smc-740-20-Exponential (definite)

Let \(f:R \to R, f(x)=e^x+e^{-x}\) and \(g:R \to R, g(x)=\dfrac{1}{2}f(2-x)\).

Complete a possible sequence of transformations to map \(f\) to \(g\). (2 marks)
• Dilation of factor \(\dfrac{1}{2}\) from the \(x\) axis.

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Two functions \(g_1\) and \(g_2\) are created, both with the same rule as \(g\) but with distinct domains, such that \(g_1\) is strictly increasing and \(g_2\) is strictly decreasing.

Give the domain and range for the inverse of \(g_1\). (2 marks)
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Shown below is the graph of \(g\), the inverse of \(g_1\) and \(g_2\), and the line \(y=x\).

The intersection points between the graphs of \(y=x, y=g(x)\) and the inverses of \(g_1\) and \(g_2\), are labelled \(P\) and \(Q\).

1. Find the coordinates of \(P\) and \(Q\), correct to two decimal places. (1 mark)
  
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1. Find the area of the region bound by the graphs of \(g\), the inverse of \(g_1\) and the inverse of \(g_2\).
  Give your answer correct to two decimal places. (2 marks)
  
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Let \(h:R\to R, h(x)=\dfrac{1}{k}f(k-x)\), where \(k\in (o, \infty)\).

The turning point of \(h\) always lies on the graph of the function \(y=2x^n\), where \(n\) is an integer.
Find the value of \(n\). (1 mark)

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Let \(h_1:[k, \infty)\to R, h_1(x)=h(x)\).

The rule for the inverse of \(h_1\) is \(y=\log_{e}\Bigg(\dfrac{1}{k}x+\dfrac{1}{2}\sqrt{k^2x^2-4}\Bigg)+k\)

What is the smallest value of \(k\) such that \(h\) will intersect with the inverse of \(h_1\)?
Give your answer correct to two decimal places. (1 mark)

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It is possible for the graphs of \(h\) and the inverse of \(h_1\) to intersect twice. This occurs when \(k=5\).

Find the area of the region bound by the graphs of \(h\) and the inverse of \(h_1\), where \(k=5\).
Give your answer correct to two decimal places. (2 marks)

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Show Answers Only

a. \(\text{Reflect in the }y\text{-axis}\)

\(\text{Then translate 2 units to the right}\)

\(\text{OR}\)

\(\text{Translate 2 units to the left}\)

\(\text{Then reflect in the }y\text{-axis}\)

\(\text{OR}\)

\(\text{Translate 2 units to the right only }(f \ \text{even function})\)

b. \(\text{See worked solution}\)

c.i. \(P(1.27, 1.27\), Q(4.09, 4.09)\)

c.ii. \(2\displaystyle \int_{1.27…}^{4.09…} (x-g(x))\,dx=5.56\)

d. \(n=-1\)

e. \(k\approx 1.27\)

f. \(A=43.91\)

Show Worked Solution

a. \(\text{Reflect in the }y\text{-axis}\)

\(\text{Then translate 2 units to the right}\)

\(\text{OR}\)

\(\text{Translate 2 units to the left}\)

\(\text{Then reflect in the }y\text{-axis}\)

\(\text{OR}\)

\(\text{Translate 2 units to the right only }(f \ \text{even function})\)

b. \(\text{Some options include:}\)

• \(\text{Domain}\to [1, \infty)\ \ \ \text{Range}\to [2, \infty)\)
\(\text{or}\)
• \(\text{Domain}\to (1, \infty)\ \ \ \text{Range}\to (2, \infty)\)

\(\text{If functions split at turning point } (2, 1)\ \text{then possible options:}\)

• \(\text{Domain}\ g_1\to [2, \infty)\ \ \ \text{Range}\to [1, \infty)\)
\(\text{and }\)
\(\text{Domain}\ g_1^{-1}\to [1, \infty)\ \ \ \text{Range}\to [2, \infty)\)

\(\text{or}\)

• \(\text{Domain}\ g_1\to (2, \infty)\ \ \ \text{Range}\to (1, \infty)\)
\(\text{and }\)
\(\text{Domain}\ g_1^{-1}\to (1, \infty)\ \ \ \text{Range}\to (2, \infty)\)

♦ Mean mark 50%.
MARKER’S COMMENT: Note: maximal domain not requested so there were many correct answers. Many students seemed to be confused by notation. Often domain and range reversed.

c.i. \(\text{By CAS:}\ P(1.27, 1.27), Q(4.09, 4.09)\)

c.ii.	\(\text{Area}\)	\(=2\displaystyle \int_{1.27…}^{4.09…} (x-g(x))\,dx\)
		\(=2\displaystyle \int_{1.27…}^{4.09…} \Bigg(x-\dfrac{1}{2}\Big(e^{2-x}+e^{x-2}\Big)\Bigg)\,dx\)
		\(\approx 5.56\)

♦♦ Mean mark (c)(ii) 30%.
MARKER’S COMMENT: Some students set up the integral but did not provide an answer.
Others did not double the integral. Some incorrectly used \(\displaystyle \int_{1.27…}^{4.09…}(g_1^-1-g(x))\,dx\).
Others had correct answer but incorrect integral.

d. \((0, 2)\ \text{turning pt of }f(x)\)

\(\text{and }h(x) \text{is the transformation from }f(x)\)

\(\therefore \Bigg(k, \dfrac{2}{k}\Bigg)\ \text{turning tp of }h(x)\)

\(\text{so the coordinates have the relationship}\)

\(y=\dfrac{2}{x}=2x^{-1}\)

\(\therefore\ n=-1\)

♦♦♦ Mean mark (d) 10%.
MARKER’S COMMENT: Question not well done. \(n=1\) a common error.

e. \(\text{Smallest }k\ \text{is when inverse of the curves }h_1\ \text{and}\ h\)

\(\text{touch with the line }y=x.\)

\(\therefore\ h(x)\)	\(=\dfrac{1}{k}\Big(e^{k-x}+e^{x-k}\Big)\)
	\(=x\)

\(\text{and}\ h'(x)\)	\(=\dfrac{1}{k}\Big(-e^{k-x}+e^{x-k}\Big)\)
	\(=1\)

\(\text{at point of intersection, where }k>0.\)

\(\text{Using CAS graph both functions to solve or solve as simultaneous equations.}\)

\(\rightarrow\ k\approx 1.2687\approx 1.27\)

♦♦♦♦ Mean mark (e) 0%.
MARKER’S COMMENT: Question poorly done with many students not attempting.

f. \(\text{When}\ k=5\)

\(h_1^{-1}(x)=\log_e\Bigg(\dfrac{5}{2}x+\dfrac{1}{2}\sqrt{25x^2-4}\Bigg)+5\)

\(\text{and }h(x)=\dfrac{1}{5}\Big(e^{5-x}+e^{x-5}\Big)\)

\(\text{Using CAS: values of }x\ \text{at of intersection of fns are}\)

\(x\approx 1.45091…\ \text{and}\ 8.78157…\)

\(\text{Area}\)	\(=\displaystyle \int_{1.45091…}^{8.78157…}h_1^{-1}(x)- h(x)\,dx\)
	\(=\displaystyle \int_{1.45091…}^{8.78157…} \log_e\Bigg(\dfrac{5}{2}x+\dfrac{1}{2}\sqrt{25x^2-4}\Bigg)+5-\dfrac{1}{5}\Big(e^{5-x}+e^{x-5}\Big)\,dx\)
	\(=43.91\)

♦♦♦ Mean mark (f) 15%.
MARKER’S COMMENT: Errors were made with terminals. Common error using \(x=2.468\), which was the \(x\) value of the pt of intersection of \(h\) and \(y=x\).

`int_0^1 e^x – e^-x\ dx`	`= [e^x + e^-x]_0^1`
	`= [e + (1)/(e) – (1 + 1)]`
	`= e + (1)/(e) – 2`

Calculus, MET2 2023 VCAA 5

Calculus, MET1-NHT 2018 VCAA 3

Calculus, MET1 2016 ADV 12d

`(dy)/(dx)`	`= x · 3e^(3x) + 1 · e^(3x)`
	`= e^(3x) (1 + 3x)`