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Calculus, MET1 2024 VCAA 7

Part of the graph of  \(f:[-\pi, \pi] \rightarrow R, f(x)=x \sin (x)\)  is shown below.

  1. Use the trapezium rule with a step size of  \(\frac{\pi}{3}\)  to determine an approximation of the total area between the graph of  \(y=f(x)\) and the \(x\)-axis over the interval  \(x \in[0, \pi]\).   (3 marks)

  2.   i. Find  \(f^{\prime}(x)\).  (1 mark)

  3.  ii. Determine the range of  \(f^{\prime}(x)\)  over the interval  \(\left[\dfrac{\pi}{2}, \dfrac{2 \pi}{3}\right]\).  (1 mark)

  4. iii. Hence, verify that \(f(x)\) has a stationary point for \(x \in\left[\dfrac{\pi}{2}, \frac{2 \pi}{3}\right]\).  (1 mark)

  5. On the set of axes below, sketch the graph of \(y=f^{\prime}(x)\) on the domain \([-\pi, \pi]\), labelling the endpoints with their coordinates.
  6. You may use the fact that the graph of \(y=f^{\prime}(x)\) has a local minimum at approximately \((-1.1,-1.4)\) and a local maximum at approximately \((1.1,1.4)\).  (3 marks)

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a.    \(\dfrac{\sqrt{3}\pi^2}{6}\)

bi.   \(x\cos(x)+\sin(x)\)

bii.  \(\left[-\dfrac{\pi}{3}+\dfrac{\sqrt{3}}{2},\ 1\right]\)

biii. \(\text{In the interval }\left[\dfrac{\pi}{2}, \dfrac{2\pi}{3}\right],\ f^{\prime}(x)\ \text{changes from positive to negative.}\)

\(\therefore\ f^{\prime}(x)=0\ \text{at some point in the interval, and hence a stationary point exists.}\)

c. 

Show Worked Solution

a.     \(A\) \(=\dfrac{\pi}{3}\times\dfrac{1}{2}\left(f(0)+2f\left(\dfrac{\pi}{3}\right)+2f\left(\dfrac{2\pi}{3}\right)+f(\pi)\right)\)
    \(=\dfrac{\pi}{3}\left(0+2.\dfrac{\pi}{3}\sin\left(\dfrac{\pi}{3}\right)+2.\dfrac{2\pi}{3}\sin\left(\dfrac{2\pi}{3}\right)+\pi\sin({\pi})\right)\)
    \(=\dfrac{\pi}{6}\left(2\times \dfrac{\pi}{3}\times\dfrac{\sqrt{3}}{2}+2\times\dfrac{2\pi}{3}\times \dfrac{\sqrt{3}}{2}+0\right)\)
    \(=\dfrac{\pi}{6}\left(\dfrac{2\pi\sqrt{3}}{6}+\dfrac{4\pi\sqrt{3}}{6}\right)\)
    \(=\dfrac{\pi}{6}\times \dfrac{6\pi\sqrt{3}}{6}\)
    \(=\dfrac{\sqrt{3}\pi^2}{6}\)

  

bi.    \(f(x)\) \(=x\sin(x)\)
  \(f^{\prime}(x)\) \(=x\cos(x)+\sin(x)\)

 

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

\(\therefore\ \text{Range of }\ f^{\prime}(x)\ \text{is}\quad\left[-\dfrac{\pi}{3}+\dfrac{\sqrt{3}}{2},\ 1\right]\)

biii. \(\text{In the interval }\left[\dfrac{\pi}{2}, \dfrac{2\pi}{3}\right],\ f^{\prime}(x)\ \text{changes from positive to negative.}\)

\(\therefore\ f^{\prime}(x)=0\ \text{at some point in the interval, and hence a stationary point exists.}\)

c.    \(f^{\prime}(\pi)=\pi\cos(\pi)+\sin(\pi)=-\pi\)

\(f^{\prime}(-\pi)=-\pi\cos(-\pi)+\sin(-\pi)=\pi\)

\(\therefore\ \text{Endpoints are }\ (-\pi,\ \pi)\ \text{and}\ (\pi,\ -\pi).\)

Calculus, MET1 2023 VCAA 1b

Let  \(f(x)=\sin(x)e^{2x}\).

Find  \(f^{'}\Big(\dfrac{\pi}{4}\Big)\).   (2 marks)

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\(\dfrac{3\sqrt{2}}{2}e^{\frac{\pi}{2}}\ \text{or}\ \dfrac{3e^{\frac{\pi}{2}}}{\sqrt{2}}\)

Show Worked Solution

\(\text{Using the product rule}\)

\(f'(x)\) \(=e^{2x}\cos(x)+2e^{2x}\sin(x)\)
  \(=e^{2x}\Big(\cos(x)+2\ \sin(x)\Big)\)
\(\therefore\ f’\Big(\dfrac{\pi}{4}\Big)\) \(=e^{2(\frac{\pi}{4})}\Bigg(\cos(\dfrac{\pi}{4})+2\ \sin(\dfrac{\pi}{4})\Bigg)\)
  \(=e^{\frac{\pi}{2}}\Bigg(\dfrac{1}{\sqrt{2}}+\sqrt{2}\Bigg)\)
  \(=e^{\frac{\pi}{2}}\Bigg(\dfrac{1+\sqrt2 \times \sqrt2}{\sqrt2} \Bigg) \)
  \(=\dfrac{3\sqrt{2}}{2}e^{\frac{\pi}{2}}\ \ \text{or}\ \ \dfrac{3e^{\frac{\pi}{2}}}{\sqrt{2}}\)

Calculus, MET1 2011 VCAA 1b

If  `g(x) = x^2 sin (2x)`,  find  `g prime (pi/6).`  (2 marks)

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`(sqrt 3 pi)/6 + pi^2/36`

Show Worked Solution

`g(x) = x^2 sin (2x)`

MARKER’S COMMENT: Incorrect determination of exact trig values lost many students marks here.

`text(Using Product Rule:)`

`(fh)′` `= f′ h + fh′`
`g prime (x)` `= 2 x sin (2x) + 2x^2 cos (2x)`
   
`:. g prime (pi/6)` `= 2 (pi/6) sin (pi/3) + 2 (pi/6)^2 cos (pi/3)`
  `= pi/3 xx sqrt 3/2 + pi^2/18 xx 1/2`
  `= (sqrt 3 pi)/6 + pi^2/36`

Calculus, MET1 2006 VCAA 3a

Let  `y = x tan(x)`. Evaluate  `(dy)/(dx)`  when `x = pi/6`.  (3 mark)

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`(3sqrt3 + 2pi)/9`

Show Worked Solution

`text(Using the product rule:)`

MARKER’S COMMENT: A common error was not knowing the exact trig function values.

`d/(dx)(uv)` `= u′v + uv′`
`(dy)/(dx)` `= tan(x) + x/(cos^2(x))`

 

`text(When)\ x = pi/6,`

`(dy)/(dx)` `= tan(pi/6) + (pi/6)/((cos(pi/6))^2)`
  `= 1/sqrt3 + pi/6 xx (2/sqrt3)^2`
  `= sqrt3/3 + (4pi)/18`
  `= (3sqrt3 + 2pi)/9`