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

To find the maximum turning point, we set the derivative equal to zero:

$-e^{-x} + 4e^{-2x} = 0$

This simplifies to: $e^{-x}(4e^{-x} - 1) = 0$

Since $e^{-x} eq 0$, we solve: $4e^{-x} - 1 = 0 \Rightarrow 4e^{-x} = 1 \Rightarrow e^{-x} = \frac{1}{4} \Rightarrow -x = \ln\left(\frac{1}{4}\right) \Rightarrow x = 2 \ln(2)$

Substituting back to find $f(2\ln(2))$: \nSo the coordinates are ((2\ln(2), f(2\ln(2)))$.

Substituting into the function:

$f(2) = e^{-2} - 2e^{-4}$

This gives us the value at $x = 2$.

As $x \to -\infty$, the term $e^{-x}$ tends to infinity and dominates the function:

$f(x) \to e^{-x} \to +\infty$

Hence, $f(x)$ approaches $+ \infty$.

The y-intercept occurs when $x = 0$:

$f(0) = e^{0} - 2e^{0} = 1 - 2 = -1$

Thus, the y-intercept is $-1$.

The sketch will include:

• The maximum turning point at $(2\ln(2), f(2\ln(2)))$.
• The y-intercept at $(0, -1)$.
• Asymptotic behaviour as $x \to -\infty$ going up. Be sure to label the axes and points accurately.

In a circle, the angle at the center ($\angle AOC$) subtended by an arc is twice the angle at the circumference subtended by the same arc ($\angle ABC$). Since $\angle ABC = x$, it follows that $\angle AOC = 2x$.

A quadrilateral is cyclic if its opposite angles sum to $180^\circ$. With $\angle AOC = 2x$ and $\angle ADB = 180^\circ - x$, we observe:

$\angle AOC + \angle ADB = 2x + (180^\circ - x) = 180^\circ$

Thus, $ACDO$ is cyclic.

To demonstrate that points $P$, $M$, and $O$ are collinear, note that if $M$ is the midpoint of $AC$, segment $PO$ connects $P$ to $O$, which also lies on the diameter through $M$. This alignment shows they are collinear.