Use a SEPARATE writing booklet - HSC - SSCE Mathematics Extension 1 - Question 11 - 2014 - Paper 1

Let ( X ) be the random variable representing the number of days it rains in November. Since rain on each day is independent, ( X ) follows a Binomial distribution:
( X \sim \text{Binomial}(n=30, p=0.1) ).

We seek the probability of fewer than 3 days of rain:

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = \sum_{k=0}^{2} \binom{30}{k} (0.1)^k (0.9)^{30-k}$

To sketch the graph, we first recognize that the function ( y = 6 \tan^{-1}x ) asymptotically approaches ( 6 \times \frac{\pi}{2} = 3\pi ) as ( x \to +\infty ) and approaches ( 0 ) as ( x \to -\infty ).
Thus, the range of the function is ( y \in (0, 3\pi) ).

• As ( x ) increases, ( y ) increases, showing a smooth curve that approaches the horizontal asymptote at ( y = 3\pi ).
• At ( x = 0 ), ( y = 3 \times 0 = 0 ).

Using the substitution ( x = u^2 + 1 ), then ( dx = 2u , du ) and the limits change from ( x=2 \Rightarrow u=1 ) to ( x=5 \Rightarrow u=2 ).

The integral becomes:

$\int_{1}^{2} \frac{u^2 + 1}{\sqrt{u^2}} (2u) \, du = \int_{1}^{2} (u^2 + 1) \cdot 2 \, du.$

This simplifies to:

$2 \int_{1}^{2} (u^2 + 1) \, du,$
which can be evaluated as:

$$2 \left[ \frac{u^3}{3} + u \right]_{1}^{2} = 2 \left[ \frac{8}{3} + 2 - \left( \frac{1}{3} + 1 \right) \right] = 2 \left[ \frac{8}{3} + 2 - \frac{4}{3} \right] = 2 \left[ \frac{4}{3} + 2 \right] = \frac{14}{3}.$

Rearranging, we have:

gl ( x^2 > 1 ). This yields two cases:

1. ( x > 1 )
2. ( x < -1 )

Therefore, the solution set is:

$x < -1 \quad \text{or} \quad x > 1.$

Using the product rule:

Let ( u = e^{x} ) and ( v = \ln x ).

The derivative is given by:

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