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

To find the domain of ( f(x) = \cos^{-1} \left( \frac{x}{2} \right) ), we must identify the values of ( x ) such that the argument of the inverse cosine function is in the interval ([-1, 1]).

Setting up the inequalities:

1. ( -1 \leq \frac{x}{2} \leq 1 )

Multiplying through by 2:

1. ( -2 \leq x \leq 2 )

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

First, we can use the properties of logarithms to rewrite the equation:

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

This means:

$x + 6 = x^2$

Rearranging gives:

$x^2 - x - 6 = 0$

Factoring the quadratic:

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

Thus, ( x = 3 ) or ( x = -2 ). Since ( x ) must be positive for the logarithm to be defined, the solution is:

$x = 3$

To solve this inequality:

1. Rewrite it as: ( 3 < 4(x + 2) )

2. Expanding gives: ( 3 < 4x + 8 ), which simplifies to ( 4x > -5 ).

Thus, dividing by 4, we have:

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

However, we must also consider any restrictions from the original denominator. The term ( x + 2 ) cannot equal zero, so:

$x \neq -2$

The final solution is:

$x > -\frac{5}{4}, \quad x \neq -2$

Using the substitution ( u = 1 - x ), then ( du = -dx ). The limits change as follows: when ( x = 0 ), ( u = 1 ) and when ( x = 1 ), ( u = 0 ).

Thus, the integral becomes:

$-\int_{1}^{0} \sqrt{u} \, du = \int_{0}^{1} \sqrt{u} \, du$

Now, we can evaluate:

$= \left[ \frac{2}{3} u^{3/2} \right]_{0}^{1} = \frac{2}{3} \cdot 1^{3/2} - 0 = \frac{2}{3}$

To find the probability of exactly two out of five dice showing a four, we can use the binomial probability formula:

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

Where:

• ( n = 5 ) (the total number of dice)
• ( k = 2 ) (the number of successes, i.e., rolling a four)
• ( p = \frac{1}{6} ) (the probability of rolling a four)

Calculating:

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

After calculating the terms:

1. Calculate ( \binom{5}{2} = 10 )
2. Then,

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

Thus, the probability that exactly two of the dice show a four is:\n $\frac{1250}{7776}$