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Date May 2019 Marks available 2 Reference code 19M.2.SL.TZ1.T_6
Level Standard Level Paper Paper 2 Time zone Time zone 1
Command term Find 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.

[5]
a.

Find the coordinates of the local minimum.

[2]
b.

Write down the interval where the gradient of the graph of f(x)f(x) is negative.

[2]
c.

Determine the equation of the normal at x=2x=2 in the form y=mx+cy=mx+c.

[5]
d.

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+x6x2+x6. 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]

a.

(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]

b.

3<x<23<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<23<y<2 award (A1)(A0). Accept equivalent notation such as (−3, 2). Award (A0)(A1)(ft) for 3x2.

[2 marks]

c.

−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)26(2)+5=493)      (A1)

Note: Award (A1) for 493  (16.3333…) seen.

493=14(2)+c  OR  y493=14(x2)       (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]

d.

Examiners report

[N/A]
a.
[N/A]
b.
[N/A]
c.
[N/A]
d.

Syllabus sections

Topic 5—Calculus » SL 5.1—Introduction of differential calculus
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Topic 5—Calculus » SL 5.2—Increasing and decreasing functions
Topic 5—Calculus » SL 5.3—Differentiating polynomials, n E Z
Topic 5—Calculus » SL 5.4—Tangents and normal
Topic 5—Calculus » SL 5.6—Stationary points, local max and min
Topic 5—Calculus

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