A-LevelMathematicsDifferentiation9709/12 · May/June 2023 · Q118 Marks

The equation of a curve is y = k√(4x + 1) − x + 5, where k is a positive constant.

(a)Find dy/dx.[2]
(b)Find the x-coordinate of the stationary point in terms of k.[2]
(c)Given that k = 10.5, find the equation of the normal to the curve at the point where the tangent to the curve makes an angle of tan⁻¹(2) with the positive x-axis.[4]

Mark scheme

Marking points from the official mark scheme (May/June 2023). The mark or mark type is shown in brackets.

[B2,1,0] dy/dx = k(1/2)(4x+1)^(-1/2)(4){-1} OE e.g. 2k (4x+1)^(-1/2)-1 B2 Three correct unsimplified { } and no others. B1 Two correct { } or three correct { } and an additional term e.g. +5. B0 More than one error.
[M1] 2k(4x+1)^(-1/2)-1=0 leading to (4x+1)^(1/2) = 2k or 2k/(4x+1)^(1/2)=1 OE Equating their dy/dx of the form ak (4x+1)^b -1 where a = 2 or 0.5, to 0 and dealing with the negative power correctly including k not multiplied by (4x+1)^(-1/2).
[A1] x = (4k²-1)/4 CAO OE simplified expression ISW.
[M1] Putting k= 10.5 into their dy/dx and equating to 2.
[A1] 7=(4x+1)^(1/2) leading to 4x+1=49 leading to x = 12 If M1 earned SCB1 available for x = 33/64 from a = 1/2.
[A1] y = [10.5√(4x+1)-x+5=]66.5 [leading to (12, 66.5)]
[A1] y-66.5 = -1/2(x-12) OE

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