SSCE HSC Mathematics Extension 1 Inverse trigonometric functions: Use a SEPARATE writing booklet.

Step 1

Solve \( (x + \frac{2}{x})^{2} - 6(x + \frac{2}{x}) + 9 = 0. \)

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Answer

To solve the equation, let ( y = x + \frac{2}{x} ). The equation simplifies to:

y^{2} - 6y + 9 = 0

This factors to:

(y - 3)^{2} = 0

Thus, ( y = 3 ). Now substituting back:

x + \frac{2}{x} = 3

Multiplying through by ( x ) gives:

x^{2} - 3x + 2 = 0

Factoring this, we find:

(x - 1)(x - 2) = 0

Thus, ( 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

Using the binomial probability formula, the expression for the probability of ( X ) being fewer than 3 can be expressed as:

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)

Where ( X ) is the number of rainy days. Using ( p = 0.1 ) and ( n = 30 ), the probabilities will be calculated as follows:

P(X = k) = \binom{n}{k} p^{k} (1-p)^{n-k}

for ( k = 0, 1, 2 ).

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) ) is a continuous function. As ( x \to -\infty ), ( y \to -3\pi ) and as ( x \to \infty ), ( y \to 3\pi ). Thus, the range is:

(-3\pi, 3\pi)

To sketch:

The curve increases from left to right, starts close to ( -3\pi ) and approaches ( 3\pi ).
The graph is symmetric about the origin, indicating its odd function characteristics.

Step 4

Evaluate \( \int_{2}^{5} \frac{x}{\sqrt{x-1}} dx \) using the substitution \( x = u^{2} + 1. \)

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Answer

Using the substitution ( x = u^{2} + 1 ), we have ( dx = 2u , du ). Changing the limits:

When ( x = 2 ), ( u = 1 ).
When ( x = 5 ), ( u = 2 ).

Transforming the integral:

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

After simplifying and evaluating:

\int_{1}^{2} (2u^{2} + 2) \, du = [\frac{2}{3}u^{3} + 2u]_{1}^{2}

The final value yields the result.

Step 5

Solve \( x^{2} + 5 > 6. \)

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Answer

To solve the inequality:

x^{2} + 5 > 6

Subtract 6 from both sides:

x^{2} > 1

Thus, ( x < -1 ) or ( x > 1 ). The solution set is:

(-\infty, -1) \cup (1, \infty)

Step 6

Differentiate \( e^{x} \ln x. \)

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Answer

Using the product rule for differentiation:

\frac{d}{dx}[e^{x} \ln x] = e^{x} \ln x + e^{x} \cdot \frac{1}{x}

Thus, the derivative is:

e^{x} \ln x + \frac{e^{x}}{x}

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

Worked Solution & Example Answer:Use a SEPARATE writing booklet - HSC - SSCE Mathematics Extension 1 - Question 11 - 2014 - Paper 1

Solve \( (x + \frac{2}{x})^{2} - 6(x + \frac{2}{x}) + 9 = 0. \)

Write an expression for the probability that it rains on fewer than 3 days in November.

Sketch the graph \( y = 6 \tan^{-1}(x) \), clearly indicating the range.

Evaluate \( \int_{2}^{5} \frac{x}{\sqrt{x-1}} dx \) using the substitution \( x = u^{2} + 1. \)

Solve \( x^{2} + 5 > 6. \)

Differentiate \( e^{x} \ln x. \)

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