Evaluate $$\lim_{x \to 0} \frac{\sin(\frac{x}{5})}{2x}.$$ Find $$\frac{d}{dx}\cos^{-1}(3x^2).$$ The line $AT$ is the tangent to the circle at $A$, and $BT$ is a secant meeting the circle at $B$ and $C$ - HSC - SSCE Mathematics Extension 1 - Question 2 - 2004 - Paper 1

Using the chain rule, we differentiate:

$\frac{d}{dx}\cos^{-1}(u) = -\frac{1}{\sqrt{1-u^2}} \cdot \frac{du}{dx}, \quad \text{where } u = 3x^2.$

Calculating ( \frac{du}{dx} = 6x ), we substitute back:

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

From the geometry, we can apply the Pythagorean theorem:

$AT^2 = AB^2 + BT^2$(which is also secant)

We have:

$12^2 = 7^2 + x^2 \implies 144 = 49 + x^2 \implies x^2 = 95 \implies x = \sqrt{95} \approx 9.746.$

To express (8 \cos x + 6 \sin x) in the form (A \cos(r - \alpha)):

We find:

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

and using $\tan \alpha = \frac{6}{8} = \frac{3}{4}$, we find ( \alpha = \tan^{-1}(\frac{3}{4})$.

Thus, it can be expressed as:

$10 \cos \left(x - \tan^{-1}\left(\frac{3}{4}\right)\right).$

Rearranging,

$10 \cos\left(x - \tan^{-1}\left(\frac{3}{4}\right)\right) = 5 \implies \cos\left(x - \tan^{-1}\left(\frac{3}{4}\right)\right) = \frac{1}{2}.$

Thus,

$x - \tan^{-1}\left(\frac{3}{4}\right) = \frac{\pi}{3} + 2k\pi ext{ or } x - \tan^{-1}\left(\frac{3}{4}\right) = -\frac{\pi}{3} + 2k\pi$ for all integers $k$.

Solving for $x$, we find specific values within (0 \leq x \leq 2\pi$.

The total number of ways to choose a team of 4 from 16 people (9 women + 7 men) is given by:

$\binom{16}{4} = \frac{16!}{4!(16-4)!} = 1820.$

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

$\binom{9}{4} = 126.$
The probability is then the ratio:

$P = \frac{\text{Ways to choose 4 women}}{\text{Total ways to choose 4}} = \frac{126}{1820} = \frac{63}{910} \approx 0.069.$