# Maths Superior Examination Paper Options

Have you ever seen the 2020 HSC Arithmetic Superior (2 Unit) examination paper but?

On this submit, we are going to work our means via the 2020 HSC Maths Superior (2 Unit) paper and provide the options, written by our Head of Arithmetic Oak Ukrit and his staff.

## Need to see how the Matrix Tutorial Head of Maths would clear up the widespread questions?

On this video, Head of Maths Oak Ukrit solves the widespread questions from the 2020 HSC Maths Superior and Maths Commonplace 2 papers.

Learn on to see the best way to reply all the 2020 questions.

## Part 1. A number of Alternative

 Query Quantity Reply Answer 1. D We will solely take the sq. root of non-negative numbers, so we want ( 2x – 3 geq 0 ). Which means that ( 2x geq 3, so x geq frac{3}{2}). 2. B The graph is translated proper as we now have ( (x – 2)) and the graph is translated up as we now have ( +5 ). 3. A start{align*} z_{French} &= frac{82-70}{8} =  1.5 z_{Commerce} &= frac{80-65}{5} = 3 z_{Music} &= frac{74-50}{12} = 2 finish{align*}Evaluating z-scores, his weakest topic is French and his strongest topic is Commerce. 4. B ( int e + e^{3x} , dx = ex + frac{1}{3} e^{3x} + c ), the place ( c ) is a continuing. 5. C Right here, ( a < 0 ) so the parabola ( y = -x^2 + bx + 1 ) must be concave down. Moreover, the axis of symmetry is start{align*} x &= – frac{b}{2a} &= -frac{b}{2 (-1)} &= frac{b}{2} &> 0 finish{align*} since it’s provided that ( b > 0 ) . 6. B Since ( -1 leq cos 3x leq 1) , and multiplying all the pieces by ( 2 ) offers ( -2 leq 2 cos 3x leq 2). Including ( 5 ) to all the pieces offers ( 5 – 2 leq 5 + 2 cos 3x leq 5 + 2 ) , so ( 3 leq 5 + 2 cos 3x leq 7) . Subsequently, the vary of ( 5 + 2 cos 3x) is ( [3,7]). 7. A start{align*} int_0^{12} fleft(xright) , dx &= textual content{space underneath \$fleft(xright)\$ from \$0\$ to \$12\$} &= 3 instances 4 + 3 instances 4 + frac{1}{2} instances pi instances left(frac{8 – 4}{2}proper)^2 + frac{3 instances left(10 – 8right)}{2} – frac{3 instances left(12 – 10right)}{2} &= 12 + 12 + 2 pi + 3 – 3 &= 24 + 2 pi ,. finish{align*} 8. A The reply is A. We’ve start{align*} f(1) > 0 finish{align*} since it’s above the ( x ) -axis. The graph is concave downwards so start{align*} f”(1) < 0 ,. finish{align*} By contemplating the tangent to ( y = f(x) ) that passes via the factors ( (0, c)) and ( (1, f(1))), we now have start{align*} f'(1) &= textual content{slope of tangent to \$f(x)\$ at \$x = 1\$} &= frac{f(1) – c}{1 – 0} &= f(1) – c &< f(1) quad textual content{(since we see that} c > 0 textual content{from the graph)} ,. finish{align*} Additionally, the slope of the tangent to ( f(x) ) at ( x = 1 ) is optimistic, so start{align*} f'(1) > 0 ,. finish{align*} In conclusion, start{align*} f”(1) < 0 < f'(1) < f(1) ,. finish{align*} 9. C We will make the next markings on the traditional curves offered: Solely C then has the decrease sure of ( 10% ) and the higher sure of ( 25% ) within the right areas. 10. D There are 2 most stationary factors at (? = 1) and since (?(?)) is steady, there have to be a minimal stationary level in between. Subsequently (3) stationary factors in complete.

## Part 2. Lengthy response

### Query 11a ### Query 11b

The quantity of water in Tank B is ( V = 30t + b ) . Substituting (  V = 0 ) and ( t = 15 ) into the equation offers

 start{align*} 0 &= 30left(15right) + b 0 &= 450 + b b &= -450 finish{align*}

Which means that the amount of water in Tank B is

 ( V = 30t – 450)

It’s provided that the amount of water in Tank A is

 ( V = 1000 – 20t )

So the 2 tanks have the identical quantity of water when

 start{align*} 30t – 450 &= 1000 – 20t 1450 &= 50t t &= frac{1450}{50} t &= 29 finish{align*}

Subsequently, at ( t = 29 ) minutes, they include the identical quantity of water.

### Query 11c

When the whole quantity of water within the two tanks is ( 1000 ) litres, we now have

 start{align*} (1000 – 20t) + (30t – 450) &= 1000 550 + 10t &= 1000 t &= frac{1000 – 550}{10} t &= 45 finish{align*}

Subsequently, at ( t = 45 ) minutes, the whole quantity of water within the two tanks is ( 1000 ) litres.

### Query 12

 start{align*} d &= T_2 – T_1 &= 10 – 4 &= 6 finish{align*}

The primary time period is

 ( a = 4 )

Utilizing

 ( T_n = a + (n – 1) d )

we now have

 start{equation*} 1354 = 4 + (n – 1) (6) ,, finish{equation*}

so

 start{align*} n &= frac{1354 – 4}{6} + 1 &= 226 finish{align*}

Subsequently, the sum of the arithmetic sequence is

 start{align*} S_{226} &= frac{226}{2} (4 + 1354) &= 153454 finish{align*}

### Query 13

We’ve

 start{align*} int_0^{pi/4} sec^2 x , dx &= left[tan xright]_0^{pi/4} &= tan left(frac{pi}{4}proper) – tan left(0right) &= 1 – 0 &= 1 finish{align*}

### Query 14a We’ve

 start{align*} n left(H cup Gright) &= n left(Hright) + n left(Gright) – nleft(H cap Gright) 33 &= 20 + 18 – n left(H cap Gright) nleft(H cap Gright) &= 5 ,. finish{align*}

Subsequently,

 start{align*} Pleft(H cap Gright) &= frac{5}{40} &= frac{1}{8} ,. finish{align*}

### Query 14b

We’ve

 start{align*} P left(overline{H} ,|, Gright) &= frac{Pleft(overline{H} cap Gright)}{Pleft(Gright)} &= frac{13}{18} ,. finish{align*}

### Query 14c

The likelihood required is

 start{align*} Pleft(Hright) instances Pleft(overline{H}proper) &= frac{20}{40} instances frac{20}{39} &= frac{10}{39} ,. finish{align*}

### Query 15a

We’ve

 start{align*} angle APB &= angle NPB – angle NPA &= 100º – 35º &= 65º finish{align*}

as required.

### Query 15b

Utilizing the cosine rule, we now have

 start{align*} AB^2 &= AP^2 + PB^2 – 2 instances AP instances PB instances cos 65º &= 7^2 + 9^2 – 2 instances 7 instances 9 instances cos 65º &= 76.75 finish{align*}

so

 start{align*} AB &= sqrt{76.75} &= 8.76 textual content{km} quad textual content{(right to 2 d.p.)} finish{align*}

### Query 15c Utilizing the cosine rule, we now have

 start{equation*} PB^2 = AP^2 + AB^2 – 2 instances AP instances AB instances cos (angle PAB) finish{equation*}

Rearranging this provides

 start{align*} cos (angle PAB) &= frac{AP^2 + AB^2 – PB^2}{2 instances AP instances AB} &= frac{7^2 + 76.75 – 9^2}{2 instances 7 instances 8.76} &= 0.365 finish{align*}

so

 start{align*} angle PAB &= cos^{-1} left(0.365right) &= 68º 36′ &approx 69º finish{align*}

Subsequently,

 start{align*} textual content{bearing} &= 180º – (69º – 35º) &= 146º quad textual content{(right to the closest diploma)} ,. finish{align*}

### Query 16

Calculating the derivatives offers

 start{align*} frac{dy}{dx} &= frac{d}{dx} (-x^3 + 3x^2 – 1) &= -3x^2 + 6x &= -3x (x – 2) finish{align*}

and

 start{align*} frac{d^2 y}{dx^2} & = frac{d}{dx} (-3x^2 + 6x) & = -6x + 6 & = -6(x – 1) finish{align*}

For the stationary factors, setting ( frac{dy}{dx} = 0 ) offers

 ( -3x (x – 2) = 0 )

so

 start{align*} x = 0 quad textual content{or} quad x = 2. finish{align*}

When ( x = 0 ),

 start{align*} y &= -0^3 + 3 (0)^2 – 1 &= -1 finish{align*}

When ( x = 2 )

 start{align*} y &= -2^3 + 3(2)^2 – 1 &= 3 finish{align*}

Subsequently, the stationary factors are ( (0,-1) ) and ( (2, 3) ).

 ( x ) ( x < 0 ) ( x = 0 ) ( 0  < x < 2 ) ( x = 2 ) ( x > 2 ) ( frac{dy}{dx} ) ( – ) ( 0 ) ( + ) ( 0 ) ( – )

For the factors of inflection, setting ( frac{d^2 y}{dx^2} = 0 ) offers

 start{align*} -6(x – 1) = 0 finish{align*}

So,

 start{align*} x = 1 finish{align*}

When ( x = 1 ),

 start{align*} y &= -1^3 + 3(1)^2 – 1 &= 1 finish{align*}

Subsequently, the purpose of inflection is ( (1,1) ).

 ( x  ) ( x < 1 ) ( x = 0 ) ( x > 1 ) ( frac{d^{2}y}{dx^{2}} ) ( + ) ( 0 ) ( – )

The ( y ) -intercept is

 start{align*} y &= -0^3 + 3(0)^2 – 1 &= -1 finish{align*} ### Query 17

We use integration by substitution. Let ( u = 4 + x^2 ). Then ( frac{du}{dx} = 2x ), so ( du = 2x dx ).

Subsequently,

 start{align*} int frac{x}{4 + x^2} , dx &= int frac{1}{2} left(frac{2x}{4 + x^2}proper) , dx &= int frac{1}{2} left(frac{du}{u}proper) &= frac{1}{2} ln left|uright| + C &= frac{1}{2} ln left|4 + x^2right| + C &= frac{1}{2} ln left(4 + x^2right) + C ,, quad textual content{since \$4 + x^2 > 0\$} ,, finish{align*}

The place ( C ) is a continuing.

### Query 18a

Utilizing the product rule offers

 start{align*} frac{d}{dx} left(e^{2x} left(2x + 1right)proper) &= e^{2x} frac{d}{dx} left(2x + 1right) + left(2x + 1right) frac{d}{dx} left(e^{2x}proper) &= e^{2x} left(2right) + left(2x + 1right) left(2e^{2x}proper) &= 2e^{2x} + 4xe^{2x} + 2e^{2x} &= 4e^{2x} + 4xe^{2x} &= 4e^{2x} left(1 + xright) ,. finish{align*}

### Query 18b

We’ve

 start{align*} int left(x + 1right) e^{2x} , dx &= frac{1}{4} int 4 left(x + 1right) e^{2x} , dx &= frac{1}{4} left[e^{2x} left(2x + 1right)right] + C ,, finish{align*}

the place ( C ) is a continuing, utilizing the consequence from half (a).

### Query 19

Ranging from the left hand aspect, we now have

 start{align*} sec theta – cos theta &= frac{1}{cos theta} – cos theta &= frac{1 – cos^2 theta}{cos theta} &= frac{sin^2 theta}{cos theta} &= sin theta cdot frac{sin theta}{cos theta} &= sin theta tan theta finish{align*}

as required.

### Query 20

Utilizing the trapezoidal rule, we discover that

 start{align*} textual content{approximate distance} &= frac{frac{1}{60}}{2} left[60 + 2 left(55 + 65 + 68 + 70right) + 67right] &= frac{643}{120} &= 5.358 &=5.4 (1text{ d.p.}) finish{align*}

### Query 21a

The temperate of the tea 4 minutes after it has been poured is discovered by substituting ( t = 4 ) into ( T ) , which supplies

 start{align*} T &= 25 + 70 left(1.5right)^{-0.4 left(4right)} &= 25 + 70 left(1.5right)^{-0.4 left(4right)} &= 25 + 70left(1.5right)^{-1.6} &approx 61.5891 textual content{levels Celsius.} finish{align*}

### Query 21b

The speed of change of the temperature of the tea is

 start{align*} frac{dT}{dt} &= frac{d}{dt} left(25 + 70left(1.5right)^{-0.4t}proper) &= 70left(1.5right)^{-0.4t} left(log_e 1.5right) frac{d}{dt} left(-0.4tright) &= 70left(1.5right)^{-0.4t} left(log_e 1.5right) left(-0.4right) &= left(-28log_e 1.5right) left(1.5right)^{-0.4t} ,. finish{align*}

When (  t = 4 ) , this turns into

 start{align*} frac{dT}{dt} &= left(-28log_e 1.5right) left(1.5right)^{-0.4 left(4right)} &approx -5.93425 ,, finish{align*}

which is destructive, so the tea is cooling on the charge of ( 5.93425 ) levels Celsius per minute.

### Query 21c

Substituting ( T = 55 ) into ( T = 25 + 70 (1.5)^{-0.4t} ) offers

 start{equation*} 55 = 25 + 70 left(1.5right)^{-0.4t} ,, finish{equation*}

so

 start{equation*} 30 = 70 left(1.5right)^{-0.4t} ,, finish{equation*}

and

 start{equation*} frac{3}{7} = left(1.5right)^{-0.4t} ,. finish{equation*}

Taking logarithms on either side of the equation offers

 start{equation*} log_e left(frac{3}{7}proper) = log_e left(1.5right)^{-0.4t} ,, finish{equation*}

so

 start{equation*} log_e left(frac{3}{7}proper) = -0.4t log_e left(1.5right) ,, finish{equation*}

and

 start{align*} t &= -frac{1}{0.4 log_e left(1.5right)} log_e left(frac{3}{7}proper) &approx 5.22423 finish{align*}

### Query 22

Utilizing the given perimeter we now have

 start{align*} AB &= frac{textual content{perimeter}}{textual content{variety of sides}} &= frac{80}{10} &= 8 , textrm{cm} finish{align*}

We even have

 start{align*} angle AOB &= 360° div 10 &= 36° finish{align*}

Utilizing the cosine rule offers

 ( AB^2 = AO^2 + BO^2 – 2 instances AO instances BO instances cos angle AOB  )

Since ( AO = BO ) and ( AB = 8 ), we now have

 start{align*} 8^2 &= AO^2 + AO^2 – 2 instances AO instances AO instances cos 36° 64 &= 2 (AO)^2 – 2 AO^2 cos 36° 64 &= AO^2  (2 – 2 cos 36°) AO^2 &= frac{64}{2 – 2 cos 36°} finish{align*}

Subsequently,

 start{align*} textual content{space of the decagon} &= 10 instances textual content{space of} triangle textual content{AOB} &= 10 instances left[frac{1}{2} times AO^2 sin 36º right] &= 10 instances frac{1}{2} instances frac{64}{2 – 2 cos 36º} instances sin 36º &= 492.429 &approx 492.4 , textrm{cm}^2 quad textual content{(right to 1 d.p.)} finish{align*}

### Query 23a

Since ( f(x) ) is a likelihood density perform, we all know that

 start{align*} int_0^ok sin x , dx &= 1 left[-cos xright]_0^ok &= 1 – cos ok – left(- cos 0right) &= 1 – cos ok + 1 &= 1 cos ok &= 0 , finish{align*}

Fixing this for ( ok ) offers

 ( ok = frac{pi}{2} )

### Query 23b

Since ( 1 < frac{pi}{2} ), we now have

 start{align*} Pleft(X leq 1right) &= int_0^1 fleft(xright) , dx &= int_0^1 sin x , dx &= left[-cos xright]_0^1 &= -cos 1 – left(- cos 0right) &= – cos 1 + 1 &= 0.4597 ,. finish{align*}

### Query 24

Beginning with the circle given, we full the sq. and acquire

 start{align*} x^2 – 6x + y^2 + 4y – 3 &= 0 x^2 – 6x + y^2 + 4y &= 3 left(x^2 – 6x + 9right) + left(y^2 + 4y + 4right) &= 3 + 9 + 4 left(x – 3right)^2 + left(y + 2right)^2 &= 16 finish{align*}

Which means that the unique circle is centred at ( (3, -2) ) and its radius is ( 4 ) models.

When it’s mirrored within the ( x ) -axis, we multiply the ( y ) -coordinate by ( -1 ) so the centre of the mirrored circle is ( (3,2) ) , and the radius remains to be ( 4 ) models. ### Query 25a

The radius of the quarter circle is ( x ) metres. We’ve

 start{align*} textual content{space of backyard mattress} &= textual content{space of rectangle} + textual content{space of quarter circle} &= xy + frac{1}{4} left(pi x^2right) ,. finish{align*}

Since this must be equal to ( 36 ), we now have

 start{align*} xy + frac{1}{4} left(pi x^2right) &= 36 xy &= 36 – tfrac{1}{4} left(pi x^2right) y &= frac{36}{x} – frac{pi x}{4} ,. finish{align*}

Subsequently, the perimeter is given by

 start{align*} P &= x + y + left(x + yright) + frac{1}{4} left(2 pi xright) &= 2 left(x + yright) + frac{pi x}{2} &= 2 left[x + left(frac{36}{x} – frac{pi x}{4}right)right] + frac{pi x}{2} &= 2x + frac{72}{x} – frac{pi x}{2} + frac{pi x}{2} &= 2x + frac{72}{x} ,, finish{align*}

as required.

### Query 25b

First, we differentiate ( P = 2x + frac{72}{x}) and acquire

 start{align*} frac{dP}{dx} &= frac{d}{dx} left(2x + frac{72}{x}proper) &= 2 – frac{72}{x^2} ,. finish{align*}

Setting ( frac{dP}{dx} = 0 ) offers ( 2 – frac{72}{x^2} = 0 ). Which means that

 ( frac{72}{x^2} = 2 )

so

 ( x^2 = 36 )

and

 ( x = 6 )

since ( x ) is the dimension of a rectangle and it have to be non-negative. When ( x = 6 ), the perimeter is given by

 start{align*} P &= 2 left(6right) + frac{72}{6} &= 12 + 12 &= 24 textual content{metres} finish{align*}
 ( x ) ( 0 < x < 6 ) ( x = 6 ) ( x > 6 ) ( frac{dP}{dx}  ) ( – ) ( 0 ) ( + )

Subsequently, (  P = 24 ) is the smallest doable perimeter.

### Query 26a

We use the recurrence relation beginning with

 ( A_0 = 60000 )

The sum of money within the account after the primary withdrawal is

 start{align*} A_1 &= A_0 left(1.005right) – 800 &= 60000 left(1.005right) – 800 ,, finish{align*}

The sum of money within the account after the second withdrawal is

 start{align*} A_2 &= A_1 left(1.005right) – 800 &= left[60000 left(1.005right) – 800right] left(1.005right) – 800 &= 60000 left(1.005right)^2 – 800 left(1 + 1.005right) ,, finish{align*}

And the sum of money within the account after the third withdrawal is

 start{align*} A_3 &= A_2 left(1.005right) – 800 &= left[60000 left(1.005right)^2 – 800 left(1 + 1.005right)right] left(1.005right) – 800 &= 60000 left(1.005right)^3 – 800 left(1 + 1.005 + 1.005^2right) &approx 58492.49 quad textual content{(right to 2 d.p.)} ,. finish{align*}

### Query 26b

The curiosity within the first month is

 start{align*} I_1 &= A_0 instances 0.5% &= 60000 instances 0.005 &= \$ 300 finish{align*}

The curiosity within the second month is

 start{align*} I_2 &= A_1 instances 0.5% &= left(60000 instances 1.005 – 800right) instances 0.005 &= \$ 297.50 finish{align*}

The curiosity within the third month is

 start{align*} I_3 &= A_2 instances 0.5% &= left(60000 instances 1.005^2 – 800 instances 1.005 – 800right) instances 0.005 &= \$ 294.99 finish{align*}

Subsequently, the whole quantity of curiosity earned within the first three months is

 start{align*} I_1 + I_2 + I_3 &= \$ 892.49 finish{align*}

### Query 26c

The sum of money within the account instantly after the 94th withdrawal is

 start{align*} A_{94} &= A_{93} left(1.005right) – 800 &= 60000 instances 1.005^{94} – 800 instances left(1 + 1.005 + 1.005^2 + dots + 1.005^{93}proper) &= 60000 instances 1.005^{94} – 800 instances left(frac{1 – 1.005^{94}}{1 – 1.005}proper) &= \$ 187.85 quad textual content{right to 2 decimal locations} finish{align*}

### Query 27

From the field plot, the median of the temperature is ( 22 )  levels Celsius. It’s provided that the imply temperature is (  0.525 )  levels beneath the median temperature, so

 start{align*} bar{x} &= 22 – 0.525 &= 21.475 textual content{levels Celsius} finish{align*}

The common variety of chirps over the 20 days is

 start{align*} bar{y} &= frac{684}{20} &= 34.2 finish{align*}

Taking the anticipated worth of either side of

 (y = -10.6063 + bx )

offers

 (bar{y} = -10.6063 + b bar{x} )

so

 ( 34.2 = -10.6063 + b left(21.475right) )

and

 start{align*} b &= frac{34.2 + 10.6063}{21.475} &= 2.08644004657 finish{align*}

Which means that

 ( y = -10.6063 + 2.08644004657x )

Subsequently, the variety of chirps anticipated in a 15-second interval when the temperature is nineteen levels Celsius is discovered by substituting ( x = 19 ), giving

 start{align*} y &= -10.6063 + 2.08644004657 left(19right) &approx 29 quad textual content{(right to the closest complete quantity)} ,. finish{align*}

### Query 28a

Let (  X ) {dollars} be the hourly charge of pay for adults who work. We’re provided that ( X sim mathcal{N} left(25, 5^2right) ) , so the likelihood that an grownup earns between ( \$ 15 ) and ( \$30 ) per hour is

 start{align*} Pleft(15 leq X leq 30right) &= Pleft(frac{15 – 25}{5} leq frac{X – 25}{5} leq frac{30 – 25}{5}proper) &= Pleft(-2 leq Z leq 1right) &= 0.84134 – 0.02275 &= 0.81859 finish{align*}

Subsequently,

 start{align*} Pleft(textual content{an grownup’s hourly charge of pay will not be between \$15 and \$30}proper) &= 1 – Pleft(15 leq X leq 30right) &= 1 – 0.81859 &= 0.18141 finish{align*}

and

 start{align*} Pleft(textual content{not less than one in all two adults earns between \$15 and \$30}proper) &= 1 – Pleft(textual content{neither earns between \$15 and \$30 an hour}proper) &= 1 – 0.18141^2 &= 0.9671 finish{align*}

### Query 28b

For the reason that variety of adults who work is the same as thrice the variety of adults who don’t work, we all know that

 start{equation*} P left(textual content{a randomly chosen grownup works}proper) = frac{1}{4} finish{equation*}

Which means that

 start{align*} &∴P left( textual content{Chosen grownup works and earns greater than} \$25 textual content{ per hour} proper) = frac{3}{4} instances frac{1}{2} & =frac{3}{8} finish{align*}

### Query 29a

Since

 start{align*} frac{dy}{dx} &= frac{d}{dx} left(c ln xright) &= frac{c}{x} ,, finish{align*}

the slope of the tangent to ( y = c ln x ) at ( x = p ) is

 (frac{dy}{dx}bigg|_{x = p} = frac{c}{p} )

When ( x = p ), the corresponding ( y )-coordinate is

 start{equation*} y = c ln p finish{equation*}

Which means that the equation of the tangent to ( y = c ln x ) at  ( x = p ) is given by

 start{equation*} frac{y – c ln p}{x – p} = frac{c}{p} finish{equation*}

Multiplying either side by ( x – p ) offers ( y – c ln p = frac{c}{p} left(x – pright) ), so

 start{align*} y &= frac{c}{p} left(x – pright) + c ln p &= frac{c}{p} x – c + c ln p ,, finish{align*}

as required.

### Query 29b

For the reason that tangent has a gradient of ( 1 ), we now have

 start{equation} label{eq:Q29bEq1} frac{c}{p} = 1 finish{equation}

which signifies that

 start{equation} label{eq:Q29bEq2} c = p finish{equation}

Moreover, because the tangent passes via the origin, substituting ( x = 0 ) and ( y = 0 ) into the equation of the tangent offers

 start{equation*} 0 = frac{c}{p} left(0right) – c + c ln p finish{equation*}

which signifies that ( 0 = -c + c ln p ), or equivalently,

 start{equation} label{eq:Q29bEq3} c left(-1 + ln pright) = 0 finish{equation}

From Eq. ( frac{c}{p} = 1), we all know that ( c neq 0 ), so Eq. ( c left(-1 + ln pright) = 0 ) implies that

 start{equation*} -1 + ln p = 0 ,, finish{equation*}

which signifies that

 start{equation*} ln p = 1 ,, finish{equation*}

and

 start{equation} label{eq:Q29bEq4} p = e finish{equation}

Substituting ( p = e ) into ( c = p ) offers

 start{equation*} c = e ,. finish{equation*}

### Query 30a

Since ( A ) is the purpose of intersection of

 start{equation*} y = 4x – x^2 finish{equation*}

and

 start{equation*} y = ax^2 finish{equation*}

its ( x ) -coordinate is discovered by equating each equations and fixing for ( x ). This offers

 start{equation*} 4x – x^2 = ax^2 finish{equation*}

so

 start{equation*} left(1 + aright) x^2 – 4x = 0 finish{equation*}

or equivalently,

 start{equation*} x left [ left(1 + a right )x – 4 right] = 0 finish{equation*}

Both ( x = 0 ) or ( left(1 + aright) x – 4 = 0 ), however from the diagram we see that the ( x ) coordinate of  ( A ) will not be ( 0 ), which signifies that we will need to have

 start{equation*} left(1 + aright) x – 4 = 0 finish{equation*}

Rearranging this provides

 start{equation*} left(1 + aright) x = 4 finish{equation*}

so

 start{equation*} x = frac{4}{1 + a} finish{equation*}

as required.

### Query 30b

The shaded space is given by

 start{align*} int_0^{4/left(1+aright)} left[left(4x – x^2right) – ax^2right] , dx &= int_0^{4/left(1+aright)} left[left(4x – x^2right) – ax^2right] , dx &= int_0^{4/left(1+aright)} left(4x – x^2 – ax^2right) , dx &= left[2x^2 – frac{x^3}{3} – frac{ax^3}{3}right]_0^{4/left(1+aright)} &= 2left(frac{4}{1 + a}proper)^2 – frac{1}{3} left(frac{4}{1 + a}proper)^3 – frac{a}{3} left(frac{4}{1 + a}proper)^3 &= frac{32}{left(1 + aright)^2} – frac{64}{3 left(1 + aright)^3} – frac{64a}{3left(1 + aright)^3} &= frac{96 left(1 + aright) – 64 -64a}{3left(1 + aright)^3} &= frac{96 + 96a – 64 -64a}{3left(1 + aright)^3} &= frac{32 left(1 + aright)}{3left(1 + aright)^3} &= frac{32}{3left(1 + aright)^2} finish{align*}

Subsequently, to ensure that the shaded space to be equal to ( frac{16}{3} ), we want

 start{equation*} frac{32}{3left(1 + aright)^2} = frac{16}{3} finish{equation*}

which implies

 start{equation*} left(1 + aright)^2 = 2 finish{equation*}

Subsequently,

 start{equation*} 1 + a = pm sqrt{2} finish{equation*}

so ( a = -1 pm sqrt{2} ). From the diagram, we want ( a > 0 ), so

 start{equation*} a = -1 + sqrt{2} finish{equation*}

### Query 31a

( a ) represents the amplitude, so

 start{align*} a &= frac{35000 – 5000}{2} &= 15000 finish{align*}

( b ) represents the vertical displacement, so

 start{align*} b &= 35000 – 15000 &= 20000 finish{align*}

### Query 31b

Differentiating ( mleft(tright) ) offers

 start{equation*} m’left(tright) = 15000left(frac{pi}{26}proper) cos left(frac{pi}{26} tright) finish{equation*}

For ( mleft(tright) ) to be growing, we assess when ( m’left(tright) > 0 ). This happens when

 start{equation*} cos left(frac{pi}{26} tright) > 0 finish{equation*}

So

 start{equation*} frac{pi}{26} t < frac{pi}{2} quad textual content{or} quad frac{pi}{26} t > frac{3 pi}{2} finish{equation*}

Which means that

 start{equation*} t < 13 quad textual content{or} quad t > 39 finish{equation*}

Likewise with ( cleft(tright) ), and we now have

 start{equation*} c’left(tright) = frac{80 pi}{26} sin left[frac{pi}{26} left(t – 10right)right] finish{equation*}

Now, ( c’left(tright) > 0 ) happens when

 start{equation*} sin left[frac{pi}{26} left(t – 10right)right] > 0 finish{equation*}

which signifies that

 start{equation*} 0 < frac{pi}{26} left(t – 10right) < pi finish{equation*}

so

 start{equation*} 10 < t < 36 finish{equation*}

Subsequently, each populations are growing when

 start{equation*} 10 < t < 13 finish{equation*}

### Query 31c

First, we discover when ( cleft(tright) ) reaches a most. This happens when ( c’left(tright) = 0 ), so

 start{equation*} frac{80 pi}{26} sin left[frac{pi}{26} left(t – 10right)right] = 0 ,, finish{equation*}

or equivalently,

 start{equation*} sin left[frac{pi}{26} left(t – 10right)right] = 0 ,. finish{equation*}

Subsequently,

 start{equation*} frac{pi}{26} left(t – 10right) = 0 quad textual content{or} quad frac{pi}{26} left(t – 10right) = pi ,, finish{equation*}

and the stationary factors are

 start{equation*} t = 10 quad textual content{or} quad t = 36 ,. finish{equation*}

To find out the character of those stationary factors, we look at ( c”left(tright) ). Now,

 start{equation*} c”left(tright) = frac{80 pi^2}{26^2} cos left(frac{pi}{26} left(t – 10right)proper) ,. finish{equation*}

At ( t = 36 ),

 start{align*} c”left(36right) &= -1.168 &< 0 finish{align*}

so ( cleft(tright) ) attains its most when ( t = 36 ). At (t = 36 ) ,

 start{align*} m’left(36right) &= frac{15000pi}{26} cos left(frac{36 pi}{26}proper) &= -642.7 finish{align*}

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