Aussie Maths & Science Teachers: Save your time with SmarterEd
Eight cars participate in a competition that lasts for four days. The probability that a car completes a day is `0.7`. Cars that do not complete a day are eliminated.
The Argand diagram shows complex numbers `w` and `z` with arguments `phi` and `theta` respectively, where `phi < theta`. The area of the triangle formed by `0, w` and `z` is `A`.
Show that `z bar w - w bar z = 4iA.` (3 marks)
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A triangle has vertices `A, B` and `C`. The point `D` lies on the interval `AB` such that `AD = 3` and `DB = 5`. The point `E` lies on the interval `AC` such that `AE = 4`, `DE = 3` and `EC = 2`.
(i) `text(In)\ \ Delta ABC and Delta AED`
`/_ BAC = /_ EAD\ \ \ \ text{(common angle)}`
| `(AB)/(AE)` | `= 8/4 = 2/1` |
| `(AC)/(AD)` | `= 6/3 = 2/1` |
| `(AB)/(AE)` | `= (AC)/(AD) = 2/1` |
| `:.\ Delta ABC\ \ text(|||)\ \ Delta AED` | `\ \ \ \ \ text{(sides about equal angles}` |
| `\ \ \ \ \ text{are in the same ratio)}` |
(ii) `/_ ABC = /_ AED\ \ \ \ \ text{(corresponding angles of similar triangles)}`
`/_ AED\ \ text(is an exterior angle of quadrilateral)\ \ BCED`
`:.\ BCED\ \ text(is a cyclic quadrilateral as an exterior)`
`text(angle equals the interior opposite angle.)`
(iii) `cos /_ AED = (3^2 + 4^2 – 3^2)/(2 xx 3 xx 4) = 2/3`
| `cos /_ CED` | `= cos(pi – /_AED)` |
| `=-cos/_AED` | |
| `=-2/3` |
`text(In)\ \ Delta CDE`
| `CD^2` | `= 2^2 + 3^2 – 2 xx 2 xx 3 xx (-2/3)` |
| `= 13 + 8` | |
| `=21` | |
| `:.CD` | `= sqrt 21` |
|
(iv) |
 |
`text(Mark the centre)\ \ O,\ \ text(draw the radii)\ \ OC, OD`
| `text(Let)\ \ /_ CBD` | `= alpha` | |
| `:. /_ COD` | `= 2 alpha` | ` \ \ \ \ text{(angles at circumference and}` |
| `\ \ \ \ text{centre on arc}\ \ CD text{)}` | ||
| `/_ DEC` | `= pi – alpha` | ` \ \ \ \ text{(opposite angles of cyclic}` |
| `\ \ \ \ text{quadrilateral}\ \ BCED text{)}` |
`cos alpha = cos /_ AED = 2/3\ \ \ \ text{(part (iii))}`
`text(Let)\ \ r= text(circle radius)`
`text(In)\ \ Delta DOC`
| `CD^2` | `=r^2 + r^2 – 2r xx r xx cos 2 alpha` |
| `21` | `=2r^2 – 2r^2 (2 cos^2 alpha – 1)` |
| `=2r^2 – 2r^2 (8/9 – 1)` | |
| `=2r^2 + (2r^2)/9` | |
| `=(20r^2)/9` | |
| `r^2` | `=(9 xx 21)/20` |
| `:.r` | `=(3 sqrt 21)/(2 sqrt 5)` |
| `=(3 sqrt 105)/10` |
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Let `z_2 = 1 + i` and, for `n > 2`, let `z_n = z_(n - 1) (1 + i/(|\ z_(n - 1)\ |)).`
Use mathematical induction to prove that `|\ z_n\ | = sqrt n` for all integers `n >= 2.` (3 marks)
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The points `A, B, C` and `D` lie on a circle of radius `r`, forming a cyclic quadrilateral. The side `AB` is a diameter of the circle. The point `E` is chosen on the diagonal `AC` so that `DE _|_ AC`. Let `alpha = /_DAC and beta = /_ACD.`
(i) `/_ ADC = (180 – (alpha + beta))`
| `/_ ABC = alpha + beta\ \ \ \ \ ` | `text{(opposite angles of cyclic}` |
| `text(quadrilateral)\ \ ABCD)` |
`/_ ACB = 90^@\ \ \ text{(angle in a semi-circle)}`
| `(AC)/(AB)` | `= sin /_ ABC` |
| `:. AC` | `= 2 r sin (alpha + beta)` |
| (ii) |  |
`text(In)\ \ Delta ABD`
`/_ ADB = 90^@\ \ \ text{(angle in a semi-circle)}`
| `(AD)/(AB)` | `= cos /_ DAB` |
| `AD` | `= AB cos /_ DAB` |
`text(In)\ \ Delta ABC`
| `/_CAB` | `= 180^@ – (90^@ +alpha + beta)\ \ \ \ text{(angle sum of}\ Delta ABC text{)}` |
| `= 90^@- (alpha + beta)` |
| `=>/_ DAB` | `= alpha + 90^@ – (alpha + beta)` |
| `= 90^@ – beta` |
| `:.AD` | `= AB cos ( 90^@ – beta)` |
| `=2r sin beta` |
| `text(In)\ \ Delta ADE` | |
| `(AE)/(AD)` | `= cos alpha` |
| `:.AE` | `= 2 r cos alpha sin beta\ \ text(… as required)` |
(iii) `text(In)\ \ Delta ADE`
| `sin alpha` | `=(DE)/(2r sin beta)` |
| `DE` | `=2r sin alpha sin beta` |
`text(In)\ \ Delta CDE`
| `tan beta` | `=(DE)/(EC)` |
| `EC` | `=(2r sin alpha sin beta)/tan beta` |
| `=2r sin alpha cos beta` |
| `text(S)text(ince)\ \ AC` | `=AE + ED` |
| `2 r sin (alpha + beta)` | `= 2 r sin beta cos alpha + 2 r sin alpha cos beta` |
| `:.sin (alpha + beta)` | `= sin alpha cos beta + sin beta cos alpha` |
Let `I_n = int_0^1 (1 - x^2)^(n/2)\ dx`, where `n >= 0` is an integer.
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Find `int(ln\ x)/((1 + ln\ x)^2)\ dx`. (3 marks)
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Suppose `n` is a positive integer.
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The diagram shows two circles `C_1` and `C_2` . The point `P` is one of their points of intersection. The tangent to `C_2` at `P` meets `C_1` at `Q`, and the tangent to `C_1` at `P` meets `C_2` at `R`.
The points `A` and `D` are chosen on `C_1` so that `AD` is a diameter of `C_1` and parallel to `PQ`. Likewise, points `B` and `C` are chosen on `C_2` so that `BC` is a diameter of `C_2` and parallel to `PR`.
The points `X` and `Y` lie on the tangents `PR` and `PQ`, respectively, as shown in the diagram.
Copy or trace the diagram into your writing booklet.
| (i) |
|
| `∠APX` | `= ∠ADP\ \ \ text{(angle in alternate segment)`βαγ |
| `∠ADP` | `= ∠DPQ\ \ \ text{(alternate angles,}\ AD\ text(||)\ PQ)` |
| `:.∠APX` | `= ∠DPQ` |
(ii) `text(Join)\ PC\ text(and)\ PB.`
`text{Let}\ \ ∠APX = ∠DPQ=α\ \ \ text{(from part (i))}`
`text{Similarly,}\ \ ∠YPB = ∠CPR=β`
`∠XPY = ∠RPQ=γ\ \ \ text{(vertically opposite angles.)}`
| `∠APD` | `= 90^@\ \ \ text{(angle in a semicircle in}\ C_1)` |
| `∠CPB` | `= 90^@\ \ \ text{(angle in a semicircle in}\ C_2)` |
| `90^@ + 90^@ + 2(α+β+γ)` | `= 360^@ \ \ text{(sum of angles at a point}\ Ptext{)}` |
| `:. (α+β+γ)` | `= 90^@` |
| `∠APC` | `=90+(α+β+γ)=180^@` |
`:.\ A, P\ text(and)\ C\ text(are collinear.)`
(iii) `:.∠APX = ∠CPR\ text{(vertically opposite angles,}\ APC\ text{is a straight line)}`
`:.α=β\ \ =>∠BCA = ∠BDA`
`text(S)text(ince chord)\ AB\ text(subtends a pair of equal angles at)\ C and D`
`=>ABCD\ text(is a cyclic quadrilateral.)`
A toy aeroplane `P` of mass `m` is attached to a fixed point `O` by a string of length `l`. The string makes an angle `ø` with the horizontal. The aeroplane moves in uniform circular motion with velocity `v` in a circle of radius `r` in a horizontal plane.
The forces acting on the aeroplane are the gravitational force `mg`, the tension force `T` in the string and a vertical lifting force `kv^2`, where `k` is a positive constant.
The point `P(a cos theta , b sin theta)` lies on the ellipse `x^2/a^2 + y^2/b^2 = 1`, where `a >b`.
The acute angle between `OP` and the normal to the ellipse at `P` is `ø`.
Let `I_n = int_0^1 (x^(2n))/(x^2 + 1)\ dx`, where `n` is an integer and `n ≥ 0`.
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The rise and fall of the tide is assumed to be simple harmonic, with the time between successive high tides being 12.5 hours. A ship is to sail from a wharf to the harbour entrance and then out to sea. On the morning the ship is to sail, high tide at the wharf occurs at 2 am. The water depths at the wharf at high tide and low tide are 10 metres and 4 metres respectively.
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Show that the earliest possible time that the ship can leave the wharf is 4:05 am. (2 marks)
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The ship takes 20 minutes to sail from the wharf to the harbour entrance and it must be out to sea by 7 am. What is the latest time the ship can leave the wharf? (2 marks)
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i. `text(Period)`
| `(2pi)/n =` | ` 12.5` |
| `12.5n =` | ` 2pi` |
| `25n =` | ` 4pi` |
| `n =` | ` (4pi)/25` |
`text(Amplitude)`
| `a` | `= 1/2(10 − 4)` |
| `= 3` |
`=>\ text(Motion centres around)\ \ x = 7\ \ text(with)`
`y = 10\ \ text(when)\ \ t= 0`
`:.\ text(Water depth is given by)`
| `y` | `= 7+a cos nt` |
| `= 7 + 3 cos\ ((4pit)/(25))\ \ …\ text(as required.)` |
| ii. |
|
`text(Find)\ \ t\ \ text(when)\ \ y = 8.5:`
| `8.5` | `= 7 + 3\ cos\ ((4pit)/25)` |
| `3\ cos\ ((4pit)/25)` | `= 1.5` |
| `cos\ ((4pit)/25)` | `= 1/2` |
| `(4pit)/25` | `=pi/3` |
| `t` | `=(25pi)/(3 xx 4pi)` |
| `=2 1/12` | |
| `= 2\ text(hrs 5 mins)` |
`:.\ text(The earliest time the ship can leave)`
`text(is 4:05 am … as required.)`
iii. `text(2 metres above low tide = 6 m)`
`text(Find)\ t\ text(when)\ y = 6:`
`6 = 7 + 3\ cos\ ((4pit)/(25))`
| `3\ cos\ ((4pit)/(25))` | `= -1` |
| `cos\ ((4pit)/(25))` | `= -1/3` |
| `(4pit)/25` | `= 1.9106…` |
| `:.t` | `= (25 xx 1.9106…)/(4pi)` |
| `= 3.801…` | |
| `= 3\ text{hr 48 min (nearest min)}` |
`:.\ text(At the harbour entrance, water depth)`
`text(of 6 m occurs when)\ \ t = 2\ text(hr 48 min.)`
`:.\ text(Given 20 mins sailing time, the latest the ship can)`
`text(leave the wharf is at)\ \ t = 2\ text(hr 28 min, or 4:28 am.)`
A small paintball is fired from the origin with initial velocity `14` metres per second towards an eight-metre high barrier. The origin is at ground level, `10` metres from the base of the barrier.
The equations of motion are
`x = 14t\ cos\ theta`
`y = 14t\ sin\ theta – 4.9t^2`
where `theta` is the angle to the horizontal at which the paintball is fired and `t` is the time in seconds. (Do NOT prove these equations of motion)
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Hence determine the maximum value of `h`. (2 marks)
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Find the other interval. (2 marks)
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| i. | `x` | `= 14t\ cos\ theta` | `\ \ …\ (1)` |
| `y` | `= 14t\ sin\ theta-4.9t^2` | `\ \ …\ (2)` |
`text(Substitute)\ \ t = x/(14\ cos\ theta)\ \ text{from (1) into (2)}`
| `y` | `= 14(x/(14\ cos\ theta))\ sin\ theta − 4.9(x/(14\ cos\ theta))^2` |
| `= x\ tan\ theta − (4.9/(14^2))((x^2)/(cos^2\ theta))` | |
| `= x\ tan\ theta − (x^2)/40\ sec^2\ theta` | |
| `= x\ tan\ theta − (x^2)/40(1 + tan^2\ theta)` | |
| `= mx − ((1 + m^2)/40)x^2\ \ \ \ \ (text(Given)\ \ m = tan\ theta)` |
ii. `text(Show paintball hits at)\ \ h\ \ text(when)`
`m = 2 ± sqrt(3 − 0.4h)`
`text(i.e.)\ \ y = h\ \ text(when)\ \ x = 10`
| `10m − ((1 + m^2)/40) · 10^2` | `= h` |
| `10m − 5/2(1 + m^2)` | `= h` |
| `20m − 5 − 5m^2` | `= 2h` |
| `5m^2 − 20m + 2h + 5` | `= 0` |
`text(Using the quadratic formula)`
| `m` | `=(20 ± sqrt((-20)^2 − 4 · 5 · (2h + 5)))/(2 · 5)` |
| `= (20 ± sqrt(400 − 40h −100))/10` | |
| `= (20 ± sqrt(300 − 40h))/10` | |
| `= (20 ± 10sqrt(3 − 0.4h))/10` | |
| `= 2 ± sqrt(3 − 0.4h)\ \ \ text(… as required)` |
`text(Find maximum)\ \ h`
| `sqrt(3 − 0.4h)` | `≥ 0` |
| `3 − 0.4h` | `≥ 0` |
| `0.4h` | `≤ 3` |
| `h` | `≤ 7.5` |
`:.\ text(Maximum)\ \ h = 7.5\ text(m)`
| iii. |  |
`text{Using part (ii)}`
`text(When)\ \ h = 3.9`
| `m` | `= 2 ± sqrt(3 − 0.4(3.9))` |
| `= 2 ± sqrt(1.44)` | |
| `= 2 ± 1.2` | |
| `= 3.2\ \ text(or)\ \ 0.8` |
`text(When)\ \ h = 5.9`
| `m` | `= 2 ± sqrt(3 − 0.4(5.9))` |
| `= 2 ± sqrt(0.64)` | |
| `= 2 ± 0.8` | |
| `= 2.8\ \ text(or)\ \ 1.2` |
`:.\ text(The other interval is)\ \ \ 0.8 ≤ m ≤ 1.2`
iv. `text(Find)\ \ x\ \ text(when)\ \ y = 0`
| `mx − ((1 + m^2)/40)x^2` | `= 0` |
| `x[m − ((1 + m^2)/40)x]` | `= 0` |
| `((1 + m^2)/40)x` | `= m,\ \ \ \ x ≠ 0` |
| `:. x` | `= (40m)/(1 + m^2)\ \ …\ text(as required)` |
`text(Consider the interval)\ \ \ 2.8 ≤ m ≤ 3.2`
`text(When)\ \ m = 2.8`
`=> x = (40(2.8))/(1 + 2.8^2) = 12.669…\ text(m)`
`text(When)\ \ m = 3.2`
`=>x = (40(3.2))/(1 + 3.2^2) = 11.387…\ text(m)`
`:.\ text(Landing width interval)`
`= 12.669… − 11.387…`
`= 1.281…`
`= 1.3\ text(m)\ \ text{to 1 d.p.}`
`text(Consider the interval)\ \ \ 0.8 ≤ m ≤ 1.2`
`text(When)\ \ m = 0.8`
`=>x = (40(0.8))/(1 + 0.8^2) = 19.512…\ text(m)`
`text(When)\ \ m = 1.2`
`=>x = (40(1.2))/(1 + 1.2^2) = 19.672…`
`text(S)text(ince interval includes)\ \ m = 1\ \ text(where the)`
`text(paintball has maximum range.)`
`x_(text(max)) = (40(1))/(1 + 1^2) = 20\ text(m)`
`:.\ text(Landing width interval)`
`= 20 − 19.512…`
`= 0.487…`
`= 0.5\ text(m)\ \ \ text{(to 1 d.p.)}`
Consider the function `f(x) = e^x − e^(-x)`.
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A particle moves in a straight line. Its displacement, `x` metres, after `t` seconds is given by
`x = sqrt3\ sin\ 2t − cos\ 2t + 3`.
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The diagram shows a point `P(2ap, ap^2)` on the parabola `x^2= 4ay`. The normal to the parabola at `P` intersects the parabola again at `Q(2aq,aq^2)`.
The equation of `PQ` is `x + py − 2ap − ap^3 = 0`. (Do NOT prove this.)
The points `P` and `Q` lie on the circle with centre `O` and radius `r`. The arc `PQ` subtends an angle `theta` at `O`. The tangent at `P` and the line `OQ` intersect at `T`, as shown in the diagram.
Show that `tan\ theta = 2theta`. (2 marks)
A gutter is to be formed by bending a long rectangular metal strip of width `w` so that the cross-section is an arc of a circle.
Let `r` be the radius of the arc and `2 theta` the angle at the centre, `O`, so that the cross-sectional area, `A`, of the gutter is the area of the shaded region in the diagram on the right.
(i) `text(Show)\ \ A = r^2(theta – sin theta cos theta)`
`text(Area of segment)\ \ OBC`
`= (2 theta)/(2 pi) xx pi r^2`
`= r^2 theta`
| `text(Area of)\ \ Delta OBC` | `= 1/2 ab sin C` |
| `= 1/2 r * r * sin 2 theta` | |
| `= 1/2 r^2 * 2 sin theta cos theta` | |
| `= r^2 sin theta cos theta` |
`:.\ text(Shaded Area (A))`
`= text(Area of segment)\ \ OBC – text(Area of)\ \ Delta OBC`
`= r^2 theta – r^2 sin theta cos theta`
`= r^2 (theta – sin theta cos theta)\ \ text(… as required.)`
(ii) `text(Consider Arc length)\ \ BC`
| `w` | `= (2 theta)/(2 pi) xx 2 pi r` |
| `= 2 theta r` | |
| `:.\ r` | `= w/(2 theta)` |
| `:. A` | `= (w/(2 theta))^2 (theta – sin theta cos theta)` |
| `= w^2/(4 theta) – (w^2 sin theta cos theta)/(4 theta^2)` |
| `:. (dA)/(d theta)` | `= (-w^2)/(4 theta^2) – (w^2/4) [((sin theta xx -sin theta + cos theta * cos theta) theta^2 – 2 theta sin theta cos theta)/theta^4]` |
| `= (-w^2)/(4 theta^2) – (w^2/(4 theta^4))[(cos^2 theta – sin^2 theta) theta^2 – 2 theta sin theta cos theta]` | |
| `= (-w^2)/(4 theta^2) – (w^2/(4 theta^3))[(2 cos^2 theta – 1) theta – 2 sin theta cos theta]` | |
| `= (-w^2)/(4 theta^2) – (w^2/(4 theta^3))[(2 cos^2 theta – 1) theta – 2 sin theta cos theta]` | |
| `= w^2/(4 theta^3) [-theta – 2 theta cos^2 theta + theta + 2 sin theta cos theta]` | |
| `= w^2/(4 theta^3) [2 sin theta cos theta – 2 theta cos^2 theta]` | |
| `= w^2/(2 theta^3) (sin theta cos theta – theta cos^2 theta)` | |
| `= (w^2 cos theta\ (sin theta – theta cos theta))/(2 theta^3)\ \ text(… as required)` |
| (iii) | `g(theta)` | `= sin theta- theta cos theta` |
| `g prime (theta)` | `= cos theta – (theta xx -sin theta + cos theta * 1)` | |
| `= cos theta + theta sin theta – cos theta` | ||
| `= theta sin theta` |
`text(S)text(ince)\ \ 0 < theta < pi,`
`=> sin theta > 0`
`:.\ g prime (theta) > 0`
`g(0) = sin 0 – 0 * cos 0 = 0`
`:.\ text(S)text(ince)\ \ g(0) = 0 and g(theta)\ \ text(is an increasing function)`
`(g prime (theta) > 0)\ \ text(for)\ \ 0 < theta < pi , text(then)\ \ g(theta) > 0.`
(iv) `text(If)\ \ (dA)/(d theta) = 0\ ,\ \ \ 0 < theta < pi`
`(w^2 cos theta\ (sin theta – theta cos theta))/(2 theta^3) = 0`
`text(Consider)`
`(w^2 cos theta)/(2 theta^3) = 0\ ,\ \ theta != 0`
`=>theta = pi/2`
`text(Consider)`
| `sin theta – theta cos theta=` | ` 0` |
| `text(i.e.)\ \ g(theta)=` | ` 0` |
`=>\ text(No solution for)\ \ 0 < theta < pi\ \ text{(using part (iii))}`
`:.\ text(There is only one value of)\ \ theta.`
| (v) `(dA)/(d theta)` | `= (w^2 cos theta\ (sin theta – theta cos theta))/(2 theta^3)` |
| `= (w^2 cos theta * g(theta))/(2 theta^3)` |
`text(If)\ \ theta = pi/4\ ,\ \ g(pi/4) > 0\ ,\ \ cos (pi/4) > 0`
`:.\ (dA)/(d theta) > 0`
`text(If)\ \ theta = (3 pi)/4\ ,\ \ g((3 pi)/4) > 0\ ,\ \ cos ((3 pi)/4) < 0`
`:.\ (dA)/(d theta) < 0`
`:.\ text(Maximum when)\ \ theta = pi/2`
`text(When)\ \ theta = pi/2`
| `A` | `= r^2 (theta – sin theta cos theta)` |
| `= (w/(2 theta))^2 (theta – sin theta cos theta)` | |
| `= w^2/(2^2 xx (pi/2)^2) (pi/2 – sin\ pi/2 cos\ pi/2)` | |
| `= w^2/pi^2 (pi/2)` | |
| `= w^2/(2 pi)\ \ text(u²)` |
In an endurance event, the probability that a competitor will complete the course is `p` and the probability that a competitor will not complete the course is `q = 1 - p.` Teams consist of either two or four competitors. A team scores points if at least half its members complete the course.
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i. `P text{(at least 3 don’t complete)}`
`= P\ text{(3 don’t complete)} + P\ text{(4 don’t complete)}`
`= \ ^4C_3 q^3 p + \ ^4C_4 q^4`
`= 4pq^3 + q^4`
ii. `P text{(4-member team scores)}`
`= 1 – P\ text{(at least 3 don’t complete)}`
`= 1 – 4pq^3 + q^4`
`= 1 – [4 (1 – q) q^3 + q^4]`
`= 1 – (4q^3 – 4q^4 + q^4)`
`= 1 – 4q^3 + 3q^4`
iii. `P text{(2-member team scores)}`
`= 1 – P\ text{(both don’t complete)}`
`= 1 – q*q`
`= 1 – q^2`
iv. `text(A 2-member team is more likely to score when)`
| `1 – q^2` | `> 1 -4q^3 + 3q^4` |
| `3q^4 – 4q^3 + q^2` | `< 0` |
| `q^2 (3q^2 – 4q + 1)` | `< 0` |
| `q^2 (3q – 1) (q – 1)` | `< 0` |
`text(Consider)\ \ (3q – 1) (q – 1) < 0`
`:.\ text(S)text(ince)\ \ q\ \ text(is positive and)\ != 1`
`q^2 (3q – 1) (q – 1) < 0\ \ text(when)`
`1/3 < q < 1.`
Two particles are fired simultaneously from the ground at time `t = 0.`
Particle 1 is projected from the origin at an angle `theta, \ \ 0 < theta < pi/2`, with an initial velocity `V.`
Particle 2 is projected vertically upward from the point `A`, at a distance `a` to the right of the origin, also with an initial velocity of `V.`
It can be shown that while both particles are in flight, Particle 1 has equations of motion:
`x = Vt cos theta`
`y = Vt sin theta -1/2 g t^2,`
and Particle `2` has equations of motion:
`x = a`
`y = Vt -1/2 g t^2.` Do NOT prove these equations of motion.
Let `L` be the distance between the particles at time `t.`
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Show that the distance between the particles in flight is smallest when
`t = (a cos theta)/(2V(1 - sin theta))` and that this smallest distance is `a sqrt ((1 - sin theta)/2).` (3 marks)
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| i. |  |
`text(Show)\ \ L^2 = 2V^2t^2 (1 – sin theta) – 2aVt cos theta + a^2`
`text(Consider)\ \ P_1`
`x_1 = Vt cos theta`
`y_1 = Vt sin theta – 1/2 g t^2`
`text(Consider)\ \ P_2`
`x_2 = a`
`y_2 = Vt -1/2 g t^2`
`text(Let)\ \ d=\ text(Vertical distance between particles)`
`d= Vt -1/2 g t^2 – (Vt sin theta – 1/2 g t^2)`
`d= Vt (1 – sin theta)`
`text(Using Pythagoras:)`
| `L^2` | `= (a – x_1)^2 + d^2` |
| `= (a – Vt cos theta)^2 + V^2t^2 (1 – sin theta)^2` | |
| `= a^2 – 2aVt cos theta + V^2t^2 cos ^2 theta + V^2t^2 (1 – 2 sin theta + sin^2 theta)` | |
| `= a^2 – 2aVt cos theta + V^2 t^2 (cos^2 theta + sin^2 theta + 1 – 2 sin theta)` | |
| `= a^2 – 2aVt cos theta + V^2 t^2 (2 – 2 sin theta)` | |
| `= 2 V^2 t^2 (1 – sin theta) – 2aVt cos theta + a^2\ \ text(… as required.)` |
ii. `L^2 = 2V^2 t^2 (1 – sin theta) – 2a Vt cos theta + a^2`
`(d(L^2))/(dt) = 4 V^2 t\ (1 – sin theta) – 2aV cos theta`
`text(Max or min when)\ \ (d(L^2))/(dt) = 0`
| `4V^2t\ (1 – sin theta)` | `= 2aV cos theta` |
| `t` | `= (2a V cos theta)/(4V^2 (1 – sin theta))` |
| `= (a cos theta)/(2V(1 – sin theta)` |
`(d^2(L^2))/(dt^2) = 4V^2(1 – sin theta) > 0\ \ text(for)\ \ V > 0,\ \ 0 < theta < pi/2`
`:.\ L^2\ \ text(is a minimum)`
`:.\ L\ \ text(is a minimum when)\ \ t = (a cos theta)/(2V (1 – sin theta)`
`text(Show minimum distance is)\ \ a sqrt {(1 – sin theta)/2}`
`text(When)\ \ t = (a cos theta)/(2V(1 – sin theta))`
| `L^2` | `= 2V^2 ((a^2 cos ^2 theta)/(4V^2 (1 – sin theta)^2)) (1 – sin theta)` |
| `\ \ – 2aV ((a cos theta)/(2V (1 – sin theta))) cos theta + a^2` | |
| `= (a^2 cos^2 theta)/(2 (1 – sin theta)) – (a^2 cos^2 theta)/((1 – sin theta)) + a^2` | |
| `= a^2[(cos^2 theta – 2 cos^2 theta + 2 (1 – sin theta))/(2(1 – sin theta))]` | |
| `= a^2[(-cos^2 theta + 2 – 2 sin theta)/(2(1 – sin theta))]` | |
| `= a^2[(-(1 – sin^2 theta) + 2 – 2 sin theta)/(2 (1 – sin theta))]` | |
| `= a^2 [(sin^2 theta – 2 sin theta + 1)/(2(1 – sin theta))]` | |
| `= a^2 [(1 – sin theta)^2/(2 (1 – sin theta))]` | |
| `= a^2 [((1 – sin theta))/2]` | |
| `:.\ L` | `= sqrt ((a^2(1 – sin theta))/2)` |
| `= a sqrt ((1 – sin theta)/2)\ \ text(… as required.)` |
iii. `text(Smallest distance occurs when)`
`t = (a cos theta)/(2V (1 – sin theta)`
`text(If)\ \ P_1\ \ text(is ascending,)\ \ dot y_1 > 0`
`y_1 = Vt sin theta – 1/2 g t^2`
`dot y_1 = V sin theta – g t`
| `:.\ V sin theta – g ((a cos theta)/(2V (1 – sin theta)))` | `> 0` |
| `2V^2 sin theta\ (1 – sin theta) – a g cos theta` | `> 0` |
| `2V^2 sin theta\ (1 – sin theta)` | `> ag cos theta` |
| `V^2` | `> (a g cos theta)/(2 sin theta\ (1 – sin theta))` |
| `V` | `> sqrt ((a g cos theta)/(2 sin theta\ (1 – sin theta)))\ \ text(… as required.)` |
The diagram shows points `A`, `B`, `C` and `D` on a circle. The lines `AC` and `BD` are perpendicular and intersect at `X`. The perpendicular to `AD` through `X` meets `AD` at `P` and `BC` at `Q`.
Copy or trace this diagram into your writing booklet.
| (i) |
|
`text(Prove)\ \ ∠QXB =∠QBX`
`∠ADX = ∠ACB = theta`
`text{(angles in the same segment on arc}\ AB)`
`text(In)\ ΔDPX`
| `∠DPX` | `= 90^@\ \ (PQ ⊥ AD)` |
| `∠PXD` | `= (90 – theta)^@\ \ text{(angle sum of}\ Δ DPX)` |
| `∠QXB` | `= (90 – theta)^@\ \ text{(vertically opposite angle)}` |
`text(In)\ Δ BXC`
| `∠BXC` | `= 90^@\ \ (∠AXC\ text{is a straight angle)}` |
| `∠QBX` | `= (90 – theta)^@\ \ text{(angle sum of}\ ΔBXC)` |
| `:.∠QXB` | `= ∠QBX` |
(ii) `text(Prove)\ \ Q\ \ text(bisects)\ \ BC`
| `BQ = QX\ \ ` | `text{(sides opposite equal angles}` |
| `text{of isosceles}\ Delta BXQ text{)}` |
| `∠QXC` | `= 180 − 90 − (90 − theta)\ \ (∠AXC\ text{is a straight angle)}` |
| `= theta` | |
| `∠XCB` | `=theta\ \ \ text{(from part (i))}` |
| `:. ΔXQC\ text(is isosceles)` | |
| `QX = QC\ \ ` | `text{(sides opposite base angles}` |
| `text{of isosceles}\ ΔQXC)` |
`:. BQ = QC`
`:. Q\ text(bisects)\ BC`
In a large city, 10% of the population has green eyes.
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A fire hose is at ground level on a horizontal plane. Water is projected from the hose. The angle of projection, `theta`, is allowed to vary. The speed of the water as it leaves the hose, `v` metres per second, remains constant. You may assume that if the origin is taken to be the point of projection, the path of the water is given by the parametric equations
`x = vt\ cos\ theta`
`y = vt\ sin\ theta − 1/2 g t^2`
where `g\ text(ms)^(−2)` is the acceleration due to gravity. (Do NOT prove this.)
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This fire hose is now aimed at a 20 metre high thin wall from a point of projection at ground level 40 metres from the base of the wall. It is known that when the angle `theta` is 15°, the water just reaches the base of the wall.
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| i. | `x` | `= vt\ cos\ theta` |
| `y` | `= vt\ sin\ theta − 1/2 g t^2` |
`text(Find)\ \ t\ \ text(when)\ \ y = 0`
| `vt\ sin\ theta − 1/2 g t^2` | `= 0` |
| `t(v\ sin\ theta − 1/2 g t)` | `= 0` |
| `v\ sin\ theta − 1/2 g t` | `= 0, \ \ t ≠ 0` |
| `1/2 g t` | `= v\ sin\ theta` |
| `t` | `= (2v\ sin\ theta)/g` |
`text(Find)\ \ x\ \ text(when)\ \ t = (2v\ sin\ theta)/g`
| `x` | `= v · (2v\ sin\ theta)/g\ cos\ theta` |
| `= (v^2 · \ 2\ sin\ theta\ cos\ theta)/g` | |
| `= (v^2\ sin\ 2theta)/g` |
`:.\ text(When)\ \ x = (v^2\ sin\ 2theta)/g,\ \ \text(the water returns)`
`text(to ground level … as required.)`
ii. `text(Show)\ \ v^2 = 80\ text(g)`
`text(When)\ \ theta = 15^@, \ x = 40`
`text{From part (i)}`
| `40` | `= (v^2\ sin\ 30^@)/g` |
| `v^2 xx 1/2` | `= 40g` |
| `v^2` | `= 80g\ \ \ …text(as required)` |
iii. `text(Show)\ \ y = x\ tan\ theta − (x^2\ sec^2\ theta)/160`
| `x` | `= vt\ cos\ theta` |
| `:.t` | `= x/(v\ cos\ theta)` |
`text(Substitute into)`
| `y` | `= vt\ sin\ theta − 1/2 g t^2` |
| `= v · x/(v\ cos\ theta) · sin\ theta −1/2 g (x/(v\ cos\ theta))^2` | |
| `= x\ tan\ theta − 1/2 g (x^2/(v^2\ cos^2\ theta))` | |
| `= x\ tan\ theta − 1/2 g · x^2/(80g\ cos^2\ theta)` | |
| `= x\ tan\ theta − (x^2\ sec^2\ theta)/160\ \ \ …text(as required.)` |
iv. `text(Water clears the top if)\ \ y = 20\ \ text(when)\ \ x = 40`
`text{Substitute into equation from (iii)}`
| `40\ tan\ theta − (40^2\ sec^2\ theta)/160` | `= 20` |
| `40\ tan\ theta − 10\ sec^2\ theta` | `= 20` |
| `40\ tan\ theta − 10(1 + tan^2\ theta)` | `= 20` |
| `40\ tan\ theta − 10 − 10\ tan^2\ theta` | `= 20` |
| `10\ tan^2\ theta − 40\ tan\ theta\ + 30` | `= 0` |
| `tan^2\ theta − 4\ tan\ theta + 3` | `= 0\ \ \ text(… as required)` |
| v. |
|
`text(Water hits the bottom of the wall when)`
`x = 40\ \ text(and)\ \ y = 0`
| `40\ tan\ theta − (40^2\ sec^2\ theta)/160` | `= 0` |
| `40\ tan\ theta − 10\ sec^2\ theta` | `= 0` |
| `4\ tan\ theta − (1 + tan^2\ theta)` | `= 0` |
| `tan^2\ theta − 4\ tan\ theta + 1` | `= 0` |
`text(Using the quadratic formula)`
| `tan\ theta` | `= (+4 ± sqrt(16 − 4 · 1 · 1))/ (2 xx 1)` |
| `= (4 ± sqrt12)/2` | |
| `= 2 ± sqrt3` | |
| `theta` | `= 15^@\ \ text(or)\ \ 75^@` |
`text(Water hits the top of the wall when)`
`x = 40\ text(and)\ \ y = 20`
| `tan^2\ theta − 4\ tan\ theta + 3` | `= 0\ \ \ text{from (iv)}` |
| `(tan\ theta − 1)(tan\ theta − 3)` | `= 0` |
| `tan\ theta` | `= 1` | `text(or)` | `tan\ theta` | `= 3` |
| `theta` | `= 45^@` | `theta` | `= tan^(−1)\ 3` | |
| `= 71.565…` | ||||
| `= 71.6^@\ \ \text{(to 1 d.p.)}` |
`:.\ text(Following the motion of the water as)\ \ theta\ \ text(increases,)`
`text(water hits the wall when)`
| `15^@ ≤ theta ≤ 45^@` | `\ text(and)` |
| `71.6^@ ≤ theta ≤ 75^@` |
The points `A, \ B, \ C` and `D` are placed on a circle of radius `r` such that `AC` and `BD` meet at `E`. The lines `AB` and `DC` are produced to meet at `F`, and `BECF` is a cyclic quadrilateral.
Copy or trace this diagram into your writing booklet.
| (i) |
|
`∠ACD = ∠ABD = theta\ \ \ text{(angles in the same segment on arc}\ AD)`
`text(Consider cyclic quad)\ \ BECF`
| `∠DBF = ∠ECD = theta` | `\ \ \ text{(exterior angle of cycle quad}` |
| `\ \ \ text{equals its interior opposite)}` |
| `2theta` | `= 180^@\ \ \ (∠ABF\ text{is a straight angle)}` |
| `theta` | `= 90^@` |
| `:.∠DBF` | `= 90^@` |
(ii) `text(S)text(ince)\ AD\ text(subtends right angles)\ \ ∠ABD\ \ text(and)\ \ ∠ACD`
`text{(from part (i))}`
`⇒ AD\ \ text(is a diameter)`
`:.AD = 2r`
The diagram below shows a sketch of the graph of `y = f(x)`, where `f(x) = 1/(1 + x^2)` for `x ≥ 0`.
| (i) |
|
(ii) `text(Domain of)\ \ f^(−1)(x)\ text(is)`
`0 < x ≤ 1`
(iii) `f(x) = 1/(1 + x^2)`
`text(Inverse: swap)\ \ x↔y`
| `x` | `= 1/(1 + y^2)` |
| `x(1 + y^2)` | `= 1` |
| `1 + y^2` | `= 1/x` |
| `y^2` | `= 1/x − 1` |
| `= (1 − x)/x` | |
| `y` | `= ± sqrt((1 − x)/x)` |
`:.y = sqrt((1 − x)/x), \ \ y >= 0`
(iv) `P\ \ text(occurs when)\ \ f(x)\ \ text(cuts)\ \ y = x`
`text(i.e. where)`
| `1/(1 + x^2)` | `= x` |
| `1` | `= x(1 + x^2)` |
| `1` | `= x + x^3` |
| `x^3 + x − 1` | `= 0` |
`=>\ text(S)text(ince)\ α\ text(is the)\ x\ text(-coordinate of)\ P,`
`text(it is a root of)\ \ \ x^3 + x − 1 = 0`
| (v) | `f(x)` | `= x^3 + x − 1` |
| `f′(x)` | `= 3x^2 + 1` | |
| `f(0.5)` | `= 0.5^3 + 0.5 − 1` | |
| `= −0.375` | ||
| `f′(0.5)` | `= 3 xx 0.5^2 + 1` | |
| `= 1.75` |
| `α_2` | `= α_1 − (f(0.5))/(f′(0.5))` |
| `= 0.5 − ((−0.375))/1.75` | |
| `= 0.5 − (−0.2142…)` | |
| `= 0.7142…` | |
| `= 0.714\ \ \ text{(to 3 d.p.)}` |
A particle is moving along the `x`-axis, starting from a position `2` metres to the right of the origin (that is, `x = 2` when `t = 0`) with an initial velocity of `5\ text(ms)^(−1)` and an acceleration given by
`ddot x = 2x^3 + 2x`.
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Katie is one of ten members of a social club. Each week one member is selected at random to win a prize.
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The two points `P(2ap, ap^2)` and `Q(2aq, aq^2)` are on the parabola `x^2 = 4ay`.
| (i) |
|
`text(Show)\ \ R(a(prq), apq)`
`text(T)text(angent equations)`
| `y` | `= px − ap^2` | `\ \ …\ (1)` |
| `y` | `= qx − aq^2` | `\ \ …\ (2)` |
`text(Substitute)\ \y = px − ap^2\ \text(into)\ (2)`
| `px − ap^2` | `= qx − aq^2` |
| `px − qx` | `= ap^2 − aq^2` |
| `x(p − q)` | `= a(p^2 − q^2)` |
| `= a(p + q)(p − q)` | |
| `:.x` | `= a(p + q)` |
`text(Substitute)\ \x = a(p + q)\ \ text(into)\ (1)`
| `y` | `= p xx a(p + q) − ap^2` |
| `= ap^2 + apq − ap^2` | |
| `= apq` |
`:.R(a(p+q), apq)\ \ \ …text(as required.)`
(ii) `text(If)\ \ ∠POQ\ text(is a right angle)`
`M_(PO) xx M_(OQ) = −1`
| `M_(PO)` | `= (ap^2 − 0)/(2ap − 0)` |
| `= p/2` | |
| `M_(OQ)` | `= (aq^2 − 0)/(2aq − 0)` |
| `= q/2` |
| `:.p/2 xx q/2` | `= −1` |
| `pq` | `= −4` |
`⇒R\ \ text(has coordinates)\ \ (a(p+q), −4a)`
`:.\ text(Locus of)\ R\ text(is)\ \ y = −4a`
The length of each edge of the cube `ABCDEFGH` is 2 metres. A circle is drawn on the face `ABCD` so that it touches all four edges of the face. The centre of the circle is `O` and the diagonal `AC` meets the circle at `X` and `Y`.
| (i) |
|
`text(S)text(ince)\ \ FA, \ AC\ \ text(and)\ \ FC\ \ text(are all)`
`text(diagonals of sides of a cube,)`
`FA = AC = FC`
`:.ΔFAC\ \ text(is equilateral)`
`:.∠FAC = 60^@`
| (ii) |
|
`text(In)\ \ ΔAEF`
| `AF^2` | `= EF^2 + EA^2` |
| `= 2^2 + 2^2` | |
| `= 8` | |
| `AF` | `= sqrt8` |
| `= 2sqrt2` |
`text(In)\ \ ΔAFO`
| `sin\ 60^@` | `= (FO)/(AF)` |
| `sqrt3/2` | `= (FO)/(2sqrt2)` |
| `FO` | `= sqrt3/2 xx 2sqrt2` |
| `= sqrt6\ text(metres … as required.)` |
| (iii) |
|
`XY\ \ text(is the diameter of a circle AND the width)`
`text(of the cube.)`
| `:.XY` | `= 2` |
| `:.OX` | `= OY = 1` |
| `tan\ ∠OFX` | `=1 /sqrt6` |
| `∠OFX` | `= 22.207…^@` |
| `:.∠XFY` | `= 2 xx 22.407…` |
| `= 44.415…` | |
| `= 44^@\ text{(nearest degree)}` |
The line `AT` is the tangent to the circle at `A`, and `BT` is a secant meeting the circle at `B` and `C`.
Given that `AT = 12`, `BC = 7` and `CT = x`, find the value of `x`. (2 marks)
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A hemispherical bowl of radius `r\ text(cm)` is initially empty. Water is poured into it at a constant rate of `k\ text(cm³)` per minute. When the depth of water in the bowl is `x\ text(cm)`, the volume, `V\ text(cm³)`, of water in the bowl is given by
`V = pi/3 x^2 (3r - x).` (Do NOT prove this)
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A particle is moving so that `ddot x = 18x^3 + 27x^2 + 9x.`
Initially `x = – 2` and the velocity, `v`, is `– 6.`
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A particle is undergoing simple harmonic motion on the `x`-axis about the origin. It is initially at its extreme positive position. The amplitude of the motion is 18 and the particle returns to its initial position every 5 seconds.
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The cubic polynomial `P(x) = x^3 + rx^2 + sx + t`. where `r, \ s` and `t` are real numbers, has three real zeros, `1, alpha` and `-alpha.`
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The points `P, Q` and `T` lie on a circle. The line `MN` is tangent to the circle at `T` with `M` chosen so that `QM` is perpendicular to `MN`. The point `K` on `PQ` is chosen so that `TK` is perpendicular to `PQ` as shown in the diagram.
| (i) |  |
`/_ QMT = 90^@\ \ \ (QM _|_ MN)`
`/_ QKT = 90^@\ \ \ (PQ _|_ TK)`
`:.\ /_ QMT + /_ QKT = 180^@`
`:.\ QKTM\ \ text(is cyclic)\ \ text{(opposite angles are supplementary)}`
`text(… as required.)`
(ii) `text(Show)\ \ /_ KMT = /_ KQT`
`/_ KMQ = /_ KTQ = theta`
`text{(angles in the same segment on arc}\ \ KQ text{)}`
| `/_ KQT` | `= 90 – theta\ \ text{(angle sum of}\ \ Delta KQT text{)}` |
| `/_ KMT` | `= /_ QMT – /_ KMQ` |
| `= 90 – theta` |
`:.\ /_ KMT = /_ KQT\ \ text(… as required.)`
(iii) `text(Show)\ \ MK\ text(||)\ TP`
| `/_ NTP` | `= /_ KQT\ \ text{(angle in alternate segment)` |
| `= 90 – theta` |
`:.\ /_ NTP = /_ KMT\ \ text{(from part (ii))}`
`:.\ MK\ text(||)\ TP\ \ text{(corresponding angles are equal)}`
Sophie has five coloured blocks: one red, one blue, one green, one yellow and one white. She stacks two, three, four or five blocks on top of one another to form a vertical tower.
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The points `P(2ap, ap^2), Q(2aq, aq^2)` and `R(2ar, ar^2)` lie on the parabola `x^2 = 4ay`. The chord `QR` is perpendicular to the axis of the parabola. The chord `PR` meets the axis of the parabola at `U`.
The equation of the chord `PR` is `y = 1/2(p + r)x - apr.` (Do NOT prove this.)
The equation of the tangent at `P` is `y = px - ap^2.` (Do NOT prove this.)
An experimental rocket is at a height of 5000 m, ascending with a velocity of ` 200 sqrt 2\ text(m s)^-1` at an angle of 45° to the horizontal, when its engine stops.
After this time, the equations of motion of the rocket are:
`x = 200t`
`y = -4.9t^2 + 200t + 5000,`
where `t` is measured in seconds after the engine stops. (Do NOT show this.)
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i.
`y = -4.9t^2 + 200t + 5000`
`dot y = -9.8t + 200`
`text(Max height occurs when)\ \ dot y = 0`
| `9.8t` | `= 200` |
| `t` | `= 20.408…` |
| `= 20.4\ text{seconds (to 1 d.p.)}` |
`text(When)\ t = 20.408…`
| `y` | `= -4.9 (20.408…)^2 + 200 (20.408…) + 5000` |
| `= 7040.816…` | |
| `= 7041\ text{m (to nearest metre)}` |
`:.\ text(The rocket will reach a maximum height of)`
`text(7041 metres when)\ \ t = 20.4\ text(seconds.)`
ii. `text(The rocket is descending at 45° at point)\ A\ text(on the graph.)`
`text(The symmetry of the parabolic motion means that)\ \ A`
`text(occurs when)\ \ t = 2 xx text(time to reach max height, or)`
`40.8\ text(seconds.)`
`text(Point)\ B\ text(occurs when the rocket is descending at 60°)`
`text(to the horizontal.)`
`text(At)\ \ B,`
`-tan 60° = (dot y)/(dot x)`
`text{(The gradient of the projectile becomes}`
`text{negative after reaching its max height.)}`
`dot y = -9.8t + 200`
`dot x = d/(dt) (200t) = 200`
| `:.\ – sqrt 3` | `= (-9.8t + 200)/200` |
| `-200 sqrt 3` | `= -9.8t + 200` |
| `9.8t` | `= 200 + 200 sqrt 3` |
| `t` | `= (200 + 200 sqrt 3)/9.8` |
| `= (546.410…)/9.8` | |
| `= 55.756…` | |
| `= 55.8\ text{seconds (nearest second)}` |
`:.\ text(The pilot can operate the ejection seat)`
`text(between)\ \ t = 40.8 and t = 55.8\ text(seconds.)`
|
iii. |
 |
`v^2 = (dot x)^2 + (dot y)^2`
`text(When)\ \ v = 350`
| `350^2` | `= 200^2 + (-9.8t + 200)^2` |
| `122\ 500` | `= 40\ 000 + (-9.8t + 200)^2` |
| `(-9.8t + 200)^2` | `= 82\ 500` |
| `-9.8t + 200` | `= +- sqrt (82\ 500)` |
| `9.8t` | `= 200 +- sqrt (82\ 500)` |
| `t` | `= (200 +- sqrt (82\ 500))/9.8` |
| `= 49.717…\ \ \ (t > 0)` | |
| `= 49.7\ text{seconds (to 1 d.p.)}` |
`:.\ text(The latest time the pilot can eject safely)`
`text(is when)\ \ t = 49.7\ text(seconds.)`
There are five matches on each weekend of a football season. Megan takes part in a competition in which she earns one point if she picks more than half of the winning teams for a weekend, and zero points otherwise. The probability that Megan correctly picks the team that wins any given match is `2/3`.
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A particle moves in a straight line and its position at time `t` is given by
`x = 5 + sqrt 3 sin3t - cos 3t.`
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Use the principle of mathematical induction to show that `4^n - 1 - 7n > 0` for all integers `n >= 2.` (3 marks)
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The points `P (2ap, ap^2)` and `Q (2aq, aq^2)` lie on the parabola `x^2 = 4ay`.
The equation of the normal to the parabola at `P` is `x + py = 2ap + ap^3` and the equation of the normal at `Q` is similarly given by `x + qy = 2aq + aq^3.`
(i) `text(Show)\ \ R\ \ text(is)\ \ (–apq [p + q], a [p^2 + pq + q^2 + 2])`
`text(Equations of normals through)\ \ P and Q`
| `x + py` | `= 2ap + ap^3` | `\ \ text{… (1)}` |
| `x + qy` | `= 2aq + aq^3` | `\ \ text{… (2)}` |
`text{Multiply (1)} xx q\ ,\ \ text{(2)} xx p`
| `qx + pqy` | `= 2apq + ap^3q` | `\ \ text{… (3)}` |
| `px + pqy` | `= 2apq + apq^3` | `\ \ text{… (4)}` |
`text{Subtract (4) – (3)}`
| `x (p – q)` | `= apq^3 – ap^3q` |
| `= apq (q^2 – p^2)` | |
| `= apq (q – p) (q + p)` | |
| `= -apq (p -q) (p + q)` | |
| `x` | `= -apq [p + q]` |
`text(Substitute)\ \ x = -apq [p + q]\ \ text{into (1)}`
| `py – apq[p + q]` | `= 2ap + ap^3` |
| `py` | `= 2ap + ap^3 + apq (p + q)` |
| `y` | `= 2a + ap^2 + aq (p + q)` |
| `= 2a + ap^2 + apq + aq^2` | |
| `= a[p^2 + pq + q^2 + 2]` |
`:.\ R\ \ text(is)\ \ (–apq [p + q], a [p^2 + pq + q^2 + 2])`
`text(… as required.)`
(ii) `PQ\ \ text(is)\ \ y = 1/2 (p + q)x – apq`
`text(Passes)\ \ (0, a)`
| `a` | `= 1/2 (p + q)0 – apq` |
| `a` | `= -apq` |
| `:. pq` | `= -1\ \ text(… as required)` |
(iii) `R\ \ text(has coordinates)`
| `x` | `= -apq [p + q]` |
| `x` | `= a(p + q)\ \ \ text{(using part (ii))}` |
| `x/a` | `=(p + q)` |
| `y` | `= a [p^2 + pq + q^2 + 2]` |
| `= a (p^2 – 1 + q^2 + 2)` | |
| `= a (p^2 + q^2 +1)` | |
| `= a [(p + q)^2 – 2pq + 1]` | |
| `= a [(p + q)^2 + 3]` | |
| `= a [(x/a)^2 + 3]` | |
| `= x^2/a + 3a` |
`:.\ text(Locus of)\ \ R\ \ text(is)\ \ y = x^2/a + 3a`
By making the substitution `t = tan\ theta/2`, or otherwise, show that
`text(cosec)\ theta + cot theta = cot\ theta/2.` (2 marks)
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`text(Show cosec)\ theta + cot\ theta = cot\ theta/2`
`tan\ theta/2=t`
| `=>sin theta` | `= 2 sin\ theta/2\ cos\ theta/2= (2t)/(t^2 +1)` |
| `=>cos theta` | `= cos^2\ theta/2 – sin^2\ theta/2= (1 – t^2)/(t^2 + 1)` |
| `=>tan theta` | `= (sin theta)/(cos theta)= (2t)/(1 – t^2)` |
| `text(LHS)` | `= 1/(sin theta) + 1/(tan theta)` |
| `= (t^2 + 1)/(2t) + (1 – t^2)/(2t)` | |
| `= (t^2 + 1 + 1 – t^2)/(2t)` | |
| `= 1/t` | |
| `= 1/(tan\ theta/2)` | |
| `= cot\ theta/2` | |
| `=\ text(RHS … as required.)` |
In the circle centred at `O` the chord `AB` has length `7`. The point `E` lies on `AB` and `AE` has length `4`. The chord `CD` passes through `E`.
Let the length of `CD` be `l` and the length of `DE` be `x`.
(i) `text(Show)\ \ x^2 – lx + 12 = 0`
| `AB` | `= 7` |
| `:.\ EB` | `= 7 – 4 = 3` |
`AE xx EB = ED xx CE`
`text{(intercepts of intersecting chords)}`
| `4 xx 3` | `= x (l – x)` |
| `12` | `= xl – x^2` |
`:.\ x^2 – lx + 12 = 0\ \ text(… as required.)`
(ii) `x^2 – lx + 12 = 0\ \ text(has a solution)`
`text(When)\ \ Delta >= 0`
| `b^2 – 4ac` | `>= 0` |
| `(-l)^2 – 4 * 1 * 12` | `>= 0` |
| `l^2 – 48` | `>= 0` |
| `l^2` | `>= 48` |
| `l` | `>= sqrt 48` |
| `>= 4 sqrt 3` |
`:.\ text(The shortest chord that could pass through)`
`E\ \ text(is)\ \ 4 sqrt 3\ text(units.)`
Use the binomial theorem to find the term independent of `x` in the expansion of
`(2x - 1/x^2)^12.` (3 marks)
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The noise level, `N`, at a distance `d` metres from a single sound source of loudness `L` is given by the formula
`N = L/d^2.`
Two sound sources, of loudness `L_1` and `L_2` are placed `m` metres apart.
The point `P` lies on the line between the sound sources and is `x` metres from the sound source with loudness `L_1.`
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Find an expression for `x` in terms of `m`, `L_1` and `L_2` if `P` is chosen to be this point. (4 marks)
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| i. |  |
`N = L/d^2`
| `text(Noise from)\ L_1` | `= L_1/x^2` |
| `text(Noise from)\ L_2` | `= L_2/(m-x)^2` |
| `:. N` | `= L_1/x^2 + L_2/(m-x)^2` |
ii. `N = L_1\ x^-2 + L_2 (m – x)^-2`
| `(dN)/(dx)` | `= -2 L_1 x^-3 + -2 L_2 (m – x)^-3 xx d/(dx) (m – x)` |
| `= (-2 L_1)/x^3 + (2 L_2)/(m – x)^3` |
`text(Max or min when)\ (dN)/(dx) = 0`
| `(2 L_1)/x^3` | `= (2 L_2)/(m – x)^3` |
| `2 L_1 (m – x)^3` | `= 2 L_2\ x^3` |
| `L_1 (m – x)^3` | `= L_2\ x^3` |
| `root 3 L_1 (m – x)` | `= root 3 L_2\ x` |
| `root 3 L_1\ m – root 3 L_1\ x` | `= root 3 L_2\ x` |
| `root 3 L_2\ x + root 3 L_1\ x` | `= root 3 L_1\ m` |
| `x (root 3 L_2 + root 3 L_1)` | `= root 3 L_1\ m` |
| `x` | `= (root 3 L_1\ m)/{(root 3 L_2 + root 3 L_1)}` |
| `(dN)/(dx)` | `= -2 L_1\ x^-3 + 2 L_2 (m – x)^-3` |
| `(d^2N)/(dx^2)` | `= 6 L_1\ x^-4 – 6 L_2 (m – x)^-4 xx -1` |
| `= (6 L_1)/x^4 + (6 L_2)/(m – x)^4 > 0` |
`:.\ text(A minimum occurs when)`
`x = (root 3 L_1\ m)/{(root 3 L_2 + root 3 L_1)}`
An object is moving on the `x`-axis. The graph shows the velocity, `(dx)/(dt)`, of the object, as a function of time, `t`. The coordinates of the points shown on the graph are `A (2, 1), B (4, 5), C (5, 0) and D (6, –5)`. The velocity is constant for `t >= 6`.
|
(i) |
 |
`text(Distance travelled)`
`= int_0^4 (dx)/(dt)\ dt`
`~~ h/3 [y_0 + 4y_1 + y_2]`
`~~ 2/3 [0 + 4 (1) + 5]`
`~~ 2/3 [9]`
`~~ 6\ \ text(units)`
(ii) `text(Displacement is reducing when the velocity is negative.)`
`:. t > 5\ \ text(seconds)`
(iii) `text(At)\ B,\ text(the displacement) = 6\ text(units)`
`text(Considering displacement from)\ B\ text(to)\ D.`
`text(S)text(ince the area below the graph from)`
`B\ text(to)\ C\ text(equals the area above the)`
`text(graph from)\ C\ text(to)\ D,\ text(there is no change)`
`text(in displacement from)\ B\ text(to)\ D.`
`text(Considering)\ t >= 6`
`text(Time required to return to origin)`
| `t` | `= d/v` |
| `= 6/5` | |
| `= 1.2\ \ text(seconds)` |
`:.\ text(The particle returns to the origin after 7.2 seconds.)`
|
(iv) |
|
Mr and Mrs Caine each decide to invest some money each year to help pay for their son’s university education. The parents choose different investment strategies.
Mr Caine makes 18 yearly contributions of $1000 into an investment fund. He makes his first contribution on the day his son is born, and his final contribution on his son’s seventeenth birthday. His investment earns 6% compound interest per annum.
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Mrs Caine makes her contributions into another fund. She contributes $1000 on the day of her son’s birth, and increases her annual contribution by 6% each year. Her investment also earns 6% compound interest per annum.
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A pack of 52 cards consists of four suits with 13 cards in each suit.
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In the diagram, `AE` is parallel to `BD`, `AE = 27`, `CD = 8`, `BD = p`, `BE = q` and `/_ABE`, `/_BCD` and `/_BDE` are equal.
Copy or trace this diagram into your writing booklet.
|
(i) |
 |
`text(Prove)\ Delta ABE\ text(|||)\ Delta BCD`
`/_ ABE = /_ BCD\ \ \ text{(given)}`
`/_ EAB = /_ DBC\ \ \ text{(corresponding angles,}\ AE\ text(||)\ BD text{)}`
`:. Delta ABE\ text(|||)\ Delta BCD\ \ \ text{(equiangular)}`
(ii) `text(Prove)\ Delta EDB\ text(|||)\ Delta BCD`
`/_ EDB = /_ BCD\ \ \ text{(given)}`
`/_ CDB = 180° – (/_ BCD + /_ DBC)\ \ \ text{(Angle sum of}\ Delta BCD text{)}`
| `/_ EBD` | `= 180° – (/_ ABE + /_ DBC)\ \ \ text{(}/_ ABC\ text{is a straight angle)}` |
| `= 180° – (/_ BCD + /_ DBC)\ \ \ text{(}/_ ABE = /_ BCD,\ text{given)}` | |
| `= /_ CDB` |
`:. Delta EDB\ text(|||)\ Delta BCD\ \ \ text{(equiangular)}`
(iii) `text(In a GP,)\ \ r = T_n/T_(n-1)`
`text(If)\ \ \ 8, p, q, 27\ \ \ text(are 1st 4 terms of a GP)`
`=> p/8 = q/p = 27/q .`
`text(S)text(ince corresponding sides of similar)`
`text(triangles are in the same ratio)`
`(BD)/(DC) = (EB)/(BD) = (AE)/(EB)`
`p/8 = q/p = 27/q .`
`:. 8, p, q, 27\ \ text(are 1st 4 terms of a GP.)`
(iv) `8, p, q, 27`
`a = 8`
`text(Using)\ \ T_n = ar^(n-1)`
`T_4=ar^3`
| `27` | `= 8 xx r^3` |
| `r^3` | `= 27/8` |
| `r` | `= 3/2` |
| `T_2` | `=ar` |
| `:.p` | `= 8 * 3/2` |
| `= 12` |
| `T_3` | `=ar^2` |
| `:.q` | `= 8 * (3/2)^2` |
| `= 18` |
One model for the number of mobile phones in use worldwide is the exponential growth model,
`N = Ae^(kt)`,
where `N` is the estimate for the number of mobile phones in use (in millions), and `t` is the time in years after 1 January 2008.
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A particle is moving on the `x`-axis and is initially at the origin. Its velocity, `v` metres per second, at time `t` seconds is given by
`v = (2t)/(16 + t^2).`
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An advertising logo is formed from two circles, which intersect as shown in the diagram.
The circles intersect at `A` and `B` and have centres at `O` and `C`.
The radius of the circle centred at `O` is 1 metre and the radius of the circle centred at `C` is `sqrt 3` metres. The length of `OC` is 2 metres.
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|
i. |
 |
`text(In)\ Delta AOC`
| `AO^2 + AC^2` | `= 1^2 + sqrt 3^2` |
| `=1 + 3` | |
| `= 4` | |
| `= OC^2` |
`:. Delta AOC\ \ text(is right-angled and)\ \ /_OAC = pi/2`
| ii. `sin\ /_ACO` | `= 1/2` |
| `:. /_ACO` | `= pi/6` |
| `sin\ /_AOC` | `= sqrt 3/2` |
| `:. /_AOC` | `= pi/3` |
iii. `text(Area)\ AOBC`
`= 2 xx text(Area)\ Delta AOC`
`= 2 xx 1/2 xx b xx h`
`= 2 xx 1/2 xx 1 xx sqrt 3`
`= sqrt 3\ \ text(m²)`
iv. `/_ACB = pi/6 + pi/6 = pi/3`
| `:. /_ACB\ text{(reflex)}` | `= 2 pi – pi/3` |
| `= (5 pi)/3` |
`text(Area of major sector)\ ACB`
`= theta/(2 pi) xx pi r^2`
`= {(5 pi)/3}/(2 pi) xx pi(sqrt 3)^2`
`= (5 pi)/6 xx 3`
`= (5 pi)/2\ text(m²)`
v. `/_AOB = pi/3 + pi/3 = (2 pi)/3`
| `:. /_AOB\ text{(reflex)}` | `= 2 pi – (2 pi)/3` |
| `= (4 pi)/3` |
`text(Area of major sector)\ AOB`
`= {(4 pi)/3}/(2 pi) xx pi xx 1^2`
`= (2 pi)/3\ text(m²)`
`:.\ text(Total area of the logo)`
`= (5 pi)/2 + (2 pi)/3 + text(Area)\ AOBC`
`= (15 pi + 4 pi)/6 + sqrt 3`
`= ((19 pi + 6 sqrt 3)/6)\ text(m²)`
Let `f(x) = Ax^3 - Ax + 1`, where `A > 0`.
(i) `f(x) = Ax^3 – Ax + 1\ \ ,\ \ \ A > 0`
`f prime (x) = 3Ax^2 – A`
`text(S.P.’s when)\ \ f prime (x) = 0`
| `3Ax^2 – A` | `= 0` |
| `A (3x^2 – 1)` | `= 0` |
| `3x^2` | `= 1` |
| `x^2` | `= 1/3` |
| `x` | `= +- 1/sqrt3 xx (sqrt 3)/(sqrt 3)` |
| `= +- (sqrt 3)/3\ \ text(… as required)` |
(ii) `text(Consider)\ \ f(x)\ \ text(with only 1 zero.)`
`f(0) = 1`
`:.\ f(x)\ \ text(passes through)\ \ (0, 1)\ \ text(and has)`
`text(turning points either side at)\ \ x = +- (sqrt 3)/3.`
`text(Given 1 zero)`
`=> f((sqrt 3)/3) > 0`
| `A ((sqrt 3)/3)^3 – A ((sqrt 3)/3) + 1` | `> 0` |
| `A((3 sqrt 3)/27) – A ((9 sqrt 3)/27)` | `> -1` |
| `A ((-6 sqrt 3)/27)` | `> -1` |
| `A` | `< 27/(6 sqrt 3)` |
| `A` | `< 9/(2 sqrt 3) xx (sqrt 3)/(sqrt 3)` |
| `A` | `< (9 sqrt 3)/6` |
| `A` | `< (3 sqrt 3)/2\ \ text(… as required.)` |
(iii) `text(S)text(ince 1 zero occurs when)\ \ A < (3 sqrt 3)/2\ \ text{(part (ii))}`
`=> text(Minimum TP when)\ \ x = (sqrt3)/3`
`=> text(Maximum TP when)\ \ x = -(sqrt 3)/3\ \ text{(see graph)}`
`text(S)text(ince)\ \ f(-1) = 1`
`:. f(x)\ \ text(cannot have a zero for)\ \ \ -1 <= x <= 1`
(iv) `g(theta) = 2 cos theta + tan theta,\ \ \ \ \ \ -pi/2 < theta < pi/2`
`g prime (theta) = -2 sin theta + sec^2 theta`
`text(S.P.’s when)\ \ g prime (theta) = 0`
| `-2 sin theta + sec^2 theta` | `= 0` |
| `-2 sin theta + 1/(cos^2 theta)` | `= 0` |
| `-2 sin theta + 1/((1 – sin^2 theta))` | `= 0` |
| `(-2 sin theta (1 – sin^2 theta) + 1)/((1 – sin^2 theta))` | `= 0` |
| `-2 sin theta (1 – sin^2 theta) + 1` | `= 0` |
| `-2 sin theta + 2 sin^3 theta + 1` | `= 0` |
| `2 sin^3 theta – 2 sin theta + 1` | `= 0\ \ text(… (1))` |
`text(Let)\ \ x = sin theta and A = 2`
`text{Equation (1) becomes}`
`Ax^3 – 2x + 1 = 0`
`text(S)text(ince)\ \ A = 2`
`0 < A < (3 sqrt 3)/2`
| `text(S)text(ince)\ -pi/2` | `< \ \ \ theta` | `< pi/2` |
| `-1` | `< sin theta` | `< 1` |
| `-1` | `< \ \ \ x` | `< 1` |
`:.\ text{Using Part (iii)} => g prime (theta)\ \ text(has no zeros and)`
`text(therefore)\ \ g (theta)\ \ text(has no S.P.’s for)\ \ -pi/2 < theta < pi/2.`
(v) `text(Given)\ \ g(theta)\ \ text(has no stationary points, it will be)`
`text(either an increasing or decreasing function in the)`
`text(range)\ \ (-pi)/2 < theta < pi/2\ \ text(and therefore it has an inverse)`
`text(function.)`
An oil tanker at `T` is leaking oil which forms a circular oil slick. An observer is measuring the oil slick from a position `P`, 450 metres above sea level and 2 kilometres horizontally from the centre of the oil slick.
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|
i. |
 |
`text(Let)\ \ Q\ \ text(be the point on the sea such)\ \ /_PQT`
`text(is a right angle.)`
`text(Consider)\ \ Delta PQT,`
`text(Using Pythagoras,)`
| `PT^2` | `= PQ^2 + QT^2` |
| `= 450^2 + 2000^2` | |
| `PT` | `= sqrt(450^2 + 2000^2)` |
| `= 2050\ text(m)` |
`text(Consider)\ \ Delta PMT`
| `tan\ /_ MPT` | `= r/(PT)` |
| `tan 0.05` | `= r/2050` |
| `:. r` | `= 2050 xx tan 0.05` |
| `= 102.585…` | |
| `= 102.6\ text{m (to 1 d.p.)}` |
ii. `text(Find)\ \ (dr)/(dt)\ \ text(when)\ \ alpha = 0.1`
| `(dr)/(dt)` | `= (dr)/(d alpha) * (d alpha)/(dt)` |
| `r` | `= 2050 xx tan\ alpha/2` |
| `:.\ (dr)/(d alpha)` | `= 2050 xx 1/2 xx sec^2\ alpha/2` |
| `= 1025 sec^2\ alpha/2` |
`text(When)\ \ (d alpha)/(dt)= 0.02\ \ text(radians per hour),`
`alpha = 0.1\ \ \ text{(given)}.`
| `:. (dr)/(dt)` | `= 1025 sec^2 0.05 xx 0.02` |
| `= 20.5513…` | |
| `= 20.6\ text{metres per hour (to 1 d.p.)}` |
Xuan and Yvette would like to meet at a cafe on Monday. They each agree to come to the cafe sometime between 12 noon and 1 pm, wait for 15 minutes, and then leave if they have not seen the other person.
Their arrival times can be represented by the point `(x, y)` in the Cartesian plane, where `x` represents the fraction of an hour after 12 noon that Xuan arrives, and `y` represents the fraction of an hour after 12 noon that Yvette arrives.
Thus `(1/3, 2/5)` represents Xuan arriving at 12:20 pm and Yvette arriving at 12:24 pm. Note that the point `(x, y)` lies somewhere in the unit square `0 ≤ x ≤ 1` and `0 ≤ y ≤ 1` as shown in the diagram.
(i) `text(Xuan and Yvette must arrive within)\ 1/4\ text(of an hour)`
`text(of each in order to meet.)`
`text(If Xuan arrives first,)\ \ y − x ≤ 1/4.`
`text(If Yvette arrives first,)\ \ x − y ≤ 1/4.`
(ii) `text{From part (i), the inequalities are}`
`y ≤ x + 1/4`
`y ≥ x − 1/4`
`text(The shaded area satisfies the inequalities)`
`text(and the conditions)`
`0 ≤ x ≤ 1`
`0 ≤ y≤ 1`
`text(Shaded Area)`
`=\ text(Area of square − 2 triangles)`
`= 1^2 −2 xx (1/2 xx b xx h)`
`= 1 − 2 xx (1/2 xx 3/4 xx 3/4)`
`= 1 − 9/16`
`= 7/16\ \ text(u²)`
`:.\ text{P(Xuan and Yvette meet)}`
`=\ text(Shaded Area)/text(Area of square)`
`= 7/16`
(iii) `text(We need the shaded area to be)\ 1/2\ text(u²)`
`text(for a 50% chance.)`
`text(If they wait)\ t\ text{minutes, the inequalities from part (i) are:}`
`y ≤ x + t/(60)`
`y ≥ x − t/(60)`
| `text(Shaded Area)\ ` | `= 1 − 2 xx (1/2 xx b xx h)` |
| `1/2` | `= 1 − 2 xx [1/2 xx (1 − t/(60))(1 − t/(60))]` |
| `1/2` | `= 1 − (1 − t/(60))^2` |
| `(1 − t/(60))^2` | `= 1/2` |
| `1 − t/(60)` | `= 1/sqrt2` |
| `t/(60)` | `= 1 − 1/sqrt2` |
| `t` | `= 60(1 − 1/sqrt2)` |
| `= 17.593…` | |
| `= text(17.6 minutes (to 1 d.p.))` |
`:.\ text(They have 50% chance of meeting if they)`
`text(wait 17.6 minutes.)`
The parabola `y = x^2` and the line `y = mx + b` intersect at the points `A(α,α^2)` and `B(β, β^2)` as shown in the diagram.
