A-LevelMathematicsDifferentiation9709/12 · Feb/Mar 2022 · Q119 Marks
It is given that a curve has equation y = k(3x – k)⁻¹ + 3x, where k is a constant.
- (a)Find, in terms of k, the values of x at which there is a stationary point.[4]
- (b)The function f has a stationary value at x = a and is defined by f(x) = 4(3x - 4)⁻¹ + 3x for x ≥ 3/2. Find the value of a and determine the nature of the stationary value.[3]
- (c)The function g is defined by g(x) = −(3x + 1)⁻¹ + 3x for x ≥ 0. Determine, making your reasoning clear, whether g is an increasing function, a decreasing function or neither.[2]
Mark scheme
Marking points from the official mark scheme (Feb/Mar 2022). The mark or mark type is shown in brackets.
- [B2, 1, 0] dy/dx = {-k(3x-k)⁻²}{×3} {+3}
- [M1] -3k / (3x-k)² + 3 = 0 leading to (3)(3x-k)² = (3)k
- [A1] leading to 3x-k = [±]√k
- [A1 OE] x = (k ± √k)/3
- [B1] a = (4 ± √4)/3 leading to a = 2
- [B1] f''(x) = [-12(3x-4)⁻²×3] = 72(3x-4)⁻³
- [B1 FT] > 0 (or 9) when x = 2 → minimum
- [M1] Substitute k = -1 leading to g'(x) = 3/(3x + 1)² + 3
- [A1] g'(x)>0 or g'(x) always positive, hence g is an increasing function
Write your answer and get it marked
On Oracle Prep you can write or photograph your own answer and have it marked by AI against this mark scheme, with feedback on what you missed.
Sign up free →