Solve $$\left( x + \frac{2}{x} \right)^2 - 6 \left( x + \frac{2}{x} \right) + 9 = 0.$$ (b) The probability that it rains on any particular day during the 30 days of November is 0.1 - HSC - SSCE Mathematics Extension 1 - Question 11 - 2014 - Paper 1

The probability of rain on any given day is ( p = 0.1 ).

Let ( X ) be the random variable representing the number of rainy days in November, which follows a binomial distribution: [ X \sim \text{Binomial}(30, 0.1) ].

The probability of rain on fewer than 3 days is: [ P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) ].

Using the binomial probability formula: [ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} ] where ( n = 30 ).

Thus, [ P(X < 3) = \binom{30}{0} (0.1)^0 (0.9)^{30} + \binom{30}{1} (0.1)^1 (0.9)^{29} + \binom{30}{2} (0.1)^2 (0.9)^{28} ].

The function ( y = 6\tan^{-1}x ) has a vertical asymptote at ( y = 6\frac{\pi}{2} = 3\pi ) as ( x \rightarrow \pm \infty ).

The range of the function is: [ \text{Range}(y) = (0, 3\pi) ]

To sketch the graph:

1. Identify key points: ( y(0) = 6\tan^{-1}(0) = 0 ).
2. As ( x ) increases or decreases, ( y ) approaches ( 3\pi ).

With the substitution ( x = u^2 + 1 ), we have: ( dx = 2u , du ).

Changing the limits:

• When ( x = 2, u = 1 )
• When ( x = 5, u = 2 )

Thus, the integral becomes: [ \int_{1}^{2} \frac{u^2 + 1}{\sqrt{u^2}} 2u , du = 2\int_{1}^{2} (u + \frac{1}{u}) , du ].

Now, evaluate: [ 2\left( \frac{u^2}{2} + \ln|u| \right) \Big|_{1}^{2} = 2\left( 2 + \ln(2) - (\frac{1}{2} + 0) \right) = 2\left( \frac{3}{2} + \ln(2) \right) = 3 + 2\ln(2) ].

First, simplify the inequality: [ x^2 + 5 > 6 ]
[ x^2 > 1 ].

Taking the square root of both sides: [ x > 1 \quad \text{or} \quad x < -1 ].

Thus, the solution set is: [ x < -1 ext{ or } x > 1 ].

Using the product rule: [ \frac{d}{dx}(e^{x} \ln x) = e^{x} \ln x + e^{x} \cdot \frac{1}{x} ].

Simplifying gives: [ e^{x} \ln x + \frac{e^{x}}{x} ].