SSCE HSC Mathematics Extension 1 Definite integrals and substitution: 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

Letting ( y = x + \frac{2}{x} ), the equation simplifies to:

(y - 3)^2 = 0

Thus, ( y = 3 ).
We substitute back to find ( x ):

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

Factoring gives:

(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

The probability of rain on any given day in November is ( p = 0.1 ).
Using the binomial probability formula:

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}

For ( n = 30 ) and ( 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 ) has a domain of all real numbers. As ( x \to -\infty ), ( y \to -6\dfrac{\pi}{2} ) and as ( x \to +\infty ), ( y \to 6\dfrac{\pi}{2} ).
Thus, the range is:

y \in \left( -6\dfrac{\pi}{2}, 6\dfrac{\pi}{2} \right) \equiv (-9.42, 9.42).

Indicate this range on the sketch.

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 substitution, let ( x = u^2 + 1 ). Then ( dx = 2u , du ) and the limits change from ( x=2 ) to ( u=1 ) and ( x=5 ) to ( u=2 ).
The integral becomes:

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

Integrating gives:

2\left[ \frac{u^2}{2} + \ln|u| \right]_{1}^{2} = [2 + \ln(2)] - [1 + 0] = 1 + \ln(2).

Step 5

Solve $$x^2 + 5 > 6.$$

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Answer

Rearranging gives:

x^2 > 1

Taking square roots results in:

x > 1 \quad \text{or} \quad x < -1.

Thus, 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, let:

y = e^x \ln x;

hence,

y' = 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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