1. Use the table of standard integrals to find $$\int \frac{1}{4 - x^2} dx.$$ 2 - HSC - SSCE Mathematics Extension 1 - Question 1 - 2010 - Paper 1

The function $f(x)$ is defined wherever the argument of the inverse cosine function is in the range of -1 to 1. Therefore, we need:

$-1 \leq \frac{x}{2} \leq 1$

Multiplying through by 2 gives:

$-2 \leq x \leq 2$

Thus, the domain of $f(x)$ is $[-2, 2]$.

Using properties of logarithms, the equation can be rewritten as:

$\ln(x + 6) = \ln(x^2)$

This leads to:

$x + 6 = x^2$

Rearranging this gives:

$x^2 - x - 6 = 0$

Factoring results in:

$(x - 3)(x + 2) = 0$

Thus, the solutions are $x = 3$ and $x = -2$. Since $x$ must be positive (from the logarithm's domain), the solution is:

$x = 3.$

To solve the inequality, first multiply both sides by $x + 2$ (noting that this is valid as long as $x + 2 > 0$, or $x > -2$):

$3 < 4(x + 2)$

Expanding gives:

$3 < 4x + 8$

Rearranging leads to:

$4x > -5$

Thus:

$x > -\frac{5}{4}$

Considering that $x > -2$, the final solution is:

$x > -\frac{5}{4}.$

With the substitution $u = 1 - x$, we have $du = -dx$, which transforms the limits: when $x = 0$, $u = 1$; when $x = 1$, $u = 0$.

Thus, the integral becomes:

$\int_1^0 \sqrt{u} (-du) = \int_0^1 \sqrt{u} du$.

This evaluates as:

$\frac{2}{3} u^{3/2} \bigg|_0^1 = \frac{2}{3}.$

The probability can be modeled using the binomial distribution:

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

where:

• $n = 5$ (total dice),
• $k = 2$ (dice showing a four), and
• $p = \frac{1}{6}$ (probability of one die showing four).

Calculating the binomial coefficient:

$\binom{5}{2} = \frac{5!}{2!(5 - 2)!} = 10$

Thus, the probability is:

$P(X = 2) = 10 \left(\frac{1}{6}\right)^2 \left(\frac{5}{6}\right)^3$

This leads to: $P(X = 2) = 10 \cdot \frac{1}{36} \cdot \frac{125}{216} = \frac{1250}{7776}$

This result represents the unsimplified probability.