The polynomial equation $2x^3 - 3x^2 - 11x + 7 = 0$ has roots $\alpha, \beta$ and $\gamma$ - HSC - SSCE Mathematics Extension 1 - Question 11 - 2013 - Paper 1

This integral can be recognized as that of the inverse sine function, where:

$\int \frac{1}{\sqrt{a^2 - x^2}} \, dx = \sin^{-1}\left(\frac{x}{a}\right) + C$

Here, $a^2 = 49$ (hence $a = 7$), leading to:

$\int \frac{1}{\sqrt{49 - 4x^2}} \, dx = \frac{1}{2} \sin^{-1}\left(\frac{2x}{7}\right) + C.$

Using the binomial probability formula, the expression for the probability of choosing the correct option for exactly $k$ questions, given $n$ total questions, is:

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

where $p = \frac{1}{4}$ (the probability of answering correctly). Therefore, for $n = 10$ and $k = 7$, we have:

$P(X = 7) = \binom{10}{7} \left(\frac{1}{4}\right)^7 \left(\frac{3}{4}\right)^3.$

To find $f'(x)$, we use the quotient rule:

$f'(x) = \frac{(4 - x^2)(-1) - (-x)(-2x)}{(4 - x^2)^2} = \frac{-4 + x^2}{(4 - x^2)^2}.$

The numerator $-4 + x^2$ is positive for $|x| > 2$, indicating that $f'(x) > 0$ for $|x| \in (0, 2)$. Thus, $f'(x) > 0$ is confirmed in its domain.

The function $f(x)$ has vertical asymptotes at $x = \pm 2$ (where $f(x)$ is undefined) and a horizontal asymptote can be determined by analyzing the limits of the function as $x \to \infty$ and $x \to -\infty$. As $|x|$ grows large, $f(x)$ approaches 0. The graph hence crosses the y-axis and has symmetry about the origin due to the odd function structure.

Using L'Hôpital's rule:

$\lim_{x \to 0} \frac{\sin \frac{x}{2}}{3x} = \lim_{x \to 0} \frac{\frac{1}{2} \cos \frac{x}{2}}{3} = \frac{1}{6}.$

Using the substitution $u = e^{3x}$, then $du = 3e^{3x}dx$, we find:

$dx = \frac{du}{3u}.$

Transforming the limits, when $x=0$, $u = 1$ and when $x=\frac{3}{2}$, $u = e^{\frac{9}{2}}$. The integral turns into:

$\int \frac{ \frac{1}{3u} }{6 - u} \, du,$ which can be solved using partial fractions.

Using the product rule and chain rule for differentiation:

$\frac{d}{dx}(x^2 \sin^{15} x) = 2x \sin^{15} x + x^2 \cdot 15 \sin^{14} x \cos x.$

Thus, the derivative is:

$\frac{d}{dx}(x^2 \sin^{15} x) = 2x \sin^{15} x + 15x^2 \sin^{14} x \cos x.$