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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?
 



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\(E\)

Show Worked Solution

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

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

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

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

\(\Rightarrow E\)

Calculus, MET2 2023 VCE SM-Bank 5 MC

The algorithm below, described in pseudocode, estimates the value of a definite integral using the trapezium rule.
 

Consider the algorithm implemented with the following inputs.

The value of the variable sum after one iteration of the while loop would be closest to

  1. 1.281
  2. 1.289
  3. 1.463
  4. 1.617
  5. 2.136
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\(C\)

Show Worked Solution

\((f(x), a, b, n)\ \rightarrow\ (\log_{e}x, 1, 3, 10) \)

\(h=\dfrac{b-a}{n}=\dfrac{3-1}{10}=\dfrac{1}{5}\)

\(\text{Sum}=f(a)+f(b)=\log_{e}1+\log_{e}3=\log_{e}3\)
  

\(\text{1st iteration of}\ \textbf{while }\text{loop:}\)

\(x\) \(=a+h=1+\dfrac{1}{5}=\dfrac{6}{5}\)
\(\text{Sum}\) \(=\text{Sum}+2\times f(x)\)
  \(=\log_{e}3+2\times f\Bigg(\dfrac{6}{5}\Bigg)\)
  \(=\log_{e}3+2\times \log_{e}{\Bigg(\dfrac{6}{5}\Bigg)}\)
  \(\approx 1.46325\dots\)

 

\(\Rightarrow C\)