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 it raining on a specific day is 0.1. Thus, the probability of it not raining is ( 1 - 0.1 = 0.9 ). Using the binomial distribution, the probability of fewer than 3 rainy days is: $P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)$ Where: $P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$ With ( n = 30 ) and ( p = 0.1 ).

The function ( y = 6 \tan^{-1} x ) has a range of ( (0, 6\pi) ). The graph approaches ( 6\pi ) asymptotically as ( x \to \infty ) and as ( x \to -\infty ) it approaches ( 0 ).

Using the substitution ( x = u^2 + 1 ), then ( dx = 2u , du ). Change the limits: when ( x = 2, u = 1 ) and when ( x = 5, u = 2 ). The integral becomes:

$\int_1^2 \frac{u^2 + 1}{\sqrt{u^2}} \cdot 2u \, du = \int_1^2 2(u^2 + 1) \, du$ Integrating gives: $[u^3 + 2u]_1^2 = [8 + 4] - [1 + 2] = 10.$

Rearranging gives ( x^2 > 1 ), thus: ( x > 1 ) or ( x < -1 ). The solution set is ( (-\infty, -1) \cup (1, \infty) ).

Using the product rule, the derivative is given by: $\frac{d}{dx}(e^{x} \ln x) = e^{x} \ln x + e^{x} \cdot \frac{1}{x} = e^{x} \left( \ln x + \frac{1}{x} \right).$