Note AcademyNote Academy

Mathematics · Oct / Nov 2021 · Paper 31 · Question 8

YMA01

As printed in the paper

The question as text

8. A curve C has equation y = f(x), where π  y  2 π f(x) = arcsin 1 –2  x  2 – 2 2 x

Read automatically from the paper; the printed page above is exact.

Show the answer and mark scheme
  • 8(a)[1]

    Shape and position

    d

    x

    and attempts to use both d

    y

    d

    x

    x

    =

    2sin

    y

    ⇒

    =

    2cos

    y

    = ÷

    1

  • 8(b)[3]

    and

    d

    y

    d

    x

    d

    y

    cos

    y

    =

    1 sin

    −

    2

    y

    =

    1 ...

    −

    x

    2

    d

    y

    1

    =

    d

    x

    x

    2

    2 1

    −

    4

    d

    y

    1

    =

    d

    x

    4

    −

    x

    2

    d

    y

    1

    2

    x =

    2

    =

    =

  • 8(c)[3]

    Substitutes

    into their

    d

    x

    2

    4

    −

    x

    2

    π

    2

    (

    )

    Finds the equation of the tangent at P

    y

    −

    =

    ''

    ''

    x

    −

    '' 2 ''

    4

    2

    2

    π

    y

    =

    x

    −+

    1

    2

    4

    Alt(b)

    x

    d

    y

    1

    1

    y

    =

    arcsin 2

    ⇒

    =

    ×

    d

    x

    2

    2

    x

    1

    −

    2

    d

    y

    1

    =

    d

    x

    4

    2

    −

    x

    (a)

    B1

    Correct shape and position: Look for a curve in quadrants 1 and 3 with non zero gradient

    at the origin and gradient →∞ at both ends. Ignore values labelled on the axes.

    If there is more than one attempt, mark the one which appears in the main body of the

    work as their fullest attempt. If there is more than one sketch on the same graph, then it

    must be clearly labelled which is the one to be marked.

    Examples of B1

    Examples of B0

    (b) Note on EPEN it is M1M1A1 which is now marked as M1A1A1

    d

    x

    x

    =

    2sin

    y

    ⇒

    = ±

    k

    cos

    y

    M1

    where k is non zero and attempts to use both

    d

    y

    d

    y

    d

    x

    = in order to obtain d

    y

    = ÷

    1

    sin

    2

    y

    cos

    2

    y

    1

    and

    +

    in terms of x. Allow slips

    d

    x

    d

    y

    d

    x

    with dealing with their “2”

    ...

    Allow working leading to

    to score M1.

    ... 1 ...x

    −

    2

    A correct expression for d

    y

    A1

    as a function of x which may be unsimplified.

    d

    x

    d

    y

    1

    1

    1

    eg

    =

    or

    or

    d

    x

    x

    2

    4

    x

    2

    4

    −

    x

    2

    2 1

    −

    −

    2

    4

    4

    4

    4

    d

    y

    1

    =

    A1cso

    d

    x

    4

    −

    x

    2

    Alt (b)

    x

    d

    y

    1

    M1

    y

    =

    arcsin 2

    ⇒

    =

    k

    ×

    (where k can equal 1) then A1A1 as above

    d

    x

    x

    2

    1

    −

    2

    x

    d

    y

    1

    SC

    with no correct intermediate working score as 100.

    y

    =

    arcsin 2

    ⇒

    =

    d

    x

    4

    −

    x

    2

    (c)

    Attempts to find the value of d

    y

    or d

    x

    y = π into

    M1

    at P. See scheme but allow sub of

    4

    d

    x

    d

    y

    their d

    x

    y = ±

    k

    cos

    y

    d

    Full method to find the equation of the tangent at P. Their d

    y

    dM1

    may not be exact. If they

    d

    x

    use y

    mx

    + they must proceed as far as

    c

    c =

    ...

    =

    A1

    Correct equation and in the correct form. The coefficient of x and the constant must be

    exact. Isw

Make a free account to save your answers and see your weak topics and ask the AI why a wrong answer is wrong.