2022TrialShort answer

Q12 (c) (i)

3 marks

Question

Find $A$ , $B$ and $C$ so that
$\frac{A}{x-1}+\frac{Bx+C}{x^2+4}=\frac{-7x-18}{(x-1)(x^2+4)}.$

Try the question before revealing the answer.

Solution

$A=-5,\qquad B=5,\qquad C=-2$

Working Out:
$\quad \frac{A}{x-1}+\frac{Bx+C}{x^2+4}=-7x-18$

$A(x^2+4)+(Bx+C)(x-1)=-7x-18$

$\text{Sub. in }x=1$

$5A=-25\Rightarrow A=-5$

$\text{Sub. in }x=0;\quad A=-5$

$-20-C=-18\Rightarrow C=-2$

$\text{Sub. in }x=2;\quad A=-5,\quad C=-2$

$-40+(2B-2)=-32$

$2B-2=8$

$B=5$

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Syllabus alignment

Calculus

Further IntegrationApply substitution, partial fractions, integration by parts and recurrence relationships to theoretical and practical problems.

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