Mathematics · Oct / Nov 2023 · Paper 41 · Question 11

3x2 – 8x – 3 [2]

B1 for two terms correct or correct answer seen

3x2 – 8x – 3 = 0 Correct method to solve their 3- term quadratic (3x + 1)(x – 3) [=0] −− ± − − − ( 8) ( 8) 2 4(3)( 3) 2(3) (3, –18) 1 14 − , 3 27 [8]

M1 FT their part (a)

M2 M1 for (3x + a)(x + b) [=0] where ab = –3 or 3b + a = –8 or for ( 8) − 2 − 4(3)( 3) − ± or for p q where p = –(–8) and r = 2(3) seen r or for a correct method for solving a 2-term quadratic

B2 B1 for one correct point or for two correct x- values,

or M1 for substitution of their x-values into = 3 − 2 − y x 4 x 3 x shown

(3, –18) minimum with reason 1 14 − , maximum with 3 27 reason [3]

Reasons could be e.g. 1. A reasonable sketch of a positive cubic 2. Correct use of 2nd derivative = 6x – 8 = 10, 10 > 0, so (3, –18) is a minimum oe. 2nd derivative = 6x – 8 = –10, –10 < 0 1 14 − so , is a maximum oe. 3 27 3. Evaluates correctly values of y on both sides of both correct stationary points 4. Finds gradient on each side of both correct stationary points.

B2 for 1 correct with a reason for that stationary point or for both x-values correct with correct conclusions and reasonable sketch of a positive cubic, or for correct substitution of both of their x-values into their second derivative shown, or substitution shown for one x-value either side of both of their stationary points to find the gradients.

Or M1 for showing [2nd derivative =] 6x – 8 or substitution shown for one x-value either side of one of their stationary points to find the gradients. or for reasonable sketch of positive cubic.