The diagram shows the curve y = √x sin 2x for 0 ≤ x ≤ π/2. The curve has a maximum point at M, where x = a. [Figure X.X]

Mark scheme

Marking points from the official mark scheme (May/June 2025). The mark or mark type is shown in brackets.

1. [M1*] Use the correct product rule to differentiate. Could be working in terms of a.
2. [A1] Obtain dy/dx = 1/(2√x) sin 2x + 2√x cos 2x. Obtain the form p/sqrt(x) sin 2x + q sqrt(x) cos 2x.
3. [DM1] Equate the derivative to zero and form an equation without surds. E.g. sin2a+4acos2a = 0. Could be working in terms of x.
4. [A1] Obtain tan 2a = -4a from correct work. AG.
5. [M1] Calculate the values of a relevant expression or pair of expressions at x = 0.9 and x = 0.95. Allow smaller interval, provided it contains the root. Must be working in radians.
6. [A1] Complete the argument correctly with correct calculated values. E.g. tan(1.8)+3.6 = -0.686... <0 and tan(1.9)+3.8=0.873... > 0.
7. [B1] State a = 1/2(π - tan⁻¹(4a)) and rearrange to π−2a = tan⁻¹(4a). Or work from right to left.
8. [B1] State tan(π – 2a) = 4a and rearrange to tan 2a = −4a. Allow working in x or a.
9. [M1] Use the iterative process correctly at least once. M0 if working in degrees.
10. [A1] Obtain final answer 0.9183.
11. [A1] Show sufficient iterations to at least 6 d.p. to justify 0.9183 to 4 d.p. or show there is a sign change in the interval (0.91825, 0.91835). E.g. 0.9, 0.920872, 0.917944, 0.918347, 0.918292...