(a)          \(R = \left( 30 – 7 \ln \left( \frac{q }{5 } \right) \right)q\)
              \(R = 30q – 7q \ln \left( \frac{q }{5 } \right)  \)        A1

(b)          \(R = 30 q  - 7q\ln(q) + 7q\ln(5)  \)             M1
\(\frac{dR}{dq} =  30 – 7q (\frac{1}{q}) – 7\ln(q) + 7\ln(5)\)     A1

OR        
              \(R = \left( 30 – 7 \ln \left( \frac{q }{5 } \right) \right)\)
\(\frac{dR}{dq} = \left( 30 – 7 \ln \left( \frac{q }{5 } \right) \right) + q(\frac{-7}{q} )    \)   M1 A1
M1 for a reasonable application of the product rule.
 \(\frac{dR}{dq} = 23 – 7 \ln \left( \frac{q }{5 } \right) .\)
Set \( \frac{dR}{dq} = 0.\)

\(23 – 7 \ln \left( \frac{q }{5 } \right) = 0 \)    M1
      \(  7 \ln \left( \frac{q }{5 } \right) = 23\)
          \(   \left( \frac{q }{5 } \right) = e^{ \frac{23}{7} }\)
             \(q = 5e^{ \frac{23}{7} }\)               A1
Since quantity of panini is an integer than \(q = 134\).     A1
        
\( p = 30 - 7 \ln \left( \frac{134 }{5 } \right)  \)       M1
              \(p \approx 6.98\) Hong Kong dollar  (3s.f.)        A1

(c)         \( p = 30 – 7 \ln \left( \frac{q }{5 } \right)\)
\(p =  30 + 7 \ln(5) – 7 \ln(q)  \)  (A1)          
\(\frac{dp}{dq} =   \frac{-7}{q}  \)     A1
 

(d)            \(  p(5) = 30 - 7 \ln \left( \frac{5 }{5 } \right)    \)   
                 \( p(5)=  30  \)    A1
We  have \(q=5\) and \(p=30\).
\(\frac{dq}{dp}  = {1 \over \frac{dp}{dq}}\)
\(\frac{dq}{dp} \vert_{p=30}  = {1 \over \left( \frac{-7}{q}  \right)}\)
\(\frac{dq}{dp} \vert_{p=30}  = \frac{q}{-7}\)           A1
Using \(q=5\) and \(p=30\).
\(y =  \frac{5}{-7} \times \frac{30}{5}   \)           A1
\(y =  \frac{-30}{7} \)     OR  -4.29 (3s.f.)                    A1

(e)   Substituting in (134, 6.98) [p and q for maximum revenue] to the price elasticity of demand gives:
\(y = \frac{134}{-7} \times \frac{6.98}{134} \)
\(y \approx -0.997\)  (3s.f.)      
When the price is adjusted to the price that maximizes the revenue, the price elasticity of demand approaches -1.   A1 
                     
Note: If \(q\) is allowed to be a real number then price that maximizes the revenue is \(p=7\) with \(q = 5e^{ \frac{23}{7} } \)and \(y = -1\). In reality, a panini seller in the school cafetaria or school fair will mostly set the panini price as HKD 7.00 rather than HKD 6.98. 

(a)          Use (0,3.32) then
\(3.32 = e^{ a + 0.2\cos (\frac{0 \pi}{b} )} \) M1
\(3.32 = e^{ a + 0.2\cos(0) } \)
\(3.32 = e^{ a + 0.2} \)        M1
\(\ln(3.32) = a + 0.2\)
\(a = \ln(3.32) – 0.2 \)          A1
Since \(a\) is an integer then \(a = 1\).  R1

(b)  \(\frac{dy}{dx} = e^{ a + 0.2\cos (\frac{x \pi}{b} )} \left( \frac{- \pi}{5b} \sin \left( \frac{x\pi}{b}  \right) \right)\)      M1A1A1
M1 for a reasonable attempt to apply chain rule. A1 for each correct term.
Let \(y = e^{u}\) and \(u = 1 + 0.2\cos (\frac{x \pi}{b} )\). 
\(\frac{du}{dx} = 0 + 0.2( \frac{\pi}{ b} )(-\sin (\frac{x\pi}{b} )  )\)  or \(\frac{du}{db} = ( \frac{ -\pi}{ 5b} )(\sin (\frac{x\pi}{b} )  )\)
\(\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}\).
\(\frac{dy}{dx} = e^{ a + 0.2\cos (\frac{x \pi}{b} )} \left(  \frac{- \pi}{ 5b} \sin (\frac{x\pi}{b} )                \right)\)

(c)          The distance between the two “peaks” is the period, and the period of the function is determined by \(\frac{- \pi}{5b} \sin \left( \frac{x\pi}{b}  \right)\).
period = \(\frac{2\pi}{ \frac{\pi}{b}}    \)        M1
period = \(2b \)
Thus, the distance between two “peaks” is \(2b \). A1

(d)          If the two peaks are at \(x=0\) and \(x=13\).
Recall that a period measures the distance between two peaks.
Thus, period is 13 in this case and using the result in (c), we have
\(2b = 13\) and \(b = \frac{13}{2}\) or \(b = 6.50\).         

If there are four peaks and there will be 3 revolutions within 13
and each period is
Thus, \(2b = \frac{13}{3}\)
and \(b = \frac{13}{6}\).
M1 for similar reasoning as above.
Thus, \(\frac{13}{6} \lt b \leq 6.50\). A1A1
Award A1A0 if the student failed to use the correct inequality signs.

(e) (i)  Let \(u = 1 + 0.2\cos (\frac{x \pi}{b} )\)  and \(\frac{du}{db} = 0 + 0.2( \frac{- x \pi}{ b^2} )(-\sin (\frac{x\pi}{b} )  )\).

We have \(y = e^{u}\) and \(\frac{du}{db} = 0 + 0.2( \frac{- x \pi}{ b^2} )(-\sin (\frac{x\pi}{b} )  )\).
\(\frac{dy}{db} = \frac{dy}{du} \times \frac{du}{db}\)
\(\frac{dy}{db} = e^{ a + 0.2\cos (\frac{x \pi}{b} )} \left(  \frac{x \pi}{ 5b^2} \sin (\frac{x\pi}{b} )                \right)\)
M1 for a reasonable attempt to apply chain rule. A1 for each correct term.

(ii)  \(\frac{dy}{db}\) is the instantaneous change in the height of a point of the fairy light decoration  from the ground in respond to a change in variable \(b\). A1

(a)   \(L = 4.41 + 7.43 \ln (0.92x) \)     M1
\(L = 4.41 + 7.43 \ln(0.92) + 7.43 \ln(x)   \)        
\(\frac{dL}{dx} = \frac{7.43}{x}  \)     A1
OR
From \( x = \frac{E}{0.92}\) we have \(E = 0.92x\) and \(\frac{dE}{dx} = 0.92\).
\(\frac{dL}{dx} = \frac{dL}{dE} \times \frac{dE}{dx}  \)           
\(\frac{dL}{dx} = \frac{7.43}{E} \times ( 0.92)  \)         M1
Since \(E = 0.92x\) then \(\frac{dL}{dx} = \frac{7.43}{0.92x} \times ( 0.92)\).
So \(\frac{dL}{dx} = \frac{7.43}{x}\) A1

(b)  Let the national income of a country doubles from \(x_o\) to \(2x_o\).
So \( \frac{dL}{dx} \vert_{x=2x_o} = \frac{7.43}{2x_o} \)or  \(\frac{1}{2} \times \frac{7.43}{x_o} \)               A1
Thus, \(\frac{dL}{dx} \vert_{x=2x_o} = \frac{1}{2} \frac{dL}{dx}\)
The rate of life expectancy at birth increases only by a half from that of the rate before the 
national income of a country doubles.  A1

(c)   \(\frac{dL}{dx} = 7.43x^{-1}\)
\(  \frac{d^2v}{dt^2} = (-1) 7.43x^{-2}    \)                   
 \( \frac{d^2v}{dt^2} = -7.43x^{-2} \)  OR      \( \frac{-7.43}{x^2} \)            A1

(d)   As national income \(x\) increases the rate at which life expectancy increases (rate of change of dL/dE) decreases  A1
The model increases indefinitey implying that life expectancy at birth, whilst the rate of increase does decline, continues to increase indefinitely (as income per capita increases). At the current time (and state of technology etc), it is not realistic to state life expectancy has no upper limit.  R1
 

(a) Since a peak is at \((h, 20)\) then \(20 = a \sin \left( 3h + \frac{\pi}{b} \right) \)and
\( \sin \left( 3h + \frac{\pi}{b} \right) = 1\).       
Thus, \(a(1) = 20\)
\(a = 20   \)       A1
\(0 = 20 \sin  \left( 3( \frac{\pi}{4}) + \frac{\pi}{b} \right) \)     M1
\(\sin (\frac{3\pi}{4}+\frac{\pi}{b}) = \sin (\pi)\)    (A1)
\(\frac{\pi}{b}= \frac{\pi}{4}\)
Thus,  \( b = 4\).           A1
 

(b)   \(v = 20 \sin \left( 3(t) + \frac{\pi}{4} \right)\)
\(\frac{dv}{dt} = 20(3)\cos \left( 3(t) + \frac{\pi}{4} \right) \)  M1
\(\frac{dv}{dt} = 60 \cos \left( 3(t) + \frac{\pi}{4} \right)\)    A1
Set \(\frac{dv}{dt} = 0\) for a stationary point
Thus,  \(\cos \left( 3(t) + \frac{\pi}{4} \right) = 0  \)        M1
\(3(t) + \frac{\pi}{4} = \frac{\pi}{2}\)
\(3(t)  = \frac{\pi}{4}\)   A1
This calculate \(t\) is our desired \(h\).
Thus, \( h = \frac{\pi}{12}\)         AG

(c) Chain rule can be used. 
Let \(u = 3t + \frac{\pi}{4}\) and \(\frac{dv}{dt} = 60 \cos (u)\). 
Then \(\frac{du}{dt} = 3\) and \(\frac{ d \frac{dv}{dt}} {du} = -60\sin(u)\)
Thus, \(\frac{d^2v}{dt^2} = \frac{ \frac{dv}{dt} }{du} \times \frac{du}{dt}\)

 \( \frac{d^2v}{dt^2} = 3(-60) \sin \left( 3t + \frac{\pi}{4} \right)  \)
OR
\(\frac{d^2v}{dt^2} = -180 \sin \left( 3t + \frac{\pi}{4} \right)  \)   M1A1

(d) Method 1:
Set \(\frac{d^2v}{dt^2} = 0\)
\(\sin \left( 3k + \frac{\pi}{4} \right)     = 0 \)   M1
\(3k + \frac{\pi}{4} = \pi \)  OR   \(3k + \frac{\pi}{4} = 2\pi \)        A1
\(k = \frac{\pi}{4} \)   OR  \(k = \frac{7\pi}{12}     \)          
Let us check the value \(k = \frac{\pi}{4}. \)  
\(v’( \frac{\pi}{4}  ) = 60 \cos \left( \frac{3\pi}{4}  + \frac{\pi}{4} \right)\)
     \(  = 60 \cos (\pi)\)
   \(    < 0\)
So \(k = \frac{\pi}{4}\)  is rejected because the first derivative at \( t= \frac{\pi}{4}\) is less than zero.
Let us check the value \(k = \frac{7\pi}{12}\).   
\(v’( \frac{7\pi}{12}  ) = 60 \cos \left( \frac{21\pi}{12}  + \frac{\pi}{4} \right)\)
      \( = 60 \cos (2\pi)\)
       \(> 0    \)   A1
Since the first derivative at \( t = \frac{7\pi}{4}\) is more than zero as required then .   
\(k = \frac{7\pi}{12}\) seconds   A1
Method 2:

M1 for graphing A1 for a correct labelled graph.
Note: We simply graph the second derivative of the original function that has been defined in f1(x) previously in the GDC. 
\(t \approx 1.83\) seconds  (3s.f.) A2 
Note: From the given graph, the gradient is positive and steepest at some point after \(t = 1.5\).
Thus, we select the second root from the left in the graph of \(\frac{d^2v}{dt^2}\).
Basically, we need a maximum of \(\frac{dv}{dt}\) and thus \(\frac{d^2v}{dt^2} \)has to go from positive to negative around the maximum point. Thus, the answer is the second root from the left of the origin in the graph of \(\frac{d^2v}{dt^2}\).

(ii)  When the voltage is increasing, the change in the increase of voltage is the largest at time \( t = \frac{7\pi}{12}\) seconds (or \(t \approx 1.83\) seconds).  A1

(a)
Values \(x\) and \(P\) are entered into L1 and L2. Note that I did the calculation for (a) in column L3.
I reported the results to 3 significant figures below. L4 is the ln(L2) and will be used in (d).

A1A0 for two correct values. A1A1 for all correct values.

(b) Since \(P = Ae^{bx} \) then \(P(x) = Ae^{bx} \)and \(P(d+k) = Ae^{b(x+k)}\).
We shall consider
\(\left( \dfrac {P(x+k)}{ P(x)} \right)^{ \frac{1}{k} }  = \left( \dfrac { Ae^{b(x+k)} }{ Ae^{bx} }  \right)^{ \frac{1}{k} }\)   A1
\( \left( \dfrac {P(x+k)}{ P(x)} \right)^{ \frac{1}{k} }  = \left( \dfrac { Ae^{bx}e^{bk}}{ Ae^{bx} }  \right)^{ \frac{1}{k} } \)  A1
\(\left( \dfrac {P(x+k)}{ P( x)} \right)^{ \frac{1}{k} }  = \left( e^{bk}  \right)^{ \frac{1}{k} }\)
\(\left( \dfrac {P(x+k)}{ P( x)} \right)^{ \frac{1}{k} }  = e^b\)  which is a constant.          A1     

(c)\( P = Ae^{bx}\)
  \( ln P = ln (Ae^{bx} ) \)  A1
 \(  ln P = lnA + ln(e^{bx} )\)
   \(ln P = lnA + bx  \)       A1       

 (d) (i) 
\(  ln P = 0.00152  + 0.115x \)  (3s.f.)  (M1)A1

(ii) \(\dfrac {dp}{dx} = Abe^{bx}\)    M1
   \(  \dfrac {dp}{dx} = 0.115e^{0.00152}e^{0.115x}      \)          
     OR        
  \(\dfrac {dp}{dx} = 0.115e^{0.115x}\)                A1      
              
(iii) \(\dfrac {dp}{dx} = Abe^{bx}\)
     \( \dfrac {dp}{dx} = bp \)   where \(p = Ae^{bx}   \) 
     \( \dfrac {dp}{dx} = 0.115p \)   where \(p = Ae^{bx} \)    A1
      Thus, \(\dfrac{dp}{dx} \left( \dfrac{1}{p} \right)= b \) or 0.115    
      The value \(\dfrac{dp}{dx} \left( \dfrac {1}{P} \right) = 0.115\) is a constant.  A1

OR
 \(  ln P= 0.00152  + 0.115x  \)
\(\frac{d(lnP)}{dx} = \frac{d(0.00152+0.115x)}{dx}\)     A1

\(\left( \frac{1}{P} \right) \frac{dP}{dx} = 0.115\)
Thus, the value of  \(\left( \dfrac{1}{P} \right) \dfrac{dP}{dx} = 0.115\)  is a constant. A1