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Question 14 (15 marks) Use a SEPARATE writing booklet - HSC - SSCE Mathematics Extension 2 - Question 14 - 2017 - Paper 1
Question 14
Question 14 (15 marks) Use a SEPARATE writing booklet. (a) It is given that $$x^4 + 4 = (x^2 + 2x + 2)(x^2 - 2x + 2).$$ (i) Find A and B so that $$\frac{16}{x^4 ... show full transcript
Worked Solution & Example Answer:Question 14 (15 marks) Use a SEPARATE writing booklet - HSC - SSCE Mathematics Extension 2 - Question 14 - 2017 - Paper 1
Step 1
Find A and B so that $$\frac{16}{x^4 + 4} = \frac{A + 2x}{x^2 + 2x + 2} + \frac{B - 2x}{x^2 - 2x + 2}$$
Answer
To find the constants A and B, first multiply both sides of the equation by (x^4 + 4):
Expanding both terms and collecting like terms, we can equate coefficients.
From this, we derive the equations:
- For (x^3): (A - B = 0)
- For (x^2): (2A + 2B - 4 = 0)
- For (x): (2A - 2B = 0)
- Constant term: (2A + 2B = 16)
Solving these equations gives us correct values for A and B.
Step 2
Hence, or otherwise, show that for any real number m, $$\int_{0}^{m} \frac{16}{x^4 + 4} dx = \ln \left( \frac{m^2 + 2m + 2}{m^2 - 2m + 2} \right) + 2\tan^{-1}(m + 1) + 2\tan^{-1}(m - 1).$$
Answer
Using the previous result, we can express the integral as:
With the substitutions required from the form derived earlier, we can integrate each fraction separately. The integral of (\frac{A + 2x}{x^2 + 2x + 2}) and (\frac{B - 2x}{x^2 - 2x + 2}) can be computed using standard calculus techniques like substitution and partial fraction decomposition.
Ultimately, after evaluating the limits, we obtain:
Step 3
Find the limiting value as m → ∞ of $$\int_{0}^{m} \frac{16}{x^4 + 4} dx.$$
Answer
As (m) approaches infinity, we analyze the behavior of each term in the derived equation:
- The logarithmic term, when (m) approaches infinity, simplifies to:
- The arctangent terms approach:
Thus, the limiting value is:
Therefore, the answer is (2\pi).