The diagram below shows a solid with volume V , obtained from a cube with edge \(a > 1\) when a smaller cube with edge \(\frac{1}{a}\) is removed.

 

 

Let \(x = a - \frac{1}{a}\)

(a)     Find V in terms of x .

(b)     Hence or otherwise, show that the only value of a for which V = 4x is \(a = \frac{{1 + \sqrt 5 }}{2}\) .

(a)     METHOD 1

\(V = {a^3} - \frac{1}{{{a^3}}}\)     A1

\({x^3} = {\left( {a - \frac{1}{a}} \right)^3}\)     M1

\( = {a^3} - 3a + \frac{3}{a} - \frac{1}{{{a^3}}}\)

\( = {a^3} - \frac{1}{{{a^3}}} - 3\left( {a - \frac{1}{a}} \right)\,\,\,\,\,\)(or equivalent)     (A1)

\( \Rightarrow {a^3} - \frac{1}{{{a^3}}} = {x^3} + 3x\)

\(V = {x^3} + 3x\)     A1     N0

METHOD 2

\(V = {a^3} - \frac{1}{{{a^3}}}\)     A1

attempt to use difference of cubes formula, \({x^3} - {y^3} = (x - y)({x^2} + xy + {y^2})\)     M1

\(V = \left( {a - \frac{1}{a}} \right)\left( {{a^2} + 1 + {{\left( {\frac{1}{a}} \right)}^2}} \right)\)

\( = \left( {a - \frac{1}{a}} \right)\left( {{{\left( {a - \frac{1}{a}} \right)}^2} + 3} \right)\)     (A1)

\( = x({x^2} + 3){\text{ or }}{x^3} + 3x\)     A1     N0

METHOD 3

diagram showing that the solid can be decomposed     M1

into three congruent \(x \times a \times \frac{1}{a}\) cuboids with volume x     A1

and a cube with edge x with volume \({x^3}\)     A1

so, \(V = {x^3} + 3x\)     A1     N0

 

(b)

Note: Do not accept any method where candidate substitutes the given value of a into \(x = a - \frac{1}{a}\) .

 

METHOD 1

\(V = 4x \Leftrightarrow {x^3} + 3x = 4x \Leftrightarrow {x^3} - x = 0\)     M1

\( \Leftrightarrow x(x - 1)(x + 1) = 0\)

\( \Rightarrow x = 1\) as \(x > 0\)     A1

so, \(a - \frac{1}{a} = 1 \Rightarrow {a^2} - a - 1 = 0 \Rightarrow a = \frac{{1 \pm \sqrt 5 }}{2}\)     M1A1

as \(a > 1\) , \(a = \frac{{1 + \sqrt 5 }}{2}\)     AG     N0

METHOD 2

\({a^3} - \frac{1}{{{a^3}}} = 4\left( {a - \frac{1}{a}} \right) \Rightarrow {a^6} - 4{a^4} + 4{a^2} - 1 = 0 \Leftrightarrow ({a^2} - 1)({a^4} - 3{a^2} + 1) = 0\)     M1A1

as \(a > 1 \Rightarrow {a^2} > 1\), \({a^2} = \frac{{3 + \sqrt 5 }}{2} \Leftrightarrow {a^2} = \sqrt {{{\left( {\frac{{1 + \sqrt 5 }}{2}} \right)}^2}} \)     M1A1

\( \Rightarrow a = \frac{{1 + \sqrt 5 }}{2}\)     AG     N0

 

[8 marks]

A fair amount of candidates had difficulties with this question. In part (a) many candidates were able to write down an expression for the volume in terms of a, but thereafter were largely unsuccessful. There is evidence that many candidates have lack of algebraic skills to manipulate the expression and obtain the volume in terms of x. In part (b) some candidates started with what they were trying to show to be true.