P2 §1.7 Review exercise for chapter 1

Pure mathematics Year 2

Table of contents

1. Edexcel P2 Review exercise 1 Q1-10

1. Edexcel P2 Review exercise 1 Q1-10

Question 1. (E/P) Prove by contradiction that there are infinitely many prime numbers. [4 marks]

 

(Relevant section: P2 §1.1 Proof by contradiction.)

 

(Edexcel 2017 Specifications, P2 Review exercise 1 Q1.)

 

Solution.

 

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Question 2. (E/P) Prove that the equation $x^2 - 2=0$ has no rational solutions. You may assume that if $n^2$ is an even integer then $n$ is also an even integer. [4 marks]

 

(Relevant section: P2 §1.1 Proof by contradiction.)

 

(Edexcel 2017 Specifications, P2 Review exercise 1 Q2.)

 

Solution.

 

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

 

(Relevant section: P2 §1.2 Algebraic fractions.)

 

(Edexcel 2017 Specifications, P2 Review exercise 1 Q3.)

 

Solution.

$$\begin{align} \frac{4x}{x^2 - 2x - 3} + \frac{1}{x^2 + x} &= \frac{4x}{ (x-3)(x+1) } + \frac{1}{ x(x+1) } \\ &= \frac{4x^2 + (x-3) }{ x(x-3)(x+1) } \\ &= \frac{4x^2 + x - 3 }{ x(x-3)(x+1) } \\ &= \frac{ (4x - 3)( x +1 ) }{ x(x-3)(x+1) } \\ &= \frac{ 4x - 3 }{ x(x-3) } \qquad \checkmark \end{align}$$

 

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Question 4. (P) $$\textrm f(x) = 1 - \frac{ 3 }{ x+2 } + \frac{ 3 }{ (x+2)^2 }, \qquad x \ne -2.$$

(a) Show that $$\textrm f(x) = \frac{ x^2 + x + 1 }{ (x+2)^2}, \qquad x \ne -2$$

 

(a) Show that $x^2 + x + 1 > 0$ for all values of $x$, $x \ne -2$.

 

(c) Show that $\textrm f(x) > 0$ for all values of $x$, $x \ne -2$.

 

(Relevant section: P2 §1.2 Algebraic fractions.)

 

(Edexcel 2017 Specifications, P2 Review exercise 1 Q4.)

 

Solution.

(a) $$\begin{align} \textrm f(x) &= 1 - \frac{ 3 }{ x+2 } + \frac{ 3 }{ (x+2)^2 } \\ &= \frac{ (x+2)^2 - 3 (x+2) + 3 }{ (x+2)^2 } \\ &= \frac{ \left( x^2 + 4x + 4 \right) - (3x+6) + 3 }{ (x+2)^2 } \\ &= \frac{ x^2 + x + 1 }{ (x+2)^2 }, \qquad x \ne -2. \qquad \square \end{align}$$

 

(b) By completing the square, $$\begin{align} x^2 + x + 1 &= \left( x + \frac12 \right)^2 - \frac14 + 1 \\ &= \left( x + \frac12 \right)^2 + \frac34 \\ &\ge \frac34 \\ &> 0. \qquad \square \end{align}$$

 

(c) This gives $$\begin{align} \textrm f(x) = \frac{ x^2 + x + 1 }{ (x+2)^2 } = \frac{ \left( x + \frac12 \right)^2 + \frac34 }{ (x+2)^2 } \end{align}$$ For $x \ne -2$, since both the numerator and denominator are positive, $\textrm f(x) > 0. \qquad \square$

 

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

 

(Relevant section: P2 §1.3 Partial fractions.)

 

(Edexcel 2017 Specifications, P2 Review exercise 1 Q5.)

 

Solution.

Let $$\begin{align} \frac{ 2x - 1 }{ (x-1) (2x-3) } &\equiv \frac{ A }{ x-1 } + \frac{ B }{ 2x - 3 } \\ &\equiv \frac{ A (2x-3) + B (x-1) }{ (x-1) (2x-3) } \end{align}$$ Compare the numerators on both sides: $$A (2x-3) + B (x-1) \equiv 2x - 1$$ By the method of substitution: $$\begin{align} \textrm{(i)}&& x &= 1 &&\Rightarrow& -A &= 1 \\ &&&&&\Rightarrow& A&=-1 \\ \\ \textrm{(ii)}&& x &= \frac32 &&\Rightarrow& \frac12 B &= 2 \\ &&&&&\Rightarrow& B&=4 \end{align}$$ and we obtain: $$\begin{align} \Rightarrow \qquad \frac{ 2x - 1 }{ (x-1) (2x-3) } \equiv - \frac{ 1 }{ x-1 } + \frac{ 4 }{ 2x - 3 } \qquad \checkmark \end{align}$$

 

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Question 6. (E) Given that $$\frac{ 3x + 7 }{ (x+1) (x+2) (x+3) } \equiv \frac{ P }{ x+1 } + \frac{ Q }{ x+2 } + \frac{ R }{ x+3 },$$ where $P,Q$ and $R$ are constants, find the values of $P,Q$ and $R$. [4 marks]

 

(Relevant section: P2 §1.3 Partial fractions.)

 

(Edexcel 2017 Specifications, P2 Review exercise 1 Q6.)

 

Solution.

It is given that: $$\begin{align} \frac{ 3x + 7 }{ (x+1) (x+2) (x+3) } &\equiv \frac{ P }{ x+1 } + \frac{ Q }{ x+2 } + \frac{ R }{ x+3 } \\ &\equiv \frac{ P (x+2)(x+3) + Q (x+1)(x+3) + R (x+1)(x+2) }{ (x+1)(x+2)(x+3) } \end{align}$$ Compare the numerators on both sides: $$P (x+2)(x+3) + Q (x+1)(x+3) + R (x+1)(x+2) \equiv 3x + 7$$ By the method of substitution: $$\begin{align} \textrm{(i)}&& x &= -1 &&\Rightarrow& (1)(2) P &= 4 \\ &&&&&\Rightarrow& P&= 2 \\ \\ \textrm{(ii)}&& x &= -2 &&\Rightarrow& (-1)(1) Q &= 1 \\ &&&&&\Rightarrow& Q&= -1 \\ \\ \textrm{(iii)}&& x &= -3 &&\Rightarrow& (-2)(-1) R &= -2 \\ &&&&&\Rightarrow& R&= -1 \end{align}$$ and we obtain: $$\begin{align} \Rightarrow \qquad \frac{ 3x + 7 }{ (x+1) (x+2) (x+3) } \equiv \frac{ 2 }{ x+1 } - \frac{ 1 }{ x+2 } - \frac{ 1 }{ x+3 } \qquad \checkmark \end{align}$$ where $$\begin{align} P &= 2 \\ Q &= -1 \\ R &= -1 \end{align}$$

 

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Question 7. (E) $$\textrm f(x) = \frac{ 2 }{ (2-x) (1+x)^2 }, \qquad x \ne -1, \quad x \ne 2.$$ Find the values of $A,B$ and $C$ such that $$\textrm f(x) = \frac{ A }{ 2-x } + \frac{ B }{ 1+x } + \frac{ C }{ (1+x)^2 }$$ [4 marks]

 

(Relevant section: P2 §1.4 Repeated factors.)

 

(Edexcel 2017 Specifications, P2 Review exercise 1 Q7.)

 

Solution.

For partial fractions with repeated factors, let $$\begin{align} \textrm f(x) = \frac{ 2 }{ (2-x) (1+x)^2 } &\equiv \frac{ A }{ 2-x } + \frac{ B }{ 1+x } + \frac{ C }{ (1+x)^2 } \\ &\equiv \frac{ A (1+x)^2 + B (2-x)(1+x) + C (2-x) }{ (2-x) (1+x)^2 } \end{align}$$ Compare the numerators on both sides: $$A (1+x)^2 + B (2-x)(1+x) + C (2-x) \equiv 2$$ By the method of substitution: $$\begin{align} \textrm{(i)}&& x &= 2 &&\Rightarrow& 9 A &= 2 \\ &&&&&\Rightarrow& A &= \frac{2}{9} \\ \\ \textrm{(ii)}&& x &= -1 &&\Rightarrow& 3 C &= 2 \\ &&&&&\Rightarrow& C &= \frac{2}{3} \\ \\ \textrm{(iii)}&& x &= 0 &&\Rightarrow& A + 2B + 2C &= 2 \\ &&&&&\Rightarrow& 2B &= 2 - A - 2C \\ &&&&&&&= 2 - \frac{2}{9} - \frac{4}{3} \\ &&&&&&&= \frac{18 - 2 - 12}{9} \\ &&&&&&&= \frac{4}{9} \\ &&&&&\Rightarrow& B &= \frac{2}{9} \\ \end{align}$$ and we obtain: $$\begin{align} \Rightarrow \qquad \textrm f(x) = \frac{ 2 }{ (2-x) (1+x)^2 } &\equiv \frac{ \frac{2}{9} }{ 2-x } + \frac{ \frac{2}{9} }{ 1+x } + \frac{ \frac{2}{3} }{ (1+x)^2 } \\ &\equiv \frac{ 2 }{ 9(2-x) } + \frac{ 2 }{ 9(1+x) } + \frac{ 2 }{ 3(1+x)^2 } \qquad \checkmark \end{align}$$ where $$\begin{align} A &= \frac{2}{9} \\ B &= \frac{2}{9} \\ C &= \frac{2}{3} \end{align}$$

 

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Question 8. $$\frac{ 14x^2 + 13x + 2 }{ (x+1) (2x+1)^2 } \equiv \frac{ A }{ x+1 } + \frac{ B }{ 2x+1 } + \frac{ C }{ (2x+1)^2 }.$$ Find the values of the constants $A,B$ and $C$.

 

(Relevant section: P2 §1.4 Repeated factors.)

 

(Edexcel 2017 Specifications, P2 Review exercise 1 Q8.)

 

Solution.

For partial fractions with repeated factors, let $$\begin{align} \frac{ 14x^2 + 13x + 2 }{ (x+1) (2x+1)^2 } &\equiv \frac{ A }{ x+1 } + \frac{ B }{ 2x+1 } + \frac{ C }{ (2x+1)^2 } \\ &\equiv \frac{ A (2x+1)^2 + B(x+1)(2x+1) + C(x+1) }{ (x+1) (2x+1)^2 } \end{align}$$ Compare the numerators on both sides: $$A (2x+1)^2 + B(x+1)(2x+1) + C(x+1) \equiv 14x^2 + 13x + 2$$ By the method of substitution: $$\begin{align} \textrm{(i)}&& x &= -1 &&\Rightarrow& (-1)^2 A &= 14(-1)^2 + 13(-1) + 2 = 3 \\ &&&&&\Rightarrow& A &= 3 \\ \\ \textrm{(ii)}&& x &= -\frac12 &&\Rightarrow& \frac12 C &= 14\left(-\frac12\right)^2 + 13\left(-\frac12\right) + 2 \\ &&&&&&&= \frac{7}{2} - \frac{13}{2} + 2 \\ &&&&&&&= -1 \\ &&&&&\Rightarrow& C &= -2 \\ \\ \textrm{(iii)}&& x &= 0 &&\Rightarrow& A + B + C &= 2 \\ &&&&&\Rightarrow& B &= 2 - A - C \\ &&&&&&&= 2 - 3 + 2 \\ &&&&&&&= 1 \end{align}$$ and we obtain: $$\begin{align} \Rightarrow \qquad \frac{ 14x^2 + 13x + 2 }{ (x+1) (2x+1)^2 } &\equiv \frac{ 3 }{ x+1 } + \frac{ 1 }{ 2x+1 } - \frac{ 2 }{ (2x+1)^2 } \qquad \checkmark \end{align}$$ where $$\begin{align} A &= 3 \\ B &= 1 \\ C &= -2 \end{align}$$

 

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Question 9. (E) Given that $$\frac{ 3x^2 + 6x -2 }{ x^2 + 4 } \equiv d + \frac{ ex + f }{ x^2 + 4 },$$ find the values of $d,e$ and $f$. [4 marks]

 

(Relevant section: P2 §1.5 Algebraic division.)

 

(Edexcel 2017 Specifications, P2 Review exercise 1 Q9.)

 

Solution.

$$\begin{align} 3x^2 + 6x - 2 &\equiv 3 \left( x^2 + 4 \right) - 12 + 6x - 2 \\ &\equiv 3 \left( x^2 + 4 \right) + 6x - 14 \\ \\ \Leftrightarrow \qquad \frac{ 3x^2 + 6x -2 }{ x^2 + 4 } &\equiv 3 + \frac{ 6x - 14 }{ x^2 + 4 } \end{align}$$

Thus, $$\begin{align} d = 3, \qquad e = 6, \qquad f = -14 \qquad \checkmark \end{align}$$

 

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Question 10. (E) $$\textrm p(x) = \frac{ 9 - 3x - 12x^2 }{ (1-x)(1+2x) }.$$ Show that $\textrm p(x)$ can be written in the form $$A + \frac{ B }{ 1-x } + \frac{ C }{ 1+2x },$$ where $A,B$ and $C$ are constants to be found. [3 marks]

 

(Relevant sections: P2 §1.3 Partial fractions and §1.5 Algebraic division.)

 

(Edexcel 2017 Specifications, P2 Review exercise 1 Q10.)

 

Solution.

(i) Step 1: Algebraic division.

The denominator reads:

$$\begin{align} (1-x)(1+2x) = 1 + x - 2x^2 \end{align}$$ and, by algebraic division, we find: $$\begin{align} && 9 - 3x - 12x^2 &= 6 \left( -2x^2 + x + 1 \right) - 6x - 6 + 9 - 3x \\ &&&= 6 \left( 1 + x - 2x^2 \right) - 9x + 3 \\ \\ &\Leftrightarrow & \frac{ 9 - 3x - 12x^2 }{ 1 + x - 2x^2 } &\equiv 6 + \frac{ -9x + 3 }{ 1 + x - 2x^2 } \\ \\ &\Leftrightarrow & \frac{ 9 - 3x - 12x^2 }{ (1-x)(1+2x) } &\equiv 6 + \frac{ -9x + 3 }{ (1-x)(1+2x) } \end{align}$$

 

(ii) Step 2: Partial fractions.

Let $$\begin{align} \frac{ -9x + 3 }{ (1-x)(1+2x) } &\equiv \frac{ B }{ 1-x } + \frac{ C }{ 1+2x } \\ &\equiv \frac{ B(1+2x) + C(1-x) }{ (1-x)(1+2x) } \end{align}$$ Compare the numerators on both sides: $$B(1+2x) + C(1-x) \equiv -9x + 3$$ By the method of substitution: $$\begin{align} \textrm{(i)}&& x &= 1 &&\Rightarrow& 3 B &= -9+3 = -6 \\ &&&&&\Rightarrow& B &= -2 \\ \\ \textrm{(ii)}&& x &= -\frac12 &&\Rightarrow& \frac32 C &= \frac{9}{2}+3=\frac{15}{2} \\ &&&&&\Rightarrow& C &= 5 \end{align}$$ and we obtain: $$\begin{align} \Rightarrow \qquad \frac{ -9x + 3 }{ (1-x)(1+2x) } &\equiv - \frac{ 2 }{ 1-x } + \frac{ 5 }{ 1+2x } \end{align}$$

 

Finally, we find: $$\begin{align} \Rightarrow \qquad \frac{ 9 - 3x - 12x^2 }{ (1-x)(1+2x) } &\equiv 6 + \frac{ -9x + 3 }{ (1-x)(1+2x) } \\ &\equiv 6 - \frac{ 2 }{ 1-x } + \frac{ 5 }{ 1+2x } \qquad \checkmark \end{align}$$ and thus: $$\begin{align} A &= 6 \\ B &= -2 \\ C &= 5 \end{align}$$