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2020 Higher Level - Paper One
Section A
Question 1
(a) \(f(x)=x^2+5x+p\) where \(x\in\mathbb{R}\), \(-3\leq p\leq8\), and \(p\in\mathbb{Z}\).
(i) Find the value of \(p\) for which \(x+3\) is a factor of \(f(x)\).
(ii) Find the value of \(p\) for which \(f(x)\) has roots which differ by \(3\).
(iii) Find the two values of \(p\) for which the graph of \(f(x)\) will not cross the \(x\)-axis.
(i) \(p=6\)
(ii) \(p=4\)
(iii) \(p=7\) or \(p=8\)
(i)
\begin{align}f(-3)=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}(-3)^2+5(-3)+p=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}p&=-9+15\\&=6\end{align}
(ii)
\begin{align}x^2+5x+p&=(x-\alpha)(x-\alpha-3)\\&=x^2-\alpha x-3x-\alpha x+\alpha^2+3\alpha\\&=x^2+(-3-2\alpha)x+(\alpha^2+3\alpha)\end{align}
We therefore have the following two equations for two unknowns:
\begin{align}5=-3-2\alpha\end{align}
\begin{align}p=\alpha^2+3\alpha\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\alpha=-4\end{align}
\begin{align}p&=\alpha^2+3\alpha\\&=(-4)^2+3(-4)\\&=4\end{align}
(iii)
\begin{align}b^2-4ac<0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}5^2-4(1)(p)<0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}25-4p<0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}-4p<-25\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\downarrow\end{align}
\begin{align}4p>25\end{align}
\begin{align}\downarrow\end{align}
\begin{align}p>6.75\end{align}
Therefore, \(p=7\) or \(p=8\) (since \(-3\leq p\leq8\)).
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(b) Find the range of values of \(x\) for which \(|2x+5|-1\leq0\), where \(x\in\mathbb{R}\).
\(-3\leq x\leq -2\)
\begin{align}|2x+5|\leq 1\end{align}
\begin{align}(2x+5)^2\leq 1^2\end{align}
\begin{align}\downarrow\end{align}
\begin{align}4x^2+20x+25\leq1\end{align}
\begin{align}\downarrow\end{align}
\begin{align}4x^2+20x+24\leq0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}x^2+5x+6\leq0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}(x+3)(x+2)\leq 0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}-3\leq x\leq -2\end{align}
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Question 2
(a) Find the two complex numbers \(z_1\) and \(z_2\) that satisfy the following simultaneous equations, where \(i^2=-1\):
\begin{align}iz_1&=-4+3i\\3z_1-z_2&=11+17i\end{align}
Write your answers in the form \(a+bi\) where \(a,b\in\mathbb{Z}\).
\(z_1=3+4i\) and \(z_2=-2-5i\)
\begin{align}iz_1=-4+3i\end{align}
\begin{align}\downarrow\end{align}
\begin{align}i(iz_1)=i(-4+3i)\end{align}
\begin{align}\downarrow\end{align}
\begin{align}-z_1=-4i+3i^2\end{align}
\begin{align}\downarrow\end{align}
\begin{align}z_1=3+4i\end{align}
and
\begin{align}3(3+4i)-z_2=11+17i\end{align}
\begin{align}\downarrow\end{align}
\begin{align}9+12i-z_2=11+17i\end{align}
\begin{align}\downarrow\end{align}
\begin{align}z_2&=-11-17i+9+12i\\&=-2-5i\end{align}
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(b) The complex numbers \(3+2i\) and \(5-i\) are the first two terms of a geometric sequence.
(i) Find \(r\), the common ratio of the sequence.
Write your answer in the form \(a+bi\) where \(a,b,\in\mathbb{Z}\).
(ii) Use de Moivre’s Theorem to find \(T_9\), the ninth term of the sequence.
Write your answer in the form \(a+bi\) where \(a,b,\in\mathbb{Z}\).
(i) \(r=1-i\)
(ii) \(48+32i\)
(i)
\begin{align}r&=\frac{T_2}{T_1}\\&=\frac{5-i}{3+2i}\\&=\frac{5-i}{3+2i}\times\frac{3-2i}{3-2i}\\&=\frac{15-10i-3i+2i^2}{9-6i+6i-4i^2}\\&=\frac{13-13i}{13}\\&=1-i\end{align}
(ii)
\begin{align}T_n=ar^{n-1}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}T_9&=ar^8\\&=(3+2i)(1-i)^8\end{align}
For \((1-i)\), the modulus is
\begin{align}r&=\sqrt{1^2+(-1)^2}\\&=\sqrt{2}\end{align}
and, since it is in the fourth quadrant, the argument is
\begin{align}\theta&=2\pi-\tan^{-1}\left(\frac{1}{1}\right)\\&=2\pi-\frac{\pi}{4}\\&=\frac{7\pi}{4}\end{align}
Therefore, we have
\begin{align}T_9&=(3+2i)(1-i)^8\\&=(3+2i)\left[\sqrt{2}\left[\cos\left(\frac{7\pi}{4}\right)+i\sin\left(\frac{7\pi}{4}\right)\right]\right]^8\\&=(3+2i)(\sqrt{2})^8\left[\cos\left(\frac{7\pi(8)}{4}\right)+i\sin\left(\frac{7\pi(8)}{4}\right)\right]\\&=(3+2i)(16)(\cos14\pi+i\sin14\pi)\\&=(3+2i)(16)(1+0i)\\&=48+32i\end{align}
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Question 3
(a) \(f(x)=6x-5\) and \(g(x)=\dfrac{x+5}{6}\). Investigate if \(f(g(x))=g(f(x))\).
\(f(g(x))=g(f(x))\)
\begin{align}f(g(x))&=6\left(\frac{x+5}{6}\right)-5\\&=x+5-5\\&=x\end{align}
and
\begin{align}g(f(x))&=\frac{(6x-5)+5}{6}\\&=\frac{6x}{6}\\&=x\end{align}
Therefore, \(f(g(x))=g(f(x))\).
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(b) The real variables \(y\) and \(x\) are related by \(y=5x^2\).
(i) The equation \(y=5x^2\) can be rewritten in the form \(\log_5y=a+b\log_5x\).
Find the value of \(a\) and the value of \(b\).
(ii) Hence, or otherwise, find the real values of \(y\) for which
\begin{align}\log_5y=2+\log_5\left(\frac{126}{25}x-1\right)\end{align}
(i) \(a=1\) and \(b=2\)
(ii) \(y=\dfrac{1}{5}\) and \(y=3{,}125\)
(i)
\begin{align}\log_5y&=\log_55x^2\\&=\log_55+\log_5x^2\\&=1+2\log_5x\end{align}
\begin{align}\downarrow\end{align}
\(a=1\) and \(b=2\)
(ii)
\begin{align}\log_5y=2+\log_5\left(\frac{126}{25}x-1\right)\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\log_5x^2=2+\log_5\left(\frac{126}{25}x-1\right)\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\log_5x^2=\log_55^2+\log_5\left(\frac{126}{25}x-1\right)\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\log_5x^2=\log_55^2+\log_5(5^2)\left(\frac{126}{25}x-1\right)\end{align}
\begin{align}\downarrow\end{align}
\begin{align}5x^2=126x-25\end{align}
\begin{align}\downarrow\end{align}
\begin{align}5x^2-126x+25=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}(5x-1)(x-25)=0\end{align}
\begin{align}\downarrow\end{align}
\(x=\dfrac{1}{5}\) or \(x=25\)
When \(x=\dfrac{1}{5}\):
\begin{align}y&=5x^2\\&=5\left(\frac{1}{5}\right)^2\\&=\frac{1}{5}\end{align}
and when \(x=25\):
\begin{align}y&=5x^2\\&=5(25^2)\\&=3{,}125\end{align}
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Question 4
The diagram below shows two functions \(f(x)\) and \(g(x)\).
The function \(f(x)\) is given by the formula \(f(x)=x^3+kx^2+15x+8\), where \(k\in\mathbb{Z}\), and \(x\in\mathbb{R}\).
(a) Given that \(f'(3)=-12\), show that \(k=-9\), where \(f'(3)\) is the derivative of \(f(x)\) at \(x=3\).
The answer is already in the question!
\begin{align}f'(x)=3x^2+2kx+15\end{align}
\begin{align}\downarrow\end{align}
\begin{align}-12=3(3^2)+2k(3)+15\end{align}
\begin{align}\downarrow\end{align}
\begin{align}-12=27+6k+15\end{align}
\begin{align}\downarrow\end{align}
\begin{align}k&=\frac{-12-27-15}{6}\\&=-9\end{align}
as required.
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(b) The function \(g(x)\) is the line that passes through the two turning points of \(f(x)=x^3-9x^2+15x+8\).
Find the equation of \(g(x)\).
\(y=-8x+23\)
\begin{align}f'(x)=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}3x^2-18x+15=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}x^2-6x+5=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}(x-1)(x-5)=0\end{align}
\begin{align}\downarrow\end{align}
\(x=1\) or \(x=5\)
\[\,\]
Points
\begin{align}f(1)&=1^3-9(1^2)+15(1)+8\\&=15\end{align}
\begin{align}\downarrow\end{align}
\begin{align}(x_1,y_1)=(1,15)\end{align}
and
\begin{align}f(5)&=5^3-9(5^2)+15(5)+8\\&=-17\end{align}
\begin{align}\downarrow\end{align}
\begin{align}(x_2,y_2)=(5,-17)\end{align}
\[\,\]
Slope
\begin{align}m&=\frac{y_2-y_1}{x_2-x_1}\\&=\frac{-17-15}{5-1}\\&=-8\end{align}
\[\,\]
Equation
\begin{align}y-y_1=m(x-x_1)\end{align}
\begin{align}\downarrow\end{align}
\begin{align}y-15=-8(x-1)\end{align}
\begin{align}\downarrow\end{align}
\begin{align}y-15=-8x+8\end{align}
\begin{align}\downarrow\end{align}
\begin{align}y=-8x+23\end{align}
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(c) Show that the graph of \(g(x)\) contains the point of inflection of \(f(x)\).
The answer is already in the question!
Setting the second derivative of \(f(x)\) to zero gives:
\begin{align}6x-18=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}x=3\end{align}
and
\begin{align}f(3)&=3^3-9(3^2)+15(3)+8\\&=-1\end{align}
Hence, the point of inflection is \((3,-1)\).
\begin{align}g(x)=-8x+23\end{align}
\begin{align}\downarrow\end{align}
\begin{align}g(3)&=-8(3)+23\\&=-1\end{align}
\begin{align}\downarrow\end{align}
\begin{align}(3,-1)\in g(x)\end{align}
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Question 5
(a) A couple agree to take out a €\(250{,}000\) mortgage in order to purchase a new home.
The loan is to be paid back monthly over \(25\) years with the repayments due at the end of each month. The bank charges an annual percentage rate (APR) which is equivalent to a monthly rate of \(0.287\%\).
Using the amortisation formula, or otherwise, find the couples’ monthly repayment on the mortgage. Give your answer in euro correct to the nearest cent.
\(1{,}244.06\mbox{ euro}\)
\begin{align}A&=P\frac{i(1+i)^t}{(1+i)^t-1}\\&=(250{,}000)\frac{(0.00287)(1+0.00287)^{300}}{(1+0.00287)^{300}-1}\\&\approx1{,}244.06\mbox{ euro}\end{align}
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(b) Another couple agree to take out a mortgage of €\(350{,}000\), at a rate of \(0.3\%\) per month, in order to purchase a new home. This loan is also to be paid back monthly over \(25\) years with the repayments due at the end of each month. The amount of each repayment is €\(1771\).
After exactly \(11\) years of repayments, the couple receive a financial windfall.
They decide to repay the remaining balance on the mortgage.
Write down a series (including the first two and last two terms) which shows the total of the present values of all the remaining monthly repayments due over the remaining \(14\) years of the mortgage (after the last monthly repayment at the end of year \(11\)).
Hence, find how much the couple will need to repay in order to clear their mortgage entirely.
Give your answer correct to the nearest cent.
Series: \(\dfrac{1771}{1.003}+\dfrac{1771}{1.003^2}+…+\dfrac{1771}{1.003^{167}}+\dfrac{1771}{1.003^{168}}\)
Repayment: \(233{,}438.25\mbox{ euro}\)
Series
\begin{align}\frac{1771}{1.003}+\frac{1771}{1.003^2}+…+\frac{1771}{1.003^{167}}+\frac{1771}{1.003^{168}}\end{align}
\[\,\]
Repayment
\begin{align}S_n=\frac{a(1-r^n)}{1-r}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}S_{168}&=\frac{\frac{1771}{1.003}\left[1-\left(\frac{1}{1.003}\right)^{168}\right]}{1-\frac{1}{1.003}}\\&\approx233{,}438.25\mbox{ euro}\end{align}
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Question 6
(a) Differentiate \((3x-5)(2x+4)\) with respect to \(x\) from first principles.
\(12x+2\)
\begin{align}f(x)&=(3x-5)(2x+4)\\&=6x^2+12x-10x-20\\&=6x^2+2x-20\end{align}
and
\begin{align}f(x+h)&=6(x+h)^2+2(x+h)-20\\&=6x^2+12hx+6h^2+2x+2h-20\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\frac{df}{dx}&=\lim_{h\rightarrow0}\frac{f(x+h)-f(x)}{h}\\&=\lim_{h\rightarrow0}\frac{(6x^2+12hx+6h^2+2x+2h-20)-(6x^2+2x-20)}{h}\\&=\lim_{h\rightarrow0}\frac{12hx+6h^2+2h}{h}\\&=\lim_{h\rightarrow0}12x+6h+2\\&=12x+2\end{align}
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(b)
(i) \(h(x)=\dfrac{1}{2}\ln(2x+3)+C\), where \(C\) is a constant.
Find \(h'(x)\), the derivative of \(h(x)\).
(ii) The diagram below shows part of the graph of the function \(h'(x)\).
The shaded region in the diagram is between the graph and the \(x\)-axis,
from \(x=0\) to \(x=A\).
This shaded region has an area of \(\mathbf{\ln 3}\) square units. Find the value of \(A\).
(i) \(h'(x)=\dfrac{1}{2x+3}\)
(ii) \(A=12\)
(i)
\begin{align}h'(x)&=\frac{1}{2}\left(\frac{1}{2x+3}\right)(2)\\&=\frac{1}{2x+3}\end{align}
(ii)
\begin{align}\int_0^A\frac{1}{2x+3}\,dx=\ln3\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\left.\frac{1}{2}\ln(2x+3)\right|_0^A=\ln3\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\frac{1}{2}[\ln(2A+3)-\ln3]=\ln3\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\frac{1}{2}\ln\left(\frac{2A+3}{3}\right)=\ln3\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\ln\left(\frac{2A+3}{3}\right)^{1/2}=\ln3\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\left(\frac{2A+3}{3}\right)^{1/2}=3\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\frac{2A+3}{3}=3^2\end{align}
\begin{align}\downarrow\end{align}
\begin{align}2A+3=27\end{align}
\begin{align}\downarrow\end{align}
\begin{align}A&=\frac{27-3}{2}\\&=12\end{align}
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Section B
Question 7
(a) A number of the form \(1+2+3+…+n\) is sometimes called a triangular number because
it can be represented as an equilateral triangle.
The diagram below shows the first three terms in the sequence of triangular numbers.
(i) Complete the table below to list the next five triangular numbers.
| Term | \(T_1\) | \(T_2\) | \(T_3\) | \(T_4\) | \(T_5\) | \(T_6\) | \(T_7\) | \(T_8\) |
|---|---|---|---|---|---|---|---|---|
|
Triangular Number |
\(1\) |
\(3\) |
\(6\) |
|
|
|
|
|
(ii) The \(n\)th triangular number can be found directly using the formula
\begin{align}T_n=\frac{n(n+1)}{2}\end{align}
Is \(1275\) a triangular number? Give a reason for your answer.
(i)
| Term | \(T_1\) | \(T_2\) | \(T_3\) | \(T_4\) | \(T_5\) | \(T_6\) | \(T_7\) | \(T_8\) |
|---|---|---|---|---|---|---|---|---|
|
Triangular Number |
\(1\) |
\(3\) |
\(6\) |
\(10\) |
\(15\) |
\(21\) |
\(28\) |
\(36\) |
(ii) Yes
(i)
| Term | \(T_1\) | \(T_2\) | \(T_3\) | \(T_4\) | \(T_5\) | \(T_6\) | \(T_7\) | \(T_8\) |
|---|---|---|---|---|---|---|---|---|
|
Triangular Number |
\(1\) |
\(3\) |
\(6\) |
\(10\) |
\(15\) |
\(21\) |
\(28\) |
\(36\) |
(ii)
\begin{align}\frac{n(n+1)}{2}=1275\end{align}
\begin{align}\downarrow\end{align}
\begin{align}n^2+n=2550\end{align}
\begin{align}\downarrow\end{align}
\begin{align}n^2+n-2550=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}(n-50)(n+51)=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}n=50\end{align}
Therefore, \(1275\) is a triangular number.
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(b)
(i) The \((n+1)\)th triangular number can be written as \(T_{n+1}=T_n+(n+1)\), where \(n\in\mathbb{N}\).
Write the expression \(\dfrac{n(n+1)}{2}+(n+1)\) as a single fraction in its simplest form.
(ii) Prove that the sum of any two consecutive triangular numbers will always be a square number (a number in the form \(k^2\), where \(k\in\mathbb{N}\)).
(iii) Two consecutive triangular numbers sum to \(12{,}544\).
Find the smaller of these two numbers.
(i) \(\dfrac{(n+1)(n+2)}{2}\)
(ii) The answer is already in the question!
(iii) \(6{,}216\)
(i)
\begin{align}T_{n+1}&=T_n+(n+1)\\&=\frac{n(n+1)}{2}+(n+1)\\&=\frac{n(n+1)+2(n+1)}{2}\\&=\frac{(n+1)(n+2)}{2}\end{align}
(ii)
\begin{align}T_{n+1}+T_n&=\frac{(n+1)(n+2)}{2}+\frac{n(n+1)}{2}\\&=\frac{(n+1)(2n+2)}{2}\\&=2\frac{(n+1)(n+1)}{2}\\&=(n+1)^2\end{align}
(iii)
\begin{align}(n+1)^2=12{,}544\end{align}
\begin{align}\downarrow\end{align}
\begin{align}n+1=112\end{align}
\begin{align}\downarrow\end{align}
\begin{align}n=111\end{align}
\begin{align}\downarrow\end{align}
\begin{align}T_{111}&=\frac{(111)(111+1)}{2}\\&=6{,}216\end{align}
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(c) Some numbers are both triangular and square, for example \(36\).
Leonhard Euler (1778) discovered the following formula for these numbers
\begin{align}N_k=\left(\frac{(3+2\sqrt{2})^k-(3-2\sqrt{2})^k}{4\sqrt{2}}\right)^2\end{align}
where \(N_k\) is the \(k\)th number that is both triangular and square.
Use Euler’s formula to find \(N_3\), the third number that is both triangular and square.
\(1{,}225\)
\begin{align}N_3&=\left(\frac{(3+2\sqrt{2})^3-(3-2\sqrt{2})^3}{4\sqrt{2}}\right)^2\\&=1{,}225\end{align}
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(d) Prove using induction that, for all \(n\in\mathbb{N}\), the sum of the first \(n\) square numbers can be found using the formula:
\begin{align}1^2+2^2+3^2+4^2+…+n^2=\frac{n(n+1)(2n+1)}{6}\end{align}
The answer is already in the question!
When \(n=1\), the right hand side is
\begin{align}\frac{1(1+1)(2(1)+1)}{6}&=\frac{1(2)(3)}{6}\\&=1\end{align}
which is indeed equal to the left hand side of \(1^2\).
Therefore, let us assume that this relation is true for \(n=k\), i.e. that
\begin{align}1^2+2^2+3^2+4^2+…+k^2=\frac{k(k+1)(2k+1)}{6}\end{align}
Using this assumption, we now wish to prove that it is true for \(n=k+1\), i.e. that
\begin{align}1^2+2^2+3^2+4^2+…+k^2+(k+1)^2=\frac{(k+1)(k+2)(2k+3)}{6}\end{align}
According to our assumption, we can rewrite the left hand side as
\begin{align}1^2+2^2+3^2+4^2+…+k^2+(k+1)^2&=\frac{k(k+1)(2k+1)}{6}+(k+1)^2\\&=\frac{k(k+1)(2k+1)+6(k+1)^2}{6}\\&=\frac{(k+1)[k(2k+1)+6(k+1)]}{6}\\&=\frac{(k+1)(2k^2+7k+6)}{6}\\&=\frac{(k+1)(k+2)(2k+3)}{6}\end{align}
which is indeed equal to the right hand side.
Therefore, the formula is indeed true for all \(n\in\mathbb{N}\).
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Question 8
A rectangle is inscribed in a circle of radius \(5\) units and centre \(O(0,0)\) as shown below.
Let \(R(x,y)\), where \(x,y\in\mathbb{R}\), be the vertex of the rectangle in the first quadrant as shown.
Let \(\theta\) be the angle between \([OR]\) and the positive \(x\)-axis, where \(0\leq\theta\leq\dfrac{\pi}{2}\).
(a)
(i) The point \(R(x,y)\) can be written as \(a\cos \theta, b\sin\theta\), where \(a,b\in\mathbb{R}\).
Find the value of \(a\) and the value of \(b\).
(ii) Show that\(A(\theta)\), the area of the rectangle, measured in square units, can be written as \(A(\theta)=50\sin2\theta\).
(iii) Use calculus to show that the rectangle with maximum area is a square.
(iv) Find this maximum area.
(i) \(a=5\) and \(b=5\)
(ii) The answer is already in the question!
(iii) The answer is already in the question!
(iv) \(50\mbox{ square units}\)
(i)
\begin{align}x=5\cos\theta&&y=5\sin\theta\end{align}
\begin{align}\downarrow\end{align}
\(a=5\) and \(b=5\)
(ii)
\begin{align}A(\theta)&=(2\times5\sin\theta)(2\times5\cos\theta)\\&=100\sin\theta\cos\theta\\&=50(2\sin\theta\cos\theta)\\&=50\sin2\theta\end{align}
(iii)
Setting the derivative to zero, we obtain:
\begin{align}(50\cos2\theta)(2)=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\cos2\theta=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}2\theta&=\cos^{-1}(0)\\&=90^{\circ}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\theta=45^{\circ}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}l&=10\cos45^{\circ}\\&=\frac{10}{\sqrt{2}}\end{align}
and
\begin{align}w&=10\sin45^{\circ}\\&=\frac{10}{\sqrt{2}}\end{align}
As these are the same, the maximum area is a square.
(iv)
\begin{align}A&=l\times w\\&=\left(\frac{10}{\sqrt{2}}\right)\times\left(\frac{10}{\sqrt{2}}\right)\\&=50\mbox{ square units}\end{align}
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(b) A person who is \(2\mbox{ m}\) tall is walking towards a streetlight of height \(5\mbox{ m}\) at a speed of \(1.5\mbox{ m/s}\).
Find the rate, in \(\mbox{m/s}\), at which the length of the person’s shadow (\(x\)), cast by the streetlight, is changing.
\(1\mbox{ m/s}\)
\begin{align}\frac{5}{2}=\frac{l+x}{x}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}5x=2(l+x)\end{align}
\begin{align}\downarrow\end{align}
\begin{align}5x=2l+2x\end{align}
\begin{align}\downarrow\end{align}
\begin{align}3x=2l\end{align}
\begin{align}\downarrow\end{align}
\begin{align}x=\frac{2}{3}l\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\frac{dx}{dl}=\frac{2}{3}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\frac{dx}{dt}&=\frac{dx}{dl}\frac{dl}{dt}\\&=\left(\frac{2}{3}\right)(1.5)\\&=1\mbox{ m/s}\end{align}
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Question 9
The number of bacteria in the early stages of a growing colony of bacteria can be approximated
using the function:
\begin{align}N(t)=450e^{0.065t}\end{align}
where \(t\) is the time, measured in hours, since the colony started to grow, and \(N(t)\) is the number of bacteria in the colony at time \(t\).
(a)
(i) Find the number of bacteria in the colony after \(4.5\) hours.
Give your answer correct to the nearest whole number.
(ii) Find the time, in hours, that it takes the colony to grow to \(790\) bacteria.
Give your answer correct to \(1\) decimal place.
(i) \(603\)
(ii) \(8.7\mbox{ h}\)
(i)
\begin{align}N(4.5)&=450e^{0.065(4.5)}\\&\approx603\end{align}
(ii)
\begin{align}450e^{0.065t}=790\end{align}
\begin{align}\downarrow\end{align}
\begin{align}e^{0.065t}=\frac{790}{450}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}0.065t=\ln\left(\frac{790}{450}\right)\end{align}
\begin{align}\downarrow\end{align}
\begin{align}t&=\frac{1}{0.065}\ln\left(\frac{790}{450}\right)\\&=8.7\mbox{ h}\end{align}
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(b) Using the function \(N(t)=450e^{0.065t}\), find the average number of bacteria in the colony during the period from \(t=3\) to \(t=12\).
Give your answer correct to the nearest whole number.
\(743\)
\begin{align}A&=\frac{1}{b-a}\int_a^bN(t)\,dt\\&=\frac{1}{12-3}\int_3^{12}450e^{0.065t}\,dt\\&=\frac{450}{9(0.065)}(e^{0.065(12)}-e^{0.065(3)})\\&\approx743\end{align}
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(c) Find the rate at which \(N(t)=450e^{0.065t}\) is changing when \(t=12\).
Give your answer correct to one decimal place.
Interpret this value in the context of the question.
In the twelfth hour, the number of bacteria is increasing at a rate of \(63.8\) bacteria per hour.
\begin{align}N'(t)&=(0.065)(450e^{0.065t})\\&=29.25e^{0.065t}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}N'(12)&=29.25e^{0.065(12)}\\&\approx63.8\end{align}
In the twelfth hour, the number of bacteria is increasing at a rate of \(63.8\) bacteria per hour.
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(d) After \(k\) hours, the rate of increase of \(N(t)\) is greater than \(90\) bacteria per hour.
Find the least value of \(k\), where \(k\in\mathbb{N}\).
\(18\)
\begin{align}N'(k)>90\end{align}
\begin{align}\downarrow\end{align}
\begin{align}29.25e^{0.065k}>90\end{align}
\begin{align}\downarrow\end{align}
\begin{align}e^{0.065k}>\frac{90}{29.25}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}0.065k>\ln\left(\frac{90}{29.25}\right)\end{align}
\begin{align}\downarrow\end{align}
\begin{align}k>\frac{1}{0.065}\ln\left(\frac{90}{29.25}\right)\end{align}
\begin{align}\downarrow\end{align}
\begin{align}k>17.29…\end{align}
Therefore, the least value of \(k\) is \(18\).
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(e) The number of bacteria in the early stages of a different colony of bacteria can be approximated using the function:
\begin{align}P(t)=220e^{0.17t}\end{align}
where \(P(t)\) is the number of bacteria and \(t\) is measured in hours.
Assume that both colonies start growing at the same time.
Find the time, to the nearest hour, at which the number of bacteria in both colonies will be equal.
\(7\mbox{ hr}\)
\begin{align}N(t)=P(t)\end{align}
\begin{align}\downarrow\end{align}
\begin{align}450e^{0.065t}=220e^{0.17t}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\frac{e^{0.065t}}{e^{0.17t}}=\frac{220}{450}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}e^{-0.105t}=\frac{220}{450}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}-0.105t=\ln\left(\frac{220}{450}\right)\end{align}
\begin{align}\downarrow\end{align}
\begin{align}t&=-\frac{1}{0.105}\ln\left(\frac{220}{450}\right)\\&\approx7\mbox{ hr}\end{align}
