{"id":389,"date":"2023-04-26T11:14:17","date_gmt":"2023-04-26T11:14:17","guid":{"rendered":"http:\/\/onlineduatease.com\/?p=389"},"modified":"2022-09-04T02:43:36","modified_gmt":"2022-09-04T02:43:36","slug":"maths-superior-examination-paper-options","status":"publish","type":"post","link":"https:\/\/onlineduatease.com\/index.php\/2023\/04\/26\/maths-superior-examination-paper-options\/","title":{"rendered":"Maths Superior Examination Paper Options"},"content":{"rendered":"

Have you ever seen the 2020 HSC Arithmetic Superior (2 Unit) examination paper but?<\/p>\n

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.<\/p>\n

 <\/p>\n

Need to see how the Matrix Tutorial Head of Maths would clear up the widespread questions?<\/h2>\n

On this video, Head of Maths Oak Ukrit solves the widespread questions from the 2020 HSC Maths Superior and Maths Commonplace 2 papers.<\/p>\n

<\/div>\n

 <\/p>\n

Learn on to see the best way to reply all the 2020 questions.<\/p>\n

 <\/p>\n

Part 1. A number of Alternative<\/h2>\n

 <\/p>\n

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
Query Quantity<\/td>\nReply<\/td>\nAnswer<\/td>\n<\/tr>\n<\/thead>\n
1.<\/strong><\/td>\n\u00a0D<\/td>\nWe will solely take the sq. root of non-negative numbers, so we want ( 2x \u2013 3 geq 0 ). Which means that ( 2x geq 3, so x geq frac{3}{2}).<\/td>\n<\/tr>\n
2.<\/strong><\/td>\n\u00a0B<\/td>\nThe graph is translated proper as we now have ( (x \u2013 2)) and the graph is translated up as we now have ( +5 ).<\/td>\n<\/tr>\n
3.<\/strong><\/td>\n\u00a0A<\/td>\nstart{align*}
\nz_{French} &= frac{82-70}{8} =\u00a0 1.5
\nz_{Commerce} &= frac{80-65}{5} = 3
\nz_{Music} &= frac{74-50}{12} = 2
\nfinish{align*}Evaluating z-scores, his weakest topic is French and his strongest topic is Commerce.<\/td>\n<\/tr>\n
4.<\/strong><\/td>\n\u00a0B<\/td>\n\u00a0( int e + e^{3x} , dx = ex + frac{1}{3} e^{3x} + c ), the place ( c\u00a0) is a continuing.<\/td>\n<\/tr>\n
5.<\/strong><\/td>\nC<\/td>\nRight here, ( a < 0 ) so the parabola ( y = -x^2 + bx + 1\u00a0) must be concave down. Moreover, the axis of symmetry is<\/p>\n

start{align*}
\nx &= \u2013 frac{b}{2a}
\n&= -frac{b}{2 (-1)}
\n&= frac{b}{2}
\n&> 0
\nfinish{align*}<\/p>\n

since it\u2019s provided that ( b > 0 ) .<\/td>\n<\/tr>\n

6.<\/strong><\/td>\n\u00a0B<\/td>\n\u00a0Since ( -1 leq cos 3x leq 1) , and multiplying all the pieces by ( 2 ) offers ( -2 leq 2 cos 3x leq 2).<\/p>\n

Including ( 5 ) to all the pieces offers ( 5 \u2013 2 leq 5 + 2 cos 3x leq 5 + 2 ) , so ( 3 leq 5 + 2 cos 3x leq 7) .<\/p>\n

Subsequently, the vary of ( 5 + 2 cos 3x) is ( [3,7]).<\/td>\n<\/tr>\n

7.<\/strong><\/td>\nA<\/td>\nstart{align*}
\nint_0^{12} fleft(xright) , dx &= textual content{space underneath $fleft(xright)$ from $0$ to $12$}
\n&= 3 instances 4 + 3 instances 4 + frac{1}{2} instances pi instances left(frac{8 \u2013 4}{2}proper)^2 + frac{3 instances left(10 \u2013 8right)}{2} \u2013 frac{3 instances left(12 \u2013 10right)}{2}
\n&= 12 + 12 + 2 pi + 3 \u2013 3
\n&= 24 + 2 pi ,.
\nfinish{align*}<\/td>\n<\/tr>\n
8.<\/strong><\/td>\nA<\/td>\n\"blog-2020-HSC-Maths-Advanced-Exam-Paper-Solutions-Question-8\"<\/p>\n

The reply is A. We\u2019ve<\/p>\n

start{align*}
\nf(1) > 0
\nfinish{align*}<\/p>\n

since it\u2019s above the ( x ) -axis.<\/p>\n

The graph is concave downwards so<\/p>\n

start{align*}
\nf\u201d(1) < 0 ,.
\nfinish{align*}<\/p>\n

By contemplating the tangent to ( y = f(x)\u00a0) that passes via the factors ( (0, c)) and ( (1, f(1))), we now have<\/p>\n

start{align*}
\nf'(1) &= textual content{slope of tangent to $f(x)$ at $x = 1$}
\n&= frac{f(1) \u2013 c}{1 \u2013 0}
\n&= f(1) \u2013 c
\n&< f(1) quad textual content{(since we see that} c > 0 textual content{from the graph)} ,.
\nfinish{align*}<\/p>\n

Additionally, the slope of the tangent to ( f(x)\u00a0) at ( x = 1\u00a0) is optimistic, so<\/p>\n

start{align*}
\nf'(1) > 0 ,.
\nfinish{align*}<\/p>\n

In conclusion,<\/p>\n

start{align*}
\nf\u201d(1) < 0 < f'(1) < f(1) ,.
\nfinish{align*}<\/td>\n<\/tr>\n

9.<\/strong><\/td>\nC<\/td>\nWe will make the next markings on the traditional curves offered:<\/p>\n

\"blog-2020-HSC-Maths-Advanced-Exam-Paper-Solutions-Question-9\"<\/p>\n

Solely C then has the decrease sure of ( 10% ) and the higher sure of ( 25%\u00a0) within the right areas.<\/td>\n<\/tr>\n

10.<\/strong><\/td>\nD<\/td>\nThere 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.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n
\n
\n
\n
\n
\n
\n
\n
\n
\n
\n
\n
\n
\n
<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

 <\/p>\n

Part 2. Lengthy response<\/h2>\n

Query 11a<\/h3>\n

\"blog-2020-HSC-Maths-Advanced-Exam-Paper-Solutions-Question-10\"<\/p>\n

 <\/p>\n

Query 11b<\/h3>\n

The quantity of water in Tank B is ( V = 30t + b ) . Substituting (\u00a0 V = 0 ) and\u00a0( t = 15 ) into the equation offers<\/p>\n

\n\n\n\n
start{align*}
\n0 &= 30left(15right) + b
\n0 &= 450 + b
\nb &= -450
\nfinish{align*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

Which means that the amount of water in Tank B is<\/p>\n

\n\n\n\n
( V = 30t \u2013 450)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

It\u2019s provided that the amount of water in Tank A is<\/p>\n

\n\n\n\n
( V = 1000 \u2013 20t )<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

So the 2 tanks have the identical quantity of water when<\/p>\n

\n\n\n\n
start{align*}
\n30t \u2013 450 &= 1000 \u2013 20t
\n1450 &= 50t
\nt &= frac{1450}{50}
\nt &= 29
\nfinish{align*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

Subsequently, at ( t = 29\u00a0) minutes, they include the identical quantity of water.<\/p>\n

 <\/p>\n

Query 11c<\/h3>\n

When the whole quantity of water within the two tanks is ( 1000 ) litres, we now have<\/p>\n

\n\n\n\n
start{align*}
\n(1000 \u2013 20t) + (30t \u2013 450) &= 1000
\n550 + 10t &= 1000
\nt &= frac{1000 \u2013 550}{10}
\nt &= 45
\nfinish{align*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

Subsequently, at ( t = 45\u00a0) minutes, the whole quantity of water within the two tanks is ( 1000 ) litres.<\/p>\n

 <\/p>\n

Query 12<\/h3>\n

The widespread distinction is<\/p>\n

\n\n\n\n
start{align*}
\nd &= T_2 \u2013 T_1
\n&= 10 \u2013 4
\n&= 6
\nfinish{align*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

The primary time period is<\/p>\n

\n\n\n\n
( a = 4 )<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

Utilizing<\/p>\n

\n\n\n\n
( T_n = a + (n \u2013 1) d )<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

we now have<\/p>\n

\n\n\n\n
start{equation*}
\n1354 = 4 + (n \u2013 1) (6) ,,
\nfinish{equation*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

so<\/p>\n

\n\n\n\n
start{align*}
\nn &= frac{1354 \u2013 4}{6} + 1
\n&= 226
\nfinish{align*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

Subsequently, the sum of the arithmetic sequence is<\/p>\n

\n\n\n\n
start{align*}
\nS_{226} &= frac{226}{2} (4 + 1354)
\n&= 153454
\nfinish{align*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

 <\/p>\n

Query 13<\/h3>\n

We\u2019ve<\/p>\n

\n\n\n\n
start{align*}
\nint_0^{pi\/4} sec^2 x , dx &= left[tan xright]_0^{pi\/4}
\n&= tan left(frac{pi}{4}proper) \u2013 tan left(0right)
\n&= 1 \u2013 0
\n&= 1
\nfinish{align*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

 <\/p>\n

Query 14a<\/h3>\n

\"blog-2020-HSC-Maths-Advanced-Exam-Paper-Solutions-Question-14\"<\/p>\n

 <\/p>\n

We\u2019ve<\/p>\n

\n\n\n\n
start{align*}
\nn left(H cup Gright) &= n left(Hright) + n left(Gright) \u2013 nleft(H cap Gright)
\n33 &= 20 + 18 \u2013 n left(H cap Gright)
\nnleft(H cap Gright) &= 5 ,.
\nfinish{align*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

Subsequently,<\/p>\n

\n\n\n\n
start{align*}
\nPleft(H cap Gright) &= frac{5}{40}
\n&= frac{1}{8} ,.
\nfinish{align*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

 <\/p>\n

Query 14b<\/h3>\n

We\u2019ve<\/p>\n

\n\n\n\n
start{align*}
\nP left(overline{H} ,|, Gright) &= frac{Pleft(overline{H} cap Gright)}{Pleft(Gright)}
\n&= frac{13}{18} ,.
\nfinish{align*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

Query 14c<\/h3>\n

The likelihood required is<\/p>\n

\n\n\n\n
start{align*}
\nPleft(Hright) instances Pleft(overline{H}proper) &= frac{20}{40} instances frac{20}{39}
\n&= frac{10}{39} ,.
\nfinish{align*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

 <\/p>\n

Query 15a<\/h3>\n

We\u2019ve<\/p>\n

\n\n\n\n
start{align*}
\nangle APB &= angle NPB \u2013 angle NPA
\n&= 100\u00ba \u2013 35\u00ba
\n&= 65\u00ba
\nfinish{align*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

as required.<\/p>\n

 <\/p>\n

Query 15b<\/h3>\n

Utilizing the cosine rule, we now have<\/p>\n

\n\n\n\n
start{align*}
\nAB^2 &= AP^2 + PB^2 \u2013 2 instances AP instances PB instances cos 65\u00ba
\n&= 7^2 + 9^2 \u2013 2 instances 7 instances 9 instances cos 65\u00ba
\n&= 76.75
\nfinish{align*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

so<\/p>\n

\n\n\n\n
start{align*}
\nAB &= sqrt{76.75}
\n&= 8.76 textual content{km} quad textual content{(right to 2 d.p.)}
\nfinish{align*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

 <\/p>\n

Query 15c<\/h3>\n

\"blog-2020-HSC-Maths-Advanced-Exam-Paper-Solutions-Question-15c\"<\/p>\n

Utilizing the cosine rule, we now have<\/p>\n

\n\n\n\n
start{equation*}
\nPB^2 = AP^2 + AB^2 \u2013 2 instances AP instances AB instances cos (angle PAB)
\nfinish{equation*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

Rearranging this provides<\/p>\n

\n\n\n\n
start{align*}
\ncos (angle PAB) &= frac{AP^2 + AB^2 \u2013 PB^2}{2 instances AP instances AB}
\n&= frac{7^2 + 76.75 \u2013 9^2}{2 instances 7 instances 8.76}
\n&= 0.365
\nfinish{align*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

so<\/p>\n

\n\n\n\n
start{align*}
\nangle PAB &= cos^{-1} left(0.365right)
\n&= 68\u00ba 36\u2032
\n&approx 69\u00ba
\nfinish{align*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

Subsequently,<\/p>\n

\n\n\n\n
start{align*}
\ntextual content{bearing} &= 180\u00ba \u2013 (69\u00ba \u2013 35\u00ba)
\n&= 146\u00ba quad textual content{(right to the closest diploma)} ,.
\nfinish{align*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

 <\/p>\n

Query 16<\/h3>\n

Calculating the derivatives offers<\/p>\n

\n\n\n\n
start{align*}
\nfrac{dy}{dx} &= frac{d}{dx} (-x^3 + 3x^2 \u2013 1)
\n&= -3x^2 + 6x
\n&= -3x (x \u2013 2)
\nfinish{align*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

and<\/p>\n

\n\n\n\n
start{align*}
\nfrac{d^2 y}{dx^2} & = frac{d}{dx} (-3x^2 + 6x)
\n& = -6x + 6
\n& = -6(x \u2013 1)
\nfinish{align*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

 <\/p>\n

For the stationary factors, setting ( frac{dy}{dx} = 0 ) offers<\/p>\n

\n\n\n\n
( -3x (x \u2013 2) = 0\u00a0)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

so<\/p>\n

\n\n\n\n
start{align*}
\nx = 0 quad textual content{or} quad x = 2.
\nfinish{align*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

When ( x = 0 ),<\/p>\n

\n\n\n\n
start{align*}
\ny &= -0^3 + 3 (0)^2 \u2013 1
\n&= -1
\nfinish{align*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

When\u00a0( x = 2 )<\/p>\n

\n\n\n\n
start{align*}
\ny &= -2^3 + 3(2)^2 \u2013 1
\n&= 3
\nfinish{align*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

Subsequently, the stationary factors are ( (0,-1)\u00a0) and ( (2, 3) ).<\/p>\n

\n\n\n\n\n\n
( x )<\/td>\n( x < 0 )<\/td>\n( x = 0 )<\/td>\n( 0\u00a0 < x < 2 )<\/td>\n( x = 2 )<\/td>\n( x > 2 )<\/td>\n<\/tr>\n<\/thead>\n
( frac{dy}{dx} )<\/td>\n( \u2013 )<\/td>\n( 0\u00a0)<\/td>\n( +\u00a0)<\/td>\n( 0\u00a0)<\/td>\n\u00a0( \u2013\u00a0)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

 <\/p>\n

For the factors of inflection, setting ( frac{d^2 y}{dx^2} = 0 ) offers<\/p>\n

\n\n\n\n
start{align*}
\n-6(x \u2013 1) = 0
\nfinish{align*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

So,<\/p>\n

\n\n\n\n
start{align*}
\nx = 1
\nfinish{align*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

When ( x = 1\u00a0),<\/p>\n

\n\n\n\n
start{align*}
\ny &= -1^3 + 3(1)^2 \u2013 1
\n&= 1
\nfinish{align*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

Subsequently, the purpose of inflection is ( (1,1)\u00a0).<\/p>\n

\n\n\n\n\n
( x\u00a0 )<\/td>\n( x < 1 )<\/td>\n( x = 0 )<\/td>\n( x > 1 )<\/td>\n<\/tr>\n
( frac{d^{2}y}{dx^{2}} )<\/td>\n( + )<\/td>\n( 0 )<\/td>\n\u00a0( \u2013 )<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

The ( y ) -intercept is<\/p>\n

\n\n\n\n
start{align*}
\ny &= -0^3 + 3(0)^2 \u2013 1
\n&= -1
\nfinish{align*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

\"blog-2020-HSC-Maths-Advanced-Exam-Paper-Solutions-Question-16\"<\/p>\n

 <\/p>\n

 <\/p>\n

Query 17<\/h3>\n

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

Subsequently,<\/p>\n

\n\n\n\n
start{align*}
\nint frac{x}{4 + x^2} , dx &= int frac{1}{2} left(frac{2x}{4 + x^2}proper) , dx
\n&= int frac{1}{2} left(frac{du}{u}proper)
\n&= frac{1}{2} ln left|uright| + C
\n&= frac{1}{2} ln left|4 + x^2right| + C
\n&= frac{1}{2} ln left(4 + x^2right) + C ,, quad textual content{since $4 + x^2 > 0$} ,,
\nfinish{align*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

The place ( C ) is a continuing.<\/p>\n

 <\/p>\n

Query 18a<\/h3>\n

Utilizing the product rule offers<\/p>\n

\n\n\n\n
start{align*}
\nfrac{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)
\n&= e^{2x} left(2right) + left(2x + 1right) left(2e^{2x}proper)
\n&= 2e^{2x} + 4xe^{2x} + 2e^{2x}
\n&= 4e^{2x} + 4xe^{2x}
\n&= 4e^{2x} left(1 + xright) ,.
\nfinish{align*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

 <\/p>\n

Query 18b<\/h3>\n

We\u2019ve<\/p>\n

\n\n\n\n
start{align*}
\nint left(x + 1right) e^{2x} , dx &= frac{1}{4} int 4 left(x + 1right) e^{2x} , dx
\n&= frac{1}{4} left[e^{2x} left(2x + 1right)right] + C ,,
\nfinish{align*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

the place ( C\u00a0) is a continuing, utilizing the consequence from half (a).<\/p>\n

 <\/p>\n

Query 19<\/h3>\n

Ranging from the left hand aspect, we now have<\/p>\n

\n\n\n\n
start{align*}
\nsec theta \u2013 cos theta &= frac{1}{cos theta} \u2013 cos theta
\n&= frac{1 \u2013 cos^2 theta}{cos theta}
\n&= frac{sin^2 theta}{cos theta}
\n&= sin theta cdot frac{sin theta}{cos theta}
\n&= sin theta tan theta
\nfinish{align*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

as required.<\/p>\n

 <\/p>\n

Query 20<\/h3>\n

Utilizing the trapezoidal rule, we discover that<\/p>\n

\n\n\n\n
start{align*}
\ntextual content{approximate distance} &= frac{frac{1}{60}}{2} left[60 + 2 left(55 + 65 + 68 + 70right) + 67right]
\n&= frac{643}{120}
\n&= 5.358
\n&=5.4 (1text{ d.p.})
\nfinish{align*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

 <\/p>\n

Query 21a<\/h3>\n

The temperate of the tea 4 minutes after it has been poured is discovered by substituting ( t = 4\u00a0) into ( T ) , which supplies<\/p>\n

\n\n\n\n
start{align*}
\nT &= 25 + 70 left(1.5right)^{-0.4 left(4right)}
\n&= 25 + 70 left(1.5right)^{-0.4 left(4right)}
\n&= 25 + 70left(1.5right)^{-1.6}
\n&approx 61.5891 textual content{levels Celsius.}
\nfinish{align*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

 <\/p>\n

Query 21b<\/h3>\n

The speed of change of the temperature of the tea is<\/p>\n

\n\n\n\n
start{align*}
\nfrac{dT}{dt} &= frac{d}{dt} left(25 + 70left(1.5right)^{-0.4t}proper)
\n&= 70left(1.5right)^{-0.4t} left(log_e 1.5right) frac{d}{dt} left(-0.4tright)
\n&= 70left(1.5right)^{-0.4t} left(log_e 1.5right) left(-0.4right)
\n&= left(-28log_e 1.5right) left(1.5right)^{-0.4t} ,.
\nfinish{align*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

When (\u00a0 t = 4 ) , this turns into<\/p>\n

\n\n\n\n
start{align*}
\nfrac{dT}{dt} &= left(-28log_e 1.5right) left(1.5right)^{-0.4 left(4right)}
\n&approx -5.93425 ,,
\nfinish{align*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

which is destructive, so the tea is cooling on the charge of ( 5.93425\u00a0) levels Celsius per minute.<\/p>\n

 <\/p>\n

Query 21c<\/h3>\n

Substituting ( T = 55\u00a0) into ( T = 25 + 70 (1.5)^{-0.4t}\u00a0) offers<\/p>\n

\n\n\n\n
start{equation*}
\n55 = 25 + 70 left(1.5right)^{-0.4t} ,,
\nfinish{equation*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

so<\/p>\n

\n\n\n\n
start{equation*}
\n30 = 70 left(1.5right)^{-0.4t} ,,
\nfinish{equation*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

and<\/p>\n

\n\n\n\n
start{equation*}
\nfrac{3}{7} = left(1.5right)^{-0.4t} ,.
\nfinish{equation*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

Taking logarithms on either side of the equation offers<\/p>\n

\n\n\n\n
start{equation*}
\nlog_e left(frac{3}{7}proper) = log_e left(1.5right)^{-0.4t} ,,
\nfinish{equation*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

so<\/p>\n

\n\n\n\n
start{equation*}
\nlog_e left(frac{3}{7}proper) = -0.4t log_e left(1.5right) ,,
\nfinish{equation*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

and<\/p>\n

\n\n\n\n
start{align*}
\nt &= -frac{1}{0.4 log_e left(1.5right)} log_e left(frac{3}{7}proper)
\n&approx 5.22423
\nfinish{align*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

 <\/p>\n

Query 22<\/h3>\n

Utilizing the given perimeter we now have<\/p>\n

\n\n\n\n
start{align*}
\nAB &= frac{textual content{perimeter}}{textual content{variety of sides}}
\n&= frac{80}{10}
\n&= 8 , textrm{cm}
\nfinish{align*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

We even have<\/p>\n

\n\n\n\n
start{align*}
\nangle AOB &= 360\u00b0 div 10
\n&= 36\u00b0
\nfinish{align*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

Utilizing the cosine rule offers<\/p>\n

\n\n\n\n
( AB^2 = AO^2 + BO^2 \u2013 2 instances AO instances BO instances cos angle AOB\u00a0\u00a0)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

Since ( AO = BO\u00a0) and\u00a0( AB = 8 ), we now have<\/p>\n

\n\n\n\n
start{align*}
\n8^2 &= AO^2 + AO^2 \u2013 2 instances AO instances AO instances cos 36\u00b0
\n64 &= 2 (AO)^2 \u2013 2 AO^2 cos 36\u00b0
\n64 &= AO^2\u00a0 (2 \u2013 2 cos 36\u00b0)
\nAO^2 &= frac{64}{2 \u2013 2 cos 36\u00b0}
\nfinish{align*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

Subsequently,<\/p>\n

\n\n\n\n
start{align*}
\ntextual content{space of the decagon} &= 10 instances textual content{space of} triangle textual content{AOB}
\n&= 10 instances left[frac{1}{2} times AO^2 sin 36\u00ba right]
\n&= 10 instances frac{1}{2} instances frac{64}{2 \u2013 2 cos 36\u00ba} instances sin 36\u00ba
\n&= 492.429
\n&approx 492.4 , textrm{cm}^2 quad textual content{(right to 1 d.p.)}
\nfinish{align*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

 <\/p>\n

Query 23a<\/h3>\n

Since ( f(x)\u00a0) is a likelihood density perform, we all know that<\/p>\n

\n\n\n\n
start{align*}
\nint_0^ok sin x , dx &= 1
\nleft[-cos xright]_0^ok &= 1
\n\u2013 cos ok \u2013 left(- cos 0right) &= 1
\n\u2013 cos ok + 1 &= 1
\ncos ok &= 0 ,
\nfinish{align*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

Fixing this for ( ok\u00a0) offers<\/p>\n

\n\n\n\n
( ok = frac{pi}{2}\u00a0)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

 <\/p>\n

Query 23b<\/h3>\n

Since ( 1 < frac{pi}{2} ), we now have<\/p>\n

\n\n\n\n
\u00a0start{align*}
\nPleft(X leq 1right) &= int_0^1 fleft(xright) , dx
\n&= int_0^1 sin x , dx
\n&= left[-cos xright]_0^1
\n&= -cos 1 \u2013 left(- cos 0right)
\n&= \u2013 cos 1 + 1
\n&= 0.4597 ,.
\nfinish{align*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

 <\/p>\n

Query 24<\/h3>\n

Beginning with the circle given, we full the sq. and acquire<\/p>\n

\n\n\n\n
start{align*}
\nx^2 \u2013 6x + y^2 + 4y \u2013 3 &= 0
\nx^2 \u2013 6x + y^2 + 4y &= 3
\nleft(x^2 \u2013 6x + 9right) + left(y^2 + 4y + 4right) &= 3 + 9 + 4
\nleft(x \u2013 3right)^2 + left(y + 2right)^2 &= 16
\nfinish{align*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

Which means that the unique circle is centred at ( (3, -2)\u00a0) and its radius is ( 4 ) models.<\/p>\n

When it\u2019s 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\u00a0) models.<\/p>\n

\"blog-2020-HSC-Maths-Advanced-Exam-Paper-Solutions-Question-25\"<\/p>\n

 <\/p>\n

Query 25a<\/h3>\n

The radius of the quarter circle is ( x\u00a0) metres. We\u2019ve<\/p>\n

\n\n\n\n
\u00a0start{align*}
\ntextual content{space of backyard mattress} &= textual content{space of rectangle} + textual content{space of quarter circle}
\n&= xy + frac{1}{4} left(pi x^2right) ,.
\nfinish{align*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

Since this must be equal to ( 36 ), we now have<\/p>\n

\n\n\n\n
start{align*}
\nxy + frac{1}{4} left(pi x^2right) &= 36
\nxy &= 36 \u2013 tfrac{1}{4} left(pi x^2right)
\ny &= frac{36}{x} \u2013 frac{pi x}{4} ,.
\nfinish{align*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

Subsequently, the perimeter is given by<\/p>\n

\n\n\n\n
start{align*}
\nP &= x + y + left(x + yright) + frac{1}{4} left(2 pi xright)
\n&= 2 left(x + yright) + frac{pi x}{2}
\n&= 2 left[x + left(frac{36}{x} \u2013 frac{pi x}{4}right)right] + frac{pi x}{2}
\n&= 2x + frac{72}{x} \u2013 frac{pi x}{2} + frac{pi x}{2}
\n&= 2x + frac{72}{x} ,,
\nfinish{align*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

as required.<\/p>\n

 <\/p>\n

Query 25b<\/h3>\n

First, we differentiate ( P = 2x + frac{72}{x}) and acquire<\/p>\n

\n\n\n\n
start{align*}
\nfrac{dP}{dx} &= frac{d}{dx} left(2x + frac{72}{x}proper)
\n&= 2 \u2013 frac{72}{x^2} ,.
\nfinish{align*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

Setting ( frac{dP}{dx} = 0 ) offers ( 2 \u2013 frac{72}{x^2} = 0 ). Which means that<\/p>\n

\n\n\n\n
( frac{72}{x^2} = 2 )<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

so<\/p>\n

\n\n\n\n
( x^2 = 36 )<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

and<\/p>\n

\n\n\n\n
( x = 6 )<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

since ( x\u00a0) is the dimension of a rectangle and it have to be non-negative. When ( x = 6 ), the perimeter is given by<\/p>\n

\n\n\n\n
start{align*}
\nP &= 2 left(6right) + frac{72}{6}
\n&= 12 + 12
\n&= 24 textual content{metres}
\nfinish{align*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n
\n\n\n\n\n\n
( x )<\/td>\n( 0 < x < 6 )<\/td>\n( x = 6 )<\/td>\n( x > 6 )<\/td>\n<\/tr>\n<\/thead>\n
( frac{dP}{dx}\u00a0 )<\/td>\n( \u2013 )<\/td>\n( 0 )<\/td>\n\u00a0( + )<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

Subsequently, (\u00a0 P = 24\u00a0) is the smallest doable perimeter.<\/p>\n

 <\/p>\n

Query 26a<\/h3>\n

We use the recurrence relation beginning with<\/p>\n

\n\n\n\n
( A_0 = 60000 )<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

The sum of money within the account after the primary withdrawal is<\/p>\n

\n\n\n\n
start{align*}
\nA_1 &= A_0 left(1.005right) \u2013 800
\n&= 60000 left(1.005right) \u2013 800 ,,
\nfinish{align*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

The sum of money within the account after the second withdrawal is<\/p>\n

\n\n\n\n
start{align*}
\nA_2 &= A_1 left(1.005right) \u2013 800
\n&= left[60000 left(1.005right) \u2013 800right] left(1.005right) \u2013 800
\n&= 60000 left(1.005right)^2 \u2013 800 left(1 + 1.005right) ,,
\nfinish{align*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

And the sum of money within the account after the third withdrawal is<\/p>\n

\n\n\n\n
start{align*}
\nA_3 &= A_2 left(1.005right) \u2013 800
\n&= left[60000 left(1.005right)^2 \u2013 800 left(1 + 1.005right)right] left(1.005right) \u2013 800
\n&= 60000 left(1.005right)^3 \u2013 800 left(1 + 1.005 + 1.005^2right)
\n&approx 58492.49 quad textual content{(right to 2 d.p.)} ,.
\nfinish{align*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

 <\/p>\n

Query 26b<\/h3>\n

The curiosity within the first month is<\/p>\n

\n\n\n\n
start{align*}
\nI_1 &= A_0 instances 0.5%
\n&= 60000 instances 0.005
\n&= $ 300
\nfinish{align*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

The curiosity within the second month is<\/p>\n

\n\n\n\n
start{align*}
\nI_2 &= A_1 instances 0.5%
\n&= left(60000 instances 1.005 \u2013 800right) instances 0.005
\n&= $ 297.50
\nfinish{align*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

The curiosity within the third month is<\/p>\n

\n\n\n\n
start{align*}
\nI_3 &= A_2 instances 0.5%
\n&= left(60000 instances 1.005^2 \u2013 800 instances 1.005 \u2013 800right) instances 0.005
\n&= $ 294.99
\nfinish{align*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

Subsequently, the whole quantity of curiosity earned within the first three months is<\/p>\n

\n\n\n\n
start{align*}
\nI_1 + I_2 + I_3 &= $ 892.49
\nfinish{align*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

 <\/p>\n

Query 26c<\/h3>\n

The sum of money within the account instantly after the 94th withdrawal is<\/p>\n

\n\n\n\n
start{align*}
\nA_{94} &= A_{93} left(1.005right) \u2013 800
\n&= 60000 instances 1.005^{94} \u2013 800 instances left(1 + 1.005 + 1.005^2 + dots + 1.005^{93}proper)
\n&= 60000 instances 1.005^{94} \u2013 800 instances left(frac{1 \u2013 1.005^{94}}{1 \u2013 1.005}proper)
\n&= $ 187.85 quad textual content{right to 2 decimal locations}
\nfinish{align*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

 <\/p>\n

Query 27<\/h3>\n

From the field plot, the median of the temperature is ( 22\u00a0)\u00a0 levels Celsius. It\u2019s provided that the imply temperature is (\u00a0 0.525\u00a0)\u00a0 levels beneath the median temperature, so<\/p>\n

\n\n\n\n
start{align*}
\nbar{x} &= 22 \u2013 0.525
\n&= 21.475 textual content{levels Celsius}
\nfinish{align*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

The common variety of chirps over the 20 days is<\/p>\n

\n\n\n\n
start{align*}
\nbar{y} &= frac{684}{20}
\n&= 34.2
\nfinish{align*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

Taking the anticipated worth of either side of<\/p>\n

\n\n\n\n
(y = -10.6063 + bx\u00a0)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

offers<\/p>\n

\n\n\n\n
(bar{y} = -10.6063 + b bar{x}\u00a0)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

so<\/p>\n

\n\n\n\n
( 34.2 = -10.6063 + b left(21.475right) )<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

and<\/p>\n

\n\n\n\n
start{align*}
\nb &= frac{34.2 + 10.6063}{21.475}
\n&= 2.08644004657
\nfinish{align*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

Which means that<\/p>\n

\n\n\n\n
( y = -10.6063 + 2.08644004657x )<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

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<\/p>\n

\n\n\n\n
start{align*}
\ny &= -10.6063 + 2.08644004657 left(19right)
\n&approx 29 quad textual content{(right to the closest complete quantity)} ,.
\nfinish{align*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

 <\/p>\n

Query 28a<\/h3>\n

Let (\u00a0 X\u00a0) {dollars} be the hourly charge of pay for adults who work. We\u2019re provided that ( X sim mathcal{N} left(25, 5^2right) ) , so the likelihood that an grownup earns between ( $ 15 ) and ( $30 ) per hour is<\/p>\n

\n\n\n\n
start{align*}
\nPleft(15 leq X leq 30right) &= Pleft(frac{15 \u2013 25}{5} leq frac{X \u2013 25}{5} leq frac{30 \u2013 25}{5}proper)
\n&= Pleft(-2 leq Z leq 1right)
\n&= 0.84134 \u2013 0.02275
\n&= 0.81859
\nfinish{align*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

Subsequently,<\/p>\n

\n\n\n\n
start{align*}
\nPleft(textual content{an grownup\u2019s hourly charge of pay will not be between $15 and $30}proper) &= 1 \u2013 Pleft(15 leq X leq 30right)
\n&= 1 \u2013 0.81859
\n&= 0.18141
\nfinish{align*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

and<\/p>\n

\n\n\n\n
start{align*}
\nPleft(textual content{not less than one in all two adults earns between $15 and $30}proper) &= 1 \u2013 Pleft(textual content{neither earns between $15 and $30 an hour}proper)
\n&= 1 \u2013 0.18141^2
\n&= 0.9671
\nfinish{align*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

 <\/p>\n

Query 28b<\/h3>\n

For the reason that variety of adults who work is the same as thrice the variety of adults who don\u2019t work, we all know that<\/p>\n

\n\n\n\n
start{equation*}
\nP left(textual content{a randomly chosen grownup works}proper) = frac{1}{4}
\nfinish{equation*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

Which means that<\/p>\n

\n\n\n\n
start{align*}
\n&\u2234P left( textual content{Chosen grownup works and earns greater than} $25 textual content{ per hour} proper) = frac{3}{4} instances frac{1}{2}
\n& =frac{3}{8}
\nfinish{align*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

 <\/p>\n

Query 29a<\/h3>\n

Since<\/p>\n

\n\n\n\n
start{align*}
\nfrac{dy}{dx} &= frac{d}{dx} left(c ln xright)
\n&= frac{c}{x} ,,
\nfinish{align*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

the slope of the tangent to ( y = c ln x ) at ( x = p ) is<\/p>\n

\n\n\n\n
(frac{dy}{dx}bigg|_{x = p} = frac{c}{p} )<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

When ( x = p ), the corresponding ( y )-coordinate is<\/p>\n

\n\n\n\n
start{equation*}
\ny = c ln p
\nfinish{equation*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

Which means that the equation of the tangent to ( y = c ln x ) at\u00a0 ( x = p\u00a0) is given by<\/p>\n

\n\n\n\n
start{equation*}
\nfrac{y \u2013 c ln p}{x \u2013 p} = frac{c}{p}
\nfinish{equation*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

Multiplying either side by ( x \u2013 p\u00a0) offers ( y \u2013 c ln p = frac{c}{p} left(x \u2013 pright) ), so<\/p>\n

\n\n\n\n
start{align*}
\ny &= frac{c}{p} left(x \u2013 pright) + c ln p
\n&= frac{c}{p} x \u2013 c + c ln p ,,
\nfinish{align*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

as required.<\/p>\n

 <\/p>\n

Query 29b<\/h3>\n

For the reason that tangent has a gradient of ( 1 ), we now have<\/p>\n

\n\n\n\n
start{equation} label{eq:Q29bEq1}
\nfrac{c}{p} = 1
\nfinish{equation}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

which signifies that<\/p>\n

\n\n\n\n
start{equation} label{eq:Q29bEq2}
\nc = p
\nfinish{equation}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

Moreover, because the tangent passes via the origin, substituting ( x = 0\u00a0) and ( y = 0 ) into the equation of the tangent offers<\/p>\n

\n\n\n\n
start{equation*}
\n0 = frac{c}{p} left(0right) \u2013 c + c ln p
\nfinish{equation*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

which signifies that ( 0 = -c + c ln p ), or equivalently,<\/p>\n

\n\n\n\n
start{equation} label{eq:Q29bEq3}
\nc left(-1 + ln pright) = 0
\nfinish{equation}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

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

\n\n\n\n
start{equation*}
\n-1 + ln p = 0 ,,
\nfinish{equation*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

which signifies that<\/p>\n

\n\n\n\n
start{equation*}
\nln p = 1 ,,
\nfinish{equation*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

and<\/p>\n

\n\n\n\n
start{equation} label{eq:Q29bEq4}
\np = e
\nfinish{equation}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

Substituting ( p = e ) into ( c = p ) offers<\/p>\n

\n\n\n\n
start{equation*}
\nc = e ,.
\nfinish{equation*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

 <\/p>\n

Query 30a<\/h3>\n

Since ( A ) is the purpose of intersection of<\/p>\n

\n\n\n\n
start{equation*}
\ny = 4x \u2013 x^2
\nfinish{equation*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

and<\/p>\n

\n\n\n\n
start{equation*}
\ny = ax^2
\nfinish{equation*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

its ( x\u00a0) -coordinate is discovered by equating each equations and fixing for ( x ). This offers<\/p>\n

\n\n\n\n
start{equation*}
\n4x \u2013 x^2 = ax^2
\nfinish{equation*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

so<\/p>\n

\n\n\n\n
start{equation*}
\nleft(1 + aright) x^2 \u2013 4x = 0
\nfinish{equation*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

or equivalently,<\/p>\n

\n\n\n\n
start{equation*}
\nx left [ left(1 + a right )x \u2013 4 right] = 0
\nfinish{equation*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

Both ( x = 0\u00a0) or ( left(1 + aright) x \u2013 4 = 0 ), however from the diagram we see that the ( x\u00a0) coordinate of\u00a0 ( A\u00a0) will not be ( 0\u00a0), which signifies that we will need to have<\/p>\n

\n\n\n\n
start{equation*}
\nleft(1 + aright) x \u2013 4 = 0
\nfinish{equation*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

Rearranging this provides<\/p>\n

\n\n\n\n
start{equation*}
\nleft(1 + aright) x = 4
\nfinish{equation*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

so<\/p>\n

\n\n\n\n
start{equation*}
\nx = frac{4}{1 + a}
\nfinish{equation*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

as required.<\/p>\n

 <\/p>\n

Query 30b<\/h3>\n

The shaded space is given by<\/p>\n

\n\n\n\n
start{align*}
\nint_0^{4\/left(1+aright)} left[left(4x \u2013 x^2right) \u2013 ax^2right] , dx &= int_0^{4\/left(1+aright)} left[left(4x \u2013 x^2right) \u2013 ax^2right] , dx
\n&= int_0^{4\/left(1+aright)} left(4x \u2013 x^2 \u2013 ax^2right) , dx
\n&= left[2x^2 \u2013 frac{x^3}{3} \u2013 frac{ax^3}{3}right]_0^{4\/left(1+aright)}
\n&= 2left(frac{4}{1 + a}proper)^2 \u2013 frac{1}{3} left(frac{4}{1 + a}proper)^3 \u2013 frac{a}{3} left(frac{4}{1 + a}proper)^3
\n&= frac{32}{left(1 + aright)^2} \u2013 frac{64}{3 left(1 + aright)^3} \u2013 frac{64a}{3left(1 + aright)^3}
\n&= frac{96 left(1 + aright) \u2013 64 -64a}{3left(1 + aright)^3}
\n&= frac{96 + 96a \u2013 64 -64a}{3left(1 + aright)^3}
\n&= frac{32 left(1 + aright)}{3left(1 + aright)^3}
\n&= frac{32}{3left(1 + aright)^2}
\nfinish{align*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

Subsequently, to ensure that the shaded space to be equal to ( frac{16}{3} ), we want<\/p>\n

\n\n\n\n
start{equation*}
\nfrac{32}{3left(1 + aright)^2} = frac{16}{3}
\nfinish{equation*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

which implies<\/p>\n

\n\n\n\n
start{equation*}
\nleft(1 + aright)^2 = 2
\nfinish{equation*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

Subsequently,<\/p>\n

\n\n\n\n
start{equation*}
\n1 + a = pm sqrt{2}
\nfinish{equation*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

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

\n\n\n\n
start{equation*}
\na = -1 + sqrt{2}
\nfinish{equation*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

 <\/p>\n

Query 31a<\/h3>\n

( a\u00a0) represents the amplitude, so<\/p>\n

\n\n\n\n
start{align*}
\na &= frac{35000 \u2013 5000}{2}
\n&= 15000
\nfinish{align*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

( b\u00a0) represents the vertical displacement, so<\/p>\n

\n\n\n\n
start{align*}
\nb &= 35000 \u2013 15000
\n&= 20000
\nfinish{align*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

 <\/p>\n

Query 31b<\/h3>\n

Differentiating ( mleft(tright)\u00a0) offers<\/p>\n

\n\n\n\n
start{equation*}
\nm\u2019left(tright) = 15000left(frac{pi}{26}proper) cos left(frac{pi}{26} tright)
\nfinish{equation*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

For ( mleft(tright) ) to be growing, we assess when ( m\u2019left(tright) > 0 ). This happens when<\/p>\n

\n\n\n\n
start{equation*}
\ncos left(frac{pi}{26} tright) > 0
\nfinish{equation*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

So<\/p>\n

\n\n\n\n
start{equation*}
\nfrac{pi}{26} t < frac{pi}{2} quad textual content{or} quad frac{pi}{26} t > frac{3 pi}{2}
\nfinish{equation*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

Which means that<\/p>\n

\n\n\n\n
start{equation*}
\nt < 13 quad textual content{or} quad t > 39
\nfinish{equation*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

Likewise with ( cleft(tright) ), and we now have<\/p>\n

\n\n\n\n
start{equation*}
\nc\u2019left(tright) = frac{80 pi}{26} sin left[frac{pi}{26} left(t \u2013 10right)right]
\nfinish{equation*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

Now, ( c\u2019left(tright) > 0\u00a0) happens when<\/p>\n

\n\n\n\n
start{equation*}
\nsin left[frac{pi}{26} left(t \u2013 10right)right] > 0
\nfinish{equation*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

which signifies that<\/p>\n

\n\n\n\n
start{equation*}
\n0 < frac{pi}{26} left(t \u2013 10right) < pi
\nfinish{equation*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

so<\/p>\n

\n\n\n\n
start{equation*}
\n10 < t < 36
\nfinish{equation*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

Subsequently, each populations are growing when<\/p>\n

\n\n\n\n
start{equation*}
\n10 < t < 13
\nfinish{equation*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

 <\/p>\n

Query 31c<\/h3>\n

First, we discover when ( cleft(tright) ) reaches a most. This happens when ( c\u2019left(tright) = 0 ), so<\/p>\n

\n\n\n\n
start{equation*}
\nfrac{80 pi}{26} sin left[frac{pi}{26} left(t \u2013 10right)right] = 0 ,,
\nfinish{equation*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

or equivalently,<\/p>\n

\n\n\n\n
start{equation*}
\nsin left[frac{pi}{26} left(t \u2013 10right)right] = 0 ,.
\nfinish{equation*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

Subsequently,<\/p>\n

\n\n\n\n
start{equation*}
\nfrac{pi}{26} left(t \u2013 10right) = 0 quad textual content{or} quad frac{pi}{26} left(t \u2013 10right) = pi ,,
\nfinish{equation*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

and the stationary factors are<\/p>\n

\n\n\n\n
start{equation*}
\nt = 10 quad textual content{or} quad t = 36 ,.
\nfinish{equation*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

To find out the character of those stationary factors, we look at ( c\u201dleft(tright) ). Now,<\/p>\n

\n\n\n\n
start{equation*}
\nc\u201dleft(tright) = frac{80 pi^2}{26^2} cos left(frac{pi}{26} left(t \u2013 10right)proper) ,.
\nfinish{equation*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

At ( t = 36 ),<\/p>\n

\n\n\n\n
start{align*}
\nc\u201dleft(36right) &= -1.168
\n&< 0
\nfinish{align*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

so ( cleft(tright)\u00a0) attains its most when ( t = 36 ). At (t = 36 ) ,<\/p>\n

\n\n\n\n
start{align*}
\nm\u2019left(36right) &= frac{15000pi}{26} cos left(frac{36 pi}{26}proper)
\n&= -642.7
\nfinish{align*}<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

 <\/p>\n

 <\/p>\n

<\/div>\n","protected":false},"excerpt":{"rendered":"

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<\/p>\n","protected":false},"author":1,"featured_media":390,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[4],"tags":[],"class_list":["post-389","post","type-post","status-publish","format-standard","has-post-thumbnail","category-math"],"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/onlineduatease.com\/index.php\/wp-json\/wp\/v2\/posts\/389","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/onlineduatease.com\/index.php\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/onlineduatease.com\/index.php\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/onlineduatease.com\/index.php\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/onlineduatease.com\/index.php\/wp-json\/wp\/v2\/comments?post=389"}],"version-history":[{"count":1,"href":"https:\/\/onlineduatease.com\/index.php\/wp-json\/wp\/v2\/posts\/389\/revisions"}],"predecessor-version":[{"id":641,"href":"https:\/\/onlineduatease.com\/index.php\/wp-json\/wp\/v2\/posts\/389\/revisions\/641"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/onlineduatease.com\/index.php\/wp-json\/wp\/v2\/media\/390"}],"wp:attachment":[{"href":"https:\/\/onlineduatease.com\/index.php\/wp-json\/wp\/v2\/media?parent=389"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/onlineduatease.com\/index.php\/wp-json\/wp\/v2\/categories?post=389"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/onlineduatease.com\/index.php\/wp-json\/wp\/v2\/tags?post=389"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}