Evaluate $$ \lim_{x \to 0} \frac{\sin(\frac{x}{5})}{2x} $$ Find $$ \frac{d}{dx} \cos^{-1}(3x^2) $$ - HSC - SSCE Mathematics Extension 1 - Question 2 - 2004 - Paper 1

Using the chain rule, we have:

$\frac{d}{dx} \cos^{-1}(u) = -\frac{1}{\sqrt{1 - u^2}} \cdot \frac{du}{dx}$

Letting $u = 3x^2$, we find:

$\frac{du}{dx} = 6x$

Then, applying the chain rule:

$\frac{d}{dx} \cos^{-1}(3x^2) = -\frac{6x}{\sqrt{1 - (3x^2)^2}} = -\frac{6x}{\sqrt{1 - 9x^4}}$

Using the relationship in a circle, we apply Pythagoras' theorem in the triangle:

$AT^2 + BT^2 = AB^2$

Here, let $BT = 7$, thus:

144 + 49 = x^2 \\ 193 = x^2 \\ \therefore x = \sqrt{193} \approx 13.89 $$

To rewrite in the required form, first calculate:

$A = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10$

Next, find $\alpha$ such that:

\sin(\alpha) = \frac{6}{10} = 0.6 $$ Thus, we can write: $$ 8\cos\alpha + 6\sin\alpha = 10\cos(\theta - \alpha) $$

We rewrite the equation using our previous result:

\cos(\theta - \alpha) = 0.5 $$ This gives solutions: $$ \theta - \alpha = \frac{\pi}{3} \quad \text{or} \quad \theta - \alpha = \frac{5\pi}{3} $$ Thus: $$ \alpha = \theta - \frac{\pi}{3} \quad \text{or} \quad \frac{5\pi}{3} $$

To choose a four-person team from nine women and seven men, we can use the combination formula $C(n, k)$:

$\text{Total ways} = C(16, 4) = \frac{16!}{4!(16-4)!} = 1820$

The number of ways to choose 4 women from 9 is:

$C(9, 4) = \frac{9!}{4!(9-4)!} = 126$

The probability is:

$P = \frac{C(9, 4)}{C(16, 4)} = \frac{126}{1820} \approx 0.06923$