P2 §1.6 Mixed exercise for chapter 1

Pure mathematics Year 2

Table of contents

1. Edexcel P2 Ch1 Mixed exercise 1

1. Edexcel P2 Ch1 Mixed exercise 1

Question 1. (E/P) Prove by contradiction that $ \sqrt{ \frac12 }$ is an irrational number. [5 marks]

 

(Edexcel 2017 Specifications, P2 Ch1 Mixed exercise 1 Q1.)

 

Solution.

 

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Question 2. (P) Prove that if $q^2$ is an irrational number then $q$ is an irrational number.

 

(Edexcel 2017 Specifications, P2 Ch1 Mixed exercise 1 Q2.)

 

Solution.

 

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Question 3. Simplify: $$ \begin{align} &\textbf{(a)} \qquad \frac{ x - 4 }{ 6 } \times \frac{ 2x + 8 }{ x^2 - 16 } \\ \\ &\textbf{(b)} \qquad \frac{ x^2 - 3x - 10 }{ 3x^2 - 21 } \times \frac{ 6x^2 + 24 }{ x^2 + 6x +8 } \\ \\ &\textbf{(c)} \qquad \frac{ 4x^2 + 12x + 9 }{ x^2 + 6x } \div \frac{ 4x^2 - 9 }{ 2x^2 + 9x - 18 } \end{align} $$

 

(Edexcel 2017 Specifications, P2 Ch1 Mixed exercise 1 Q3.)

 

Solution.

 

(a) $$ \begin{align} \frac{ x - 4 }{ 6 } \times \frac{ 2x + 8 }{ x^2 - 16 } &= \frac{ x - 4 }{ 6 } \times \frac{ 2(x + 4) }{ (x+4)(x-4) } \\ &= \frac13 \qquad \checkmark \end{align} $$

 

(b) $$ \begin{align} \frac{ x^2 - 3x - 10 }{ 3x^2 - 21 } \times \frac{ 6x^2 + 24 }{ x^2 + 6x +8 } &= \frac{ (x-5)(x+2) }{ 3 \left( x^2 - 7 \right) } \times \frac{ 6 \left( x^2 + 4 \right) }{ (x+2)(x+4) } \\ &= \frac{ (x-5) }{ \left( x^2 - 7 \right) } \times \frac{ 2 \left( x^2 + 4 \right) }{ (x+4) } \\ &= \frac{ 2 \left( x^2 + 4 \right) ( x - 5 ) }{ \left( x^2 - 7 \right) ( x + 4 ) } \qquad \checkmark \end{align} $$

 

(c) $$ \begin{align} \frac{ 4x^2 + 12x + 9 }{ x^2 + 6x } \div \frac{ 4x^2 - 9 }{ 2x^2 + 9x - 18 } &= \frac{ (2x+3)^2 }{ x (x + 6) } \times \frac{ (2x - 3) (x + 6) }{ (2x + 3)(2x - 3) } \\ &= \frac{ 2x + 3 }{ x } \qquad \checkmark \end{align} $$

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Question 4. (E/P)

(a) Simplify fully $$ \begin{align} \frac{ 4x^2 - 8x }{ x^2 - 3x - 4 } \times \frac{ x^2 + 6x + 5 }{ 2x^2 + 10x } \end{align} $$ [3 marks]

 

(b) Given that $$ \begin{align} \ln \Big[ \left( 4x^2 - 8x \right) \left( x^2 + 6x + 5 \right) \Big] = 6 + \ln \Big[ \left( x^2 - 3x - 4 \right) \left( 2x^2 + 10x \right) \Big], \end{align} $$ find $x$ in terms of $ \textrm e $. [4 marks]

 

(Edexcel 2017 Specifications, P2 Ch1 Mixed exercise 1 Q4.)

 

Solution.

(a) $$ \begin{align} \frac{ 4x^2 - 8x }{ x^2 - 3x - 4 } \times \frac{ x^2 + 6x + 5 }{ 2x^2 + 10x } &= \frac{ 4x (x - 2) }{ (x - 4) (x + 1) } \times \frac{ (x + 1) (x + 5) }{ 2x ( x + 5 ) } \\ &= \frac{ 2 (x - 2) }{ x - 4 } \qquad \checkmark \end{align} $$

 

(b) $$ \begin{align} && \ln \Big[ \left( 4x^2 - 8x \right) \left( x^2 + 6x + 5 \right) \Big] &= 6 + \ln \Big[ \left( x^2 - 3x - 4 \right) \left( 2x^2 + 10x \right) \Big] \\ \\ &\Rightarrow & \ln \left[ \frac{ \left( 4x^2 - 8x \right) \left( x^2 + 6x + 5 \right) }{ \left( x^2 - 3x - 4 \right) \left( 2x^2 + 10x \right) } \right] &= 6 \\ &\Rightarrow & \ln \left[ \frac{ 2 (x - 2) }{ x - 4 } \right] &= 6 \\ &\Rightarrow & \frac{ 2 (x - 2) }{ x - 4 } &= \textrm e^6 \\ &\Rightarrow & 2x - 4 &= \textrm e^6 ( x - 4 ) \\ &\Rightarrow & \left( 2 - \textrm e^6 \right) x &= 4 \left( 1 - \textrm e^6 \right) \\ \\ &\Rightarrow & x &= \frac{ 4 \left( 1 - \textrm e^6 \right) }{ \left( 2 - \textrm e^6 \right) } = \frac{ 4 \left( \textrm e^6 - 1 \right) }{ \left( \textrm e^6 - 2 \right) } \qquad \checkmark \end{align} $$

 

Note: A similar question: P2 Ch1.2 Exercise 1B Q5.

 

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Question 5. (E/P) $$ \textrm g(x) = \frac{ 4x^3 - 9x^2 - 9x }{ 32x + 24 } \div \frac{ x^2 + 6x + 5 }{ 6x^2 - 13x - 5 } $$

(a) Show that $ \textrm g(x) $ can be written in the form $ ax^2 + bx + c $, where $a,b$ and $c$ are constants to be found. [4 marks]

 

(b) Hence differentiate $ \textrm g(x) $ and find $ \textrm g'(-2) $. [3 marks]

 

(Edexcel 2017 Specifications, P2 Ch1 Mixed exercise 1 Q5.)

 

Solution.

(a) $$ \begin{align} \textrm g(x) &= \frac{ 4x^3 - 9x^2 - 9x }{ 32x + 24 } \div \frac{ x^2 - 3x }{ 6x^2 - 13x - 5 } \\ &= \frac{ x(4x + 3) (x - 3) }{ 8 (4x + 3) } \times \frac{ (3x + 1)(2x - 5) }{ x(x-3) } \\ &= \frac{ (3x + 1)(2x - 5) }{ 8 } \\ &= \frac{ 6x^2 - 13x - 5 }{ 8 } \\ &= \frac34 x^2 - \frac{13}{8}x - \frac{5}{8} \qquad \checkmark \end{align} $$

 

(b) $$ \begin{align} && \textrm g(x) &= \frac34 x^2 - \frac{13}{8}x - \frac{5}{8} \\ \\ &\Rightarrow & \textrm g'(x) &= \frac32 x - \frac{13}{8} \\ &\Rightarrow & \textrm g'(-2) &= \frac32 (-2) - \frac{13}{8} \\ &&&= - 3 - \frac{13}{8} \\ &&&= - \frac{37}{8} \qquad \checkmark \end{align} $$

 

Note: A similar question: P2 Ch1.2 Exercise 1B Q6.

 

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Question 6. (E) Express $$ \frac{ 6x + 1 }{ x - 5 } + \frac{ 5x + 3 }{ x^2 - 3x - 10 } $$ as a single fraction in its simplest form. [4 marks]

 

(Edexcel 2017 Specifications, P2 Ch1 Mixed exercise 1 Q6.)

 

Solution.

$$ \begin{align} \frac{ 6x + 1 }{ x - 5 } + \frac{ 5x + 3 }{ x^2 - 3x - 10 } &= \frac{ 6x + 1 }{ x - 5 } + \frac{ 5x + 3 }{ (x-5)(x+2) } \\ &= \frac{ (6x + 1)(x+2) + 5x + 3 }{ (x-5)(x+2) } \\ &= \frac{ \left( 6x^2 + 13x + 2 \right) + 5x + 3 }{ (x-5)(x+2) } \\ &= \frac{ 6x^2 + 18x + 5 }{ (x-5)(x+2) } \qquad \checkmark \\ &= \frac{ 6x^2 + 18x + 5 }{ x^2 - 3x - 10 } \qquad \checkmark \end{align} $$

 

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Question 7. (E) $$ \textrm f(x) = x + \frac{ 3 }{ x - 1 } - \frac{ 12 }{ x^2 + 2x - 3 }, \qquad x \in \mathbb R, \quad x > 1 $$ Show that $$ \textrm f(x) = \frac{ x^2 + 3x + 3 }{ x + 3 } $$ [4 marks]

 

(Edexcel 2017 Specifications, P2 Ch1 Mixed exercise 1 Q7.)

 

Solution.

$$ \begin{align} \textrm f(x) &= x + \frac{ 3 }{ x - 1 } - \frac{ 12 }{ x^2 + 2x - 3 } \\ &= x + \frac{ 3 }{ x - 1 } - \frac{ 12 }{ (x+3)(x-1) } \\ &= \frac{ x(x+3)(x-1) + 3 (x+3) - 12 }{ (x+3)(x-1) } \\ &= \frac{ \left( x^3 + 2x^2 - 3x \right) + (3x+9) - 12 }{ (x+3)(x-1) } \\ &= \frac{ x^3 + 2x^2 - 3 }{ (x+3)(x-1) } \end{align} $$ For the numerator, let $ \textrm g(x) = x^3 + 2x^2 - 3 $ and note that: $$ \begin{align} g(1) = 1 + 2 - 3 = 0 \end{align} $$ By the factor theorem, we find: $$ \begin{align} \textrm g(x) = x^3 + 2x^2 - 3 = (x-1) \left( x^2 + 3x + 3 \right) \end{align} $$ This gives $$ \begin{align} \Rightarrow \qquad \textrm f(x) = \frac{ \textrm g(x) }{ (x+3)(x-1) } &= \frac{ (x-1) \left( x^2 + 3x + 3 \right) }{ (x+3)(x-1) } \\ &= \frac{ x^2 + 3x + 3 }{ x+3 } \qquad \checkmark \end{align} $$

 

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Question 8. (E) $$ \textrm f(x) = \frac{ x - 3 }{ x (x-1) } $$ Show that $ \textrm f(x) $ can be written in the form $$ \frac{ A }{ x } + \frac{ B }{ x - 1 } $$ where $A$ and $B$ are constants to be found. [3 marks]

 

(Edexcel 2017 Specifications, P2 Ch1 Mixed exercise 1 Q8.)

 

Solution.

Let $$ \begin{align} \textrm f(x) = \frac{ x - 3 }{ x (x-1) } &\equiv \frac{ A }{ x } + \frac{ B }{ x - 1 } \\ &\equiv \frac{ A (x-1) + B x }{ x (x - 1) } \end{align} $$ Compare the numerators on both sides: $$ A (x-1) + B x \equiv x - 3. $$ By the method of substitution: $$ \begin{align} \textrm{(i)}&& x &= 0 &&\Rightarrow& -A&=-3 \\ &&&&&\Rightarrow& A&=3 \\ \\ \textrm{(ii)}&& x &= 1 &&\Rightarrow& B&=-2 \end{align} $$ and we find: $$ \Rightarrow \qquad \textrm f(x) = \frac{ x - 3 }{ x (x-1) } \equiv \frac{ 3 }{ x } - \frac{ 2 }{ x - 1 } \qquad \checkmark $$ where $$ A = 3, \qquad B = -2 \qquad \checkmark $$

 

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Question 9. (E) $$ \frac{ - 15x + 21 }{ (x-2) (x+1) (x-5) } \equiv \frac{ P }{ x - 2 } + \frac{ Q }{ x + 1 } + \frac{ R }{ x - 5 }. $$ Find the values of the constants $P,Q$ and $R$. [4 marks]

 

(Edexcel 2017 Specifications, P2 Ch1 Mixed exercise 1 Q9.)

 

Solution.

Let $$ \begin{align} \frac{ - 15x + 21 }{ (x-2) (x+1) (x-5) } &\equiv \frac{ P }{ x - 2 } + \frac{ Q }{ x + 1 } + \frac{ R }{ x - 5 } \\ &\equiv \frac{ P (x + 1) (x - 5) + Q (x - 2) (x - 5) + R (x - 2) (x + 1) }{ (x - 2) (x + 1) (x - 5) } \\ \end{align} $$ Compare the numerators on both sides: $$ P (x + 1) (x - 5) + Q (x - 2) (x - 5) + R (x - 2) (x + 1) \equiv - 15x + 21. $$ By the method of substitution: $$ \begin{align} \textrm{(i)}&& x &= 2 &&\Rightarrow& (3)(-3) P &= -30 + 21\\ &&&&&\Rightarrow& -9 P&= - 9 \\ &&&&&\Rightarrow& P&= 1 \\ \\ \textrm{(ii)}&& x &= -1 &&\Rightarrow& (-3)(-6) Q &= 15 + 21 \\ &&&&&\Rightarrow& 18 Q &= 36 \\ &&&&&\Rightarrow& Q &= 2 \\ \\ \textrm{(iii)}&& x &= 5 &&\Rightarrow& (3)(6) R &= - 75 + 21 \\ &&&&&\Rightarrow& 18 R &= - 54 \\ &&&&&\Rightarrow& R &= - 3 \end{align} $$ and we find: $$ \Rightarrow \qquad \frac{ - 15x + 21 }{ (x-2) (x+1) (x-5) } \equiv \frac{ 1 }{ x - 2 } + \frac{ 2 }{ x + 1 } - \frac{ 3 }{ x - 5 } \qquad \checkmark $$ where $$ P = 1, \qquad Q = 2, \qquad R = -3 \qquad \checkmark $$

 

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Question 10. (E) Show that $$ \frac{ 16x - 1 }{ (3x+2) (2x-1) } $$ can be written in the form $$ \frac{ D }{ 3x + 2 } + \frac{ E }{ 2x - 1 } $$ and find the values of the constants $D$ and $E$. [4 marks]

 

(Edexcel 2017 Specifications, P2 Ch1 Mixed exercise 1 Q10.)

 

Solution.

Let $$ \begin{align} \frac{ 16x - 1 }{ (3x+2) (2x-1) } &\equiv \frac{ D }{ 3x + 2 } + \frac{ E }{ 2x - 1 } \\ &\equiv \frac{ D (2x - 1) + E (3x + 2) }{ (3x + 2) (2x - 1) } \end{align} $$ Compare the numerators on both sides: $$ D (2x - 1) + E (3x + 2) \equiv 16x - 1. $$ By the method of substitution: $$ \begin{align} \textrm{(i)}&& x &= - \frac23 &&\Rightarrow& \left( -\frac43 - 1 \right) D &= - \frac{32}{3} - 1 \\ &&&&&\Rightarrow& - \frac{7}{3} D &= - \frac{35}{3} \\ &&&&&\Rightarrow& D &= 5 \\ \\ \textrm{(ii)}&& x &= \frac12 &&\Rightarrow& \left( \frac32 + 2 \right) E &= 8 - 1 \\ &&&&&\Rightarrow& \frac{7}{2} E &= 7 \\ &&&&&\Rightarrow& E &= 2 \end{align} $$ and we find: $$ \Rightarrow \qquad \frac{ 16x - 1 }{ (3x+2) (2x-1) } \equiv \frac{ 5 }{ 3x + 2 } + \frac{ 2 }{ 2x - 1 } \qquad \checkmark $$ where $$ D = 5, \qquad E = 2 \qquad \checkmark $$

 

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Question 11. (E) $$ \frac{ 7x^2 + 2x - 2 }{ x^2 (x+1) } \equiv \frac{ A }{ x } + \frac{ B }{ x^2 } + \frac{ C }{ x + 1 } $$ Find the values of the constants [4 marks]

 

(Edexcel 2017 Specifications, P2 Ch1 Mixed exercise 1 Q11.)

 

Solution.

We have: $$ \begin{align} \frac{ 7x^2 + 2x - 2 }{ x^2 (x+1) } &\equiv \frac{ A }{ x } + \frac{ B }{ x^2 } + \frac{ C }{ x + 1 } \\ &\equiv \frac{ A x (x+1) + B (x+1) + C x^2 }{ x^2 (x + 1) } \end{align} $$ Compare the numerators on both sides: $$ A x (x+1) + B (x+1) + C x^2 \equiv 7x^2 + 2x - 2. $$ By the method of substitution: $$ \begin{align} \textrm{(i)}&& x &= 0 &&\Rightarrow& B &= -2 \\ \\ \textrm{(ii)}&& x &= -1 &&\Rightarrow& C &= 7(-1)^2 + 2(-1) -2 = 3 \\ \\ \textrm{(iii)}&& x &= 1 &&\Rightarrow& 2A + 2B + C &= 7 + 2 - 2 = 7 \\ &&&&&\Rightarrow& 2A &= 7 - 2B - C \\ &&&&&&&= 7 + 4 - 3 \\ &&&&&&&= 8 \\ &&&&&\Rightarrow& A &= 4 \end{align} $$ and we find: $$ \Rightarrow \qquad \frac{ 7x^2 + 2x - 2 }{ x^2 (x+1) } \equiv \frac{ 4 }{ x } - \frac{ 2 }{ x^2 } + \frac{ 3 }{ x + 1 } \qquad \checkmark $$ where $$ A = 4, \qquad B = -2, \qquad C = 3 \qquad \checkmark $$

 

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Question 12. (E) $$ \textrm h(x) = \frac{ 21x^2 - 13 }{ (x+5) (3x - 1)^2 } $$ Show that $ \textrm h(x) $ can be written in the form $$ \frac{ D }{ x+5 } + \frac{ E }{ (3x - 1) } + \frac{ F }{ (3x - 1)^2 }, $$ where $D,E$ and $F$ are constants to be found. [5 marks]

 

(Edexcel 2017 Specifications, P2 Ch1 Mixed exercise 1 Q12.)

 

Solution.

Let $$ \begin{align} \textrm h(x) = \frac{ 21x^2 - 13 }{ (x+5) (3x - 1)^2 } &\equiv \frac{ D }{ x+5 } + \frac{ E }{ (3x - 1) } + \frac{ F }{ (3x - 1)^2 } \\ &\equiv \frac{ D (3x - 1)^2 + E (x+5) (3x - 1) + F (x+5) }{ (x+5) (3x - 1)^2 } \end{align} $$ Compare the numerators on both sides: $$ D (3x - 1)^2 + E (x+5) (3x - 1) + F (x+5) \equiv 21x^2 - 13. $$ By the method of substitution: $$ \begin{align} \textrm{(i)}&& x &= -5 &&\Rightarrow& (-16)^2 D &= 21(-5)^2 - 13 \\ &&&&&\Rightarrow& 256 D &= 512 \\ &&&&&\Rightarrow& D &= 2 \\ \\ \textrm{(ii)}&& x &= \frac13 &&\Rightarrow& \left( \frac13 + 5 \right) F &= 21 \left( \frac13 \right)^2 - 13 \\ &&&&&\Rightarrow& \frac{16}{3} F &= \frac{7}{3} -13 \\ &&&&&&&= \frac{7-39}{3} \\ &&&&&&&= -\frac{32}{3} \\ &&&&&\Rightarrow& F &= -2 \\ \\ \textrm{(iii)}&& x &= 0 &&\Rightarrow& D - 5E + 5F &= -13 \\ &&&&&\Rightarrow& -5E &= -13 - D - 5F \\ &&&&&&&= -13 - 2 + 10 \\ &&&&&&&= -5 \\ &&&&&\Rightarrow& E &= 1 \end{align} $$ and we find: $$ \Rightarrow \qquad \textrm h(x) = \frac{ 21x^2 - 13 }{ (x+5) (3x - 1)^2 } \equiv \frac{ 2 }{ x+5 } + \frac{ 1 }{ (3x - 1) } - \frac{ 2 }{ (3x - 1)^2 } \qquad \checkmark $$ where $$ D = 2, \qquad E = 1, \qquad F = -2 \qquad \checkmark $$

 

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Question 13. (E) Find the values of the constants $A,B,C$ and $D$ in the following identity: $$ x^3 - 6x^2 + 11x + 2 \equiv (x-2) \left( Ax^2 + Bx + C \right) + D $$ [5 marks]

 

(Edexcel 2017 Specifications, P2 Ch1 Mixed exercise 1 Q13.)

 

Solution.

By algebraic long division (see the YouTube clip below for more details), $$ \begin{align} x^3 - 6x^2 + 11x + 2 \equiv (x-2) \left( x^2 - 4x + 3 \right) + 7 \end{align} $$

 

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Question 14. (E) Show that $$ \frac{ 4x^3 - 6x^2 + 8x - 5 }{ 2x + 1 } $$ can be put in the form $$ Ax^2 + Bx + C + \frac{ D }{ 2x + 1 }. $$ Find the values of the constants $A,B,C$ and $D$. [5 marks]

 

(Edexcel 2017 Specifications, P2 Ch1 Mixed exercise 1 Q14.)

 

Solution.

By algebraic long division (see the YouTube clip below for more details), $$ \begin{align} 4x^3 - 6x^2 + 8x - 5 &\equiv \left( 2x^2 - 4x + 6 \right) (2x + 1) - 11 \\ \\ \Leftrightarrow \qquad \frac{ 4x^3 - 6x^2 + 8x - 5 }{ 2x + 1 } &\equiv 2x^2 - 4x + 6 - \frac{ 11 }{ 2x + 1 } \end{align} $$

 

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Question 15. (E) Show that $$ \frac{ x^4 + 2 }{ x^2 - 1 } \equiv Ax^2 + Bx + C + \frac{ D }{ x^2 - 1 }, $$ where $A,B,C$ and $D$ are constants to be found. [5 marks]

 

(Edexcel 2017 Specifications, P2 Ch1 Mixed exercise 1 Q15.)

 

Solution.

By algebraic long division (see the YouTube clip below for more details), $$ \begin{align} x^4 + 2 &\equiv \left( x^2 + 1 \right) \left( x^2 - 1 \right) + 3 \\ \\ \Leftrightarrow \qquad \frac{ x^4 + 2 }{ x^2 - 1 } &\equiv x^2 + 1 + \frac{ 3 }{ x^2 - 1 } \end{align} $$

 

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Question 16. (E) $$ \frac{ x^4 }{ x^2 - 2x + 1 } \equiv Ax^2 + Bx + C + \frac{ D }{ x-1 } + \frac{ E }{ (x-1)^2 }. $$ Find the values of the constants $A,B,C,D$ and $E$. [5 marks]

 

(Edexcel 2017 Specifications, P2 Ch1 Mixed exercise 1 Q16.)

 

Solution.

(i) Step 1: Algebraic division.

By algebraic long division (see the YouTube clip below for more details), $$ \begin{align} x^4 &\equiv \left( x^2 + 2x + 3 \right) \left( x^2 - 2x + 1 \right) + 4x - 3 \\ \\ \Leftrightarrow \qquad \frac{ x^4 }{ x^2 - 2x + 1 } &\equiv x^2 + 2x + 3 + \frac{ 4x - 3 }{ x^2 - 2x + 1 } \end{align} $$

 

(ii) Step 2: Partial fractions.

Factorise the denominator first: $$ \begin{align} x^2 - 2x + 1 = (x-1)^2 \end{align} $$ and this gives partial fractions with repeated factors. Let $$ \begin{align} \frac{ 4x - 3 }{ x^2 - 2x + 1 } &\equiv \frac{ 4x - 3 }{ (x - 1)^2 } \\ &\equiv \frac{ D }{ x - 1 } + \frac{ E }{ (x - 1)^2 } \\ &\equiv \frac{ D (x-1) + E }{ 2x - 1 } \end{align} $$ Compare the numerators on both sides: $$ D (x-1) + E \equiv 4x - 3 $$ By the method of equating coefficients: $$ \begin{align} \left\{ \begin{array}{l} D = 4 \\ - D + E = - 3 \end{array} \right. \qquad \Rightarrow \qquad \left\{ \begin{array}{l} D = 4 \\ E = 1 \end{array} \right. \end{align} $$ and we obtain: $$ \begin{align} \Rightarrow \qquad \frac{ 4x - 3 }{ x^2 - 2x + 1 } \equiv \frac{ 4x - 3 }{ (x - 1)^2 } \equiv \frac{ 4 }{ x - 1 } + \frac{ 1 }{ (x - 1)^2 } \end{align} $$

 

Finally, we find: $$ \begin{align} \Rightarrow \qquad \frac{ x^4 }{ x^2 - 2x + 1 } &\equiv x^2 + 2x + 3 + \frac{ 4x - 3 }{ x^2 - 2x + 1 } \\ &\equiv x^2 + 2x + 3 + \frac{ 4 }{ x - 1 } + \frac{ 1 }{ (x - 1)^2 } \qquad \checkmark \end{align} $$ and thus: $$ \begin{align} A &= 1 \\ B &= 2 \\ C &= 3 \\ D &= 4 \\ E &= 1 \end{align} $$

 

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Question 17. (E) $$ \textrm h(x) = \frac{ 2x^2 + 2x - 3 }{ x^2 + 2x - 3 } $$ Show that $ \textrm h(x) $ can be written in the form $$ A + \frac{ B }{ x + 3 } + \frac{ C }{ x - 1 }, $$ where $A,B$ and $C$ are constants to be found. [5 marks]

 

(Edexcel 2017 Specifications, P2 Ch1 Mixed exercise 1 Q17.)

 

Solution.

(i) Step 1: Algebraic division.

By algebraic long division (see the YouTube clip below for more details), $$ \begin{align} 2x^2 + 2x - 3 &\equiv 2 \left( x^2 + 2x - 3 \right) - 2x + 3 \\ \\ \Leftrightarrow \qquad \textrm h(x) = \frac{ 2x^2 + 2x - 3 }{ x^2 + 2x - 3 } &\equiv 2 + \frac{ - 2x + 3 }{ x^2 + 2x - 3 } \end{align} $$

 

(ii) Step 2: Partial fractions.

Factorise the denominator first: $$ \begin{align} x^2 + 2x - 3 = (x+3) (x-1) \end{align} $$ Let $$ \begin{align} \frac{ - 2x + 3 }{ x^2 + 2x - 3 } &\equiv \frac{ - 2x + 3 }{ (x+3) (x-1) } \\ &\equiv \frac{ B }{ x + 3 } + \frac{ C }{ x - 1 } \\ &\equiv \frac{ B (x-1) + C (x+3) }{ (x+3) (x-1) } \end{align} $$ Compare the numerators on both sides: $$ B (x-1) + C (x+3) \equiv -2x + 3 $$ By the method of substitution: $$ \begin{align} \textrm{(i)}&& x &= -3 &&\Rightarrow& -4B&=6+3=9 \\ &&&&&\Rightarrow& B&=-\frac{9}{4} \\ \\ \textrm{(ii)}&& x &= 1 &&\Rightarrow& 4C&=-2+3=1 \\ &&&&&\Rightarrow& C&=\frac{1}{4} \end{align} $$ and we obtain: $$ \begin{align} \Rightarrow \qquad \frac{ - 2x + 3 }{ x^2 + 2x - 3 } &\equiv \frac{ - 2x + 3 }{ (x+3) (x-1) } \\ &\equiv \frac{ -\frac{9}{4} }{ x + 3 } + \frac{ \frac14 }{ x - 1 } \\ &\equiv \frac14 \left( - \frac{ 9 }{ x + 3 } + \frac{ 1 }{ x - 1 } \right) \end{align} $$

 

Finally, we find: $$ \begin{align} \Rightarrow \qquad \textrm h(x) = \frac{ 2x^2 + 2x - 3 }{ x^2 + 2x - 3 } &\equiv 2 + \frac{ - 2x + 3 }{ x^2 + 2x - 3 } \\ &\equiv 2 + \frac14 \left( - \frac{ 9 }{ x + 3 } + \frac{ 1 }{ x - 1 } \right) \qquad \checkmark \end{align} $$ and thus: $$ \begin{align} A &= 2 \\ B &= -\frac{9}{4} \\ C &= \frac14 \end{align} $$

 

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Question 18. (E) Given that $$ \frac{ x^2 + 1 }{ x (x - 2) } \equiv P + \frac{ Q }{ x } + \frac{ R }{ x - 2 }, $$ find the values of the constants $P,Q$ and $R$. [5 marks]

 

(Edexcel 2017 Specifications, P2 Ch1 Mixed exercise 1 Q18.)

 

Solution.

(i) Step 1: Algebraic division.

$$ \begin{align} x^2 + 1 &\equiv x^2 - 2x + 2x + 1 \\ &\equiv x(x-2) + 2x + 1 \\ \\ \Leftrightarrow \qquad \frac{ x^2 + 1 }{ x (x - 2) } &\equiv 1 + \frac{ 2x + 1 }{ x (x - 2) } \end{align} $$

 

(ii) Step 2: Partial fractions.

Let $$ \begin{align} \frac{ 2x + 1 }{ x (x - 2) } &\equiv \frac{ Q }{ x } + \frac{ R }{ x - 2 } \\ &\equiv \frac{ Q (x-2) + R x }{ x (x-2) } \end{align} $$ Compare the numerators on both sides: $$ (Q+R)x - 2Q \equiv 2x + 1 $$ By the method of equating coefficients: $$ \begin{align} \left\{ \begin{array}{l} Q+R = 2 \\ -2Q = 1 \end{array} \right. \qquad \Rightarrow \qquad \left\{ \begin{array}{l} Q = -\frac12 \\ R = \frac52 \end{array} \right. \end{align} $$ and we obtain: $$ \begin{align} \Rightarrow \qquad \frac{ 2x + 1 }{ x (x - 2) } &\equiv \frac{ Q }{ x } + \frac{ R }{ x - 2 } \\ &\equiv -\frac{ 1 }{ 2x } + \frac{ 5 }{ 2(x - 2) } \end{align} $$

 

Finally, we find: $$ \begin{align} \Rightarrow \qquad \frac{ x^2 + 1 }{ x (x - 2) } &\equiv 1 + \frac{ 2x + 1 }{ x (x - 2) } \\ &\equiv 1 -\frac{ 1 }{ 2x } + \frac{ 5 }{ 2(x - 2) } \qquad \checkmark \end{align} $$ and thus: $$ \begin{align} P &= 1 \\ Q &= -\frac{1}{2} \\ R &= \frac52 \end{align} $$

 

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Question 19. (P) Given that $$ \textrm f(x) = 2x^3 + 9x^2 + 10x + 3, $$

(a) show that $-3$ is a root of $ \textrm f(x) $;

 

(b) express $ \frac{ 10 }{ \textrm f(x) } $ as partial fractions.

 

(Edexcel 2017 Specifications, P2 Ch1 Mixed exercise 1 Q19.)

 

Solution.

(a) $$ \begin{align} \textrm f(-3) &= 2(-3)^3 + 9(-3)^2 + 10(-3) + 3 \\ &= -54 + 81 - 30 + 3 \\ &= 0. \qquad \square \end{align} $$

 

(b) We factorise the denominator $ \textrm f(x) $ first. By the factor theorem, $ (x+3) $ is a factor. $$ \begin{align} \textrm f(x) &= 2x^3 + 9x^2 + 10x + 3 \\ &= (x+3) \left( 2x^2 + 3x + 1 \right) \\ &= (x+3)( 2x + 1 )(x + 1) \end{align} $$ For partial fractions, let $$ \begin{align} \frac{ 10 }{ \textrm f(x) } = \frac{ 10 }{ (x+3)( 2x + 1 )(x + 1) } &\equiv \frac{ A }{ x+3 } + \frac{ B }{ 2x+1 } + \frac{ C }{ x+1 } \\ &\equiv \frac{ A (2x+1)(x+1) + B (x+3)(x+1) + C (x+3)(2x+1) }{ (x+3)( 2x + 1 )(x + 1) } \end{align} $$ Compare the numerators on both sides: $$ A (2x+1)(x+1) + B (x+3)(x+1) + C (x+3)(2x+1) \equiv 10 $$ By the method of substitution: $$ \begin{align} \textrm{(i)}&& x &= -3 &&\Rightarrow& (-5)(-2)A &= 10 \\ &&&&&\Rightarrow& A&=1 \\ \\ \textrm{(ii)}&& x &= -\frac12 &&\Rightarrow& \left( \frac52 \right) \left( \frac12 \right) B &= 10 \\ &&&&&\Rightarrow& B&=8 \\ \\ \textrm{(iii)}&& x &= -1 &&\Rightarrow& (2)(-1)C &= 10 \\ &&&&&\Rightarrow& C&=-5 \end{align} $$ and we obtain: $$ \begin{align} \Rightarrow \qquad \frac{ 10 }{ \textrm f(x) } = \frac{ 10 }{ (x+3)( 2x + 1 )(x + 1) } \equiv \frac{ 1 }{ x+3 } + \frac{ 8 }{ 2x+1 } - \frac{ 5 }{ x+1 } \qquad \checkmark \end{align} $$

 

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Challenge. The line $L$ meets the circle $C$ with centre $O$ at exactly one point, $A$.

 

Prove by contradiction that the line $L$ is perpendicular to the radius $OA$. [Hint: In a right-angled triangle, the side opposite the right-angle is always the longest side.]

 

(Edexcel 2017 Specifications, P2 Ch1 Mixed exercise 1 Challenge.)

 

Solution.

 

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