A-LevelMathematicsDifferentiation9709/12 · May/June 2017 · Q99 Marks

The equation of a curve is y = 8√x − 2x.

(i)Find the coordinates of the stationary point of the curve.[3]
(ii)Find an expression for d²y/dx² and hence, or otherwise, determine the nature of the stationary point.[2]
(iii)Find the values of x at which the line y = 6 meets the curve.[3]
(iv)State the set of values of k for which the line y = k does not meet the curve.[1]

Mark scheme

Marking points from the official mark scheme (May/June 2017). The mark or mark type is shown in brackets.

dy/dx = 4x^-1/2 - 2
[B1] Accept unsimplified.
= 0 when √x = 2
x = 4, y = 8
[B1B1]
d²y/dx² = -2x^-3/2
[B1FT] FT providing -ve power of x
(-2/4^(3/2)) → Maximum
[B1] Correct d²y/dx² and x=4 in (i) are required. Followed by "< 0 or negative" is sufficient" but d²y/dx² must be correct if evaluated.
EITHER: Recognises a quadratic in √x
[M1] Eg √x=u→ 2u² - 8u + 6 = 0
1 and 3 as solutions to this equation
[A1]
→ x = 9, x = 1.
[A1]
OR: Rearranges then squares
(M1)
x²-10x+9=0 oe
[A1]
→ x = 9, x = 1.
[A1] Both correct by trial and improvement gets 3/3
k > 8
[B1]

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