `text(Limitations can include:)`
`•\ n\ text(must be a whole number)`
`•\ C > 0`
v. `text(If)\ C = 94:`
`94` | `= (120\ 000)/n` |
`94n` | `= 120\ 000` |
`n` | `= (120\ 000)/94` |
`= 1276.595…` |
`:.\ text(C)text(ost cannot be $94 per person,)`
`text(because)\ n\ text(isn’t a whole number.)`
i.
\begin{array} {|l|c|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex}\text{Number of people} (n) \rule[-1ex]{0pt}{0pt} & \ 500\ & \ 1000 \ & 1500 \ & 2000 \ & 2500\ & 3000 \ \\
\hline
\rule{0pt}{2.5ex}\text{Cost per person} (C)\rule[-1ex]{0pt}{0pt} & 240 & 120 & 80 & 60 & 48\ & 40 \ \\
\hline
\end{array}
ii. |
iii. `C = (120\ 000)/n`
iv. `text(Limitations can include:)`
`•\ n\ text(must be a whole number)`
`•\ C > 0`
v. `text(If)\ C = 94`
`=> 94` | `= (120\ 000)/n` |
`94n` | `= 120\ 000` |
`n` | `= (120\ 000)/94` |
`= 1276.595…` |
`:.\ text(C)text(ost cannot be $94 per person,)`
`text(because)\ n\ text(isn’t a whole number.)`
An aerial diagram of a swimming pool is shown.
The swimming pool is a standard length of 50 metres but is not in the shape of a rectangle.
(i) Given `AB=8\ text(cm)`, determine the scale of the diagram such that
1 cm = `x` m (1 mark)
(ii) If the length of a carpark next to the pool measured 5 cm (not shown), how long would it be in real life? (1 mark)
(iii) In the diagram of the swimming pool, the five widths are measured to be:
`CD = 21.88\ text(m)`
`EF = 25.63\ text(m)`
`GH = 31.88\ text(m)`
`IJ = 36.25\ text(m)`
`KL = 21.88\ text(m)`
The average depth of the pool is 1.2 m
Calculate the approximate volume of the swimming pool, in cubic metres. In your calculations, use TWO applications of Simpson’s Rule. (3 marks)
(i) `x=6.25\ text(m)`
(ii) `31.25\ text(m)`
(iii) `1775\ text(m³)`
(i) | `\ \ \ \ 8\ text(cm)` | `=50\ text(m)` |
`1\ text(cm)` | `=50/8` | |
`=6.25\ text(m)` |
`:.x=6.25\ text(m)`
(ii) `text{Using scale from (i)}`
`5\ text(cm)` | `=5 xx 6.25` |
`=31.25\ text(m)` |
`:.\ text(The carpark would be 31.25 m long)`
(iii) |
`h = 50/4 = 12.5\ text(m)`
`A` | `~~ h/3 [y_0 + 4(y_1) + y_2]\ \ text(… applied twice)` |
`~~ 12.5/3 [21.88 + 4(25.63) + 31.88]` | |
`+ 12.5/3 [31.88 + 4(36.25) + 21.88]` | |
`~~ 12.5/3 [156.28] + 12.5/3 [198.76]` | |
`~~ 1,479.33\ text(m²)` |
`V` | `= Ah` |
`~~ 1479.33 xx 1.2` | |
`~~ 1775.2` | |
`~~ 1775\ text(m³)` |
A fair coin is tossed three times. Using a tree diagram, or otherwise, calculate the probability of obtaining two heads and a tail in any order. (2 marks)
--- 4 WORK AREA LINES (style=lined) ---
`3/8`
A radial compass survey of a sports centre is shown in the diagram.
--- 2 WORK AREA LINES (style=lined) ---
--- 4 WORK AREA LINES (style=lined) ---
--- 6 WORK AREA LINES (style=lined) ---
i. |
`text(Let)\ D\ text(be directly north of)\ O`
`/_AOD = 360\ – 320 = 40^@`
`:.\ /_AOB = 40 + 74 = 114^@\ \ \ text(… as required)`
ii. `text(Using cosine rule)`
`AB^2` | `= AO^2 + BO^2\ – 2 xx AO xx BO xx cos /_AOB` |
`= 287^2 + 211^2\ – 2 xx 287 xx 211 xx cos 114^@` | |
`= 126\ 890\ – 121\ 114 (-0.4067…)` | |
`= 176\ 151.50…` | |
`AB` | `= 419.704…` |
`= 420\ text(m)\ text{(nearest m)}` |
iii. `text(Using)\ A = 1/2 ab sin C,`
`text(Area)\ Delta AOB` | `= 1/2 xx 287 xx 211 xx sin 114^@` |
`= 27\ 660.786…\ text(m²)` | |
`= 2.7660…\ text(ha)\ \ \ \ text{(1 ha = 10 000 m²)}` | |
`= 2.8\ text(ha)\ text{(2 sig figures)}` |
Alex is buying a used car which has a sale price of $13 380. In addition to the sale price there are the following costs:
(i) | `($13\ 380)/100 = 133.8` |
`:.\ text(Stamp duty)` | `= 134 xx $3` |
`= $402` |
(ii) | `text(Total loan)` | `= $13\ 380 + 30 + 402` |
`= $13\ 812` |
`text(Total interest)\ (I)` | `= Prn` |
`= 13\ 812 xx 7.5/100 xx 3` | |
`= 3107.70` |
`text(Total to repay)` | `= 3107.70 + 13\ 812` |
`= 16\ 919.70` |
`text(# Repayments) = 3 xx 12 = 36`
`:.\ text(Monthly repayment)` | `= (16\ 919.70)/36` |
`= 469.9916…` | |
`= $470\ text{(nearest dollar)}` |
(iii) | `text(Base rate) = $845` |
`text(FSL) =\ text(1%) xx 845 = $8.45`
`text(Stamp)` | `=\ text(5.5%) xx(845 + 8.45)` |
`= 46.9397…` | |
`= $46.94\ text{(nearest cent)}` |
`text(GST)` | `= 10 text(%) xx(845 + 8.45)` |
`= 85.345` | |
`= $85.35` |
`:.\ text(Total cost)` | `= 845 + 8.45 + 46.94 + 85.35` |
`= $985.74` |
(iv) | `text(Comprehensive insurance covers Alex)` |
`text(for damage done to his own car as well.)` |
The weight of an object on the moon varies directly with its weight on Earth. An astronaut who weighs 84 kg on Earth weighs only 14 kg on the moon.
A lunar landing craft weighs 2449 kg when on the moon. Calculate the weight of this landing craft when on Earth. (2 marks)
--- 4 WORK AREA LINES (style=lined) ---
`14\ 694\ text(kg)`
`W_text(moon) prop W_text(earth)`
`=> W_text(m) = k xx W_text(e)`
`text(Find)\ k\ text{given}\ W_text(e) = 84\ text{when}\ W_text(m) = 14`
`14` | `= k xx 84` |
`k` | `= 14/84 = 1/6` |
`text(If)\ W_text(m) = 2449\ text(kg),\ text(find)\ W_text(e):`
`2449` | `= 1/6 xx W_text(e)` |
`W_text(e)` | `= 14\ 694\ text(kg)` |
`:.\ text(Landing craft weighs)\ 14\ 694\ text(kg on earth)`
Solve the equation `(5x + 1)/3-4 = 5-7x`. (3 marks)
`x = 1`
`(5x + 1)/3-4` | `= 5-7x` |
`5x + 1-3(4)` | `= 3(5-7x)` |
`5x + 1-12` | `= 15-21x` |
`26x` | `= 26` |
`:. x` | `= 1` |
Expand `4x(7x^4 - x^2)`. (1 mark)
`28x^5 – 4x^3`
`4x(7x^4 – x^2)`
`= 28x^5 – 4x^3`
Heather’s car uses fuel at the rate of 6.6 L per 100 km for long-distance driving and 8.9 L per 100 km for short-distance driving.
She used the car to make a journey of 560 km, which included 65 km of short-distance driving.
Approximately how much fuel did Heather’s car use on the journey?
(A) 37 L
(B) 38 L
(C) 48 L
(D) 50 L
`B`
`text(Fuel used in short distance)`
`= 65/100 xx 8.9\ text(L) = 5.785\ text(L)`
`text(Fuel used in long distance)`
`= 495/100 xx 6.6\ text(L) = 32.67\ text(L)`
`:.\ text(Total Fuel)` | `= 5.785 + 32.67` |
`= 38.455\ text(L)` |
`=> B`
Jaz has 2 bags of apples.
Bag A contains 4 red apples and 3 green apples.
Bag B contains 3 red apples and 1 green apple.
Jaz chooses an apple from one of the bags.
Which tree diagram could be used to determine the probability that Jaz chooses a red apple?
`A`
`text(The tree diagram needs to identify 2 separate events.)`
`text(1st event – which bag is chosen)`
`text(2nd event – choosing a red apple from a particular bag)`
`=> A`
A child who weighs 14 kg needs to be given 15 mg of paracetamol for every 2 kg of body weight.
Every 10 mL of a particular medicine contains 120 mg of paracetamol.
What is the correct dosage of this medicine for the child?
(A) 5.6 mL
(B) 8.75 mL
(C) 11.43 mL
(D) 17.5 mL
`B`
`text(Paracetamol needed)`
`= 14/2 xx 15\ text(mg)`
`= 105\ text(mg)`
`text(S)text(ince 120 mg is contained in 10 mL,)`
`=> 105\ text(mg is contained in)`
`105/120 xx 10\ text(mL)= 8.75\ text(mL)`
`=> B`
Which expression will give the shortest distance, in kilometres, between Mount Isa `(20^@ text(S)\ 140^@ text(E))` and Tokyo `(35^@ text(N)\ 140^@ text(E))`?
(A) `15/360 xx 2 xx pi xx 6400`
(B) `55/360 xx 2 xx pi xx 6400`
(C) `140/360 xx 2 xx pi xx 6400`
(D) `305/360 xx 2 xx pi xx 6400`
`B`
Simplify `6w^4 xx 1/3 w^2`.
`A`
`6w^4 xx 1/3w^2` | `= (6w^4w^2)/3` | |
`= 2w^6` |
`=> A`
Use mathematical induction to prove that
`sum_(r=1)^n r^3 = 1/4 n^2 (n + 1)^2`
`text(for integers)\ n>=1` (3 marks)
--- 10 WORK AREA LINES (style=lined) ---
`text(Proof)\ \ text{(See Worked Solutions)}`
`text(Need to prove)\ sum_(r=1)^n r^3 = 1/4 n^2 (n + 1)^2\ \ text(integral)\ n>=1 `
`text(i.e.)\ 1^3 + 2^3 + 3^3 + … + n^3 = 1/4 n^2 (n + 1)^2`
`text(When)\ n = 1`
`text(LHS) = 1^3 = 1`
`text(RHS) = 1/4 1^2 (1 + 1)^2 = 1`
`:.\ text(True for)\ n = 1`
`text(Assume true for)\ n = k`
`text(i.e.)\ 1^3 + 2^3 + … + k^3 = 1/4 k^2 (k + 1)^2`
`text(Need to prove true for)\ n = k + 1`
`1^3 + 2^3 + … + k^3 + (k + 1)^3 = 1/4 (k + 1)^2 (k + 2)^2`
`text(LHS)` | `= 1/4 k^2 (k + 1)^2 + (k + 1)^3` |
`= 1/4 (k + 1)^2 [k^2 + 4(k + 1)]` | |
`= 1/4 (k + 1)^2 (k^2 + 4k + 4)` | |
`= 1/4 (k + 1)^2 (k + 2)^2` | |
`=\ text(RHS)` |
`=>text(True for)\ n = k + 1`
`:.text(S)text(ince true for)\ n = 1,\ text(true for integral)\ n >= 1`
The point `P(2at, at^2)` lies on the parabola `x^2 = 4ay` with focus `S`.
The point `Q` divides the interval `PS` internally in the ratio `t^2 :1`.
(i) `text(Show)\ \ Q = ((2at)/(1 + t^2), (2at^2)/(1 + t^2))`
`P (2at, at^2),\ S (0, a)`
`PS\ text(is divided internally in ratio)\ t^2: 1`
`Q` | `= ((nx_1 + mx_2)/(m + n), (ny_1 + my_2)/(m + n))` |
`= ((1(2at) + t^2(0))/(t^2 + 1), (1(at^2) + t^2 (a))/(t^2 + 1))` | |
`= ((2at)/(1 + t^2), (2at^2)/(1 + t^2))\ \ text(… as required.)` |
(ii) | `m_(OQ)` | `= (y_2\ – y_1)/(x_2\ – x_1)` |
`= ((2at^2)/(1 + t^2))/((2at)/(1 + t^2)) xx (1+t^2)/(1+t^2)` | ||
`= (2at^2)/(2at)` | ||
`= t` |
(iii) | `text(Show)\ Q\ text(lies on a fixed circle radius)\ a` |
`text(S)text(ince)\ Q\ text(passes through)\ (0, 0)` |
`=>\ text(If locus of)\ Q\ text(is a circle, it has)`
`text(diameter)\ QT\ text(where)\ T(0, 2a)`
`text(Show)\ \ QT _|_ OQ`
`text{(} text(angles on circum. subtended by)`
`text(a diameter are)\ 90^@ text{)}`
`m_(OQ) = t\ \ \ \ text{(see part (ii))}`
`text(Find)\ m_(QT),\ \ text(where:)`
`Q((2at)/(1 + t^2), (2at^2)/(1 + t^2)),\ \ \ \ \ T (0,2a)`
`m_(QT)` | `= (y_2\ – y_1)/(x_2\ – x_1)` |
`= ((2at^2)/(1 + t^2)\ – 2a)/((2at)/(1 + t^2)\ – 0)` | |
`= (2at^2\ – 2a (1 + t^2))/(2at)` | |
`= – (2a)/(2at)` | |
`= – 1/t` |
`m_(QT) xx m_(OT) = -1/t xx t = -1`
`=> QT _|_ OQ`
`=>O,\ T,\ Q\ text(lie on a circle.)`
`:.\ text(Locus of)\ Q\ text(is a fixed circle,)`
`text(centre)\ (0, a),\ text(radius)\ a`
The diagram shows points `P(2t, t^2)` and `Q(4t, 4t^2)` which move along the parabola `x^2 = 4y`. The tangents to the parabola at `P` and `Q` meet at `R`.
(i) | `x^2` | `= 4y` |
`y` | `= (x^2)/4` | |
`dy/dx` | `= x/2` |
`text(At)\ \ x = 2t,`
`dy/dx = (2t)/2 = t`
`:.\ text(T)text(angent has)\ \ m = t\ \ text(and passes)`
`text(through)\ \ (2t, t^2).`
`y – t^2` | `= t (x – 2t)` |
`y` | `= tx\ – 2t^2 + t^2` |
`= tx\ – t^2\ \ text(… as required)` |
(ii) | `text(T)text(angent at)\ \ P (2t, t^2)\ \ text(is)\ \ y = tx\ – t^2` |
`=>Q(4t, 4t^2) -= Q(2(2t), (2t)^2)`
`:.\ text(Equation of the tangent at)\ \ Q,`
`y` | `= 2tx\ – (2t)^2` |
`= 2tx\ – 4t^2` |
`text(T)text(angents intersect at)\ \ R`
`y` | `= tx\ – t^2\ \ \ \ \ …\ text{(1)}` |
`y` | `= 2tx\ – 4t^2\ \ \ \ \ …\ text{(2)}` |
`text(Intersection when)\ text{(1)} = text{(2)}`
`tx\ – t^2` | `= 2tx\ – 4t^2` |
`tx` | `= 3t^2` |
`x` | `= 3t` |
`text(Substitute)\ \ x = 3t\ \ text(into)\ text{(1)}`
`y` | `= t (3t)\ – t^2` |
`= 2t^2` |
`:.\ R (3t, 2t^2)`
(iii) | `text(Find locus of)\ \ R` |
`x` | `= 3t` | `\ \ \ \ \ …\ text{(1)}` |
`y` | `= 2t^2\ \ \ \ \ …\ text{(2)}` |
`text{From (1),}\ \ \ \ t = x/3`
`text(Substitute)\ \ t = x/3\ text(into)\ text{(2)}`
`y` | `= 2 (x/3)^2` |
`= 2/9 x^2` |
`:.\ text(Locus of)\ \ R\ \ text(is)\ \ y = 2/9 x^2`
The equation `e^t = 1/t` has an approximate solution `t_0 = 0.5`
(i) | `e^t` | `= 1/t` |
`e^t\ – 1/t` | `= 0` |
`text(Let)\ \ f(t)` | `= e^t\ – 1/t` |
`f prime (t)` | `= e^t + 1/(t^2)` |
`f(0.5)` | `=e^0.5-1/0.5` |
`=–0.3512…` | |
`f′(0.5)` | `=e^0.5 +1/0.5^2` |
`=5.6487…` |
`text(Applying Newton where)\ \ t_0 = 0.5`
`t_1` | `= 0.5\ – (f(0.5))/(f prime (0.5)` |
`= 0.5\ – (–0.3512…)/(5.6487…)` | |
`= 0.5\ – (- 0.062)` | |
`= 0.56\ text{(2 d.p.)}\ \ text(… as required.)` |
(ii) | `y_1` | `= e^(rx),` | `\ \ \ \ \ (dy_1)/(dx) = re^(rx)` |
`y_2` | `= log_e x,` | `\ \ \ \ \ (dy_2)/(dx) = 1/x` |
`text(S)text(ince tangent is common),\ \ (dy_1)/(dx)=(dy_2)/(dx)`
`re^(rx)` | `= 1/x` |
`e^(rx)` | `= 1/(rx)` |
`text(Using part)\ text{(i)},\ rx ~~ 0.56`
`text(At intersection of curves)\ \ y_1 = y_2`
`e^(rx)` | `= log_e x` |
`e^0.56` | `= log_e x` |
`x` | `= e^(e^0.56)` |
`= 5.758 …` |
`text(S)text(ince)\ rx` | `~~ 0.56` |
`=> r` | `~~ 0.56/(5.758 … )` |
`~~ 0.0972 …` | |
`~~ 0.097\ text{(to 3 d.p.)}` |
(i) `text(Find co-efficient of)\ \ x^(2n)`
`text(Expanding)\ \ (1+x)^(4n)`
`((4n),(0)) + ((4n),(1))x + ((4n),(2))x^2 + … + ((4n),(2n))x^(2n) + …`
`:.\ text(Co-efficient of)\ \ x^(2n)\ text(is)\ ((4n),(2n))`
(ii) `text(Show)\ (1 + x^2 + 2x)^(2n) = sum_(k=0)^(2n) ((2n),(k)) x^(2n\ – k) (x + 2)^(2n\ – k)`
`text(Using)\ (1 + x^2 + 2x)^(2n) = [x(x + 2) + 1]^(2n)`
`[x (x + 2) + 1]^(2n)` |
`= ((2n),(0)) (x(x + 2))^(2n) + ((2n),(1)) (x(x + 2))^(2n\ – 1) + … + ((2n),(2n))` |
`= ((2n),(0)) x^(2n)(x + 2)^(2n) + ((2n),(1)) x^(2n\ – 1) (x + 2)^(2n\ – 1) + … + ((2n),(2n))` |
`= sum_(k=0)^(2n) ((2n),(k)) x^(2n\ – k) (x + 2)^(2n\ – k)\ text(… as required.)` |
(iii) `((4n),(2n))\ text(is the co-eff of)\ x^(2n)\ text(in expansion)\ (1+x)^(4x)`
`text(S)text(ince)\ (1 + x^2 + 2x)^(2n) = ((x+1)^2)^(2n) = (1 + x)^(4n)`
`=> ((4n),(2n))\ text(is co-efficient of)\ x^(2n)\ text(in expansion)\ (1 + x^2 + 2x)^(2n)`
`text(Using part)\ text{(ii)}`
`(1 + x^2 + 2x)^(2n) = sum_(k=0)^(2n) ((2n),(k))\ x^(2n\ – k) (x + 2)^(2n\ – k)`
`text(Using the given identity,)\ x^(2n)\ text(co-efficients are)`
`k = 0,` | `\ ((2n),(0))((2n\ – 0),(0)) 2^(2n\ – 0)` |
`k = 1,` | `\ ((2n),(1))((2n\ – 1),(1)) 2^(2n\ – 1\ – 1)` |
`vdots` | |
`k = n,` | `\ ((2n),(n))((2n\ – n),(n)) 2^(2n\ – n\ – n)` |
`:.\ ((4n),(2n))` |
`= ((2n),(0))((2n),(0))2^(2n) + ((2n),(1))((2n\ – 1),(1))2^(2n\ – 2) + … + ((2n),(n))((n),(n)) 2^0` |
` = sum_(k=0)^(n)\ ((2n),(k))((2n\ – k),(k)) 2^(2n\ – 2k)\ \ \ \ text(… as required)` |
A spherical raindrop of radius `r` metres loses water through evaporation at a rate that depends on its surface area. The rate of change of the volume `V` of the raindrop is given by
`(dV)/(dt) = -10^(-4) A`,
where `t` is time in seconds and `A` is the surface area of the raindrop. The surface area and the volume of the raindrop are given by `A = 4pir^2` and `V = 4/3 pi r^3` respectively.
--- 5 WORK AREA LINES (style=lined) ---
--- 6 WORK AREA LINES (style=lined) ---
i. | `text(Show)\ \ (dr)/(dt)\ \ text(is a constant)` |
`(dV)/(dt) = (dV)/(dr) * (dr)/(dt)\ \ \ …\ text{(1)}`
`V` | `= 4/3 pi r^3` |
`:. (dV)/(dr)` | `= 4 pi r^2` |
`(dV)/(dt)` | `= -10^(-4) A\ \ text{(given)}` |
`text(Substituting into)\ text{(1)}`
`-10^(-4) A` | `= 4 pi r^2 xx (dr)/(dt)` |
`= A xx (dr)/(dt)` | |
`:.\ (dr)/(dt)` | `= -10^(-4)\ \ text(… as required)` |
ii. | `V` | `= 10^(-6)\ text(m³)` |
`4/3 pi r^3` | `= 10^(-6)` | |
`r^3` | `= (3 xx 10^(-6))/(4pi)` | |
`r` | `= root(3)((3 xx 10^(-6))/(4pi))` |
`text(S)text(ince the radius decreases at a constant rate,)`
`t=(root(3)((3 xx 10^(-6))/(4pi)))/(10^(-4))`
`\ \ =62.035 …`
`\ \ =62\ text(seconds)\ text{(nearest whole)}`
`:.\ text(It takes 62 seconds for the raindrop)`
`text(to evaporate.)`
A particle moves along a straight line. The displacement of the particle from the origin is `x`, and its velocity is `v`. The particle is moving so that `v^2 + 9x^2 = k`, where `k` is a constant.
Show that the particle moves in simple harmonic motion with period `(2pi)/3`. (2 marks)
--- 5 WORK AREA LINES (style=lined) ---
`text(Proof)\ \ text{(See Worked Solutions)}`
`v^2 + 9x^2` | `= k` |
`v^2` | `= k\ – 9x^2` |
`1/2 v^2` | `= 1/2k\ – 9/2 x^2` |
`text(For SHM,)\ \ ddot x = -n^2x`
`ddot x` | `= d/(dx) (1/2v^2)` |
`= -9x` | |
`= -3^2 x \ \ \ text(… as required)` |
`text(Period)\ (T)\ text(of SHM) = (2pi)/n`
`text(Here,)\ \ n=3`
`:.T= (2pi)/3\ \ \ text(… as required)`
The point `P(t, t^2 + 3)` lies on the curve `y = x^2 + 3`. The line `l` has equation `y = 2x\ – 1`. The perpendicular distance from `P` to the line `l` is `D(t)`.
(i) | `text(Show)\ \ D(t) = (t^2\ – 2t + 4)/sqrt5` |
`D(t)` | `= _|_\ text(dist of)\ (t, t^2 + 3)\ text(from)\ 2x\ – y\ – 1 = 0` |
`D` | `= |(ax_1 + by_1 + c)/sqrt(a^2 + b^2)|` |
`= |(2(t)\ – 1 (t^2 + 3)\ – 1)/sqrt(2^2 + (-1)^2)|` | |
`= |(2t\ – t^2\ – 3\ – 1)/sqrt5|` | |
`= |(-(t^2\ – 2t + 4))/sqrt5|` |
`text(S)text(ince)\ \ t^2\ – 2t + 4\ \ text(is positive definite)`
`text{(i.e.} \ Delta < 0\ text(and co-efficient of)\ t^2 > 0 text{)},`
`t^2\ – 2t + 4\ \ text(is positive for all)\ t.`
`:.\ D(t) = (t^2\ – 2t + 4)/sqrt5\ \ text(… as required)`
(ii) | `text(MAX or MIN when)\ \ (dD)/(dt) = 0` |
`(dD)/(dt) = 1/sqrt5 (2t\ – 2)` | `= 0` |
`2t\ – 2` | `= 0` |
`t` | `= 1` |
`(d^2D)/(dt^2) = 2/sqrt5 > 0\ =>\ text(MIN)`
`:.\ P\ text(is closest to)\ l\ text(when)\ t = 1`
(iii) | `y` | `= x^2 + 3` |
`dy/dx` | `= 2x` | |
`text(When)\ x = t` |
`dy/dx = 2t`
`text(At)\ t = 1`
`dy/dx = 2,`
`=> m_text(tangent) = 2\ \ and\ \ m_l = 2`
`:.\ text(T)text(angent is parallel to)\ \ l\ \ \ \ text(… as required)`
The region bounded by the graph `y = 3 sin\ x/2` and the `x`-axis between `x = 0` and `x = (3pi)/2` is rotated about the `x`-axis to form a solid.
Find the exact volume of the solid. (3 marks)
--- 6 WORK AREA LINES (style=lined) ---
`(9pi)/2 ((3pi)/2 + 1) text(u³)`
`y` | `= 3 sin\ x/2` |
`y^2` | `= 9 sin^2\ x/2` |
`text(Using:)\ \ sin^2x= 1/2 (1 – cos 2x)`
`:. V` | `= pi int_0^((3pi)/2) 9 sin^2\ x/2\ dx` |
`= (9pi)/2 int_0^((3pi)/2) (1\ – cosx)\ dx` | |
`= (9pi)/2 [x\ – sinx]_0^((3pi)/2)` | |
`= (9pi)/2 [((3pi)/2\ – sin\ (3pi)/2)\ – 0]` | |
`= (9pi)/2 ((3pi)/2 + 1)\ text(u³)` |
--- 6 WORK AREA LINES (style=lined) ---
--- 6 WORK AREA LINES (style=lined) ---
i. `text(Write)\ sqrt3 cosx\ – sinx\ text(in form)`
`2cos (x + alpha),\ \ \ 0 < alpha < pi/2`
`2 (cosx cos alpha\ – sinx sin alpha)` | `= sqrt 3 cosx\ – sinx` |
`cosx cos alpha\ – sinx sin alpha` | `= sqrt3/2 cos x\ – 1/2 sinx` |
`=> cos alpha` | `= sqrt3/2` |
`=> sin alpha` | `= 1/2` |
`alpha` | `= pi/6` |
`:.\ 2cos (x + pi/6) = sqrt3 cosx\ – sinx`
ii. `text(Solve)\ sqrt3 cosx = 1 + sinx,\ \ \ 0 < x < 2pi`
`sqrt3 cosx\ – sinx` | `= 1` |
`2 cos (x + pi/6)` | `= 1\ \ \ text{(from part (i))}` |
`cos (x + pi/6)` | `= 1/2` |
`cos(pi/3)` | `=1/2` |
`text(S)text(ince cos is positive in)\ 1^text(st) // 4^text(th)\ text(quadrants)`
`x + pi/6` | `= pi/3,\ 2pi\ – pi/3\ \ \ \ \ (0 < x < 2pi)` |
`:.\ x` | `= pi/6,\ (3pi)/2` |
Use the substitution `u = e^(3x)` to evaluate `int_0^(1/3) (e^(3x))/(e^(6x) + 1)\ dx`. (3 marks)
--- 8 WORK AREA LINES (style=lined) ---
`1/3 (tan^(-1)e\ – pi/4)`
`text(Let)\ \ u = e^(3x)`
`(du)/(dx)` | `= 3e^(3x)` |
`:.dx` | `= (du)/(3e^(3x))` |
`text(When)` | `\ x = 1/3,` | `\ u = e^(3 xx 1/3) = e` |
`\ x = 0,` | `\ u = e^0 = 1` |
`:.int_0^(1/3) (e^(3x))/(e^(6x) + 1)\ dx`
`=int_1^e (e^(3x))/(u^2 + 1) xx (du)/(3e^(3x))`
`= 1/3 int_1^e 1/(u^2 + 1)\ du`
`= 1/3 [tan^(-1)u]_1^e`
`= 1/3 [tan^(-1) e\ – tan^(-1) 1]`
`= 1/3 (tan^(-1)e\ – pi/4)`
Find `lim_(x -> 0) (sin\ x/2)/(3x)`. (1 mark)
`1/6`
`lim_(x -> 0) (sin\ x/2)/(3x)`
`= lim_(x -> 0) (sin\ x/2)/(6 xx x/2)`
`= 1/6 xx lim_(x -> 0) (sin\ x/2)/(x/2)`
`= 1/6`
Consider the function `f(x) = x/(4\ - x^2)`.
(i) `f(x) = x/(4\ – x^2)`
`u` | `= x` | `\ \ \ \ \ v` | `= 4\ – x^2` |
`u prime` | `= 1` | `\ \ \ \ \ \ \ \ \ \ v prime` | `= -2x` |
`f prime (x)` | `= (u prime v\ – u v prime)/(v^2)` |
`= (1* (4\ – x^2)\ – x(-2x))/((4\ – x^2)^2)` | |
`= (4 + x^2)/((4\ – x^2)^2)` |
`text(Domain is all)\ x,\ x != +- 2`
`text(Consider)\ f prime (x)`
`4 + x^2` | `> 0\ \ \ text{(in domain)}` |
`(4\ – x^2)^2` | `> 0\ \ \ text{(in domain)}` |
`:.\ f prime (x) > 0\ text(for all)\ x,\ x != +- 2`
(ii) |
`text(Asymptotes at)\ x = +- 2`
`text(Passes through)\ (0,0)`
`text(As)` | `\ x -> 2 (-),` | `\ y -> oo` |
`\ x -> 2 (+),` | `\ y -> – oo` |
`text(As)` | `\ x -> -2 (-),` | `\ y -> oo` |
`\ x -> -2 (+),` | `\ y -> -oo` |
`text(As)` | `\ x -> oo,` | `\ y -> 0` |
`\ x -> – oo,` | `\ y -> 0` |
Find `int 1/sqrt (49 - 4x^2)\ dx`. (2 marks)
`1/2 sin^(-1) ((2x)/7) + c`
`int 1/sqrt(49 – 4x^2)\ dx`
`= int 1/(2 sqrt(49/4 – x^2))\ dx`
`= 1/2 int 1/sqrt((7/2)^2 – x^2)\ dx`
`= 1/2 sin^(-1) ((2x)/7) + c`
The diagram shows the graph of a function.
Which function does the graph represent?
(A) `y = cos^(-1) x`
(B) `y = pi/2 + sin^(-1) x`
(C) `y = - cos^(-1) x`
(D) `y = - pi/2\ - sin^(-1) x`
`B`
`text(By elimination,)`
`text(The graph passes through)\ \ (1, pi)`
`text(The only equation to satisfy this point is)`
`y = pi/2 + sin^(-1) x`
`=> B`
A family of eight is seated randomly around a circular table.
What is the probability that the two youngest members of the family sit together?
(A) `(6!\ 2!)/(7!)`
(B) `(6!)/(7!\ 2!)`
(C) `(6!\ 2!)/(8!)`
(D) `(6!)/(8!\ 2!)`
`A`
`text(Fix youngest person in 1 seat,)`
`text(Total combinations around table) = 7!`
`text(Combinations with youngest side by side) =2!6!`
`:.\ text{P(sit together)} = (6!\ 2!)/(7!)`
`=> A`
Which integral is obtained when the substitution `u = 1 + 2x` is applied to `int x sqrt(1 + 2x)\ dx`?
(A) `1/4 int (u - 1) sqrt u\ du`
(B) `1/2 int (u - 1) sqrt u\ du`
(C) `int (u - 1) sqrt u\ du`
(D) `2 int (u - 1) sqrt u\ du`
`A`
`text(Let)\ \ u` | `= 1 + 2x` |
`:.x` | `= 1/2 (u – 1)` |
`(du)/(dx)` | `= 2` |
`:.dx` | `= 1/2\ du` |
`int x sqrt(1 + 2x)\ dx`
`=int 1/2 (u – 1) xx u^(1/2) xx 1/2\ du`
`= 1/4 int (u – 1) sqrt u\ du`
`=> A`
`D`
`y = x(1 – x)^3 (3 – x)^2`
`text(By elimination)`
`text(Consider when)\ \ x < 0,`
`y = text{(–ve)} xx text{(+ve)} xx text{(+ve)} < 0`
`:.\ text(Cannot be)\ A\ text(or)\ C`
`text(Consider the cubic factor)\ \ (1 – x)^3,`
`text(The graph must have a stationary point at)\ x = 1`
`:.\ text(Cannot be)\ B`
`=> D`
The binomial theorem states that
`(1 + x)^n = ((n),(0)) + ((n),(1))x + ((n),(2))x^2 + ((n),(3))x^3 + ... + ((n),(n))x^n.`
(i) `(1 + x)^n = ((n),(0)) + ((n),(1))x + … + ((n),(n))x^n`
`text(Let)\ \ x = 1`
`(1 + 1)^n` | `= ((n),(0)) + ((n),(1))1 + ((n),(2))1^2 + … + ((n),(n))1^n` |
`2^n` | `= ((n),(0)) + ((n),(1)) + … + ((n),(n))` |
`= sum_(k=0)^n ((n),(k))\ text(… as required)` |
(ii) `((100),(0)) + ((100),(1)) + ((100),(2)) + … + ((100),(100)) = 2^100`
`text{(} text(from part)\ text{(i)} text{)}`
(iii) `(1 + x)^n = ((n),(0)) + ((n),(1))x + ((n),(2))x^2 + … + ((n),(n))x^n`
`text(Differentiate both sides)`
`n(1 + x)^(n\ – 1) = 1((n),(1)) + 2x ((n),(2)) + 3x^2 ((n),(3)) + … + nx^(n\ – 1) ((n),(n))`
`text(Let)\ \ x = 1`
`n (1 + 1)^(n\ – 1)` | `= 1 ((n),(1)) + 2 ((n),(2)) + 3 ((n),(3)) + … + n ((n),(n))` |
`n 2^(n\ – 1)` | `= sum_(k = 1)^n k ((n),(k))\ text(… as required)` |
--- 3 WORK AREA LINES (style=lined) ---
--- 4 WORK AREA LINES (style=lined) ---
i. `text(Show)\ \ cos(A\ – B) = cosA cosB (1 + tanA tanB)`
`text(RHS)` | `= cosA cosB (1 + (sinA sinB)/(cosA cosB))` |
`= cosA cos B + sinA sin B` | |
`= cos(A – B)\ text(… as required)` |
ii. `text(Given)\ \ tanA tanB = -1`
`cos (A – B)` | `= cosA cosB (1\ – 1)` |
`cos (A – B)` | `= 0` |
`A – B` | `= cos^(-1) 0` |
`= pi/2, (3pi)/2, …` |
`text(S)text(ince)\ \ \ 0 < B < pi/2\ \ text(and)\ \ \ B < A < pi,`
`=> A\ – B = pi/2`
A boat is sailing due north from a point `A` towards a point `P` on the shore line.
The shore line runs from west to east.
In the diagram, `T` represents a tree on a cliff vertically above `P`, and `L` represents a landmark on the shore. The distance `PL` is 1 km.
From `A` the point `L` is on a bearing of 020°, and the angle of elevation to `T` is 3°.
After sailing for some time the boat reaches a point `B`, from which the angle of elevation to `T` is 30°.
(i) `text(Show)\ \ BP = (sqrt3 tan 3°)/(tan 20°)`
`text(In)\ Delta ATP` |
`tan 3°` | `= (TP)/(AP)` |
`=> AP` | `= (TP)/(tan 3)` |
`text(In)\ Delta APL`
`tan 20°` | `= 1/(AP)` |
`=> AP` | `= 1/tan 20` |
`:. (TP)/(tan3)` | `= 1/(tan20)` |
`TP` | `= (tan3°)/(tan20°)\ \ \ text(…)\ text{(} text(1)text{)}` |
`text(In)\ \ Delta BTP`
`tan 30°` | `= (TP)/(BP)` |
`1/sqrt3` | `= (TP)/(BP)` |
`BP` | `= sqrt3 xx TP\ \ \ \ \ text{(using (1) above)}` |
`= (sqrt3 tan3°)/(tan20°)\ \ \ text(… as required)` |
(ii) | `AB` | `= AP\ – BP` |
`AP` | `= 1/(tan20°)\ \ \ text{(} text(from part)\ text{(i)} text{)}` |
`:.\ AB` | `= 1/(tan 20°)\ – (sqrt3 tan3°)/(tan20°)` |
`= (1\ – sqrt3 tan 3)/(tan20°)` | |
`= 2.4980…` | |
`= 2.5\ text(km)\ text{(to 1 d.p.)` |
The diagram shows the parabola `x^2 = 4ay`. The point `P(2ap, ap^2)`, where `p != 0`, is on the parabola.
The tangent to the parabola at `P`, `y = px − ap^2`, meets the `y`-axis at `L`.
The point `M` is on the directrix, such that `PM` is perpendicular to the directrix.
Show that `SLMP` is a rhombus. (3 marks)
`text(Proof)\ \ text{(See Worked Solutions)}`
`text(Solution 1)`
`text(Show)\ \ SLMP\ \ text(is a rhombus)`
`L\ \ text(occurs when tangent meets)\ x text(-axis)`
`y = 0 – ap^2`
`:.L(0,–ap^2),\ \ M(2ap, –a)`
`m_(PS)` | `= (y_2\ – y_1)/(x_2\ – x_1)` |
`= (ap^2\ – a)/(2ap\ – 0)` | |
`= (p^2\ – 1)/(2p)` | |
`m_(ML)` | `= (-a + ap^2)/(2ap\ – 0` |
`= (p^2\ – 1)/(2p)` |
`text(S)text(ince)\ \ ` | `PS \ text(||)\ ML\ \ \ text{(same gradient)}` |
`text(and)\ \ ` | `PM \ text(||)\ SL\ \ \ text{(vertical)}` |
`=>` | `\ \ SLMP\ text(is a parallelogram)` |
`text(S)text(ince)\ \ ` | `PS = PM\ \ \ text{(definition of a parabola)}` |
`=>` | `\ \ SLMP\ text(is a rhombus … as required.)` |
`text(Alternate Solution)`
`L(0,–ap^2),\ \ M(2ap, –a)`
`m_(SM)` | `= (y_2\ – y_1)/(x_2\ – x_1) = (-a\ -a)/(2ap\ – 0) = – 1/p` |
`m_(LP)` | `= p` |
`text(S)text(ince)\ \ m_(SM) xx m_(LP) = p xx – 1/p = -1`
`=>\ text(Diagonals are perpendicular)`
`text(Midpoint)\ LP` | `= ( (0 + 2ap)/2,\ (-ap^2 + ap^2)/2)` |
`= (ap, 0)` | |
`text(Midpoint)\ MS` | `= ((0 + 2ap)/2,\ (a\ – a)/2)` |
`= (ap,\ 0)` |
`=>\ text(Diagonals bisect)`
`:.\ text(S)text(ince diagonals are)\ _|_\ text(bisectors,)`
`SLMP\ text(is a rhombus … as required.)`
--- 12 WORK AREA LINES (style=lined) ---
--- 8 WORK AREA LINES (style=lined) ---
i. | `2 cos theta + 2 cos (theta + pi/3)` |
`= 2 cos theta + 2 (cos theta cos (pi/3)\ – sin theta sin (pi/3))` | |
`= 2 cos theta + 2 cos theta xx 1/2\ – 2 sin theta xx sqrt3/2` | |
`= 2 cos theta + cos theta\ – sqrt3 sin theta` | |
`= 3 cos theta\ – sqrt3 sin theta` |
`R cos (theta + alpha) = R cos theta cos alpha – R sin theta sin alpha`
`R cos alpha` | `= 3\ \ \ \ \ ` | `R sin alpha` | `= sqrt3` |
`cos alpha` | `= 3/R\ \ \ \ \ ` | `sin alpha` | `= sqrt3/R` |
`tan alpha` | `= sin alpha/cos alpha = sqrt3/3 = 1/sqrt3` |
`tan\ pi/6` | `=1/sqrt3` |
`:. alpha` | `=pi/6\ \ \ \ \ (0 < alpha < pi/2)` |
`R^2` | `= 3^2 + (sqrt3)^2` |
`= 9 + 3` | |
`R` | `= sqrt 12 = 2 sqrt3` |
`:.\ 2 cos theta + 2 cos (theta + pi/3) = 2 sqrt 3 cos (theta + pi/6)`
ii. | `2 cos theta + 2 cos(theta + pi/3)` | `= 3` |
`2 sqrt 3 cos (theta + pi/6)` | `= 3` | |
`cos (theta + pi/6)` | `= 3/(2sqrt3) = sqrt3/2` | |
`cos^(-1) (sqrt3/2)` | `= pi/6` |
`text(S)text(ince cos is positive in 1st and 4th quadrants,)`
`theta + pi/6` | `= pi/6,\ 2pi\ – pi/6` |
`:. theta` | `= (5pi)/3\ \ \ \ \ (0 < theta < 2pi)` |
A particle is moving in simple harmonic motion along the `x`-axis.
Its velocity `v`, at `x`, is given by `v^2 = 24 − 8x − 2x^2`.
--- 4 WORK AREA LINES (style=lined) ---
--- 2 WORK AREA LINES (style=lined) ---
--- 6 WORK AREA LINES (style=lined) ---
(i) | `v^2 = 24\ – 8x\ – 2x^2` |
`text(Find)\ \ x\ \ text(when)\ \ v=0`
`24 – 8x – 2x^2` | `= 0` |
`x^2 + 4x – 12` | `= 0` |
`(x + 6)(x – 2)` | `= 0` |
`x= -6\ \ text(or)\ \ 2` |
`:.\ text(Particle at rest when)\ \ x = –6\ \ text(or)\ \ 2`
(ii) | `ddot x` | `= d/(dx) (1/2 v^2)` |
`= d/(dx) (12 – 4x – x^2)` | ||
`= -4 – 2x` |
(iii) `text(Solution 1)`
`text(Max speed when)\ \ ddot x = 0`
`-4 – 2x` | `= 0` |
`2x` | `= -4` |
`x` | `= -2` |
`text(At)\ \ x = -2`
`v^2` | `= 24 – 8(–2) – 2(–2)^2` |
`= 24 + 16 – 8` | |
`= 32` | |
`v` | `= +- sqrt32` |
`= +- 4 sqrt2` |
`:.\ text(Maximum speed is)\ \ 4 sqrt2`
`text(Alternate Answer)`
`text(Maximum speed occurs when)`
`x = (-6+2)/2 = –2`
`text(At)\ \ x = –2`
`v^2` | `= 32\ \ \ text{(see working above)}` |
`v` | `= +- 4 sqrt 2` |
`:.\ text(Maximum speed is)\ 4sqrt2`
Let `f(x) = e^(-x^2)`. The diagram shows the graph `y = f(x)`.
(i) | `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` |
`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`
(ii) | `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).` |
(iii) | `y = e^(-x^2)` |
`text(Inverse function can be written)`
`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))`
(iv) | `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`
(v) |
(vi)(1) `x = e^(-x^2)`
`text(Let)\ g(x) = x\ – e^(-x^2)`
`g(0.6)` | `=0.6\ – e^(-0.6^2)` |
`=0.6\ – 0.6977 < 0` | |
`g(0.7)` | `=0.7\ – e^(-0.7^2)` |
`=0.7\ – 0.6126 > 0` | |
`=>g(x)\ text(changes sign)` |
`:.\ g(x)\ \ text(has a root between 0.6 and 0.7)`
`:.\ x = e^(-x^2)\ \ text(has a solution between 0.6 and 0.7)`
(vi)(2) | `g(0.65)` | `=0.65\ – e^(-0.65^2)` |
`=0.65\ – 0.655 < 0` |
`:.\ text(A solution lies between 0.65 and 0.7)`
`:.\ x = 0.7\ \ text{(1 d.p.)}`
At the front of a building there are five garage doors. Two of the doors are to be painted red, one is to be painted green, one blue and one orange.
--- 1 WORK AREA LINES (style=lined) ---
--- 2 WORK AREA LINES (style=lined) ---
i. | `text(# Arrangements)` | `= (5!)/(2!)` |
`= 60` |
ii. | `text(When 2 red doors are side-by-side,)` |
`text(# Arrangements)` | `= 4!` |
`= 24` |
A radio transmitter `M` is situated 6 km from a straight road. The closest point on the road to the transmitter is `S`.
A car is travelling away from `S` along the road at a speed of `text(100 km h)`−1. The distance from the car to `S` is `x\ text(km)` and from the car to `M` is `r\ text(km)`.
Find an expression in terms of `x` for `(dr)/(dt)`, where `t` is time in hours. (3 marks)
--- 8 WORK AREA LINES (style=lined) ---
`(100x)/sqrt(x^2 + 36)\ \ text(km/hr)`
`text(Using Pythagoras,)`
`r^2` | `= x^2 + 6^2` |
`r` | `= sqrt(x^2 + 36),\ r > 0` |
`(dr)/(dt)` | `= (dx)/(dt) * (dr)/(dx)\ \ \ \ …\ (1)` |
`(dx)/(dt)` | `= 100\ \ \ text{(given)}` |
`(dr)/(dx)` | `= 1/2 (x^2 + 36)^(-1/2) xx d/(dx) (x^2 + 36)` |
`= x/sqrt(x^2 + 36)` | |
`text{Substituting into (1)}` | |
`:.\ (dr)/(dt)` | `= (100x)/sqrt(x^2 + 36)\ \ text(km/hr)` |
Let `P(x) = (x + 1)(x-3) Q(x) + ax + b`,
where `Q(x)` is a polynomial and `a` and `b` are real numbers.
The polynomial `P(x)` has a factor of `x-3`.
When `P(x)` is divided by `x + 1` the remainder is `8`.
--- 6 WORK AREA LINES (style=lined) ---
--- 2 WORK AREA LINES (style=lined) ---
i. `P(x) = (x+1)(x-3)Q(x) + ax + b`
`(x-3)\ \ text{is a factor (given)}`
`:. P (3)` | `= 0` |
`3a + b` | `= 0\ \ \ …\ text{(1)}` |
`P(x) ÷ (x+1)=8\ \ \ text{(given)}`
`:.P(-1)` | `= 8` |
`-a + b` | `= 8\ \ \ …\ text{(2)}` |
`text{Subtract (1) – (2)}`
`4a` | `= -8` |
`a` | `= -2` |
`text(Substitute)\ \ a = -2\ \ text{into (1)}`
`-6 + b` | `= 0` |
`b` | `= 6` |
`:. a= – 2, \ b=6`
ii. `P(x) -: (x + 1)(x-3)`
`= ((x+1)(x-3)Q(x)-2x + 6)/((x+1)(x-3))`
`= Q(x) + (-2x + 6)/((x+1)(x-3))`
`:.\ text(Remainder is)\ \ -2x + 6`
The derivative of a function `f(x)` is given by
`f^{′}(x) = sin^2 x`.
Find `f(x)`, given that `f(0) = 2`. (2 marks)
--- 5 WORK AREA LINES (style=lined) ---
`1/2x-1/4 sin 2x + 2`
`f^{′}(x)` | `= sin^2 x` |
`f(x)` | `= int sin^2 x\ dx` |
`= int 1/2(1-cos 2x)\ dx` | |
`= int 1/2-1/2 cos 2x\ dx` | |
`= 1/2 x-1/4 sin 2 x + c` |
`text(Given)\ \ f(0) = 2,`
`2` | `= 1/2 xx 0-1/4 sin 0 + c` |
`:. c= 2` |
`:.\ f(x) = 1/2 x-1/4 sin 2x + 2`
Five ordinary six-sided dice are thrown.
What is the probability that exactly two of the dice land showing a four?
Leave your answer in unsimplified form. (1 mark)
`\ ^5C_2 (1/6)^2 (5/6)^3`
`P(4) = 1/6`
`P(bar4) = 5/6`
`text(# Combinations of two 4’s) =\ ^5C_2`
`:.\ P text{(exactly two 4s)} =\ ^5C_2 (1/6)^2 (5/6)^3`
Solve `ln(x + 6) = 2 ln x`. (3 marks)
`x = 3`
`ln(x + 6)` | `= 2 lnx` |
`ln(x + 6)` | `= ln x^2` |
`x^2` | `= x + 6` |
`x^2\ – x\ – 6` | `= 0` |
`(x\ – 3)(x + 2)` | `= 0` |
`:. x = 3\ \ text(or)\ \ -2` |
`text(S)text(ince RHS)\ = 2lnx,\ \ x>0`
`:. x=3`
Let `f(x) = cos^(-1) (x/2)`. What is the domain of `f(x)`? (1 mark)
`-2 <= x <= 2`
`f(x) = cos^(-1) (x/2)`
`–1` | `<= x/2` | `<= 1` |
`–2` | `<= x` | `<= 2` |
`:.\ text(Domain of)\ \ f(x)\ \ text(is)\ \ \ –2 <= x <= 2`
The diagram shows two identical circular cones with a common vertical axis. Each cone has height `h` cm and semi-vertical angle 45°.
The lower cone is completely filled with water. The upper cone is lowered vertically into the water as shown in the diagram. The rate at which it is lowered is given by
`(dl)/(dt) = 10`,
where `l` cm is the distance the upper cone has descended into the water after `t` seconds.
As the upper cone is lowered, water spills from the lower cone. The volume of water remaining in the lower cone at time `t` is `V` cm³.
--- 4 WORK AREA LINES (style=lined) ---
--- 5 WORK AREA LINES (style=lined) ---
--- 6 WORK AREA LINES (style=lined) ---
i. `text(Show that)\ V = pi/3 (h^3\ – l^3)`
`text(S)text(ince)\ \ tan45° = r/h=1`
`=>r=h`
`=>\ text(Radius of lower cone) = h`
`:.\ V text{(lower cone)}` | `= 1/3 pi r^2 h` |
`= 1/3 pi h^3` |
`text(Similarly,)`
`V text{(submerged upper cone)} = 1/3 pi l^3`
`V text{(water left)}` | `= 1/3 pi h^3\ – 1/3 pi l^3` |
`= pi/3 (h^3\ – l^3)\ \ \ text(… as required)` |
ii. `text(Find)\ (dV)/(dt)\ text(at)\ l = 2`
`(dV)/(dt)= (dV)/(dl) xx (dl)/(dt)\ …\ text{(1)}`
`=>(dl)/(dt)` | `= 10\ text{(given)}` |
`text(Using)\ \ V` | `= pi/3 (h^3\ – l^3)\ \ \ \ text(from part)\ text{(i)}` |
`=>(dV)/(dl)` | `= -3 xx pi/3 l^2` |
`= -pi l^2` |
`text(At)\ \ l = 2,`
`text{Substitute into (1) above}`
`(dV)/(dt)` | `= -pi xx 2^2 xx 10` |
`= -40 pi\ \ \ text(cm³)//text (sec)` |
iii. `text(Find)\ \ (dV)/(dt)\ \ text(when lower cone has lost)\ 1/8 :`
`text(Find)\ \ l\ \ text(when)\ \ V = 7/8 xx 1/3 pi h^3`
`pi/3 (h^3\ – l^3)` | `= 7/8 xx 1/3 pi h^3` |
`h^3 -l^3` | `= 7/8 h^3` |
`l^3` | `= 1/8 h^3` |
`l` | `= h/2` |
`text(When)\ \ l = h/2 ,`
`(dV)/(dt)` | `= -pi (h/2)^2 xx 10\ \ …\ text{(*)}` |
`= (-5pih^2)/2\ text(cm³)// text(sec)` |
`:. V\ text(is decreasing at the rate of)\ \ (5 pi h^2)/2\ text(cm³)//text(sec).`
Consider the function `f(x) = e^(-x)\ - 2e^(-2x)`.
(i) | `f(x)` | `= e^(-x)\ – 2e^(-2x)` |
`f prime (x)` | `= -e^(-x) + 4e^(-2x)` |
(ii) | `text(T.P. when)\ f prime (x) = 0` |
`-e^(-x) + 4e^(-2x) = 0`
`text(Let)\ e^(-x) = X`
`- X + 4X^2` | `= 0` |
`X (4X\ – 1)` | `= 0` |
`X = 0\ \ text(or)\ \ 1/4`
`text(If)\ \ e^(-x) = 0,\ \ \ text(No solution)`
`text(If)\ \ e^(-x) = 1/4`
`ln e^(-x)` | `= ln (1/4)` |
`-x` | `= ln 4^(-1)` |
`= -ln 4` | |
`x` | `= ln4` |
`f″(x)` | `= e^(-x)\ – 8e^(-2x)` |
`f″(ln4)` | `< 0 => text(MAX)` |
`f (ln4)` | `= e^(-ln4)\ – 2e^(-2ln4)` |
`= e^(ln(1/4))\ – 2 e^(ln(1/16))` | |
`= 1/4\ – 2(1/16)` | |
`= 1/8` |
`:.\ text(Maximum T.P. at)\ \ (ln4, 1/8)`
(iii) | `f(ln2)` | `= e^(-ln2)\ – 2e^(-2ln2)` |
`= e^(ln(1/2))\ – 2e^(ln(1/4))` | ||
`= 1/2\ – 2 xx 1/4` | ||
`= 0` |
(iv) | `f(x)` | `= e^(-x)\ – 2e^(-2x)` |
`= e^(-x) (1\ – 2e^(-x))` | ||
`text(As)\ x -> oo, \ e^(-x) ->0,\ f(x) -> 0` |
(v) | `y text(-intercept at)\ \ f(0)` |
`f(0)` | `= e^0\ – 2e^0` |
`= -1` |
(vi)
The diagram shows two distinct points `P(t, t^2)` and `Q(1\ - t, (1\ - t)^2)` on the parabola `y = x^2`. The point `R` is the intersection of the tangents to the parabola at `P` and `Q`.
(i) |
`text(Show tangent at)\ P\ text(is)\ y = 2tx\ – t^2`
`y` | `= x^2` |
`dy/dx` | `= 2x` |
`x=t\ \ \ \ text(at)\ P`
`dy/dx = 2t`
`text(Find equation with)\ \ m = 2t\ \ text(through)\ \ P(t, t^2)`
`y\ – y_1` | `= m(x\ – x_1)` |
`y\ – t^2` | `= 2t ( x\ – t)` |
`y` | `= 2tx\ – 2t^2 + t^2` |
`= 2tx\ – t^2\ text(… as required)` |
(ii) `text(T)text(angent at)\ Q\ text(has equation)`
`y = 2(1\ – t)x\ – (1\ – t)^2`
(iii) `text(Need to show)\ R(1/2,\ t\ – t^2)`
`R\ text(is at intersection of tangents)`
`2tx\ – t^2` | `= 2(1\ – t)x\ – (1\ – t)^2` |
`2tx\ – t^2` | `= 2x\ – 2tx\ – 1 + 2t\ – t^2` |
`4tx\ – 2x` | `= -1 + 2t\ – t^2 + t^2` |
`2x(2t\ – 1)` | `= 2t\ – 1` |
`2x` | `= 1` |
`x` | `= 1/2` |
`text(Using)\ \ y = 2tx – t^2\ \ text(when)\ \ x = 1/2`
`y` | `= 2t(1/2)\ – t^2` |
`= t\ – t^2` |
`:.\ R(1/2, t\ – t^2)\ text(… as required)`
(iv) `text(Locus of)\ R`
`text(S)text(ince)\ x = 1/2\ text(is a constant)`
`R\ text(is a vertical line)`
`text(Now,)\ y = t\ – t^2 = t(1\ – t)`
`text(Graphically,)\ \ y\ \ text(has a maximum at)\ \ t = 1/2`
`text(Max)\ \ y = 1/2\ – (1/2)^2 = 1/4`
`=> y < 1/4\ \ text{(tangents can’t meet on parabola)}`
`:.\ text(Locus of)\ R\ text(is vertical line)\ x = 1/2,\ \ y<1/4`
The equation of motion for a particle undergoing simple harmonic motion is
`(d^2x)/(dt^2) = -n^2 x`,
where `x` is the displacement of the particle from the origin at time `t`, and `n` is a positive constant.
--- 5 WORK AREA LINES (style=lined) ---
Find the values of `A` and `B` in the solution `x = A cos nt + B sin nt`. (2 marks)
--- 6 WORK AREA LINES (style=lined) ---
--- 5 WORK AREA LINES (style=lined) ---
--- 6 WORK AREA LINES (style=lined) ---
i. | `x` | `= A cos nt + B sin nt` |
`(dx)/(dt)` | `= – An sin nt + Bn cos nt` | |
`(d^2x)/(dt^2)` | `= – An^2 cos nt\ – Bn^2 sin nt` | |
`= -n^2 (A cos nt + B sin nt)` | ||
`= -n^2 x\ \ \ text(… as required)` |
ii. `text(At)\ \ t=0, \ x=0, \ v=2n:`
`x` | `= Acosnt + Bsinnt` |
`0` | `= A cos 0 + B sin 0` |
`:.A` | `= 0` |
`text(Using)\ \ (dx)/(dt) = Bn cos nt`
`2n` | `= Bn cos 0` |
`Bn` | `= 2n` |
`:.B` | `= 2` |
iii. `text(Max distance from origin when)\ (dx)/(dt) = 0`
`(dx)/(dt)` | `= 2n cos nt` |
`0` | `= 2n cos nt` |
`cos nt` | `= 0` |
`nt` | `= pi/2,\ (3pi)/2,\ (5pi)/2` |
`t` | `= pi/(2n),\ (3pi)/(2n), …` |
`:.\ text(Particle is first at greatest distance)`
`text(from)\ O\ text(when)\ t = pi/(2n).`
iv. `text(Solution 1)`
`text(Find the distance travelled from)\ \ t=0\ \→\ \ t=(2pi)/n`
`text{(i.e. 1 full period)}`
`text(S)text(ince)\ \ x=2 sin (nt)`
`=> text(Amplitude)=2`
`:.\ text(Distance travelled)=4 xx2=8\ text(units)`
`text(Solution 2)`
`text(At)\ t = 0,\ x = 0`
`text(At)\ t` | `= pi/(2n)` |
`x` | `= 2 sin (n xx pi/(2n)) = 2` |
`text(At)\ t` | `= (3pi)/(2n)\ \ \ text{(i.e. the next time}\ \ (dx)/(dt) = 0 text{)}` |
`x` | `= 2 sin (n xx (3pi)/(2n)) = -2` |
`text(At)\ t` | `= (2pi)/n` |
`x` | `= 2 sin (n xx (2pi)/n) = 0` |
`:.\ text(Total distance travelled)`
`= 2 + 4 + 2` |
`= 8\ \ text(units)` |
Alex’s playlist consists of 40 different songs that can be arranged in any order.
--- 1 WORK AREA LINES (style=lined) ---
--- 2 WORK AREA LINES (style=lined) ---
i. | `#\ text(Arrangements) = 40!` |
ii. | `#\ text(Arrangements)` | `= 3! xx 37!` |
`= 6 xx 37!` |