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Use a SEPARATE writing booklet - HSC - SSCE Mathematics Extension 1 - Question 11 - 2014 - Paper 1
Question 11
Use a SEPARATE writing booklet. (a) Solve \(\left( x + \frac{2}{x} \right)^2 - 6 \left( x + \frac{2}{x} \right) + 9 = 0.\) (b) The probability that it rains on... show full transcript
Worked Solution & Example Answer:Use a SEPARATE writing booklet - HSC - SSCE Mathematics Extension 1 - Question 11 - 2014 - Paper 1
Step 1
Solve \(\left( x + \frac{2}{x} \right)^2 - 6 \left( x + \frac{2}{x} \right) + 9 = 0.\)
Answer
Let (y = x + \frac{2}{x}). Then the equation simplifies to ((y - 3)^2 = 0), leading to (y = 3).
Substituting back gives: [x + \frac{2}{x} = 3] Multiplying through by (x) results in: [x^2 - 3x + 2 = 0] Factoring the quadratic yields: [(x - 1)(x - 2) = 0] Thus, (x = 1) or (x = 2).
Step 2
Write an expression for the probability that it rains on fewer than 3 days in November.
Answer
Using the binomial distribution, the probability (P(X < 3)) for (X) being the number of days it rains is given by: [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} ] Here, (n = 30), (p = 0.1).
Step 3
Sketch the graph \(y = 6 \tan^{-1}(x),\) clearly indicating the range.
Answer
The function (y = 6 \tan^{-1}(x)) has horizontal asymptotes at (y = 6 \times \frac{\pi}{2} = 3\pi). The range is therefore: ((0, 3\pi)). To sketch:
- The graph approaches this upper limit as (x \to \infty) and (y = 0) as (x \to -\infty).
Step 4
Evaluate \(\int_{2}^{5} \frac{x}{\sqrt{x-1}} \ dx\) using the substitution \(x = u^2 + 1.\)
Answer
Using the substitution (x = u^2 + 1), then (dx = 2u , du). Changing the limits:
- When (x = 2), (u = 1); and when (x = 5), (u = 2).
The integral becomes: [\int_{1}^{2} \frac{u^2 + 1}{u} (2u) , du = \int_{1}^{2} 2(u + \frac{1}{u}) , du] This evaluates to: [2\left(\frac{u^2}{2} + \ln|u|\right) \bigg|_{1}^{2}].
Step 5
Step 6