Let f x = x 2 - 8 x + 9 ( 1 - x ) ( 2 - x ) 2 . Express f ( x ) in partial fractions.

Given, f x = x 2 - 8 x + 9 ( 1 - x ) ( 2 - x ) 2 Since the given proper fraction has repeated a linear factor. So, we can split the given fraction into partial fractions as follows: x 2 - 8 x + 9 ( 1 - x ) ( 2 - x ) 2 = A 1 - x + B 2 - x + C 2 - x 2 ⇒ x 2 - 8 x + 9 ( 1 - x ) ( 2 - x ) 2 = A 2 - x 2 + B 1 - x 2 - x + C 1 - x ( 1 - x ) ( 2 - x ) 2 Comparing the numerators of the above equation, we have x 2 - 8 x + 9 = A 2 - x 2 + B 1 - x 2 - x + C 1 - x . . . . . i Substituting x = 1 in the above equation i , we have 1 2 - 8 1 + 9 = A 2 - 1 2 ⇒ A = 1 - 8 + 9 ⇒ A = 2 . . . . i i Now, substituting x = 2 in equation i , we have 2 2 - 8 × 2 + 9 = A 2 - 2 2 + B 1 - 2 2 - 2 + C 1 - 2 ⇒ 4 - 16 + 9 = - C ⇒ - C = - 3 ⇒ C = 3 . . . . . i i i Putting x = 0 in equation i , we have 9 = 4 A + 2 B + C ⇒ 4 × 2 + 2 B + 3 = 9 ⇒ 11 + 2 B = 9 ⇒ 2 B = - 2 ⇒ B = - 1 . . . . i v Hence, x 2 - 8 x + 9 ( 1 - x ) ( 2 - x ) 2 = 2 1 - x - 1 2 - x + 3 2 - x 2 [From i i , i i i , and i v ]

Let f x = 5 x - 2 ( x - 1 ) 2 x 2 - 1 . Express f x in partial fractions.

Given, f x = 5 x - 2 ( x - 1 ) 2 x 2 - 1 Since the denominator of a proper algebraic fraction has a quadratic factor of the form c x 2 + d that cannot be factorised. So, we can split the given fraction into partial fractions as follows: 5 x - 2 ( x - 1 ) 2 x 2 - 1 = A x - 1 + B x + C 2 x 2 - 1 ⇒ 5 x - 2 ( x - 1 ) 2 x 2 - 1 = A 2 x 2 - 1 + B x + C x - 1 ( x - 1 ) 2 x 2 - 1 Comparing the numerators of the above equation, we have 5 x - 2 = A 2 x 2 - 1 + B x + C x - 1 ⇒ 5 x - 2 = A 2 x 2 - 1 + B x x - 1 + C x - 1 . . . . i Substituting x = 1 in the above equation i , we have 5 1 - 2 = A 2 - 1 ⇒ A = 3 . . . . i i Now, substituting x = 0 in equation i , we have ⇒ 5 0 - 2 = A 2 0 2 - 1 + 0 + C 0 - 1 ⇒ - A - C = - 2 ⇒ - 3 - C = - 2 [Using i i ] ⇒ C = - 1 . . . . . i i i Putting x = - 1 in equation i , we have ⇒ 5 × - 1 - 2 = A 2 × - 1 2 - 1 + B × - 1 - 1 - 1 + C - 1 - 1 ⇒ - 7 = A + 2 B - 2 C ⇒ - 7 = 3 + 2 B - 2 - 1 ⇒ 2 B + 5 = - 7 ⇒ B = - 6 . . . . i v Hence, 5 x - 2 ( x - 1 ) 2 x 2 - 1 = 3 x - 1 + - 6 x - 1 2 x 2 - 1 [From i i , i i i , and i v ].

E M B I B E

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 16 - Cross-Topic Review Exercise 3>CROSS-TOPIC REVIEW EXERCISE 3

Sue Pemberton, Julianne Hughes and, Julian Gilbey Solutions for Chapter: Cross-Topic Review Exercise 3, Exercise 1: CROSS-TOPIC REVIEW EXERCISE 3

Author:Sue Pemberton, Julianne Hughes & Julian Gilbey

Sue Pemberton Mathematics Solutions for Exercise - Sue Pemberton, Julianne Hughes and, Julian Gilbey Solutions for Chapter: Cross-Topic Review Exercise 3, Exercise 1: CROSS-TOPIC REVIEW EXERCISE 3

Attempt the free practice questions on Chapter 16: Cross-Topic Review Exercise 3, Exercise 1: CROSS-TOPIC REVIEW EXERCISE 3 with hints and solutions to strengthen your understanding. Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book solutions are prepared by Experienced Embibe Experts.

Questions from Sue Pemberton, Julianne Hughes and, Julian Gilbey Solutions for Chapter: Cross-Topic Review Exercise 3, Exercise 1: CROSS-TOPIC REVIEW EXERCISE 3 with Hints & Solutions

HARD

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 16 - Cross-Topic Review Exercise 3>CROSS-TOPIC REVIEW EXERCISE 3>Q 8

Expand $\frac{1}{\sqrt{(1 - 4 x)}}$ in ascending powers of $x$ , up to and including the term in $x^{2}$ , simplifying the coefficients.

HARD

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 16 - Cross-Topic Review Exercise 3>CROSS-TOPIC REVIEW EXERCISE 3>Q 8

Find the coefficient of $x^{2}$ in the expansion of $\frac{1 + 2 x}{\sqrt{(4 - 16 x)}}$ .

HARD

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 16 - Cross-Topic Review Exercise 3>CROSS-TOPIC REVIEW EXERCISE 3>Q 9

When $(1 + a x)^{- 2}$ , where $a$ is a positive constant, is expanded in ascending powers of $x$ , the coefficients of $x$ and $x^{3}$ are equal. Find the exact value of $a$ .

MEDIUM

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 16 - Cross-Topic Review Exercise 3>CROSS-TOPIC REVIEW EXERCISE 3>Q 9

Expand ${(1 + \frac{1}{\sqrt{2}} x)}^{- 2}$ in ascending powers of $x$ , up to and including the term in $x^{2}$ , simplifying the coefficients.

MEDIUM

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 16 - Cross-Topic Review Exercise 3>CROSS-TOPIC REVIEW EXERCISE 3>Q 18

Let $f (x) = \frac{x^{2} - 8 x + 9}{(1 - x) (2 - x)^{2}}$ . Express $f (x)$ in partial fractions.

HARD

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 16 - Cross-Topic Review Exercise 3>CROSS-TOPIC REVIEW EXERCISE 3>Q 18

Let $f (x) = \frac{x^{2} - 8 x + 9}{(1 - x) (2 - x)^{2}}$ . Express $f (x)$ in partial fractions. Hence obtain the expansion of $f (x)$ in ascending powers of $x$ , up to and including the term in $x^{2} .$

MEDIUM

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 16 - Cross-Topic Review Exercise 3>CROSS-TOPIC REVIEW EXERCISE 3>Q 19

Let $f (x) = \frac{5 x - 2}{(x - 1) (2 x^{2} - 1)}$ . Express $f (x)$ in partial fractions.

HARD

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 16 - Cross-Topic Review Exercise 3>CROSS-TOPIC REVIEW EXERCISE 3>Q 19

Let $f (x) = \frac{5 x - 2}{(x - 1) (2 x^{2} - 1)}$ . Express $f (x)$ in partial fractions. Hence obtain the expansion of $f (x)$ in ascending powers of $x$ , up to and including the term in $x^{2}$ .

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