(a) The point $P$ divides the interval from $A(-4,-4)$ to $B(1,6)$ internally in the ratio $2:3$ - HSC - SSCE Mathematics Extension 1 - Question 11 - 2017 - Paper 1

Let $y = \tan^{-1}(x^2)$. To differentiate, we apply the chain rule:

$\frac{dy}{dx} = \frac{1}{1 + (x^2)^2} \cdot 2x = \frac{2x}{1 + x^4}$

To solve the inequality, we first rearrange it:

$\frac{2x}{x+1} - 1 > 0 \Rightarrow \frac{2x - (x + 1)}{x+1} > 0 \Rightarrow \frac{x - 1}{x + 1} > 0$

Next, we find the critical points:

• $x = 1$
• $x = -1$

We can test the intervals defined by these points:

1. For $x < -1$, $\frac{x-1}{x+1} < 0$
2. For $-1 < x < 1$, $\frac{x-1}{x+1} < 0$
3. For $x > 1$, $\frac{x-1}{x+1} > 0$

Thus, the solution is: $x > 1$

The function $y = 2 \cos^{-1} x$ is defined for $x \in [-1, 1]$ and gives a range of $[0, 2\pi]$.

• The critical points include:
  • At $x = -1$, $y = 2\pi$.
  • At $x = 0$, $y = \pi$.
  • At $x = 1$, $y = 0$.

The graph will start at $(1, 0)$, reach $(0, \pi)$, and end at $(-1, 2\pi)$.

Using the substitution $x = u^2 - 1$, we have:

• $dx = 2u \, du$
• Change limits: when $x = 0$, $u = 1$; when $x = 3$, $u = 2$.

So we rewrite the integral:

$\int_0^3 \frac{x}{\sqrt{x+1}} \ dx = \int_{1}^{2} \frac{u^2 - 1}{u} \cdot 2u \, du = 2 \int_{1}^{2} (u - \frac{1}{u}) \, du$

This simplifies to: $2\left[ \frac{u^2}{2} - \ln |u| \right]_{1}^{2} = 2\left[(2 - \ln 2) - (\frac{1}{2}) \right] = 3 - 2\ln 2$

We can use substitution. Let:

• $u = \sin x$
• $du = \cos x \, dx$

Thus:

$\int \sin^2 x \cos x \, dx = \int u^2 \, du = \frac{u^3}{3} + C = \frac{\sin^3 x}{3} + C$

Let $p = \frac{1}{5}$ be the probability of a seedling producing red flowers. The probability of exactly three out of eight seedlings producing red flowers can be expressed using the binomial probability formula:

$P(X = k) = {n \choose k} p^k (1 - p)^{n-k}$

Therefore:

$P(X = 3) = {8 \choose 3} \left(\frac{1}{5}\right)^3 \left(\frac{4}{5}\right)^{5}$

The probability that none of the eight seedlings produces red flowers is:

$P(X = 0) = {8 \choose 0} \left(\frac{1}{5}\right)^0 \left(\frac{4}{5}\right)^{8} = \left(\frac{4}{5}\right)^{8}$

The probability that at least one of the eight seedlings produces red flowers can be found by taking the complement of the probability that none produce:

$P(X \geq 1) = 1 - P(X = 0) = 1 - \left(\frac{4}{5}\right)^{8}$