O-LevelAdditional MathematicsTrigonometry4037/22 · Oct/Nov 2017 · Q1011 Marks
Without using a calculator, solve the equation 6c³ - 7c² + 1 = 0.
- (i)Without using a calculator, solve the equation 6c³ - 7c² + 1 = 0.[5]
- (ii)It is given that y = tan x + 6 sin x. Find dy/dx.[2]
- (iii)If dy/dx = 7 show that 6cos³x - 7cos²x + 1 = 0.[2]
- (iv)Hence solve the equation dy/dx = 7 for 0 ≤ x ≤ π radians.[2]
Mark scheme
Marking points from the official mark scheme (Oct/Nov 2017). The mark or mark type is shown in brackets.
- [B1] c=1 → 6(1)³-7(1)² +1=0 → (c-1) is a factor.; Or correct division. Finding or using one correct factor.
- [M1] Attempt to factorise or use long division to obtain 6c²...±1 or 6c² °c... respectively
- [A1] (c−1)(6c² -c-1)=0
- [A1] (c-1)(2c-1)(3c+1) =0; FT From three different linear factors
- [A1] c=1, 1/2, -1/3
- [B2] dy/dx = sec²x +6cosx; B1 for each term
- [B1] sec²x + 6cosx = 7; B1dep Replaces sec²x by 1/cos²x
- [B1] →6cos³x-7cos²x+1=0; B1dep Answer given so all steps must be correct.
- [A2] cosx=1, 1/2, -1/3; A1 for 2 values awrt A1 for third value and no others in range. No credit for answers in degrees
- →x = 0, 1.05 or (2π/3), 1.91
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