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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 ) = 1 3 x 3 + 1 2 x 2 + k x + 5  has a local maximum and a local minimum. The local maximum is at x = 3 .

Show that k = 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 ) is negative.

[2]
c.

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

[5]
d.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

x 2 + 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 x 2 + 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    (M1)(M1)

Note: Award (M1) for substituting in x = 3 into their derivative and (M1) for setting it equal to zero. Substituting k = 6 invalidates the process, award at most (A1)(A1)(A1)(M0)(M0).

( k = ) 6       (AG)

Note: For the final (M1) to be awarded, no incorrect working must be seen, and must lead to the conclusion k = 6 . The final (AG) must be seen.

[5 marks]

a.

(2, −2.33)  OR  ( 2 , 7 3 )       (A1)(A1)

Note: Award (A1) for each correct coordinate. Award (A0)(A1) if parentheses are missing. Accept x = 2 , y = 2.33 . Award (M1)(A0) for their derivative, a quadratic expression with –6 substituted for k , equated to zero but leading to an incorrect answer.

[2 marks]

b.

3 < x < 2       (A1)(ft)(A1)

Note: Award (A1) for x > 3 , (A1)(ft) for x < 2 . Follow through for their "2" in part (b). It is possible to award (A0)(A1). For 3 < y < 2 award (A1)(A0). Accept equivalent notation such as (−3, 2). Award (A0)(A1)(ft) for 3 x 2 .

[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  1 4       (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.

49 3     ( f ( 2 ) = 1 3 ( 2 ) 3 + 1 2 ( 2 ) 2 6 ( 2 ) + 5 = 49 3 )       (A1)

Note: Award (A1) for  49 3   (16.3333…) seen.

49 3 = 1 4 ( 2 ) + c   OR   y 49 3 = 1 4 ( x 2 )        (M1)

Note: Award (M1) for substituting their normal gradient into equation of line formula.

y = 1 4 x + 101 6   OR  y = 0.25 x + 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 = 49 3 and x = 2 .

[5 marks]

d.

Examiners report

[N/A]
a.
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b.
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c.
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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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