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Calculus, MET2 2025 VCAA 4

Consider the function  \(f:\left[0, \dfrac{5 \pi}{2}\right] \rightarrow R, f(x)=\sin (x)+1\).

The graph of  \(y=f(x)\)  is shown below.
 

   

  1. Evaluate  \(f\left(\dfrac{2 \pi}{3}\right)\).   (1 mark)

  2. Find the exact values of \(x\) for which  \(f(x)=\dfrac{3}{2}\).   (1 mark)

  3. There exist real numbers \(a\) and \(k\) in the interval \(\left(0, \dfrac{5 \pi}{2}\right)\), such that  \(f(x+k)=f(x)\) for all  \(x \in[0, a]\).
  4. Find the value of \(k\) and the largest possible value of \(a\).   (2 marks)

  5. Consider the tangent to the graph of  \(y=f(x)\) at the point \(A\) where  \(x=\dfrac{2 \pi}{3}\), as shown on the axes below.
     

  1. Find the equation of the tangent to the graph of \(y=f(x)\) at the point where  \(x=\dfrac{2 \pi}{3}\).   (1 mark)

  2. Apply two iterations of Newton's method to \(f\) with  \(x_0=\dfrac{2 \pi}{3}\).
    1. Write down \(x_2\), correct to one decimal place.   (1 mark)

    2. On the axes in part d, draw the tangent to the graph of  \(y=f(x)\) at the point where  \(x=x_1\).
    3. Answer on the graph in part d.   (1 mark)

  3. Now consider the line \(y=t(x)\), which is the tangent to the graph of  \(y=f(x)\) at the point  \((p, f(p))\), where  \(p \in\left(0, \dfrac{5 \pi}{2}\right)\).

      1. Show that  \(t(x)=\cos (p)(x-p)+\sin (p)+1\).   (2 marks)

      2. Determine the minimum and maximum possible values for the \(y\)-intercept of  \(y=t(x)\), for  \(p \in\left(0, \dfrac{5 \pi}{2}\right)\).   (2 marks)

      3. Determine the values of \(p\) for which  \(y=t(x)\) has a unique \(x\)-intercept that is equal to the \(x\)-intercept of  \(y=f(x)\).
      4. Give your answers correct to two decimal places.   (2 marks)

  1. Let  \(g:\left[0, \dfrac{5 \pi}{2}\right] \rightarrow R, g(x)=a x^3+b x^2+c x+d\)  be a polynomial function, where \(a, b, c, d \in R\).
  2. Suppose  \(g(0)=f(0)\)  and  \(g^{\prime}(0)=f^{\prime}(0)\).
    1. Show that  \(c=1\)  and  \(d=1\).   (2 marks)

    2. If  \(g(2 \pi)=f(2 \pi)\) and  \(g^{\prime}(2 \pi)=f^{\prime}(2 \pi)\), determine the area bounded by the graphs of  \(y=f(x)\)  and  \(y=g(x)\), for  \(x \in[0,2 \pi]\).
    3. Give your answer correct to two decimal places.    (2 marks)

    4. Let  \(a=0, c=1, d=1\).
    5. Find \(b\) and \(r\), such that  \(g(r)=f(r)\) and  \(g^{\prime}(r)=f^{\prime}(r)\), where  \(b \in R\) and  \(r \in\left(0, \dfrac{5 \pi}{2}\right)\).   (2 marks)

Show Answers Only

a.    \(f\left(\dfrac{2 \pi}{3}\right)=\dfrac{2+\sqrt{3}}{2}\)
 

b.   \(x=\dfrac{\pi}{6}, \dfrac{5 \pi}{6}, \dfrac{13 \pi}{6}\)
 

c.    \(k=2 \pi, \ \ a=\dfrac{\pi}{2}\)
 

d.    \(y=-\dfrac{x}{2}+\dfrac{\pi}{3}+\dfrac{\sqrt{3}}{2}+1\)
 

e.i.  \(x_2=5.2\)
 

e.ii.

f.i.    \(f^{\prime}(p)=\cos (p)\)

\(\text{Equation of tangent at}\ (p, \sin (p)+1):\)

\(y-(\sin (p)+1)\) \(=\cos (p)(x-p)\)
\(t(x)\) \(=\cos (p)(x-p)+\sin (p)+1\)

 

f.ii.  \(y \text{-int (min)}=1-2 \pi \ \text { (at } p=2 \pi)\)

\(y \text{-int (max)}=1+\pi \ \text { (at } p= \pi)\)
 

f.iii. \(p=2.38 \ \text{or} \ 7.04\)
 

g.i.  \(g(x)=a x^3+b x^2+c x+d, \ f(x)=\sin (x)+1\)

\(\text{Given} \ \ f(0)=g(0):\)

\(d=\sin (0)+1=1\)
 

\(g^{\prime}(x)=3 a x^2+2 b x+c, \ f^{\prime}(x)=\cos (x)\)

\(\text{Given} \ \ f^{\prime}(0)=g^{\prime}(0):\)

\(c=1\)
 

g.ii.  \(\text {Area}=1.53\)
  

g.iii. \(r=\pi, \ b=-\dfrac{1}{\pi}\)

Show Worked Solution

a.    \(f(x)=\sin (x)+1\)

\(f\left(\dfrac{2 \pi}{3}\right)=\sin \left(\dfrac{2 \pi}{3}\right)+1=\dfrac{2+\sqrt{3}}{2}\)
 

b.   \(\text{Solve \(\ f(x)=\dfrac{3}{2} \ \) for \(x\) (by CAS):}\)

\(\sin (x)+1=\dfrac{3}{2} \ \Rightarrow \ \sin (x)=\dfrac{1}{2}\)

\(\text{Solve} \ \ \sin (x)=\dfrac{1}{2} \ \text { for } \ x \in\left[0, \dfrac{5 \pi}{2}\right]:\)

\(x=\dfrac{\pi}{6}, \dfrac{5 \pi}{6}, \dfrac{13 \pi}{6}\)
 

c.    \(f(x+k)=f(x) \ \ \text{for} \ \ x \in[0, a]\)

\(\text{By inspection of graph:}\)

\(k=2 \pi, \ \ a=\dfrac{\pi}{2}\)

♦♦ Mean mark (c) 28%.

d.    \(f(x)=\sin (x)+1 \ \Rightarrow \ f^{\prime}(x)=\cos (x)\)

\(f^{\prime}\left(\dfrac{2 \pi}{3}\right)=-\dfrac{1}{2}\)

\(\text{Find equation of line} \ \ m_2=-\dfrac{1}{2} \ \ \text{through}\ \ \left(\dfrac{2 \pi}{3}, \dfrac{2+\sqrt{3}}{2}\right):\)

\(y=-\dfrac{x}{2}+\dfrac{\pi}{3}+\dfrac{\sqrt{3}}{2}+1\)
 

e.i.  \(x_0=\dfrac{2 \pi}{3}\)

\(x_1=\dfrac{2 \pi}{3}-\dfrac{f\left(\dfrac{2 \pi}{3}\right)}{f^{\prime}\left(\dfrac{2 \pi}{3}\right)}=5.8264 \ldots\)

\(x_2=5.8264 \ldots-\dfrac{f(5.8264)}{f^{\prime}(5.8264)}=5.2 \ \text{(1 d.p.)}\)
 

e.ii.

♦♦ Mean mark (e.ii) 28%.

f.i.    \(f^{\prime}(p)=\cos (p)\)

\(\text{Equation of tangent at}\ (p, \sin (p)+1):\)

\(y-(\sin (p)+1)\) \(=\cos (p)(x-p)\)
\(t(x)\) \(=\cos (p)(x-p)+\sin (p)+1\)

 

f.ii.  \(t(x)=\cos (p) x+\sin (p)+1-p \times \cos (p)\)

\(y\text{-intercept}=\sin (p)+1-p \times \cos (p)\)

\(\text{Find max/min of} \ y\text{-int for} \ p \in\left[0, \dfrac{5 \pi}{2}\right] \ \ \text{(by CAS):}\)

\(y \text{-int (min)}=1-2 \pi \ \text { (at } p=2 \pi)\)

\(y \text{-int (max)}=1+\pi \ \text { (at } p= \pi)\)

♦♦ Mean mark (f.ii) 26%.
♦♦♦ Mean mark (f.iii) 23%.

f.iii. \(x \text{-intercept of} \ f(x) \ \text{occurs at} \ \ x=\dfrac{3 \pi}{2}\)

\(\text{Solve} \ \ t\left(\dfrac{3 \pi}{2}\right)=0 \ \ \text {for}\  p:\)

\(p=2.38 \ \text{or} \ 7.04\)
 

g.i.  \(g(x)=a x^3+b x^2+c x+d, \ f(x)=\sin (x)+1\)

\(\text{Given} \ \ f(0)=g(0):\)

\(d=\sin (0)+1=1\)
 

\(g^{\prime}(x)=3 a x^2+2 b x+c, \ f^{\prime}(x)=\cos (x)\)

\(\text{Given} \ \ f^{\prime}(0)=g^{\prime}(0):\)

\(c=1\)
 

g.ii.  \(g(x)=a x^3+b x^2+x+1 \ \Rightarrow \ g^{\prime}(x)=3 a x^2+2 b x+1\)

\(\text{Given \(\ g(2 \pi)=f(2 \pi)=1\ \) and \(\ \ g^{\prime}(2 \pi)=f^{\prime}(2 \pi)=1\)}\)

\(\text{Solve \(\ g(2 \pi)=1 \ \) and \(\ g^{\prime}(2 \pi)=1\)  simultaneously for \(a, b\):}\)

\(a=\dfrac{1}{2 \pi^2}, \ b=-\dfrac{3}{2 \pi}\)

♦♦ Mean mark (g.ii) 33%.
♦♦♦ Mean mark (g.iii) 24%.

\(\text{Find intersection of}\ f(x)\ \text{and}\ g(x)\ \text{(by CAS)}:\)

\(f(x)=g(x)\ \ \Rightarrow\ \ x=\pi\)

\(\text {Area}=\displaystyle \int_0^\pi f(x)-g(x)\, d x+\int_\pi^{2 \pi} g(x)-f(x)\, d x=1.53\)
  

g.iii. \(a=0, c=1, d=1\)

\(g(x)=b x^2+x+1, \ f(x)=\sin (x)+1\)

\(g^{\prime}(x)=2 b x+1, \ f^{\prime}(x)=\cos (x)\)

\(\text{Solve simultaneous equations for \(b\) and \(r\):}\)

\(br^2+r=\sin (r)\ \ldots\ (1)\)

\(2 b r+1=\cos (r)\ \ldots\ (2)\)

\(r=\pi, \ b=-\dfrac{1}{\pi}\)

Calculus, MET1 2023 VCAA SM-Bank 6

Newton's method is used to estimate the \(x\)-intercept of the function  \(f(x)=\dfrac{1}{3} x^3+2 x+4\).

  1. Verify that  \(f(-1)>0\)  and  \(f(-2)<0\).   (1 mark)

  2. Using an initial estimate of  \(x_0=-1\), find the value of \(x_1\).   (2 marks)

Show Answers Only

a.    \(\text{See worked solutions}\)

b.    \(-\dfrac{14}{9}\)

Show Worked Solution
a.     \(f(x)\) \(=\dfrac{1}{3}x^3+2x+4\)
  \(f(-1)\) \(=\dfrac{1}{3}.(-1)^3+2.(-1)+4=\dfrac{5}{3}>0\)
  \(f(-2)\) \(=\dfrac{1}{3}.(-2)^3+2.(-2)+4=-\dfrac{8}{3}<0\)

 

b.     \(x_1\) \(=x_0-\dfrac{f(x_0)}{f^{\prime}(x_0)},\quad\ \text{where }x_0=-1\)
   

\(=-1-\dfrac{-\dfrac{1}{3}+2.(-1)+4}{(-1)^2+2}\)
    \(=-\dfrac{14}{9}\)

Calculus, MET2 2023 SM-Bank 1

The function \(g\) is defined as follows.

\(g:(0,7] \rightarrow R, g(x)=3\, \log _e(x)-x\)

  1. Sketch the graph of \(g\) on the axes below. Label the vertical asymptote with its equation, and label any axial intercepts, stationary points and endpoints in coordinate form, correct to three decimal places.   (3 marks)

     

     

  2.  i. Find the equation of the tangent to the graph of \(g\) at the point where \(x=1\).   (1 mark)

  3. ii. Sketch the graph of the tangent to the graph of \(g\) at \(x=1\) on the axes in part a.   (1 mark)

Newton's method is used to find an approximate \(x\)-intercept of \(g\), with an initial estimate of \(x_0=1\).

  1. Find the value of \(x_1\).   (1 mark)

  2. Find the horizontal distance between \(x_3\) and the closest \(x\)-intercept of \(g\), correct to four decimal places.   (1 mark)

  3.  i. Find the value of \(k\), where \(k>1\), such that an initial estimate of  \(x_0=k\)  gives the same value of  \(x_1\)  as found in part \(c\). Give your answer correct to three decimal places.   (2 marks)

  4. ii. Using this value of \(k\), sketch the tangent to the graph of \(g\) at the point where  \(x=k\)  on the axes in part a.   (1 mark)

Show Answers Only
a.    
  
b.i

 \(y=2x-3\)
  
b.ii
  
c. \(\dfrac{3}{2}=1.5\)
  
d. \(0.0036\)
  
e.i \(k=2.397\)
  
e.ii
Show Worked Solution

a.

b.i    \(g(x)\) \(=3\log_{e}x-x\)
  \(g(1)\) \(=3\log_{e}1-1=-1\)
  \(g^{\prime}(x)\) \(=\dfrac{3}{x}-1\)
  \(g^{\prime}(1)\) \(=\dfrac{3}{1}-1=2\)

  
\(\text{Equation of tangent at }(1, -1)\ \text{with }m=2\)

\(y+1=2(x-1)\ \ \rightarrow \ \ y=2x-3\)

b.ii

 
c.    \(\text{Newton’s Method}\)

\(x_1\) \(=x_0-\dfrac{g(x)}{g'(x)}\)
  \(=1-\left(\dfrac{-1}{2}\right)\)
  \(=\dfrac{3}{2}=1.5\)

\(\text{Using CAS:}\)
  
 

d.    \(\text{Using CAS}\)

\(x\text{-intercept}:\ x=1.85718\)

\(\therefore\ \text{Horizontal distance}=1.85718-1.85354=0.0036\)

 

e.i.  \(\text{Using CAS}\)

\(k-\dfrac{3\log_{e}x-x}{\dfrac{3}{x}-1}\) \(=1.5\)
\(k>1\ \therefore\ \ k\) \(=2.397\)

 
e.ii

Calculus, MET2 2023 VCE SM-Bank 7 MC

One way of implementing Newton's method using pseudocode, with a tolerance level of 0.001 , is shown below.

The pseudocode is incomplete, with two missing lines indicated by an empty box.
 

  

Which one of the following options would be most appropriate to fill the empty box?
 



Show Answers Only

\(E\)

Show Worked Solution

\(\text{The tolerance}=\pm 0.001\)

\(\Big|\text{next_ x}-\text{prev_ x}\Big|<\ \text{tolerance}\)

\(\therefore\ \textbf{If }\ \ -0.001<\ \text{next_ x}-\text{prev_ x}\ <0.001\ \textbf{Then }\)

\(\text{If true }\textbf{Return }\text{next_ x}\)

\(\Rightarrow E\)