1. Find $ \int \frac{dx}{49 + x^2} $ 2. Using the substitution $ u = x^2 + 8 $, or otherwise, find $ \int x \sqrt{4 + 8} \, dx $. 3. Evaluate $ \lim_{x \to 0} \frac{\sin 5x}{3x} $. 4. Using the sum of cubes, simplify: $ \frac{\sin^3 \theta + \cos^3 \theta}{\sin \theta + \cos \theta} - 1 $, for $ 0 < \theta < \frac{\pi}{2} $. 5. For what values of $ b $ is the line $ y = 12x + b $ tangent to $ y = x^3 $?

SSCE HSC Mathematics Extension 1 Integration involving inverse trigonometric functions: 1. Find \( \int \frac{dx}{49 + x

1. Find $ \int \frac{dx}{49 + x^2} $ 2 - HSC - SSCE Mathematics Extension 1 - Question 1 - 2006 - Paper 1

Question 1

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1. Find $ \int \frac{dx}{49 + x^2} $ 2. Using the substitution $ u = x^2 + 8 $, or otherwise, find $ \int x \sqrt{4 + 8} \, dx $. 3. Evaluate \( \lim_{x \to ... show full transcript

Worked Solution & Example Answer:1. Find $ \int \frac{dx}{49 + x^2} $ 2 - HSC - SSCE Mathematics Extension 1 - Question 1 - 2006 - Paper 1

Step 1

Find $ \int \frac{dx}{49 + x^2} $

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Answer

To determine the integral, we utilize the formula for the integral of the form ( \int \frac{dx}{a^2 + x^2} = \frac{1}{a} \tan^{-1}(\frac{x}{a}) + C ). Here, ( a^2 = 49 ) implies ( a = 7 ). Thus,

\int \frac{dx}{49 + x^2} = \frac{1}{7} \tan^{-1}(\frac{x}{7}) + C.

Step 2

Using the substitution $ u = x^2 + 8 $, or otherwise, find $ \int x \sqrt{4 + 8} \, dx $

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Answer

Here, the integral implies:

\int x \sqrt{12} \, dx = \sqrt{12} \int x \, dx = \sqrt{12} \cdot \frac{x^2}{2} + C = \frac{\sqrt{12}}{2} x^2 + C.

Step 3

Evaluate $ \lim_{x \to 0} \frac{\sin 5x}{3x} $

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Answer

Using L'Hôpital's Rule, we take the derivative of the numerator and denominator:

\lim_{x \to 0} \frac{\sin 5x}{3x} = \lim_{x \to 0} \frac{5 \cos 5x}{3} = \frac{5(1)}{3} = \frac{5}{3}.

Step 4

Using the sum of cubes, simplify: $ \frac{\sin^3 \theta + \cos^3 \theta}{\sin \theta + \cos \theta} - 1 $

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Answer

The sum of cubes can be simplified using:

\frac{\sin^3 \theta + \cos^3 \theta}{\sin \theta + \cos \theta} = \sin^2 \theta + \cos^2 \theta = 1.

Thus,

1 - 1 = 0.

Step 5

For what values of $ b $ is the line $ y = 12x + b $ tangent to $ y = x^3 $?

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Answer

To find the tangent condition, set the equations equal:

12x + b = x^3.

Taking derivatives,

For the y-coordinates,

At ( x = 2 ): ( y = 12(2) + b = 2^3 = 8 \Rightarrow 24 + b = 8 \Rightarrow b = -16.)

At ( x = -2 ): ( y = 12(-2) + b = (-2)^3 = -8 \Rightarrow -24 + b = -8 \Rightarrow b = 16.)

Thus, the values of ( b ) are ( -16 ) and ( 16.)

1. Find \( \int \frac{dx}{49 + x^2} \) 2 - HSC - SSCE Mathematics Extension 1 - Question 1 - 2006 - Paper 1

Worked Solution & Example Answer:1. Find \( \int \frac{dx}{49 + x^2} \) 2 - HSC - SSCE Mathematics Extension 1 - Question 1 - 2006 - Paper 1

Find \( \int \frac{dx}{49 + x^2} \)

Using the substitution \( u = x^2 + 8 \), or otherwise, find \( \int x \sqrt{4 + 8} \, dx \)

Evaluate \( \lim_{x \to 0} \frac{\sin 5x}{3x} \)

Using the sum of cubes, simplify: \( \frac{\sin^3 \theta + \cos^3 \theta}{\sin \theta + \cos \theta} - 1 \)

For what values of \( b \) is the line \( y = 12x + b \) tangent to \( y = x^3 \)?

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