Consider the function $f(x) = e^{-x} - 2e^{-2x}$. (i) Find $f'(x)$. (ii) The graph $y = f(x)$ has one maximum turning point. Find the coordinates of the maximum turning point. (iii) Evaluate $f(2)$. (iv) Describe the behaviour of $f(x)$ as $x \to -\infty$. (v) Find the $y$-intercept of the graph $y = f(x)$. (vi) Sketch the graph $y = f(x)$ showing the features from parts (ii)-(v). You are not required to find any points of inflexion. In the diagram, the vertices of $\triangle ABC$ lie on the circle with centre $O$. The point $D$ lies on $BC$ such that $\triangle ABD$ is isosceles and $\angle ABC = x$. (i) Explain why $\angle AOC = 2x$. (ii) Prove that $ACDO$ is a cyclic quadrilateral. (iii) Let $M$ be the midpoint of $AC$ and $P$ the centre of the circle through $A, C, D$ and $O$. Show that $P, M$ and $O$ are collinear.

SSCE HSC Mathematics Extension 1 Integrals involving trigonometric substitution: Consider the function $f(x) = e^

Consider the function $f(x) = e^{-x} - 2e^{-2x}$ - HSC - SSCE Mathematics Extension 1 - Question 4 - 2011 - Paper 1

Question 4

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Consider the function $f(x) = e^{-x} - 2e^{-2x}$. (i) Find $f'(x)$. (ii) The graph $y = f(x)$ has one maximum turning point. Find the coordinates of the maxi... show full transcript

Worked Solution & Example Answer:Consider the function $f(x) = e^{-x} - 2e^{-2x}$ - HSC - SSCE Mathematics Extension 1 - Question 4 - 2011 - Paper 1

Step 1

(i) Find $f'(x)$.

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Answer

To find the derivative of the function $f(x) = e^{-x} - 2e^{-2x}$ , we apply the rules of differentiation.

Using the chain rule: $f'(x) = -e^{-x} + 4e^{-2x}$

Step 2

(ii) The graph $y = f(x)$ has one maximum turning point. Find the coordinates of the maximum turning point.

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Answer

To find the maximum turning point, we set the first derivative equal to zero:
$-e^{-x} + 4e^{-2x} = 0$
Solving for $x$ , we get:
$e^{-x} = 4e^{-2x} \Rightarrow e^{x} = 4 \Rightarrow x = ext{ln}(4).$

Next, substituting $x = ext{ln}(4)$ back into $f(x)$ to find the corresponding $y$ -coordinate:
$f( ext{ln}(4)) = e^{- ext{ln}(4)} - 2e^{-2 ext{ln}(4)} = \frac{1}{4} - 2 \cdot \frac{1}{16} = \frac{1}{4} - \frac{1}{8} = \frac{1}{8}.$

Hence, the coordinates of the maximum turning point are $( ext{ln}(4), \frac{1}{8})$ .

Step 3

(iii) Evaluate $f(2)$.

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Answer

To evaluate $f(2)$ , we substitute $2$ into the function:
$f(2) = e^{-2} - 2e^{-4} = \frac{1}{e^2} - 2 \cdot \frac{1}{e^4} = \frac{1}{e^2} - \frac{2}{e^4} = \frac{e^2 - 2}{e^4}.$

Step 4

(iv) Describe the behaviour of $f(x)$ as $x \to -\infty$.

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Answer

As $x \to -\infty$ , both terms in the function $f(x) = e^{-x} - 2e^{-2x}$ will tend to infinity since $e^{-x}$ grows larger. Hence,
$$f(x) \to \infty \text{ as } x \to -\infty.$

Step 5

(v) Find the $y$-intercept of the graph $y = f(x)$.

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Answer

To find the $y$ -intercept, we set $x = 0$ :
$f(0) = e^{0} - 2e^{0} = 1 - 2 = -1.$
Thus, the $y$ -intercept is at $(0, -1)$ .

Step 6

(vi) Sketch the graph $y = f(x)$ showing the features from parts (ii)-(v).

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Answer

When sketching the graph, we include:

The maximum turning point at $( ext{ln}(4), \frac{1}{8})$ ,
The $y$ -intercept at $(0, -1)$ ,
The behavior as $x \to -\infty$ (growing towards infinity).
Ensure to represent the curve smoothly, connecting these key features while observing the general shape dictated by the derivative.

Step 7

(i) Explain why $\angle AOC = 2x$.

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Answer

In the circle, the angle at the center ( $\angle AOC$ ) is twice the angle at the circumference ( $\angle ABC$ ) based on the inscribed angle theorem. Therefore,
$\angle AOC = 2\angle ABC = 2x.$

Step 8

(ii) Prove that $ACDO$ is a cyclic quadrilateral.

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Answer

For quadrilateral $ACDO$ to be cyclic, the opposite angles must sum to $180^{\circ}$ .
We have $\angle AOC + \angle ADC = 2x + (180^{\circ} - 2x) = 180^{\circ}$ , hence $ACDO$ is indeed a cyclic quadrilateral.

Step 9

(iii) Let $M$ be the midpoint of $AC$ and $P$ the center of the circle through $A, C, D$ and $O$. Show that $P, M$ and $O$ are collinear.

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Answer

Since $M$ is the midpoint of $AC$ , it lies on the segment connecting $A$ and $C$ . Point $P$ , being in the circle of $A$ and $C$ , also lies on the line joining $O$ and $M$ due to the properties of a circumcircle.
By the nature of lines and midpoints in a cyclic structure, it can be shown that $P$ , $M$ , and $O$ are collinear.

Consider the function $f(x) = e^{-x} - 2e^{-2x}$ - HSC - SSCE Mathematics Extension 1 - Question 4 - 2011 - Paper 1

Worked Solution & Example Answer:Consider the function $f(x) = e^{-x} - 2e^{-2x}$ - HSC - SSCE Mathematics Extension 1 - Question 4 - 2011 - Paper 1

(i) Find $f'(x)$.

(ii) The graph $y = f(x)$ has one maximum turning point. Find the coordinates of the maximum turning point.

(iii) Evaluate $f(2)$.

(iv) Describe the behaviour of $f(x)$ as $x \to -\infty$.

(v) Find the $y$-intercept of the graph $y = f(x)$.

(vi) Sketch the graph $y = f(x)$ showing the features from parts (ii)-(v).

(i) Explain why $\angle AOC = 2x$.

(ii) Prove that $ACDO$ is a cyclic quadrilateral.

(iii) Let $M$ be the midpoint of $AC$ and $P$ the center of the circle through $A, C, D$ and $O$. Show that $P, M$ and $O$ are collinear.

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