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Mathematics · Oct / Nov 2023 · Paper 41 · Question 2

YMA01

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The question as text

2. x cm Figure 1 Figure 1 shows a cube which is increasing in size. At time t seconds, • the length of each edge of the cube is x cm 2 • the surface area of the cube is S cm 3 • the volume of the cube is V cm 2 s –1 Given that the surface area of the cube is increasing at a constant rate of 4 cm

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  • 2(a)[4]

    or d

    S

    States or uses

    S

    =

    6

    x

    2

    t =

    4

    d

    and d

    S

    States or uses

    S

    =

    6

    x

    2

    t =

    4

    d

    Attempts to use d

    S

    d

    S

    d

    x

    d

    x

    d

    x

    4

    12

    ...

    =

    ×

    =

    x

    ×

    =

    d

    t

    d

    x

    d

    t

    d

    t

    d

    t

    d

    x

    1

    oe

    =

    d

    t

    3

    x

    (a) Alt

    or d

    S

    States or uses

    S

    =

    6

    x

    2

    t =

    4

    d

    and d

    S

    States or uses

    S

    =

    6

    x

    2

    4

    t =

    d

    d

    S

    d

    x

    d

    x

    =

    4

    S

    =

    4

    t

    +

    c

    S

    =

    6

    x

    2

    6

    x

    2

    =

    4

    t

    +

    c

    12

    x

    =

    4

    =

    ...

    d

    t

    d

    t

    d

    t

    or

    d

    t

    d

    x

    S

    6

    x

    2

    6

    x

    2

    4

    t

    c

    12

    x

    4

    ...

    =

    =

    +

    =

    =

    d

    x

    d

    t

    d

    x

    1

    oe

    =

    d

    t

    3

    x

  • 2(b)[3]

    d

    V

    States or uses

    3

    x

    2

    x =

    d

    d

    V

    d

    V

    d

    x

    d

    V

    1

    Attempts to use

    =

    ×

    =

    3

    x

    2

    ×

    =

    x

    d

    t

    d

    x

    d

    t

    d

    t

    3

    x

    1

    d

    V

    =

    V

    3

    d

    t

    (b) Alt

    2

    d

    S

    1

    −

    V

    =

    x

    3

    ,

    S

    =

    6

    x

    2

    S

    =

    6

    V

    3

    =

    4

    V

    3

    d

    V

    d

    V

    d

    V

    d

    S

    1

    Attempts to use

    =

    ×

    =

    ×

    4

    d

    t

    d

    S

    d

    t

    1

    −

    4

    V

    3

    d

    V

    1

    =

    V

    3

    d

    t

    This will most likely be the use of A rather than S for surface area.

    or d

    S

    B1: States or uses

    S

    6

    x

    2

    (ignore any units associated with either)

    =

    t =

    4

    d

    may be implied by d

    S

    Note that

    S

    =

    6

    x

    2

    x =

    12

    x

    d

    and d

    S

    B1: States or uses

    (ignore any units associated with either)

    S

    =

    6

    x

    2

    t =

    4

    d

    may be implied by d

    S

    Note that

    S

    =

    6

    x

    2

    12

    x

    x =

    d

    M1: Attempts to use d

    d

    d

    or equivalent with their d

    x and d

    t where d

    and d

    S

    S

    x

    S

    S

    S

    =

    ×

    x =

    kx

    d

    t

    d

    x

    d

    t

    d

    d

    d

    constant, in an attempt to find d

    x

    t . May be implied by their working.

    d

    1

    A1: d

    x

    1

    4

    3 , 12

    oe e.g.

    x from correct work and apply isw once a correct answer is seen.

    =

    d

    t

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