The constant a is such that ∫(from 2 to a) 6x ln x dx = 4.

Mark scheme

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

1. [M1*] Commence integration and reach px² ln x + q ∫ x dx Allow M1 with q ∫ (x²/x) dx.
2. [A1] Obtain 3x² ln x - 3 ∫ x dx x²/x must be simplified.
3. [A1] Complete integration and obtain 3x² ln x - (3/2)x²
4. [DM1] Use limits correctly in an expression of the form ax² ln x – bx² and equate to 4, having integrated twice 3a² lna-a²-0+3/2 = 4
5. [A1] Obtain a = exp( (1/6)(5/a² + 3) ) correctly AG
6. [M1] Calculate the values of a relevant expression or pair of expressions at a = 2 and a = 2.1 Not from using calculator to evaluate the original integral.
7. [A1] Justify the given statement with correct calculated values E.g., using f(x) = 6x² ln x − 3x² f(2) = 4.636 < 5 and f(2.1) = 6.402 > 5, or using the exponential, 2 < 2.0306 and 2.1 > 1.9917 or 0.0306 > 0 and -0.1083 <0..
8. [M1] Use the iterative process a_(n+1) = exp( (1/6)(5/a_n² + 3) ) correctly at least once.
9. [A1] Obtain final answer 2.02
10. [A1] Show sufficient iterations to at least 4 d.p. to justify 2.02 to 2 d.p. or show that there is a sign change in (2.015, 2.025)