The diagram shows the curve y = sin 3x cos x for 0 ≤ x ≤ ½π and its minimum point M. The shaded region R is bounded by the curve and the x-axis. [Figure 10.1]

Mark scheme

Marking points from the official mark scheme (May/June 2019). The mark or mark type is shown in brackets.

1. [B1] State correct expansion of sin(3x + x) or sin(3x - x) (B0 If their formula retains ± in the middle)
2. [M1] Substitute expansions in ½(sin 4x + sin 2x)
3. [A1] Obtain sin 3x cos x = ½(sin 4x + sin 2x) correctly (AG) (Must see the sin 4x and sin 2x or reference to LHS and RHS for A1)
4. [B1 B1 B1] Integrate and obtain -¼cos 4x - ¼cos 2x (FT in their expression)
5. [M1] Substitute limits x = 0 and x = ½π correctly (In their expression)
6. [A1] Obtain answer 9/16 (From correct working seen.)
7. [B1] State correct derivative 2 cos 4x + cos 2x
8. [M1] Using correct double angle formula, express derivative in terms of cos 2x and equate the result to zero
9. [A1] Obtain 4cos²2x + cos2x - 2 = 0
10. [M1] Solve for x or 2x (could be labelled x) (cos 2x = (-1 ± √33)/8). Must see working if solving an incorrect quadratic. The roots of the correct quadratic are -0.843 and 0.593. Need to get as far as x = ... The wrong value of x is 0.468 and can imply M1 if correct quadratic seen. Could be working from a quartic in cos x: 16cos⁴x - 14cos²x + 1 = 0)
11. [A1] Obtain answer x ≈ 1.29 only