The diagram shows part of the curve y = x-x²/4 and the line y =-4. The curve and the line intersect at the point A.
- (a)The maximum point on the curve is at a perpendicular distance h from the line y =-4. Find the value of h.[4]
- (b)Find the exact x-coordinate of A.[3]
- (c)Find the acute angle between the tangent to the curve at A and the line y =-4.[4]
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Mark scheme
Marking points from the official mark scheme (Oct/Nov 2024). The mark or mark type is shown in brackets.
- [B4] Maximum point at (2, 1) oe, nfww and h = 5 nfww [B3] for maximum point at (2, 1) oe
- [B2] for maximum point when x = 2
- [B1] for a correct method which could be used to find the maximum point x(1-x/4)=0 when x=0 and x=4 or [differentiating and equating to 0:] 1 - 2x/4 = 0 or [completing the square to find:] 1 - 1/4 (x-2)^2
- If 0 scored, [SC1] for h = 5 with incorrect or no method shown
- [B1] x² - 4x - 16 [= 0] oe or -1/4 (x-2)² = -4-1 oe [FT] their attempt to complete the square, if already seen and of the form a+b(x+c)² where a, b and c are constants and b is negative
- [M1] Solves their 3-term quadratic using the formula or completing the square
- [A1] x = 2 + 2√5 oe [FT] their rearrangement of -4 = x - x²/4 oe
- [M1] First derivative 1 - x/2 and substitution of their 2+2√5 or 6.47 [FT] their attempt to differentiate x - x²/4 if already seen, and their 2+2√5
- [A1] 1 - 2(2+2√5)/4 or -√5 or awrt –2.24 soi [dep] on correct derivative and correct x-coordinate of A
- [M1] Correct method to find the angle e.g. -tan⁻¹(1 - 2(2+2√5)/4) soi or tan⁻¹√5 or -tan⁻¹(-√5) soi [dep] on previous M1 A1
- [A1] Degrees: awrt 65.9 or Radians: awrt 1.15
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