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 \longrightarrow 1

1 = 2C - 5B \longrightarrow 2

-13 = 9A - 5C \longrightarrow 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!

    pingbacks / trackbacks

    Leave a Comment

    16 − two =

    Contact Us

    CONTACT US We would love to hear from you. Contact us, or simply hit our personal page for more contact information

    Not readable? Change text. captcha txt

    Start typing and press Enter to search