2013 A-level H2 Mathematics (9740) Paper 1 Question 4 Suggested Solutions

By KS Teng

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

(i)
(1+2i)^{3}

=(1+2i)^{2}(1+2i)

=[1+4i+(2i)^{2}](1+2i)

=(1+4i-4)(1+2i)

=(4i-3)(1+2i)

=4i-3+4i(2i)-3(2i)

=4i-3-8-6i

=-11-2i

(ii)
Since w=1+2i is a root,
a(1+2i)^{3}+5(1+2i)^{2}+17(1+2i)+b=0

a(-11-2i)+5(-3+4i)+17(1+2i)+b=0

-11a-2ai-15+20i+17+34i+b=0

-11a+2+b-2ai+54i=0

Comparing the real and imaginary coefficients,

-2a+54 = 0 ~\Rightarrow ~ a=27,

-11(27)+2+b= 0 ~\Rightarrow ~ b=295,

\therefore a=27, b=295

(iii)
Since all coefficient of the polynomial is real, by conjugate root theorem, if 1+2i is root, then 1-2i is also a root.
(z-1-2i)(z-1+2i)=z^{2}-2z+5

27z^{3}+5z^{2}+17z+295 \equiv (z^{2}-2z+5)(cz+d),

by comparing coefficients, c=27, d=59

\mathrm{Roots}=1+2i, 1-2i, -\frac{59}{27}

KS Comments:

Since they want to see all workings, let us not skip any steps here and do dilligently. Do check your answers with the GC though. It is good if students emphasise that the coefficients of polynomial is real here, to indicate your understanding. It is shocking that some students confused a root with a factor though.

Comments
    pingbacks / trackbacks

    Leave a Comment

    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
    0

    Start typing and press Enter to search

    2013 A-level H2 Mathematics (9740) Paper 1 Question 3 Suggested SolutionsJC Mathematics
    2013 A-level H2 Mathematics (9740) Paper 1 Question 5 Suggested SolutionsJC Mathematics