Solve $$rac{(x + rac{2}{x})^2}{6} - rac{(x + rac{2}{x})}{9} = 0.$$ (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

This scenario can be modeled using the binomial distribution. The probability of rain on a day is 0.1, and we have 30 days.

The expression for the probability that it rains on fewer than 3 days is: $P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = \sum_{k=0}^{2} \binom{30}{k} (0.1)^k (0.9)^{30-k}$

To sketch the graph, we note that:

• The function ( \tan^{-1}x ) approaches ( \frac{\pi}{2} ) as ( x ) approaches ( \infty ) and approaches ( -\frac{\pi}{2} ) as ( x ) approaches ( -\infty ).
• Thus, ( y = 6\tan^{-1}x ) will approach values of ( 3\pi ) and ( -3\pi ).

The range of ( y ) is therefore:
$(-3\pi, 3\pi).$

Applying the substitution ( x = u^2 + 1 ) gives ( dx = 2u , du ). When ( x = 2 ), ( u = 1 ), and when ( x = 5 ), ( u = 2 ).

The integral becomes: $\int_{1}^{2} \frac{u^2 + 1}{\sqrt{u^2}} \, 2u \, du = 2 \int_{1}^{2} (u + \frac{1}{u}) \, 2u \, du$

This can be computed to yield the final answer.

To solve the inequality, we rearrange it: $x^2 > 1$

Taking square roots gives: $x > 1 \, \text{ or } \, x < -1$. Thus, the solution set is: [(-∞, -1) U (1, ∞)].

Using the product rule for differentiation: Let ( u = e^{x} ) and ( v = \ln x ).

( u' = e^{x} ); ( v' = \frac{1}{x} )

Then, $\frac{d}{dx}(e^{x} \ln x) = u'v + uv' = e^{x} \ln x + e^{x} \cdot \frac{1}{x}.$

This can be simplified to: $e^{x} \ln x + \frac{e^{x}}{x}.$