If `f^{′}(x)= (x^2)/(x^3 + 1)` and `f(1)= log_e 2,` find `f(x)`. (3 marks)
--- 6 WORK AREA LINES (style=lined) ---
Aussie Maths & Science Teachers: Save your time with SmarterEd
If `f^{′}(x)= (x^2)/(x^3 + 1)` and `f(1)= log_e 2,` find `f(x)`. (3 marks)
--- 6 WORK AREA LINES (style=lined) ---
`f(x) = 1/3 log_e(x^3 + 1) + 2/3 log_e 2`
`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`
Evaluate `int_1^4(sqrt x + 1)\ dx`. (3 marks)
`23/3`
`int_1^4 (x^(1/2) + 1)\ dx`
`= [2/3 x^(3/2) + x]_1^4`
`= (2/3 xx (sqrt4)^3 + 4) – (2/3 + 1)`
`=(2/3 xx 8 +4) – 5/3`
`= 23/3`
Let `y=xsinx.` Evaluate `dy/dx` for `x=pi`. (3 marks)
--- 6 WORK AREA LINES (style=lined) ---
`-pi`
`y = x sin x`
`(dy)/(dx)` | `= x xx d/(dx) (sin x) + d/(dx) (x) xx sin x` |
`= x cos x + sin x` |
`text(When)\ \ x = pi,`
`(dy)/(dx)` | `= pi xx cos pi + sin pi` |
`= pi (-1) + 0` | |
`=-pi` |
`f(x) = 1/sqrt2 sqrtx`, where `x in [0,2]`
--- 4 WORK AREA LINES (style=lined) ---
The graph of `y = f(x)`, where `x ∈ [0, 2]`, is shown on the axes below.
--- 0 WORK AREA LINES (style=lined) ---
`f^(-1)(x) = 2x^2`
a. `text(Domain)\ \ f^(-1)(x)= text(Range)\ \ f(x)=[0,1]`
`y = 1/sqrt2 x`
`text(Inverse: swap)\ \ x ↔ y`
`x` | `= 1/sqrt2 sqrty` | |
`sqrty` | `= sqrt2 x` | |
`y` | `= 2x^2` |
`:. f^(-1)(x) = 2x^2`
b. | |
Let `f:[0,2] -> R`, where `f(x) = 1/sqrt2 sqrtx`.
--- 5 WORK AREA LINES (style=lined) ---
The graph of `y = f(x)`, where `x ∈ [0, 2]`, is shown on the axes below.
--- 0 WORK AREA LINES (style=lined) ---
--- 8 WORK AREA LINES (style=lined) ---
`f^(-1)(x) = 2x^2`
a. | `text(Domain)\ \ f^(-1)(x)` | `= text(Range)\ \ f(x)=[0,1]` |
`y = 1/sqrt2 x`
`text(Inverse: swap)\ \ x ↔ y`
`x` | `= 1/sqrt2 sqrty` |
`sqrty` | `= sqrt2 x` |
`y` | `= 2x^2` |
`:. f^(-1)(x) = 2x^2`
b. | |
c. | |
`A` | `= int_0^(1/2) 1/sqrt2 sqrtx-2x^2 dx + int_(1/2)^1 2x^2-1/sqrt2 sqrtx\ dx` |
`= [sqrt2/3 x^(3/2)-2/3 x^3]_0^(1/2) + [2/3 x^3-sqrt2/3 x^(3/2)]_(1/2)^1` | |
`= [sqrt2/3 (1/sqrt2)^3-2/3(1/2)^3] + [(2/3-sqrt2/3)-(2/24-sqrt2/3 · 1/(2sqrt2))]` | |
`= (1/6-1/12) + 2/3-sqrt2/3-(1/12-1/6)` | |
`= 1/12 + 2/3-sqrt2/3 + 1/12` | |
`= (1 + 8-4sqrt2 + 1)/12` | |
`= (5-2sqrt2)/6\ \ text(u²)` |
For a certain population the probability of a person being born with the specific gene SPGE1 is `3/5`.
The probability of a person having this gene is independent of any other person in the population having this gene.
In a randomly selected group of four people, what is the probability that three or more people have the SPGE1 gene? (2 marks)
`297/625`
`text(Let)\ \ X =\ text(number of people with gene)`
`X\ ~\ text(Bin) (4, 3/5)`
`P(X >= 3)` | `= P(X = 3) + P(X = 4)` |
`= \ ^4C_3(3/5)^3(2/5) + \ ^4C_4(3/5)^4` | |
`= (4 xx 27 xx 2)/625 + 81/625` | |
`= 297/625` |
For a certain population the probability of a person being born with the specific gene SPGE1 is `3/5`. The probability of a person having this gene is independent of any other person in the population having this gene. --- 5 WORK AREA LINES (style=lined) --- --- 5 WORK AREA LINES (style=lined) ---
a. `text(Let)\ \ X =\ text(number of people with gene)` `X\ ~\ text(Bi) (4, 3/5)`
`P(X >= 3)`
`= P(X = 3) + P(X = 4)`
`= \ ^4C_3(3/5)^3(2/5) + \ ^4C_4(3/5)^4`
`= (4 xx 27 xx 2)/625 + 81/625`
`= 297/625`
b.
`P(X = 2 | X >= 1)`
`= (P(X = 2))/(1 – P(X = 0))`
`= (\ ^4C_2(3/5)^2(2/5)^2)/(1 – (2/5)^4)`
`= (6 xx (6^2)/(5^4))/((5^4 – 2^4)/(5^4))`
`= (6^3)/(5^4 – 2^4)`
A car manufacturer is reviewing the performance of its car model X. It is known that at any given six-month service, the probability of model X requiring an oil change is `17/20`, the probability of model X requiring an air filter change is `3/20` and the probability of model X requiring both is `1/20`.
--- 5 WORK AREA LINES (style=lined) ---
--- 6 WORK AREA LINES (style=lined) ---
a. |
`text(Pr)(F ∩ O′)` | `= text(Pr)(F) – text(Pr)(F∩ O)` | |
`= 3/20 – 1/20` | ||
`= 1/10` |
b. |
`text(Pr)(F ∩ O′)` | `= n/(m + n) – 1/(m + n)` |
`1/20` | `= (n – 1)/(m + n)` |
`m + n` | `= 20n – 20` |
`m` | `= 19n – 20` |
A car manufacturer is reviewing the performance of its car model X. It is known that at any given six-month service, the probability of model X requiring an oil change is `17/20`, the probability of model X requiring an air filter change is `3/20` and the probability of model X requiring both is `1/20`. --- 5 WORK AREA LINES (style=lined) --- --- 5 WORK AREA LINES (style=lined) ---
a.
`text(Pr)(F ∩ O′)`
`= text(Pr)(F) – text(Pr)(F∩ O)`
`= 3/20 – 1/20`
`= 1/10`
b.
`text(Pr)(F ∩ O′)`
`= n/(m + n) – 1/(m + n)`
`1/20`
`= (n – 1)/(m + n)`
`m + n`
`= 20n – 20`
`m`
`= 19n – 20`
Solve the equation `2 log_2(x + 5) - log_2(x + 9) = 1`. (3 marks)
`x = text{−1}`
`2 log_2(x + 5) – log_2(x + 9)` | `= 1` |
`log_2(x + 5)^2 – log_2(x + 9)` | `= 1` |
`log_2(((x + 5)^2)/(x + 9))` | `= 1` |
`((x + 5)^2)/(x + 9)` | `= 2` |
`x^2 + 10x + 25` | `= 2x + 18` |
`x^2 + 8x + 7` | `= 0` |
`(x + 7)(x + 1)` | `= 0` |
`:. x = −1\ \ \ \ (x != text{−7}\ \ text(as)\ \ x > text{−5})`
Shown below is part of the graph of a period of the function of the form `y = tan(ax + b)`.
Find the value of `a` and the value of `b`, where `a > 0` and `0 < b < 1`. (3 marks)
--- 6 WORK AREA LINES (style=lined) ---
`a = (7pi)/24, b = pi/24`
`y = tan(ax + b)`
`text(Substitute)\ \ (1, sqrt3), (−1, −1)\ \ text(into equation:)`
`tan(a + b)` | `= sqrt3` |
`tan(b – a)` | `= −1` |
`a + b` | `= pi/3 \ …\ (1)` |
`b – a` | `= −pi/4 \ …\ (2)` |
`text{Add (1) + (2):}`
`2b` | `= pi/3 – pi/4` |
`b` | `= pi/24` |
`text{Substitute into (1):}`
`a + pi/24` | `= pi/3` |
`a` | `= (7pi)/24` |
Shown below is part of the graph of a period of the function of the form `y = tan(ax + b)`.
The graph is continuous for `x = ∈ [-1, 1]`.
Find the value of `a` and the value of `b`, where `a > 0` and `0 < b < 1`. (3 marks)
--- 10 WORK AREA LINES (style=lined) ---
`a = (7pi)/24, b = pi/24`
`y = tan(ax + b)`
`text(Substitute)\ \ (1, sqrt3), (-1,-1)\ \ text(into equation:)`
`tan(a + b)` | `= sqrt3` |
`tan(b-a)` | `=-1` |
`a + b` | `= pi/3 \ …\ (1)` |
`b-1` | `=-pi/4 \ …\ (2)` |
`text{Add (1) + (2):}`
`2b` | `= pi/3-pi/4` |
`b` | `= pi/24` |
`text{Substitute into (1):}`
`a + pi/24` | `= pi/3` |
`a` | `= (7pi)/24` |
A study on the mobile phone usage of NSW high school students is to be conducted.
Data is to be gathered using a questionnaire.
The questionnaire begins with the three questions shown.
--- 1 WORK AREA LINES (style=lined) ---
--- 2 WORK AREA LINES (style=lined) ---
i. `text(Categorical, nominal)`
ii. `text(How many outgoing calls do you make per day?)`
`text{(Ensure it can be answered with a numerical score.)}`
Prove that `log_3 7` is irrational. (2 marks)
--- 5 WORK AREA LINES (style=lined) ---
`text(See Worked Solution)`
`text(Proof by contradiction:)`
`text(Assume that)\ \ log_3 7\ \ text(is rational.)`
`text(i.e.)\ \ log_3 7 = p/q\ text(where)\ \ p, q ∈ ZZ\ \ text(with no common factor except 1.)`
`log_3 7` | `= p/q` |
`q log_3 7` | `= p` |
`log_3 7^q` | `= p` |
`7^q` | `= 3^p` |
`text(S)text(ince 7 and 3 are prime numbers), 7^q != 3^p`
`:.\ text(Contradiction: integer values)\ \ p, q\ \ text(do not exist.)`
`:. log_3 7\ text(is irrational.)`
A light inextensible string hangs over a frictionless pulley connecting masses of 3 kg and 7 kg, as shown below.
Particles of mass 3 kg and `m` kg are attached to the ends of a light inextensible string that passes over a smooth pulley, as shown.
If the acceleration of the 3 kg mass is `4.9\ text(m/s)^2` upwards, then
A. `m` = 4.5
B. `m` = 6.0
C. `m` = 9.0
D. `m` = 13.5
E. `m` = 18.0
`C`
A light inextensible string passes over a smooth pulley, as shown below, with particles of mass 1 kg and `m` kg attached to the ends of the string.
If the acceleration of the 1 kg particle is 4.9 `text(ms)^(-2)` upwards, then `m` is equal to
`C`
Particles of mass 3 kg and 5 kg are attached to the ends of a light inextensible string that passes over a fixed smooth pulley, as shown above. The system is released from rest.
Assuming the system remains connected, the speed of the 5 kg mass after two seconds is
A. 4.0 m/s
B. 4.9 m/s
C. 9.8 m/s
D. 19.6 m/s
`B`
`sum F` | `= 5g – 3g = 2g` |
`=2 xx 9.8` | |
`=19.6\ \ text(N)` | |
`(5+3)a` | `= 19.6` |
`a` | `=2.45` |
`u = 0, quad t = 2, quad a = 2.45`
`v` | `= u + at` |
`= 0 + 2.45 xx 2` | |
`=4.9\ \ text(m/s)` |
`=> B`
Find all solutions for `z`, in exponential form, given `z^4 = -2 sqrt3 - 2 i`. (3 marks)
--- 8 WORK AREA LINES (style=lined) ---
`sqrt2 e^(-frac{i 17 pi}{24}) \ , \ sqrt2 e^(-frac{i 5 pi}{24}) \ , \ sqrt2 e^(frac{i 7 pi}{24}) \ , \ sqrt2 e^(frac{i 19 pi}{24})`
`text{Convert} \ \ z^4\ \ text{to Mod/Arg:}`
`| z^4 | = sqrt{(2 sqrt3)^2 + 2^2} = 4`
`tan \ theta` | `= frac{2 sqrt3}{2} = sqrt3` |
`theta` | `= frac{pi}{3}` |
`text{arg} (z^4) = – (frac{pi}{2} + frac{pi}{3}) = – frac{5 pi}{6}`
`text{By De Moivre:}`
`| z | = root4 (4) = sqrt2`
`text{arg}(z) = -frac{5 pi}{24} + frac{ k pi}{2} \ , \ k = 0 , ± 1 , ± 2`
`k = 0:\ text{arg} (z)=-frac{5 pi}{24}`
`k = 1:\ text{arg} (z)= -frac{5 pi}{24} + frac{pi}{2} = frac{7 pi}{24}`
`k = – 1:\ text{arg} (z)= -frac{5 pi}{24} – frac{pi}{2} = frac{-17 pi}{24}`
`k = 2:\ text{arg} (z)= -frac{5 pi}{24} + pi = frac{19 pi}{24}`
`therefore \ text{Solutions are:} \ sqrt2 e^(- frac{i 17 pi}{24}) \ , \ sqrt2 e^(frac{ -i 5 pi}{24}) \ , \ sqrt2 e^(frac{i 7 pi}{24}) \ , \ sqrt2 e^(frac{i 19 pi}{24})`
If `(x + iy)^3 = e^( - frac{i pi}{2}),\ \ x, y ∈ R`, find a solution in the form `x + i y, \ x, y ≠ 0`. (2 marks)
--- 5 WORK AREA LINES (style=lined) ---
`frac{sqrt3}{2} – i frac{1}{2}`
`text{Let}` | `(x + i y)` | `= cos theta + i sin theta` |
`(x + i y)^3` | `= cos (3 theta) + isin (3 theta)` |
`e^(- frac{i pi}{2}) = cos (frac{-pi}{2}) + i sin (frac{-pi}{2})`
`text{Equating real parts:}`
`3 theta` | `= frac{-pi}{2}` |
`theta` | `= frac{-pi}{6}` |
`therefore \ x + i y` | `= cos (frac{-pi}{6}) + i sin (frac{-pi}{6})` |
`= frac{sqrt3}{2} – i frac{1}{2}` |
Let `z = sqrt3 - 3 i`
--- 6 WORK AREA LINES (style=lined) ---
--- 6 WORK AREA LINES (style=lined) ---
i. | `z` | `= sqrt3 – 3 i` |
`|z|` | `= sqrt((sqrt3)^2 + 3^2) = 2 sqrt3` |
`tan theta` | `= frac{3}{sqrt3}=sqrt3` |
`theta` | `= frac{pi}{3}` |
`text{arg} (z)` | `= – frac{pi}{3}` |
`therefore z = 2 sqrt3 \ text{cis} (frac{-pi}{3})`
ii. `z^n + (overset_z)^n = 0`
`[2 sqrt3 \ cos (frac{-pi}{3}) + i sin (frac{-pi}{3})]^n + [ 2 sqrt3 \ cos (frac{-pi}{3}) – i sin (frac{-pi}{3}) ]^n = 0`
`(2 sqrt3)^n [cos (frac{-n pi}{3}) + i sin (frac{-n pi}{3}) + cos (frac{-n pi}{3}) – i sin (frac{-n pi}{3}) = 0`
`2 \ cos (frac{-n pi}{3})` | `= 0` |
`cos (frac{n pi}{3})` | `= 0` |
`frac{n pi}{3}` | `= frac{pi}{2} + k pi \ , \ k = 0, ± 1, ± 2, …` |
`frac{n}{3}` | `= frac{(2k + 1)}{2}` |
`n` | `= frac{3 (2k + 1)}{2}` |
`text{Numerator will always be odd ⇒ no solution exists}`
Sketch the region in the complex plane where the inequalities
`| z + overset_z | ≤ 1` and `| z - i | ≤ 1`
hold simultaneously. (3 marks)
`| z + overset_z | ≤ 1 `
`| x + i y + x – iy |` | `≤ 1` |
`| 2x |` | `≤ 1` |
`| x |` | `≤ frac{1}{2}` |
`| z – i | ≤ 1 \ => \ text{Circle, radius = 1 , centre (0, 1)`
Sketch the region on the Argand diagram where the inequalities
`| z - overset_z | < 2` and `| z - 1 | >=1`
hold simultaneously. (3 marks)
`z = sqrt2 e^((ipi)/15)` is a root of the equation `z^5 = alpha(1 + isqrt3), \ alpha ∈ R`.
--- 4 WORK AREA LINES (style=lined) ---
--- 4 WORK AREA LINES (style=lined) ---
--- 6 WORK AREA LINES (style=lined) ---
i. `beta = 1 + isqrt3`
`|beta| = sqrt(1 + (sqrt3)^2) = 2`
`text(arg)(beta) = tan^(−1) (sqrt3/1) = pi/3`
`beta = 2e^((ipi)/3)`
ii. | `z` | `= sqrt2 e^((ipi)/15)` |
`z^5` | `= (sqrt2 e^((ipi)/15))^5` | |
`= (sqrt2)^5 e^((ipi)/15 xx 5)` | ||
`= 4sqrt2 e^((ipi)/3)` |
`:. alpha = 2sqrt2`
iii. `text(arg)(z^5) = pi/3 + 2kpi, \ \ k = 0, ±1, ±2, …`
`text(arg)(z) = pi/15 + (2kpi)/5`
`k = 1:\ text(arg)(z) = pi/15 + (2pi)/5 = (11pi)/15`
`k = text(−1):\ text(arg)(z) = pi/15 – (2pi)/5 = −pi/3`
`k = 2:\ text(arg)(z) = pi/15 + (4pi)/5 = (13pi)/15`
`k =text(−2):\ text(arg)(z) = pi/15 – (4pi)/5 = −(11pi)/15`
`:. 4\ text(other roots are:)`
`e^((i11pi)/15), e^(−(ipi)/3), e^((i13pi)/15), e^(−(i11pi)/15)`
Given \(z=\dfrac{-1-i \sqrt{3}}{1+i}\), calculate \(z^2\) in exponential form. (3 marks)
--- 10 WORK AREA LINES (style=lined) ---
\(2 e^{\small{\dfrac{i \pi}{6}}}\)
\(z\) | \(=\dfrac{-1-i \sqrt{3}}{1+i} \times \dfrac{1-i}{1-i}\) |
\(=\dfrac{(-1-i \sqrt{3})(1-i)}{1-i^2}\) | |
\(=\dfrac{-1+i-i \sqrt{3}+i^2 \sqrt{3}}{2}\) | |
\(=\dfrac{-1-\sqrt{3}}{2}+i\left(\dfrac{1-\sqrt{3}}{2}\right)\) |
\(\abs{z}\) | \(=\sqrt{\left(\dfrac{-1-\sqrt{3}}{2}\right)^2+\left(\dfrac{1-\sqrt{3}}{2}\right)^2}\) |
\(=\sqrt{\dfrac{1+2 \sqrt{3}+3+1-2 \sqrt{3}+3}{4}}\) | |
\(=\sqrt{2}\) |
\(\text{Find}\ \ \arg (z):\)
\(\tan \theta\) | \(=\dfrac{\frac{1-\sqrt{3}}{2}}{\frac{-1-\sqrt{3}}{2}}=\dfrac{\sqrt{3}-1}{\sqrt{3}+1}\) |
\(\theta\) | \(=15^{\circ}=\dfrac{\pi}{12}\) |
\(\therefore \arg (z)=-\dfrac{11 \pi}{12}\)
\(z\) | \(=\sqrt{2} \operatorname{cis}\left(-\dfrac{11 \pi}{12}\right)\) |
\(z^2\) | \(=(\sqrt{2})^2 \operatorname{cis}\left(-\dfrac{11 \pi}{12} \times 2+2 \pi\right)\) |
\(=2 \operatorname{cis}\left(\dfrac{\pi}{6}\right)\) | |
\(=2 e^{\small{\dfrac{i \pi}{6}}}\) |
Let `t=tan(theta/2).`
--- 3 WORK AREA LINES (style=lined) ---
--- 6 WORK AREA LINES (style=lined) ---
--- 5 WORK AREA LINES (style=lined) ---
i. | `t` | `= tan frac{theta}{2}` |
`frac{dt}{d theta}` | `= frac{1}{2} text{sec}^2 frac{theta}{2}` | |
`= frac{1}{2} (1 + tan^2 frac{theta}{2})` | ||
`= frac{1}{2} (1 + t^2)` |
ii. `text{Show} \ \ sin theta = frac{2t}{1 + t^2} :`
`sin theta` | `= 2 \ sin frac{theta}{2} cos frac{theta}{2}` |
`= 2 * frac{t}{sqrt(1 + t^2)} * frac{1}{sqrt(1 + t^2)}` | |
`= frac{2t}{1 + t^2}` |
iii. `int \ text{cosec} \ theta \ d theta`
`t = tan frac {theta}{2}`
`frac{dt}{d theta} = frac{1}{2} text{sec}^2 frac{theta}{2} \ , \ d theta = frac{2dt}{sec^2 frac{theta}{2}} = frac{2}{1 + t^2} dt`
`int \ text{cosec} \ theta\ d theta` | `= int frac{1 + t^2}{2t} xx frac{2}{1 + t^2} dt` |
`= int frac{1}{t}\ dt` | |
`= log_e | t | + c` | |
`= log_e | tan frac{theta}{2} | + c` |
--- 2 WORK AREA LINES (style=lined) ---
--- 8 WORK AREA LINES (style=lined) ---
--- 3 WORK AREA LINES (style=lined) ---
--- 2 WORK AREA LINES (style=lined) ---
i. | `frac{1 + i sqrt3}{1 + i} xx frac{1 – i}{1 – i}` | `= frac{(1 + i sqrt3)(1 – i)}{1 – i^2}` |
`= frac{1 – i + i sqrt3 – sqrt3 i^2}{2}` | ||
`= frac{1 + sqrt3}{2} – i ( frac{1 – sqrt3}{2} )` |
ii. `z_1 = 1 + i sqrt3`
`| z_1 | = sqrt(1 + ( sqrt3)^2) = 2`
`text{arg} (z_1) = tan^-1 (sqrt3) = frac{pi}{3}`
`z_1 = 2 (cos frac{pi}{3} + i sin frac{pi}{3})`
`z_2 = 1 + i`
`| z_2 | = sqrt(1^2 + 1^2) = sqrt2`
`text{arg} (z_2) = tan^-1 (1) = frac{pi}{4}`
`z_2 = sqrt2 (cos frac{pi}{4} + i sin frac{pi}{4})`
`frac{1 + i sqrt3}{1 + i}` | `= frac{z_1}{z_2}` |
`= frac{2}{sqrt2} ( cos ( frac{pi}{3} – frac{pi}{4} ) + i sin ( frac{pi}{3} – frac{pi}{4} ) )` | |
`= sqrt2 ( cos (frac{pi}{12}) + i sin (frac{pi}{12}) )` |
iii. `text{Equating real parts of i and ii:}`
`sqrt2 cos (frac{pi}{12})` | `= frac{1 + sqrt3}{2}` |
`cos(frac{pi}{12})` | `= frac{1 + sqrt3}{2 sqrt2} xx frac{sqrt2}{sqrt2}` |
`= frac{sqrt2 + sqrt6}{4}` |
iv. | `(frac{1 + i sqrt2}{1 + i})^12` | `= (sqrt2)^12 (cos (frac{pi}{12} xx 12) + i sin (frac{pi}{12} xx 12))` |
`= 64 (cos pi + i sin pi)` | ||
`= – 64` |
Which of the following is the complex number \(-3-\sqrt{3}i\)?
\(C\)
\(\abs{z}\) | \(=\sqrt{3^2+(\sqrt{3})^2}\) |
\(=\sqrt{12}\) | |
\(=2 \sqrt{3}\) |
\(\tan \theta\) | \(=\dfrac{3}{\sqrt{3}} \times \dfrac{\sqrt{3}}{\sqrt{3}}=\sqrt{3}\) |
\(\theta\) | \(=\dfrac{\pi}{3}\) |
\(\arg (z)\) | \(=-\left(\dfrac{\pi}{3}+\dfrac{\pi}{2}\right)=-\dfrac{5 \pi}{6}\) |
\(\text {In exponential form:}\)
\(z=2 \sqrt{3} e^{-\small{\dfrac{i5\pi}{6}}}\)
\(\Rightarrow C\)
What is the maximum value of `|e^(i theta) - 2| + |e^(i theta) + 2|` for `0 ≤ theta ≤ 2 pi`?
`C`
`e^(i theta) = a + ib \ \ text{where}\ \ a^2 + b^2 =1`
`X` | `= |e^(i theta) – 2| + |e^(i theta) + 2|` |
`= | (a – 2) + ib | + | (a + 2) + i b |` | |
`= sqrt((a-2)^2 + b^2) + sqrt((a+2)^2 + b^2)` | |
`= sqrt(a^2 – 4a + 4 + b^2) + sqrt(a^2 + 4a + 4 + b^2` | |
`= sqrt(5 – 4a) + sqrt(5 + 4a)` |
`frac{dX}{da}` | `= -frac{1}{2} xx 4 xx frac{1}{sqrt(5 – 4a)} + frac{1}{2} xx 4 xx frac{1}{sqrt(5 + 4a)}` |
`= frac{2}{sqrt(5 + 4a)} – frac{2}{sqrt(5 – 4a)}` |
`text(When)\ \ frac{dX}{da} = 0:`
`5 + 4a` | `= 5 – 4a` |
`a` | `= 0` |
`:. X_text(max) = 2 sqrt5`
`=> \ C`
Two masses, `2m` kg and `4m` kg, are attached by a light string. The string is placed over a smooth pulley as shown.
The two masses are at rest before being released and `v` is the velocity of the larger mass at time `t` seconds after they are released.
The force due to air resistance on each mass has magnitude `kv`, where `k` is a positive constant.
--- 6 WORK AREA LINES (style=lined) ---
--- 8 WORK AREA LINES (style=lined) ---
i.
`text{Taking} \ v \ text{downwards as positive.}`
`text{Forces acting on}\ 2m\ text{mass:}`
`kv + 2 mg-T = -2m * frac{dv}{dt}\ …\ (1)`
`text{Forces acting our} \ 4m \ text{mass:}`
`4mg-kv-T = 4m * frac{dv}{dt}\ …\ (2)`
`text{Subtract} \ \ (2)-(1)`
`2 mg-2 kv` | `= 6 m * frac{dv}{dt}` |
`:. frac{dv}{dt}` | `= frac{2mg-2 kv}{6 m}` |
`= frac{gm-kv}{3m}` |
ii. | `frac{dv}{dt}` | `= frac{gm-kv}{3m}` |
`frac{dt}{dv}` | `= frac{3m}{gm-kv}` | |
`t` | `= int frac{3m}{gm-kv}\ dv` | |
`= -frac{3m}{k} log_e |gm-kv | + c` |
`text{When} \ \ t = 0, v = 0:`
`0` | `= -frac{3m}{k} log_e \ | gm | + c` |
`c` | `= frac{3m}{k} log_e \ | gm | ` |
`t` | `= frac{3m}{k} log_e \ | gm | \-frac{3m}{k} log_e \ | gm -kv |` |
`= frac{3m}{k} log_e \ | frac{mg}{gm-kv} |` |
`text{Find} \ \ v\ \ text{when} \ \ t = frac{3m}{k} log_e 2 :`
`frac{3m}{k} log_e 2` | `= frac{3m}{k} log_e | frac{gm}{gm-kv} |` |
`2` | `= frac{gm}{gm-kv}` |
`2gm-2kv` | `= gm` |
`2kv` | `= gm` |
`therefore \ v` | `= frac{gm}{2k}` |
Prove that for any integer `n > 1, log_n (n + 1)` is irrational. (3 marks)
--- 8 WORK AREA LINES (style=lined) ---
`text{See Worked Solutions}`
`text{Proof by contradiction:}`
`text{Assume} \ log_n(n + 1) \ text{is rational}`
`therefore \ log_n (n + 1) = frac{p}{q} \ \ text{where} \ \ p,q ∈ ZZ \ text{with no common factor except 1}`
`n^(frac{p}{q}` | `= n + 1` |
`n^p` | `= (n + 1)^q` |
`text{Strategy 1}`
`n^p = (n + 1)^q \ \ text{when} \ \ p = q = 0\ \ text{only}`
`q ≠ 0`
`:.\ text{By contradiction}, log_n (n + 1) \ \ text{is irrational.}`
`text{Strategy 2}`
`n^p = (n + 1)^q`
`text{If} \ \ n\ \ text{is odd, LHS is odd and RHS is even.}`
`text{If} \ \ n\ \ text{is even LHS is even and RHS is odd.}`
`text{Statement is true for} \ \ p = q = 0 , text{but} \ \ q ≠ 0`
`therefore \ text{By contradiction,} \ log_n (n + 1) \ text{is irrational.}`
The point `C` divides the interval `AB` so that `frac{CB}{AC} = frac{m}{n}`. The position vectors of `A` and `B` are `underset~a` and `underset~b` respectively, as shown in the diagram.
--- 6 WORK AREA LINES (style=lined) ---
--- 5 WORK AREA LINES (style=lined) ---
Let `OPQR` be a parallelogram with `overset->(OP) = underset~p` and `overset->(OR) = underset~r`. The point `S` is the midpoint of `QR` and `T` is the intersection of `PR` and `OS`, as shown in the diagram.
--- 8 WORK AREA LINES (style=lined) ---
--- 3 WORK AREA LINES (style=lined) ---
i.
`frac{overset->(AC)}{overset->(AB)}` | `= frac{n}{m + n}` |
`overset->(AC)` | `= frac{n}{m + n} * overset->(AB)` |
`= frac{n}{m + n} (underset~b – underset~a)` |
ii. | `overset->(OC)` | `= overset->(OA) + overset->(AC)` |
`= underset~a + frac{n}{m + n} (underset~b – underset~a)` | ||
`= underset~a – frac{n}{m + n} underset~a + frac{n}{m + n} underset~b` | ||
`= (1 – frac{n}{m + n}) underset~a + frac{n}{m + n} underset~b` | ||
`= (frac{m + n – n}{m + n}) underset~a + frac{n}{m + n} underset~b` | ||
`= frac{m}{m + n} underset~a + frac{n}{m + n} underset~b` |
iii. `text{Show} \ \ overset->(OT) = frac{2}{3} underset~r + frac{1}{3} underset~p`
`text{Consider} \ \ Delta PTO \ \ text{and} \ \ Delta RTS:`
`angle PTO = angle RTS \ (text{vertically opposite})`
`angle OPT = angle SRT \ (text{vertically opposite})`
`therefore \ Delta PTO \ text{|||} \ Delta RTS\ \ text{(equiangular)}`
`OT : TS = OP : SR = 2 : 1`
`(text{corresponding sides in the same ratio})`
`frac{overset->(OT)}{overset->(OS)}` | `= frac{2}{3}` |
`overset->(OT)` | `= frac{2}{3} overset->(OS)` |
`= frac{2}{3} ( underset~r + frac{1}{2} underset~p)` | |
`= frac{2}{3} underset~r + frac{1}{3} underset~p` |
iv. `text{Let} \ \ overset->(OT) \ text{divide} \ PR\ text{so that}\ \ frac{TR}{PT} = frac{m}{n}`
`text{Using part (ii):}`
`overset->(OT)` | `= frac{m}{m + n} underset~p + frac{n}{m + n} c` |
`overset->(OT)` | `= frac{1}{3} underset~p + frac{2}{3} underset~r \ \ \ (text{part (iii)})` |
`frac{m}{m + n}` | `= frac{1}{3} , frac{n}{m + n} = frac{2}{3}` |
`=> \ m = 1 \ , \ n = 2`
`therefore \ T \ text{divides} \ PR \ text{in ratio 2 : 1}.`
In the set of integers, let `P` be the proposition:
'If `k + 1` is divisible by 3, then `k^3 + 1` divisible by 3.'
--- 5 WORK AREA LINES (style=lined) ---
--- 2 WORK AREA LINES (style=lined) ---
--- 8 WORK AREA LINES (style=lined) ---
i. `text{Let} \ \ k + 1 = 3N, \ N∈ Z`
`=> k = 3N – 1`
`k^3 + 1` | `= (3N -1)^3 + 1` |
`= (3N)^3 + 3(3N)^2 (-1) + 3(3N)(-1)^2 + (-1)^3 + 1` | |
`= 27N^3 – 27N^2 + 9N – 1 + 1` | |
`= 3 (9N^3 – 9N^2 + 3N)` | |
`= 3Q \ , \ Q ∈ Z` |
`therefore \ text{If} \ \ k+ 1 \ \ text{is divisible by 3}, text{then} \ \ k^3 + 1 \ \ text{is divisible by 3.}`
ii. `text{Contrapositive}`
`text{If} \ \ k^3 + 1 \ \ text{is not divisible by 3, then}\ \ k + 1\ \ text{is not divisible by 3.}`
iii. `text{Converse:}`
`text{If} \ \ k^3 + 1\ \ text{is divisible by 3, then}\ \ k + 1\ \ text{is divisible by 3.}`
`text(Contrapositive of converse:)`
`text{If}\ \ k + 1\ \ text{is not divisible by 3, then}\ \ k^3 + 1\ \ text{is not divisible by 3.}`
`text(i.e.)\ \ k + 1 \ \ text{is not divisible by 3 when}\ \ k + 1 = 3Q + 1\ \ text{or}\ \ k + 1 = 3Q + 2, text{where}\ Q ∈ Z`
`text{If} \ \ k + 1` | `= 3Q + 1\ \ => \ k=3Q` |
`k^3 + 1` | `= (3Q)^3 + 1` |
`= 27Q^3 + 1` | |
`= 3(9Q^3) + 1` | |
`= 3M + 1 \ \ (text{not divisible by 3,}\ M ∈ Z)` |
`text{If} \ \ k + 1` | `= 3Q + 2\ \ => \ k=3Q+1` |
`k^3 + 1` | `= (3Q + 1)^3 + 1` |
`= (3Q)^3 + 3(3Q)^2 + 3(3Q) + 1 + 1` | |
`= 27Q^3 + 27Q^2 + 9Q + 2` | |
`= 3(9Q^3 + 9Q^2 + 3Q) + 2` | |
`= 3M + 2 \ (text{not divisible by 3,}\ M ∈ Z) ` |
`therefore \ text{By contrapositive, if}\ \ k^3 + 1\ \ text {is divisible by 3, k + 1 is divisible by 3.}`
Prove by mathematical induction that, for `n ≥ 2`,
`frac{1}{2^2} + frac{1}{3^2} + ... + frac{1}{n^2} < frac{n - 1}{n}` (4 marks)
--- 10 WORK AREA LINES (style=lined) ---
`text{See Worked Solutions}`
`text{Prove} \ frac{1}{2^2} + frac{1}{3^2} + … + frac{1}{n^2} < frac{n – 1}{n}\ \ \ text(for)\ \ n>=2`
`text(If)\ \ n=2:`
`text{LHS} = frac{1}{2^2} = frac{1}{4}`
`text{RHS} = frac{2-1}{2} = frac{1}{2} > text{LHS}`
`therefore \ text{True for} \ \ n = 2`
`text{Assume true for} \ \ n =k`
`frac{1}{2^2} + frac{1}{3^2} + … + frac{1}{k^2} < frac{k – 1}{k}`
`text{Prove true for} \ \ n = k + 1`
`frac{1}{2^2} + frac{1}{3^2} + … + frac{1}{k^2} + frac{1}{(k + 1)^2}< frac{k}{k+1}`
`text{LHS}` | `= frac{k – 1}{k} + frac{1}{(k + 1}^2}` |
`= frac{(k – 1)(k + 1)^2 + k}{k (k + 1)^2}` | |
`= frac{(k – 1)(k^2 + 2 k +1) + k}{k(k +1)^2}` | |
`= frac{k^3 + 2k^2 + k – k^2 – 2k – 1 + k}{k(k + 1)^2}` | |
`= frac{k^3 + k^2 -1}{k(k + 1)^2}` | |
`= frac{k^2 (k + 1 – frac{1}{k^2})}{k(k + 1)^2}` | |
`= frac{k}{k +1} xx frac{(k + 1 – frac{1}{k^2})}{k +1}` | |
`< frac{k}{k +1} \ \ \ (text{since} \ frac{k + 1 – frac{1}{k^2}}{k +1} < 1 )` |
`=> \ text{True for} \ n = k + 1`
`therefore \ text{S}text{ince true for} \ n = 2, \ text{by PMI, true integral} \ n ≥ 2.`
A particle starts from rest and falls through a resisting medium so that its acceleration, in m/s2, is modelled by
`a = 10 (1 - (kv)^2)`,
where `v` is the velocity of the particle in m/s and `k = 0.01`.
Find the velocity of the particle after 5 seconds. (4 marks)
`46.21 \ text{ms}^-1`
`a` | `= 10 (1 – (kv)^2)` |
`frac{dv}{dt}` | `= 10 – 10 xx 0.01^2 xx v^2` |
`= 10 – 0.001 v^2` | |
`frac{dt}{dv}` | `= frac{1}{10 – 0.001 \ v^2}` |
`= frac{1000}{10 \ 000 – v^2}` | |
`t` | `= int frac{1000}{100^2 – v^2}\ dv` |
`text{Using partial fractions}:`
`frac{1}{100^2 – v^2}` | ` = frac{A}{100 + v} + frac{B}{100 – v}` |
`1` | `= A (100 – v) + B(100 + v)` |
`text{If} \ \ v = 100 \ , \ 1 = 200 B \ => \ B = frac{1}{200}`
`text{If} \ \ v = -100 \ , \ 1 = 200 A \ => \ A = frac{1}{200}`
`t` | `= 1000 int frac{1}{200 (100 + v)}\ dv + 1000 int frac{1}{200(100 -v)}\ dv` |
`= 5 int frac{1}{100 + v}\ dv + 5 int frac{1}{100 – v}\ dv` | |
`= 5 ln \ | 100 + v | – 5 ln \|100 – v | + c` | |
`= 5 ln \ | frac{100 + v}{100 – v} | + c` |
`text{When} \ \ t = 0 , \ v = 0`
`0 = 5 ln 1 + c \ => \ c = 0`
`:. t = 5 ln \ | frac{100 + v}{100 – v} |`
`text{Find} \ \ v \ \ text{when} \ \ t = 5 :`
`5` | `= 5 ln | frac{100 + v}{100 – v} |` |
`1` | `= ln | frac{100 + v}{100 – v} | ` |
`e` | `= frac{100 + v}{100 – v}` |
`100 e – ve` | `= 100 + v` |
`v + ve` | `= 100 e – 100` |
`v(1 + e)` | `= 100 e – 100` |
`:. v` | `= frac{100 e – 100}{1 + e}` |
`= 46.21 \ text{ms}^-1 \ \ (text{2 d.p.})` |
Let `z_1` be a complex number and let `z_2 = e^(frac{i pi}{3}) z_1`
The diagram shows points `A` and `B` which represent `z_1` and `z_2`, respectively, in the Argand plane.
--- 6 WORK AREA LINES (style=lined) ---
--- 8 WORK AREA LINES (style=lined) ---
i.
`text{Let}` | `z_1` | `= r(cos theta + i sin theta)` |
`z_2` | `= e^(i frac{pi}{3}) z_1` | |
`= r (cos ( theta + frac{pi}{3} ) + i sin (theta + frac{pi}{3}))` |
`| z_1 | = | z_2 | => OA = OB`
` angle AOB = frac{pi}{3} \ ( z_2 \ text{is a} \ frac{pi}{3} \ text{anti-clockwise rotation of} \ z_1 )`
`=> angle OBA = angle BAO = pi/3\ \ \ text{(angles opposite equal sides)}`
`therefore \ OAB \ text{is equilateral}`
ii. | `z_1` | `= z_2 e^(i frac{pi}{3})` |
`frac{z_1}{z_2}` | `= e^(i frac{pi}{3})` | |
`(frac{z_1}{z_2})^3` | `= e^((3 xx i frac{pi}{3})) \ \ (text{by De Moivre})` | |
`frac{z_1^3}{z_2^3}` | `= e^(i pi)` | |
`z_1^3` | `= -z_2^3` | |
`z_1^3 + z_2^3` | `= 0` |
`(z_1 + z_2)(z_1^2 – z_1 z_2 + z_2^2)` | `= 0` |
`z_1^2 – z_1 z_2 + z_2^2` | `= 0` |
`z_1^2 + z_2^2` | `= z_1 z_2` |
A particle is undergoing simple harmonic motion with period `frac{pi}{3}`. The central point of motion of the particle is at `x = sqrt(3)`. When `t = 0` the particle has its maximum displacement of `2 sqrt(3)` from the central point of motion.
Find an equation for the displacement, `x`, of the particle in terms of `t`. (3 marks)
`x = 2 sqrt(3) cos (6t) + sqrt(3)`
`text{Period}` | `= frac{pi}{3}` |
`frac{2 pi}{n}` | `= frac{pi}{3}` |
`n` | `= 6` |
`text{Amplitude} = 2 sqrt(3)`
`text{Centre of motion} = sqrt(3)`
`text{S} text{ince maximum displacement at}\ \ t = 0,`
`x = 2 sqrt(3) cos (6t) + sqrt(3)`
A particle is projected from the origin with initial velocity `u` m/s at an angle `theta` to the horizontal. The particle lands at `x = R` on the `x`-axis. The acceleration vector is given by `underset~a = ((0),(-g))`, where `g` is the acceleration due to gravity. (Do NOT prove this.)
--- 8 WORK AREA LINES (style=lined) ---
--- 6 WORK AREA LINES (style=lined) ---
--- 7 WORK AREA LINES (style=lined) ---
i. `underset~a = ((0),(-g))`
`ddotx = 0`
`dotx = int ddotx \ dt = c`
`text{When} \ \ t = 0, \ dotx = u cos theta \ => \ c = u cos theta`
`=> dotx = u cos theta`
`x = int dotx \ dt =u t cos theta + c`
`text{When} \ \ t = 0 , \ x = 0, \ c = 0`
`therefore \ x = ut cos theta`
`ddoty = -g`
`doty = int – g \ dt = -g t + c`
`text{When} \ \ t = 0 , \ doty = u sin theta`
`=> doty = u sin theta – g t`
`y = int doty \ dt = ut sin theta – frac {1}{2} g t^2 + c`
`text{When} \ \ t = 0, \ y = 0 \ => \ c = 0`
`therefore y = ut sin theta – frac(1)(2) g t^2`
`:. underset~r = ((x),(y)) = ((ut cos theta),(ut sin theta – frac{1}{2} g t^2))`
ii. `x = ut \ cos theta`
`t = frac{x}{u \ cos theta}`
`text{Substitute into} \ y:`
`y` | `= u * frac{x}{u \ cos theta}\ sin theta – frac{1}{2} g ( frac{x}{u \ cos theta} )^2` |
`= x tan theta – frac{gx^2}{2 u^2 cos^2 theta}` | |
`= frac{-gx^2}{2u^2} ( frac{1}{cos^2 theta} – frac{2u^2}{gx} tan theta )` | |
`= frac{-gx^2}{2u^2} ( sec^2 theta – frac{2u^2}{gx} tan theta )` | |
`= frac{-gx^2}{2u^2} ( tan^2 theta – frac{2u^2}{gx} tan theta + 1 )` |
iii. `text{When} \ \ x = R \ , \ y = 0`
`frac{-gR^2}{2 u^2}` | `( tan^2 theta – frac{2u^2}{gR} tan theta + 1 ) = 0` |
`tan^2 theta – frac{2u^2}{gR} tan theta + 1 = 0` | |
`Delta` |
`= ( frac{-2u^2}{gR} )^2 – 4 * 1 * 1` |
`= frac{4u^2}{g^2 R^2} – 4` |
`u^2` | `> gR\ \ \ text{(given)}` | |
`u^4` | `> g^2 R^2` | |
`frac{u^4}{g^2 R^2}` | `> 1` | |
`frac{4u^4}{g^2 R^2}` | `> 4` | |
`frac{4u^4}{g^2 R^2} – 4` | `> 0` | |
`Delta` | `> 0` |
`therefore \ 2 \ text{distinct values of} \ \ theta\ \ text{satisfy} \ \ x = R.`
A 50-kilogram box is initially at rest. The box is pulled along the ground with a force of 200 newtons at an angle of 30° to the horizontal. The box experiences a resistive force of `0.3R` newtons, where `R` is the normal force, as shown in the diagram.
Take the acceleration `g` due to gravity to be 10m/s2.
--- 5 WORK AREA LINES (style=lined) ---
--- 4 WORK AREA LINES (style=lined) ---
--- 6 WORK AREA LINES (style=lined) ---
i.
`text{Resolving forces vertically:}`
`R + 200 \ sin 30^@` | `= 50g` |
`R + 200 xx frac{1}{2}` | `= 50 xx 10` |
`R + 100` | `= 500` |
`therefore \ R` | `= 400 \ text(N)` |
ii. `text{Resolving forces horizontally:}`
`text{Net Force}` | `= 200 \ cos 30^@ – 0.3 R` |
`= 200 xx frac{sqrt3}{2} – 0.3 xx 400` | |
`= 100 sqrt3 – 120` | |
`= 53.2 \ text{N (to 1 d. p.)}` |
iii. | `F` | `=ma` |
`50 a` | `=100 sqrt300 – 120` | |
`a` | `= frac{100 sqrt3 – 120}{50}\ text(ms)^(-2)` |
`text{Initially} \ \ u = 0,`
`v` | `= u + at` |
`v_(t=3)` | `= 0 + frac{100 sqrt3 – 120}{50} xx 3` |
`= 3.1923 \ …` | |
`= 3.19 \ text{ms}^-1 \ text{(to 2 d.p.)}` |
A particle undergoing simple harmonic motion has a maximum acceleration of 6 m/s2 and a maximum velocity of 4 m/s.
What is the period of the motion?
`D`
`ddotx = -n^2 x`
`text{Find} \ n :`
`ddotx_text{max} = 6 \ \ text{occurs when} \ \ x = – a`
`6 = n^2 a\ …\ (1)`
`v^2 = n^2 (a^2 – x^2)`
`v_text{max} = 4 \ \ text{occurs when} \ \ x = 0`
`4^2` | `= n^2 (a^2-0)` | |
`16` | `= n^2 a^2` | |
`4` | `= n a\ …\ (2)` |
`text{Substitute} \ \ na = 4 \ \ text{from (2) into (1):}`
`6` | `= 4n` |
`n` | `= frac{3}{2}` |
`therefore \ text{Period} = frac{2pi}{n} = frac{4pi}{3}`
`=> \ D`
The diagram shows the complex number `z` on the Argand diagram.
Which of the following diagrams best shows the position of `frac{z^2}{|z|}`?
|
`A`
`text{Let} \ \ z = r\ text(cis)\ theta`
`z^2` | `= r^2 text(cis)\ (2 theta)` |
`|z|` | `= r` |
`therefore frac{z^2}{|z|}` | `= frac{r^2 \ text(cis)\ (2 theta)}{r}` |
`= r\ text(cis)\ (2 theta)` |
`text{On Argand diagram, it lies on the dotted line`
`text{(modulus the same) with an argument that is}`
`text{doubled.}`
`=> \ A`
Barbara plays a game of chance, in which two unbiased six-sided dice are rolled. The score for the game is obtained by finding the difference between the two numbers rolled. For example, if Barbara rolls a 2 and a 5, the score is 3.
The table shows some of the scores.
--- 3 WORK AREA LINES (style=lined) ---
a.
b. | `Ptext{(not zero)}` | `= frac{text(numbers) ≠ 0}{text(total numbers)}` |
`= frac{30}{36}` | ||
`= frac{5}{6}` |
\(\text{Alternate solution (b)}\)
b. | `Ptext{(not zero)}` | `= 1 – Ptext{(zero)}` |
`= 1 – frac{6}{36}` | ||
`= frac{5}{6}` |
A group of students sat a test at the end of term. The number of lessons each student missed during the term and their score on the test are shown on the scatterplot.
--- 5 WORK AREA LINES (style=lined) ---
--- 1 WORK AREA LINES (style=lined) ---
Use the line of the best fit drawn in part (c) to estimate Meg's score on this test. (1 mark)
--- 1 WORK AREA LINES (style=lined) ---
Is it appropriate to use the line of the best fit to estimate his score on the test? Briefly explain your answer. (1 mark)
--- 2 WORK AREA LINES (style=lined) ---
a. \(\text{Strength : strong}\)
\(\text{Direction : negative} \)
b. \(\text{Range}\ = \text{high}-\text{low}\ = 100-80=20\)
c.
d.
e. \(\text{John’s missed days are too extreme and the LOBF is not}\)
\(\text{appropriate. The model would estimate a negative score for}\)
\(\text{John which is impossible.}\)
a. \(\text{Strength : strong}\)
\(\text{Direction : negative} \)
b. \(\text{Range}\ = \text{high}-\text{low}\ = 100-80=20\)
c.
d.
\(\therefore\ \text{Meg’s estimated score = 40}\)
e. \(\text{John’s missed days are too extreme and the LOBF is not}\)
\(\text{appropriate. The model would estimate a negative score for}\)
\(\text{John which is impossible.}\)
The weight of a bundle of A4 paper (`W` kg) varies directly with the number of sheets (`N`) of A4 paper that the bundle contains.
This relationship is modelled by the formula `W = kN`, where `k` is a constant.
The weight of a bundle containing 500 sheets of A4 paper is 2.5 kilograms.
--- 2 WORK AREA LINES (style=lined) ---
--- 4 WORK AREA LINES (style=lined) ---
a. `W = 2.5\ text{kg when} \ N = 500:`
`2.5` | `= k xx 500` |
`therefore \ k` | `= frac{2.5}{500}` |
`= 0.005` |
b. `text{Find}\ \ N \ \ text{when} \ \ W = 1.2\ text{kg:}`
`1.2` | `= 0.005 xx N` |
`therefore N` | `= frac{1.2}{0.005}` |
`= 240 \ text{sheets}` |
Matilda has a weekly net income of $ 510. She has created a budget where she allocates this income to rent, car expenses, personal expenses, phone, and rest to savings.
Her budget is shown below, with some details missing.
Matilda allocates 20% of her weekly net income to personal expenses.
How many weeks will it take Matilda to save $4930? (3 marks)
`85 \ text{weeks}`
`text{Personal expenses} = 20text(%) xx 510 = $102`
`text{Savings}` | `= 510 – (115 + 210 + 102 + 25)` |
`= $58` |
`therefore \ text{Weeks to save} \ $4930`
`= frac{4930}{58}`
`= 85 \ text{weeks}`
Consider the equation `m = 6 - frac{3R}{2R - 5}`.
Find the value of `m` when `R = 10`. (2 marks)
`4`
`m` | `= 6 – frac{3 xx 10}{2 xx 10 – 5}` |
`= 6 – frac{30}{15}` | |
`= 6 – 2` | |
`= 4` |
Adam travels on a straight road away from his home. His journey is shown in the distance – time graph.
--- 5 WORK AREA LINES (style=lined) ---
On the above, complete the distance-time using the information provided. (2 marks)
--- 0 WORK AREA LINES (style=lined) ---
`text{Speed: Adam increases speed until approximately}\ \ t=2,`
`text{and then decreases speed until he stops when}\ \ t=4.`
`text{Distance travelled: Adam’s distance from home increases}`
`text{at an increasing rate until}\ \ t=2, \ text{and then continues}`
`text{to increase but at a decreasing rate until}\ \ t=4, \ text(when the)`
`text(distance from home remains the same.)`
a. `text{Speed: Adam increases speed until approximately}\ \ t=2,`
`text{and then decreases speed until he stops when}\ \ t=4.`
`text{Distance travelled: Adam’s distance from home increases}`
`text{at an increasing rate until}\ \ t=2, \ text{and then continues}`
`text{to increase but at a decreasing rate until}\ \ t=4, \ text(when the)`
`text(distance from home remains the same.)`
b. `text(S)text(ince Adam travels home at a constant speed, the graph is)`
`text{is a straight line and ends at (10, 0).}`
Two painters each provide a quote for painting an area of 1500 square metres. Painter A charges $100 per 30 square metres. Painter B charges $80 per hour and bases their quote on painting 25 square metres per hour.
Calculate how much will be saved by choosing the cheaper quote. (3 marks)
--- 6 WORK AREA LINES (style=lined) ---
`$200`
`text{Painter A cost:}`
`C_A = frac{1500}{30} xx 100 = $5000`
`text{Painter B cost:}`
`text(Hours)\ = frac{1500}{25} = 60`
`C_B = 60 xx 80 = $ 4800`
`therefore \ text{Savings by using Painter B}`
`= 5000 – 4800`
`= $200`
The table shows the average brain weight (in grams) and average body weight (in kilograms) of nine different mammals.
\begin{array} {|l|c|c|c|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} \textit{Brain weight (g)} \rule[-1ex]{0pt}{0pt} & 0.7 & 0.4 & 1.9 & 2.4 & 3.5 & 4.3 & 5.3 & 6.2 & 7.8 \\
\hline
\rule{0pt}{2.5ex} \textit{Body weight (kg)} \rule[-1ex]{0pt}{0pt} & 0.02 &0.06 & 0.05 & 0.34 & 0.93 & 0.97 & 0.43 & 0.33 & 0.22 \\
\hline
\end{array}
Which of the following is the correct scatterplot for this dataset?
|
|
`C`
`text{Consider data point} \ (1.9, 0.05)`
`→ \ text{Eliminate} \ A \ text{(too high)}`
`→ \ text{Eliminate} \ D \ text{(should be below 2nd data point)}`
`text{Consider data point} \ (2.4, 0.34)`
`→ \ text{Eliminate} \ B \ text{(not on graph)}`
`=> \ C`
--- 7 WORK AREA LINES (style=lined) ---
Show that `sin (3theta) = 1/2`. (2 marks)
--- 7 WORK AREA LINES (style=lined) ---
--- 10 WORK AREA LINES (style=lined) ---
i. `text(Prove:)\ \ sin^3 theta-3/4 sin theta + (sin(3theta))/4 = 0`
`text(LHS)` | `= sin^3 theta-3/4 sin theta + 1/4 (sin 2thetacostheta + cos2thetasintheta)` |
`= sin^3 theta-3/4 sintheta + 1/4(2sinthetacos^2theta + sintheta(1 – 2sin^2theta))` | |
`= sin^3theta-3/4 sintheta + 1/4(2sintheta(1-sin^2theta) + sintheta – 2sin^3theta)` | |
`= sin^3theta-3/4 sintheta + 1/4(2sintheta-2sin^3theta + sintheta-2sin^3theta)` | |
`= sin^3theta-3/4sintheta + 3/4sintheta-sin^3theta` | |
`= 0` |
ii. `text(Show)\ \ sin(3theta) = 1/2`
`text{Using part (i):}`
`(sin(3theta))/4` | `= 3/4 sintheta-sin^3 theta` |
`sin(3theta)` | `= 3sintheta-4sin^3theta\ …\ (1)` |
`x^3-12x + 8 = 0`
`text(Let)\ \ x = 4 sin theta`
`(4sintheta)^3-12(4sintheta) + 8` | `= 0` |
`64sin^3theta-48sintheta` | `= 0` |
`−16underbrace{(3sintheta-4sin^2theta)}_text{see (1) above}` | `= −8` |
`−16 sin(3theta)` | `= −8` |
`sin(3theta)` | `= 1/2` |
iii. `text(Prove:)\ \ sin^2\ pi/18 + sin^2\ (5pi)/18 + sin^2\ (25pi)/18 = 3/2`
`text(Solutions to)\ \ x^3-12x + 8 = 0\ \ text(are)`
`x = 4sintheta\ \ text(where)\ \ sin(3theta) = 1/2`
`text(When)\ \ sin3theta = 1/2,`
`3theta` | `= pi/6, (5pi)/6, (13pi)/6, (17pi)/6, (25pi)/6, (29pi)/6, …` |
`theta` | `= pi/18, (5pi)/18, (13pi)/18, (17pi)/18, (25pi)/18, (29pi)/18, …` |
`:.\ text(Solutions)`
`x = 4sin\ pi/18 \ \ \ (= 4sin\ (17pi)/18)`
`x = 4sin\ (5pi)/18 \ \ \ (= 4sin\ (13pi)/18)`
`x = 4sin\ (25pi)/18 \ \ \ (= 4sin\ (29pi)/18)`
`text(If roots of)\ \ x^3-12x + 8 = 0\ \ text(are)\ \ α, β, γ:`
`α + β + γ = −b/a = 0`
`αβ + βγ + αγ = c/a = −12`
`(4sin\ pi/18)^2 + (4sin\ (5pi)/18)^2 + (4sin\ (25pi)/18)^2` | `= (α + β + γ)^2 – 2(αβ + βγ + αγ)` |
`16(sin^2\ pi/18 + sin^2\ (5pi)/18 + sin^2\ (25pi)/18)` | `= 0-2(−12)` |
`:. sin^2\ pi/18 + sin^2\ (5pi)/18 + sin^2\ (25pi)/18` | `= 24/16=3/2` |
A random sample of students was taken from each of two universities, and their ages were recorded. The boxplots of their ages are shown.
For the given samples of students' ages, which of the following statements is FALSE?
`A`
`text{Consider} \ A :`
`text{Range of} \ A \ ≈ 40 – 17.5 ≈ 22.5`
`text{Range of} \ B \ ≈ 30 – 17.5 ≈ 12.5`
`therefore \ text{Range for} \ A > text{Range for}\ B`
`=> A`
Consider the triangle shown.
--- 4 WORK AREA LINES (style=lined) ---
--- 4 WORK AREA LINES (style=lined) ---
a. | `tan theta` | `= frac{8}{10}` |
`theta` | `= tan ^(-1) frac{8}{10}` | |
`= 38.659…` | ||
`= 39^@ \ text{(nearest degree)}` |
b. `text{Using Pythagoras:}`
`x` | `= sqrt{8^2 + 10^2}` |
`= 12.806…` | |
`= 12.8 \ \ text{(to 1 d.p.)}` |
The distance between Bricktown and Koala Creek is 75 km. A person travels from Bricktown to Koala Creek at an average speed of 50 km/h.
How long does it take the person to complete the journey?
`C`
`text(Time)` | `= frac(text(Distance))(text(Speed))` |
`= frac(75)(50)` | |
`=1.5 \ text(hours)` | |
`= 1 \ text(hour) \ 30 \ text(minutes)` |
`=> \ C`
Mr Ali, Ms Brown and a group of students were camping at the site located at `P`. Mr Ali walked with some of the students on a bearing of 035° for 7 km to location `A`. Ms Brown, with the rest of the students, walked on a bearing of 100° for 9 km to location `B`.
--- 2 WORK AREA LINES (style=lined) ---
--- 4 WORK AREA LINES (style=lined) ---
--- 4 WORK AREA LINES (style=lined) ---
a. | `angle APB` | `= 100 – 35` |
`= 65^@` |
b. `text(Using cosine rule:)`
`AB^2` | `= AP^2 + PB^2 – 2 xx AP xx PB cos 65^@` |
`= 49 + 81 – 2 xx 7 xx 9 cos 65^@` | |
`= 76.750…` | |
`:.AB` | `= 8.760…` |
`= 8.76\ text{km (to 2 d.p.)}` |
c.
`anglePAC = 35^@\ (text(alternate))`
`text(Using cosine rule, find)\ anglePAB:`
`cos anglePAB` | `= (7^2 + 8.76 – 9^2)/(2 xx 7 xx 8.76)` | |
`= 0.3647…` | ||
`:. angle PAB` | `= 68.61…^@` | |
`= 69^@\ \ (text(nearest degree))` |
`:. text(Bearing of)\ B\ text(from)\ A\ (theta)`
`= 180 – (69 – 35)`
`= 146^@`
The region `R` is bounded by the `y`-axis, the graph of `y = cos(2x)` and the graph of `y = sin x`, as shown in the diagram.
Find the volume of the solid of revolution formed when the region `R` is rotated about the `x`-axis. (4 marks)
--- 8 WORK AREA LINES (style=lined) ---
`(3sqrt3 pi)/16\ text(u)³`
`text(Find intersection:)`
`sin x = cos 2x`
`sin x = 1 – 2sin^2 x`
`2sin^2 x + sinx – 1` | `= 0` |
`(2 sinx – 1)(sinx + 1)` | `= 0` |
`sin x` | `= 1/2` | `text(or)` | `sin x` | `= −1` |
`x` | `= pi/6` | `x` | `= (3pi)/2` |
`V` | `= pi int_0^(pi/6) (cos 2x)^2\ dx – pi int_0^(pi/6)(sin x)^2\ dx` |
`= pi int_0^(pi/6) cos^2 2x – sin^2 x\ dx` | |
`= pi int_0^(pi/6) 1/2 (1 + cos 4x) – 1/2 (1 – cos 2x)\ dx` | |
`= pi/2 int_0^(pi/6) cos 4x + cos 2x\ dx` | |
`= pi/2 [1/4 sin 4x + 1/2 sin 2x]_0^(pi/6)` | |
`= pi/8 [sin\ (2pi)/3 + 2sin\ pi/3]` | |
`= pi/8 (sqrt3/2 + 2 xx sqrt3/2)` | |
`= (3sqrt3 pi)/16\ text(u)³` |
--- 1 WORK AREA LINES (style=lined) ---
--- 12 WORK AREA LINES (style=lined) ---
i. `d/(d theta) (sin^3 theta) = 3 cos theta sin^2 theta`
ii. `text(Let)\ x = tan theta`
`(dx)/(d theta) = sec^2 theta \ => \ dx = sec^2 theta\ d theta`
`text(When)\ x = 1, \ theta = pi/4`
`text(When)\ x = 0, \ theta = 0`
`int_0^1 (x^2)/(1 + x^2)^(5/2) dx` | `= int_0^(pi/4) (tan^2 theta)/((1 + tan^2 theta)^(5/2)) xx sec^2 theta\ d theta` |
`= int_0^(pi/4) (tan^2 theta)/((sec^2 theta)^(5/2)) xx sec^2 theta\ d theta` | |
`= int_0^(pi/4) (sin^2 theta)/(cos^2 theta) · 1/((sec^2 theta)^(3/2))\ d theta` | |
`= int_0^(pi/4) (sin^2 theta)/(cos^2 theta) · 1/(sec^3 theta)\ d theta` | |
`= int_0^(pi/4) sin^2 theta cos theta\ d theta` | |
`= 1/3[sin^3 theta]_0^(pi/4)` | |
`= 1/3(sin^3\ pi/4 – 0)` | |
`= 1/3 (1/sqrt2)^3` | |
`= 1/(6sqrt2)` | |
`= sqrt2/12` |
Find `int_0^(pi/2) cos 5x\ sin 3x\ dx`. (3 marks)
`−1/2`
`int_0^(pi/2) cos 5x\ sin 3x\ dx` | `= 1/2 int_0^(pi/2) 2cos 5x\ sin 3x\ dx` |
`= 1/2 int_0^(pi/2) sin 8x-sin 2x\ dx` | |
`= 1/2[−1/8 cos 8x + 1/2 cos 2x]_0^(pi/2)` | |
`= 1/2[(−1/8 cos 4pi + 1/2 cospi)-(−1/8 cos0 + 1/2 cos0)]` | |
`= 1/2(−1/8-1/2 + 1/8-1/2)` | |
`= −1/2` |
To complete a course, a student must choose and pass exactly three topics.
There are eight topics from which to choose.
Last year 400 students completed the course.
Explain, using the pigeonhole principle, why at least eight students passed exactly the same three topics. (2 marks)
--- 6 WORK AREA LINES (style=lined) ---
`text(See Worked Solution)`
`\ ^8C_3 = 56\ text(ways of choosing 3 topics)`
`text(400 students pass)`
`text(Pigeonholes)\ (k)= 56`
`text(Pigeons)\ (n) = 400`
`n/k` | `= 400/56` |
`= 7.14…` |
`:.\ text(By PHP, at least 8 students passed the)`
`text(same 3 subjects.)`
When a particular biased coin is tossed, the probability of obtaining a head is `3/5`.
This coin is tossed 100 times.
Let `X` be the random variable representing the number of heads obtained. This random variable will have a binomial distribution.
--- 2 WORK AREA LINES (style=lined) ---
--- 3 WORK AREA LINES (style=lined) ---
--- 1 WORK AREA LINES (style=lined) ---
i. `X = text(number of heads)`
`X\ ~\ text(Bin) (n, p)\ ~\ text(Bin) (100, 3/5)`
`E(X)` | `= np` |
`= 100 xx 3/5` | |
`= 60` |
ii. | `text(Var)(X)` | `= np(1 – p)` |
`= 60 xx 2/5` | ||
`= 24` |
`sigma(x)` | `= sqrt24` |
`~~ 5` |
iii. | `P(55 <= x <=65)` | `~~ P(−1 <= z <= 1)` |
`~~ 68text(%)` |
A composite solid consists of a triangular prism which fits exactly on top of a cube, as shown.
Find the surface area of the composite solid. (3 marks)
--- 6 WORK AREA LINES (style=lined) ---
`424 \ text{cm}^2`
`text{S.A. of 1 face of cube} = 8 xx 8 = 64 \ text{cm}^2`
`text{Height of triangle} = 11 – 8 = 3 \ text{cm}`
`therefore \ text{S.A. (triangular prism)}` | `= 2 xx ( frac{1}{2} xx 8 xx 3 ) + 2 xx (5 xx 8)` |
`= 24 + 80` | |
`= 104 \ text{cm}^2` |
`therefore \ text{Total S.A.}` | `= 5 xx 64 + 104` |
`= 424 \ text{cm}^2` |