Find the inverse of the function $y = x^3 - 2$ - HSC - SSCE Mathematics Extension 1 - Question 11 - 2016 - Paper 1

Using the substitution $u = x - 4$, we have:

1. Differentiate to find $dx$: [ dx = du ]
2. The integral becomes [ \int \sqrt{u} , du ]
3. Evaluate the integral: [ \int \sqrt{u} , du = \frac{2}{3} u^{3/2} + C ]
4. Substitute back: [ = \frac{2}{3} (x - 4)^{3/2} + C ]

Using the chain rule:

1. Let [ y = 3 \tan^{-1}(2x) ]
2. Differentiate: [ \frac{dy}{dx} = 3 \cdot \frac{1}{1 + (2x)^2} \cdot 2 = \frac{6}{1 + 4x^2} ]

First, recognize that (2 \sin x \cos x = \sin(2x)), so:

1. The limit can be rewritten as [ \lim_{x \to 0} \frac{\sin(2x)}{3x} = \frac{2}{3} \lim_{x \to 0} \frac{\sin(2x)}{2x} ]
2. Evaluating gives [ = \frac{2}{3} \cdot 1 = \frac{2}{3} ]

1. Rearranging gives: [ \frac{3}{2x + 5} > x ]
2. Multiply through by $2x + 5$ (assuming $2x + 5 > 0$): [ 3 > x(2x + 5) ]
3. This leads to the quadratic: [ 2x^2 + 5x - 3 < 0 ]
4. Factor to find the roots, then test intervals to solve the inequality.

Let $p = \frac{2}{3}$, then the probability of hitting exactly once is:

1. Probability formula: [ P(X = 1) = \binom{3}{1} p^1 (1-p)^2 = 3 \cdot \left(\frac{2}{3}\right)^1 \cdot \left(\frac{1}{3}\right)^2 = 3 \cdot \frac{2}{3} \cdot \frac{1}{9} = \frac{6}{27} = \frac{2}{9} ]

To find the probability of at least two hits:

1. Calculate probabilities for zero and one hits, then subtract from 1:
   • [ P(X = 0) = \left(\frac{1}{3}\right)^6 ]
   • [ P(X = 1) = \binom{6}{1} \left(\frac{2}{3}\right)^1 \left(\frac{1}{3}\right)^5 ]
   • Evaluate and subtract these from 1 to find $P(X \geq 2) = 1 - (P(X = 0) + P(X = 1))$.