Solve \(\left( x + \frac{2}{x} \right)^2 - 6 \left( x + \frac{2}{x} \right) + 9 = 0\) - HSC - SSCE Mathematics Extension 1 - Question 11 - 2014 - Paper 1

Let ( X ) be the random variable representing the number of rainy days in November. ( X ) follows a binomial distribution ( X \sim B(30, 0.1) ).

We want to find ( P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) ).

The probability mass function is given by: ( P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k} ) where ( n = 30 ) and ( p = 0.1 ). Thus, we write: [ P(X < 3) = \sum_{k=0}^{2} \binom{30}{k} (0.1)^k (0.9)^{30-k} ]

The function ( y = 6 \tan^{-1}(x) ) has horizontal asymptotes. As ( x ) approaches +( \infty ), ( y ) approaches 6( \frac{\pi}{2} ). As ( x ) approaches -( \infty ), ( y ) approaches -6( \frac{\pi}{2} ).

The range of the function is therefore ( \left(-6\frac{\pi}{2}, 6\frac{\pi}{2}\right) ).

Make the substitution ( x = u^2 + 1 ) so that ( dx = 2u , du ).

Change the limits: When ( x = 2, u = \sqrt{1} = 1 ); When ( x = 5, u = \sqrt{4} = 2 ).

Thus, the integral becomes: [ \int_1^2 \frac{u^2 + 1}{u} 2u , du = \int_1^2 (2u^2 + 2) , du = \left[ \frac{2}{3} u^3 + 2u \right]_1^2 ] Evaluating from 1 to 2 will yield the final answer.

This simplifies to ( x^2 > 1 ). The solutions are ( x > 1 ) or ( x < -1 ).

Using the product rule: [ \frac{d}{dr}(e^{r} \ln x) = e^{r} \ln x + e^{r} \frac{1}{x} \frac{dx}{dr} ] Thus, the derivative is:\n( e^r \ln x + e^r \frac{1}{x} \frac{dx}{dr} ).