2014 A-level H2 Mathematics (9740) Paper 2 Question 2 Suggested Solutions

All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

Using partial fractions formula found in MF15,

Let $\frac{9x^{2}+x-13}{(2x-5)(x^{2}+9)} = \frac{A}{2x-5} + \frac{Bx+C}{x^{2}+9}$

$9x^{2}+x-13 = A(x^{2}+9)+(Bx+C)(2x-5)$

$9 = A + 2B - - - (1)$

$1 = 2C - 5B - - - (2)$

$-13 = 9A - 5C - - - (3)$

Using GC, $A = 3, B = 3, C = 8$

$\int_{0}^{2} \frac{9x^{2}+x-13}{(2x-5)(x^{2}+9)} dx$

$= \int_{0}^{2} \frac{3}{2x-5} + \frac{3x+8}{x^{2}+9} dx$

$= \int_{0}^{2} \frac{3}{2} \frac{2}{2x-5} + \frac{3x}{x^{2}+9} + \frac{8}{x^{2}+9}dx$

$= \int_{0}^{2} \frac{3}{2} \frac{2}{2x-5} + \frac{3}{2} \frac{2x}{x^{2}+9} + 8 \frac{1}{x^{2}+9}dx$

$= [\frac{3}{2} \mathrm{ln}|2x-5|+ \frac{3}{2} \mathrm{ln}(x^{2}+9) + \frac{8}{3} tan^{-1} \frac{8}{3}]\biggl|_0^2$

$= \frac{3}{2} \mathrm{ln} 13 + \frac{8}{3} tan^{-1} \frac{2}{3} - (\frac{3}{2} \mathrm{ln}5 + \frac{3}{2} \mathrm{ln}9]$

$= \frac{3}{2} \mathrm{ln} \frac{13}{45} + \frac{8}{3} tan^{-1} \frac{2}{3}$

$a = \frac{3}{2}, b = \frac{13}{45}, c = \frac{8}{3}, d = \frac{2}{3}$

Personal Comments:
Firstly, this question is a lot of marks! It is quite a standard tutorial question that tests students on all their integration techniques. I think that they combined system of linear equations here too, which is really neat. Students could also have solved the partial fractions using the substitution (cover-up) method. Do use the MF15 for the partial fractions and integration formula!
Finally, the answers are to be left in rational form, which means to be expressed as a fraction of two integers!

About

KS Teng

KS has been teaching H2/H1 Mathematics and IB mathematics for the more than 10 years. Having taught students from all various Junior Colleges, KS adapts to students' abilities and help them better understand the topics. As someone who loves to teach mathematics and sees it as a truly useful tool in life, KS seeks to enable students to appreciate math. Therefore, his tuition mission is to motivate and cultivate students to be independent and confident thinkers.

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