A test consists of five multiple-choice questions - HSC - SSCE Mathematics Extension 1 - Question 4 - 2009 - Paper 1

To find the probability that Huong selects three or more correct answers, we must consider the probabilities for selecting 3, 4, and 5 correct answers.

1. For 3 correct answers, we already calculated this as ( \frac{45}{512} ).

2. For 4 correct answers:

   $P(X = 4) = C(5, 4) \left(\frac{1}{4}\right)^{4} \left(\frac{3}{4}\right)^{1} = 5 \cdot \frac{1}{256} \cdot \frac{3}{4} = \frac{15}{1024}$

3. For 5 correct answers:

   $P(X = 5) = C(5, 5) \left(\frac{1}{4}\right)^{5} = 1 \cdot \frac{1}{1024} = \frac{1}{1024}$

Combining these, the total probability:

$P(X \geq 3) = P(X = 3) + P(X = 4) + P(X = 5) = \frac{45}{512} + \frac{15}{1024} + \frac{1}{1024}$

To combine these fractions, we convert ( \frac{45}{512} ) to a common denominator:

$\frac{45}{512} = \frac{90}{1024}$

Thus,

$P(X \geq 3) = \frac{90}{1024} + \frac{15}{1024} + \frac{1}{1024} = \frac{106}{1024} = \frac{53}{512}$

To determine the probability that Huong selects at least one incorrect answer, we use the complementary probability method. First, we calculate the probability of selecting no incorrect answers (i.e., selecting all correct answers).

The probability of picking all 5 correct answers is:

$P(X = 5) = C(5, 5) \left(\frac{1}{4}\right)^{5} = 1 \cdot \frac{1}{1024} = \frac{1}{1024}$

Thus,

$P(X \geq 1) = 1 - P(X = 5) = 1 - \frac{1}{1024} = \frac{1023}{1024}$

To show that the function ( f(x) ) is even, we need to prove that ( f(-x) = f(x) ).

First, we calculate ( f(-x) ):

$f(-x) = \frac{(-x)^4 + 3(-x)^2}{(-x)^3 + 3} = \frac{x^4 + 3x^2}{-x^3 + 3}$

Since the signs are different in the denominator, we inspect its structure and find that:

$-f(-x) = -\frac{x^4 + 3x^2}{-x^3 + 3} = \frac{x^4 + 3x^2}{x^3 + 3} = f(x)$

Thus, since ( f(-x) = f(x) ), we conclude that ( f(x) ) is indeed an even function.

To find the horizontal asymptote of the function ( y = f(x) ), we look at the degrees of the numerator and denominator as ( x ) approaches infinity.

The degree of the numerator (( x^4 )) is greater than the degree of the denominator (( x^3 )). Hence, there is no horizontal asymptote. Therefore, the horizontal asymptote of the graph is:

$y = \infty$. If considering the leading coefficients, one can also say that tendencies lead to increasing functions without bound.

To find stationary points, we need to differentiate ( f(x) ) and set the derivative equal to zero:

$f'(x) = \frac{d}{dx} \left( \frac{x^4 + 3x^2}{x^3 + 3} \right)$

Using the quotient rule:

$f'(x) = \frac{ (3x^3 + 0)(x^4 + 3x^2) - (x^4 + 3x^2)(4x^3 + 6x) }{(x^3 + 3)^2} = 0$

Setting the numerator equal to zero and solving for ( x ) yields the stationary points. You should simplify and factor as necessary to find the exact values.

To sketch the graph of ( y = f(x) ), observe:

• The function is even, so it is symmetric about the y-axis.
• As previously stated, there's no horizontal asymptote, implying unbounded behavior.
• Analyze critical points and the behavior near those points. Set the x-coordinates we previously found, and consider limits to determine the behavior towards \\ ( +, and -\infty). A smooth curve can be drawn reflecting these relationships.