Date | May 2019 | Marks available | 5 | Reference code | 19M.2.SL.TZ1.T_6 |
Level | Standard Level | Paper | Paper 2 | Time zone | Time zone 1 |
Command term | Show that | Question number | T_6 | Adapted from | N/A |
Question
The function f(x)=13x3+12x2+kx+5f(x)=13x3+12x2+kx+5 has a local maximum and a local minimum. The local maximum is at x=−3x=−3.
Show that k=−6k=−6.
Find the coordinates of the local minimum.
Write down the interval where the gradient of the graph of f(x)f(x) is negative.
Determine the equation of the normal at x=−2x=−2 in the form y=mx+cy=mx+c.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
x2+x+kx2+x+k (A1)(A1)(A1)
Note: Award (A1) for each correct term. Award at most (A1)(A1)(A0) if additional terms are seen or for an answer x2+x−6x2+x−6. If their derivative is seen in parts (b), (c) or (d) and not in part (a), award at most (A1)(A1)(A0).
(−3)2+(−3)+k=0(−3)2+(−3)+k=0 (M1)(M1)
Note: Award (M1) for substituting in x=−3x=−3 into their derivative and (M1) for setting it equal to zero. Substituting k=−6k=−6 invalidates the process, award at most (A1)(A1)(A1)(M0)(M0).
(k=)−6(k=)−6 (AG)
Note: For the final (M1) to be awarded, no incorrect working must be seen, and must lead to the conclusion k=−6k=−6. The final (AG) must be seen.
[5 marks]
(2, −2.33) OR (2,−73)(2,−73) (A1)(A1)
Note: Award (A1) for each correct coordinate. Award (A0)(A1) if parentheses are missing. Accept x=2x=2, y=−2.33y=−2.33. Award (M1)(A0) for their derivative, a quadratic expression with –6 substituted for kk, equated to zero but leading to an incorrect answer.
[2 marks]
−3<x<2−3<x<2 (A1)(ft)(A1)
Note: Award (A1) for x>−3x>−3 , (A1)(ft) for x<2x<2. Follow through for their "2" in part (b). It is possible to award (A0)(A1). For −3<y<2−3<y<2 award (A1)(A0). Accept equivalent notation such as (−3, 2). Award (A0)(A1)(ft) for −3⩽x⩽2.
[2 marks]
−4 (A1)(ft)
Note: Award (A1)(ft) for the gradient of the tangent seen. If an incorrect derivative was used in part (a), then working for their f′(−2) must be seen. Follow through from their derivative in part (a).
gradient of normal is 14 (A1)(ft)
Note: Award (A1)(ft) for the negative reciprocal of their gradient of tangent. Follow through within this part. Award (G2) for an unsupported gradient of the normal.
493 (f(−2)=13(−2)3+12(−2)2−6(−2)+5=493) (A1)
Note: Award (A1) for 493 (16.3333…) seen.
493=14(−2)+c OR y−493=14(x−−2) (M1)
Note: Award (M1) for substituting their normal gradient into equation of line formula.
y=14x+1016 OR y=0.25x+16.8333… (A1)(ft)(G4)
Note: Award (G4) for the correct equation of line in correct form without any prior working. The final (A1)(ft) is contingent on y=493 and x=−2.
[5 marks]