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 solved using the substitution ( x = \frac{7}{2} \sin \theta ) which leads to:

[ \int \frac{1}{\sqrt{49 - 4x^2}} \ dx = \frac{1}{4} \arcsin\left(\frac{2x}{7}\right) + C ]

The expression for the probability of choosing exactly ( k ) correct answers out of ( n ) questions follows a binomial distribution:

[ P(X = k) = \binom{n}{k} p^k (1 - p)^{n-k} ] where ( n = 10, k = 7, p = \frac{1}{4} ):

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

First, we find the derivative:

[ f'(x) = \frac{d}{dx}\left(\frac{-x}{4 - x^2}\right) ] Applying the quotient rule:

[ f'(x) = \frac{(4 - x^2)(-1) - (-x)(-2x)}{(4 - x^2)^2} = \frac{-4 + x^2}{(4 - x^2)^2} ] This numerator, ( x^2 - 4 ) is positive for all ( x ) where ( |x| < 2 ). Therefore, ( f'(x) > 0 ).

To determine the asymptotes, we note the vertical asymptotes occur at the points where the denominator is zero, namely at ( x = 4 ) and ( x = -4 ). We should sketch the graph of the function keeping these asymptotes in mind.

1. Vertical asymptotes at ( x = 4 ) and ( x = -4 ).
2. The function approaches zero as ( x \to \infty ) or ( x \to -\infty ).

By applying L'Hôpital's Rule or the small-angle approximation for sine, we can confirm:

[ \lim_{x \to 0} \frac{\sin \frac{x}{2}}{3x} = \lim_{x \to 0} \frac{\frac{1}{2}\cos\frac{x}{2}}{3} = \frac{1/2}{3} = \frac{1}{6} ]

Substituting ( u = e^{3x} ), gives us:

[ \int_{0}^{1} \frac{3}{e^{2x}} \ dx = \int_{1}^{e^{3}} \frac{1}{u^{\frac{2}{3}}} du = \frac{3}{1} [u^{\frac{1}{3}}]_{1}^{e^{3}} = 3\left(e - 1\right) ]

Using the product rule:

[ \frac{d}{dx}(x^2 \sin 5x) = 2x \sin 5x + x^2 \cdot 5\cos 5x = 2x \sin 5x + 5x^2 \cos 5x ]