A-LevelMathematicsDifferentiation9709/12 · Feb/Mar 2023 · Q1010 Marks
At the point (4, −1) on a curve, the gradient of the curve is -3/2. It is given that dy/dx = x^(-1/2) + k, where k is a constant.
- (a)Show that k = -2.[1]
- (b)Find the equation of the curve.[4]
- (c)Find the coordinates of the stationary point.[3]
- (d)Determine the nature of the stationary point.[2]
Mark scheme
Marking points from the official mark scheme (Feb/Mar 2023). The mark or mark type is shown in brackets.
- [B1] -3/2 = 4^(-1/2) + k leading to k = -2 AG Need to see 4^(-1/2) evaluated as 1/2 or better.
- [M1 A1] [y] = (2/3)x^(3/2) -2x [+c] Allow (1/ (1/2 + 1))x^(1/2+1) - 2x.
- [M1] -1 = (2/3)(4)^(3/2) -2(4)+c Substitute x = 4, y = −1 (c present) Expect c = 3.
- [A1] y = (2/3)x^(3/2) - 2x + 3 or y = 2√x - 2x + 3 Allow if f(x) = or y = anywhere in the solution.
- [M1] x^(-1/2) - 2 = 0 Set their dy/dx to zero.
- [A1] x = 1/4
- [A1] (1/4, 3/2) If (1/4, 3/2) seen. A1 = max of M1A1 if (1/4, 3/2) seen.
- [B1] d²y/dx² = -1/2 x^(-3/2)
- [DB1] < 0 (or -4) hence Maximum WWW Ignore extra solutions from x = -1/4.
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