Using integration by parts, find the exact value of $$\(\int_{0}^{2} \tan ^{-1}\left(\frac{1}{2} x\right) \mathrm{d} x\)$$. ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................
Exam No：9709_s21_qp_32 Year：2021 Question No：4
Answer:
Square \(a+\mathrm{i} b\), use \(\mathrm{i}^{2}=-1\) and equate real and imaginary parts to 10 and \(-4 \sqrt{6}\) respectively
Obtain \(a^{2}-b^{2}=10\) and \(2 a b=-4 \sqrt{6}\)
Eliminate one unknown and find an equation in the other
Obtain \(a^{4}-10 a^{2}-24[=0]\), or \(b^{4}+10 b^{2}-24[=0]\), or 3-term equivalent
Obtain final answers \(\pm(2 \sqrt{3}-\sqrt{2} i)\), or exact equivalents
Alternative method for Question 4
Use the correct method to find the modulus and argument of \(\sqrt{u}\)
Obtain modulus \(\sqrt{14}\)
Obtain argument \(\tan ^{-1} \frac{-1}{\sqrt{6}}\) using an exact method
Convert to the required form
Obtain answers \(\pm(2 \sqrt{3}-\sqrt{2} \mathrm{i})\), or exact equivalents
Obtain \(a^{2}-b^{2}=10\) and \(2 a b=-4 \sqrt{6}\)
Eliminate one unknown and find an equation in the other
Obtain \(a^{4}-10 a^{2}-24[=0]\), or \(b^{4}+10 b^{2}-24[=0]\), or 3-term equivalent
Obtain final answers \(\pm(2 \sqrt{3}-\sqrt{2} i)\), or exact equivalents
Alternative method for Question 4
Use the correct method to find the modulus and argument of \(\sqrt{u}\)
Obtain modulus \(\sqrt{14}\)
Obtain argument \(\tan ^{-1} \frac{-1}{\sqrt{6}}\) using an exact method
Convert to the required form
Obtain answers \(\pm(2 \sqrt{3}-\sqrt{2} \mathrm{i})\), or exact equivalents
Knowledge points:
3.4.1 use the derivatives of together with constant multiples, sums, differences and composites (Derivatives of are not required.)
3.5.1 extend the idea of ‘reverse differentiation’ to include the integration of (Including examples such as.)
3.5.5 recognise when an integrand can usefully be regarded as a product, and use integration by parts
Solution:
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