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Question 1 (12 marks) Use a SEPARATE writing booklet - HSC - SSCE Mathematics Extension 1 - Question 1 - 2006 - Paper 1
Question 1
Question 1 (12 marks) Use a SEPARATE writing booklet. (a) Find \( \int \frac{dx}{49 + x^2} \) (b) Using the substitution \( u = x^2 + 8 \), or otherwise, find \( \... show full transcript
Worked Solution & Example Answer:Question 1 (12 marks) Use a SEPARATE writing booklet - HSC - SSCE Mathematics Extension 1 - Question 1 - 2006 - Paper 1
Step 1
Find \( \int \frac{dx}{49 + x^2} \)
Answer
To solve this integral, we can use the formula for integrals of the form ( \int \frac{dx}{a^2 + x^2} = \frac{1}{a} \tan^{-1}\left( \frac{x}{a} \right) + C ).
In our case, ( a^2 = 49 ) implies ( a = 7 ). Therefore, we have:
[ \int \frac{dx}{49 + x^2} = \frac{1}{7} \tan^{-1}\left( \frac{x}{7} \right) + C. ]
Step 2
Using the substitution \( u = x^2 + 8 \), or otherwise, find \( \int x \sqrt{4 + 8} dx \)
Answer
We first simplify ( \sqrt{4 + 8} = \sqrt{12} = 2\sqrt{3} ). Hence, the integral becomes:
[ \int x \cdot 2\sqrt{3} dx = 2\sqrt{3} \int x dx = 2\sqrt{3} \cdot \frac{x^2}{2} + C = \sqrt{3} x^2 + C. ]
Step 3
Evaluate \( \lim_{x \to 0} \frac{\sin 5x}{3x} \)
Answer
To evaluate the limit, we can apply L'Hôpital's Rule since both the numerator and denominator approach 0 as ( x \to 0 ):
[ \lim_{x \to 0} \frac{\sin 5x}{3x} = \lim_{x \to 0} \frac{5 \cos 5x}{3} = \frac{5}{3} \cdot \cos(0) = \frac{5}{3}. ]
Step 4
Using the sum of cubes, simplify: \( \frac{\sin^3\theta + \cos^3\theta}{\sin\theta + \cos\theta} - 1 \)
Answer
Using the identity for the sum of cubes, we have:
[ a^3 + b^3 = (a + b)(a^2 - ab + b^2) ]
Where ( a = \sin\theta ) and ( b = \cos\theta ). Thus,
[ \frac{(\sin\theta + \cos\theta)(\sin^2\theta - \sin\theta \cos\theta + \cos^2\theta)}{\sin\theta + \cos\theta} - 1 = \sin^2\theta - \sin\theta \cos\theta + \cos^2\theta - 1 = 0 - \sin\theta \cos\theta. ]
Step 5
For what values of \( b \) is the line \( y = 12x + b \) tangent to \( y = x^3 \)?
Answer
To find the tangent point, we need the slopes to be equal. The derivative of ( y = x^3 ) is ( y' = 3x^2 ).
Setting the slope of the line to the derivative, we have:
[ 12 = 3x^2 \Rightarrow x^2 = 4 \Rightarrow x = 2 \text{ or } x = -2. ]
Next, we substitute these values back to find ( b ):
For ( x = 2 ): [ y = 12(2) + b = (2)^3 \Rightarrow 24 + b = 8 \Rightarrow b = -16. ]
For ( x = -2 ): [ y = 12(-2) + b = (-2)^3 \Rightarrow -24 + b = -8 \Rightarrow b = 16. ]
Thus, the values of ( b ) are ( -16 ) and ( 16 ).