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

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

Starting with the inequality:

$\frac{2x}{x + 1} > 1$

Multiply both sides by $(x + 1)^2$ to clear the denominator:

$2x(x + 1) > (x + 1) \implies 2x^2 + 2x > x + 1$

Rearranging gives:

$2x^2 + x - 1 > 0$

Factoring:

$(2x - 1)(x + 1) > 0$

The critical points are $x = \frac{1}{2}$ and $x = -1$. Testing intervals, we find the solution is $x < -1 \text{ or } x > \frac{1}{2}$.

To sketch $y = 2 \cos^{-1}(x)$, we note that:

• The domain is $[-1, 1]$ (from $\cos^{-1}(x)$).
• The range is $[0, 2\pi]$ because multiplying by 2 scales the range.

Plot key points, such as $(x, y) = (-1, 2\pi)$, $(0, \pi)$, and $(1, 0)$, and connect them smoothly. The shape is a decreasing curve starting from $(1, 0)$ down to $(-1, 2\pi)$.

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

$dx = 2u \, du$

The limits change: when $x = 0, u = 1$ and when $x = 3, u = 2$.

The integral transforms to:

$\int_{1}^{2} \frac{(u^2 - 1)}{\sqrt{u^2}} (2u) \, du = \int_{1}^{2} (2u^2 - 2) \, du$

Calculating gives:

$= [\frac{2}{3}u^3 - 2u]_{1}^{2} = \frac{2}{3}(8) - 4 - (\frac{2}{3} - 2) = \frac{16}{3} - 4 + 2 - \frac{2}{3} = \frac{16 - 12 + 2}{3} = \frac{6}{3} = 2$.

We can use the substitution $u = ext{sin}(x)$. Then $du = ext{cos}(x)dx$:

$\int \sin^2(x)\cos(x)dx = \int u^2 \; du = \frac{u^3}{3} + C = \frac{ ext{sin}^3(x)}{3} + C$.

Let X be the number of seedlings producing red flowers, modeled by the binomial distribution: $P(X = 3) = \binom{8}{3} \left(\frac{1}{5}\right)^3 \left(\frac{4}{5}\right)^{5}$

Similarly, for none of the seedlings to produce red flowers: $P(X = 0) = \binom{8}{0} \left(\frac{1}{5}\right)^0 \left(\frac{4}{5}\right)^{8} = \left(\frac{4}{5}\right)^{8}$.

For at least one seedling: $$P(X \geq 1) = 1 - P(X = 0) = 1 - \left(\frac{4}{5}\right)^{8}.$