SSCE HSC Mathematics Extension 1 Binomial distribution: Use a SEPARATE writing booklet.

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 $

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Answer

Let ( y = x + \frac{2}{x} ). Then the equation transforms to:

(y - 3)^2 = 0.

Hence, ( y - 3 = 0 ) implies ( y = 3 ). Substituting back:

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

Multiplying through by ( x ):

3x = x^2 + 2 \implies x^2 - 3x + 2 = 0.

Factoring gives:

(x - 1)(x - 2) = 0 \implies x = 1 \text{ or } 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

Let ( X ) be the random variable representing the number of rainy days in November, which follows a binomial distribution:

X \sim \text{Binomial}(n = 30, p = 0.1).

The probability of rain on fewer than 3 days is:

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = \sum_{k=0}^{2} {30 \choose k} (0.1)^k (0.9)^{30 - 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 ) is an increasing function. The range can be calculated by evaluating the limits: As ( x \to -\infty, \tan^{-1} x \to -\frac{\pi}{2} \implies y \to -3\pi ); As ( x \to \infty, \tan^{-1} x \to \frac{\pi}{2} \implies y \to 3\pi.\n$$ Thus, the range is ( (-3\pi, 3\pi) ).

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 ) gives ( dx = 2u , du ) and changes the limits:

When ( x = 2, u = 1 )
When ( x = 5, u = 2 ) Thus, the integral becomes:

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

Calculating the limits gives the evaluated result.

Step 5

Solve $ x^2 + 5 > 6 $.

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Answer

Rearranging the inequality gives:

x^2 > 1.

Taking the square root results in:\n1. ( x > 1 ) 2. ( x < -1 ) Thus, the solution is ( x < -1 ) or ( x > 1 ).

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} = e^{x} \left( \ln x + \frac{1}{x} \right).

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 \( \left( x + \frac{2}{x} \right)^2 - 6 \left( x + \frac{2}{x} \right) + 9 = 0 \)

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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