Question 11 (15 marks) Use a SEPARATE writing booklet - HSC - SSCE Mathematics Extension 1 - Question 11 - 2014 - Paper 1
Question 11
Question 11 (15 marks) Use a SEPARATE writing booklet.
(a) Solve
$$ (x + rac{2}{x})^2 - 6igg(x + rac{2}{x}igg) + 9 = 0 $$
(b) The probability that it rains on... show full transcript
Worked Solution & Example Answer:Question 11 (15 marks) Use a SEPARATE writing booklet - HSC - SSCE Mathematics Extension 1 - Question 11 - 2014 - Paper 1
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Answer
Let ( y = x + \frac{2}{x} ). Then, the equation simplifies to ( y^2 - 6y + 9 = 0 ). Factoring gives ( (y - 3)^2 = 0 ), thus ( y = 3 ).
Substituting back, we have:
[ x + \frac{2}{x} = 3 ]
This can be rearranged to:
[ x^2 - 3x + 2 = 0 ]
Factoring yields ( (x - 1)(x - 2) = 0 ), giving solutions ( x = 1 ) and ( x = 2 ).
Step 2
Write an expression for the probability that it rains on fewer than 3 days in November.
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Answer
The number of days it rains, ( X ), follows a binomial distribution ( X \sim Binomial(n = 30, p = 0.1) ).
The expression for the probability of it raining on fewer than 3 days is:
[ P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) ]
Using the binomial formula:
[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} ]
Step 3
Sketch the graph $y = 6 \tan^{-1} x$, clearly indicating the range.
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Answer
The function ( y = 6 \tan^{-1} x ) has a horizontal asymptote. As ( x \to -\infty ), ( y \to -6\frac{\pi}{2} = -6\pi ) and as ( x \to \infty ), ( y \to 6\frac{\pi}{2} = 6\pi ).
Thus, the range is ( (-6\pi, 6\pi) ). Sketch must indicate these asymptotic behaviors.
Step 4
Evaluate $\int \frac{x}{2\sqrt{x} - 1} dx$ using the substitution $x = u^2 + 1$.
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Answer
Using the substitution ( x = u^2 + 1 ), we find ( dx = 2u , du ).
Substituting into the integral:
[ \int \frac{u^2 + 1}{2u - 1} (2u , du) = 2 \int \frac{u(u^2 + 1)}{2u - 1} , du ]
Proceed to simplify and evaluate this integral using appropriate techniques.
Step 5
Solve $x^2 + 5 > 6$
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Answer
First, simplify the inequality:
[ x^2 > 1 ]
This leads to two cases:
[ x > 1 \text{ or } x < -1 ]
Thus, the solution set is ( x < -1 ) or ( x > 1 ).
Step 6
Differentiate $e^x \text{ln } x$.
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Answer
Using the product rule:
[ \frac{d}{dx}(e^x \text{ln } x) = e^x \text{ln } x + e^x \cdot \frac{1}{x} = e^x \left(\text{ln } x + \frac{1}{x}\right) ]