At the front of a building there are five garage doors - HSC - SSCE Mathematics Extension 1 - Question 3 - 2010 - Paper 1

If the two red doors are next to each other, we can treat them as a single item or block. This simplifies our arrangement:

Let the red doors be denoted as RR, so we have:

• RR, G, B, O

Now, we have four items to arrange: RR, G, B, O.

The number of arrangements is: $4! = 24$

To find the x-coordinates of the points of inflexion, we need to calculate the second derivative of the function $f(x) = e^{x^2}$ and set it equal to zero.

First, we find the first derivative: $f'(x) = 2xe^{x^2}$

Now, we find the second derivative using the product rule: $f''(x) = 2e^{x^2} + 4x^2 e^{x^2} = 2e^{x^2}(1 + 2x^2)$

Setting the second derivative equal to zero gives: $2e^{x^2}(1 + 2x^2) = 0$ Since $e^{x^2}$ is never zero, we solve:

The function $f(x) = e^{x^2}$ is not one-to-one over its entire domain, due to the fact that it is symmetric about the y-axis. For a function to have an inverse, it must be one-to-one (i.e., each y-value corresponds to exactly one x-value). Therefore, we must restrict the domain to either $x \geq 0$ or $x \leq 0$.

Given the function $y = e^{x^2}$ with the restricted domain of $x \geq 0$, we solve for $x$ as follows:

Let $y = e^{x^2}$ then, taking natural logarithms, we have: $ext{ln}(y) = x^2$ Thus, $x = \sqrt{ln(y)}$

Therefore, the inverse function is given by: $f^{-1}(x) = \sqrt{ln(x)}$

The domain of $f^{-1}(x) = \sqrt{ln(x)}$ is determined by the restriction that $ln(x)$ must be non-negative. Thus, we have: $x > 1$

Consequently, the domain of $f^{-1}(x)$ is: $x \in (1, \infty)$

To sketch the curve of $y = f^{-1}(x) = \sqrt{ln(x)}$, we note:

• The function is defined for $x > 1$.
• As $x$ approaches 1, $y$ approaches 0.
• As $x$ increases, $y$ continues to increase without bound.

The graph starts at the point (1, 0) and increases, giving the general shape of a curve that grows gradually.

To show there is a solution for the equation $x = e^{x^2}$ in the interval [0.6, 0.7], calculate both sides at the endpoints:

• For $x = 0.6$: $e^{(0.6)^2} = e^{0.36} \approx 1.4333$
• For $x = 0.7$: $e^{(0.7)^2} = e^{0.49} \approx 1.6323$

Since $0.6 < 1.4333$ and $0.7 < 1.6323$, and because of the Intermediate Value Theorem, we conclude there is at least one root within the interval.