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Date May 2017 Marks available 1 Reference code 17M.2.SL.TZ1.T_6
Level Standard Level Paper Paper 2 Time zone Time zone 1
Command term Hence and Justify Question number T_6 Adapted from N/A

Question

Consider the function g ( x ) = x 3 + k x 2 15 x + 5 .

The tangent to the graph of y = g ( x ) at x = 2 is parallel to the line y = 21 x + 7 .

Find g ( x ) .

[3]
a.

Show that k = 6 .

[2]
b.i.

Find the equation of the tangent to the graph of y = g ( x ) at x = 2 . Give your answer in the form y = m x + c .

[3]
b.ii.

Use your answer to part (a) and the value of k , to find the x -coordinates of the stationary points of the graph of y = g ( x ) .

[3]
c.

Find g ( 1 ) .

[2]
d.i.

Hence justify that g is decreasing at x = 1 .

[1]
d.ii.

Find the y -coordinate of the local minimum.

[2]
e.

Markscheme

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

3 x 2 + 2 k x 15     (A1)(A1)(A1)

 

Note:     Award (A1) for 3 x 2 , (A1) for 2 k x and (A1) for 15 . Award at most (A1)(A1)(A0) if additional terms are seen.

 

[3 marks]

a.

21 = 3 ( 2 ) 2 + 2 k ( 2 ) 15     (M1)(M1)

 

Note:     Award (M1) for equating their derivative to 21. Award (M1) for substituting 2 into their derivative. The second (M1) should only be awarded if correct working leads to the final answer of k = 6 .

Substituting in the known value, k = 6 , invalidates the process; award (M0)(M0).

 

k = 6     (AG)

[2 marks]

b.i.

g ( 2 ) = ( 2 ) 3 + ( 6 ) ( 2 ) 2 15 ( 2 ) + 5   ( = 7 )     (M1)

 

Note:     Award (M1) for substituting 2 into g .

 

7 = 21 ( 2 ) + c     (M1)

 

Note:     Award (M1) for correct substitution of 21, 2 and their 7 into gradient intercept form.

 

OR

 

y 7 = 21 ( x 2 )     (M1)

 

Note:     Award (M1) for correct substitution of 21, 2 and their 7 into gradient point form.

 

y = 21 x 35     (A1)     (G2)

[3 marks]

b.ii.

3 x 2 + 12 x 15 = 0 (or equivalent)     (M1)

 

Note:     Award (M1) for equating their part (a) (with k = 6 substituted) to zero.

 

x = 5 ,   x = 1     (A1)(ft)(A1)(ft)

 

Note:     Follow through from part (a).

 

[3 marks]

c.

3 ( 1 ) 2 + 12 ( 1 ) 15     (M1)

 

Note:     Award (M1) for substituting 1 into their derivative, with k = 6 substituted. Follow through from part (a).

 

= 24     (A1)(ft)     (G2)

[2 marks]

d.i.

g ( 1 ) < 0 (therefore g is decreasing when x = 1 )     (R1)

[1 marks]

d.ii.

g ( 1 ) = ( 1 ) 3 + ( 6 ) ( 1 ) 2 15 ( 1 ) + 5     (M1)

 

Note:     Award (M1) for correctly substituting 6 and their 1 into g .

 

= 3     (A1)(ft)     (G2)

 

Note:     Award, at most, (M1)(A0) or (G1) if answer is given as a coordinate pair. Follow through from part (c).

 

[2 marks]

e.

Examiners report

[N/A]
a.
[N/A]
b.i.
[N/A]
b.ii.
[N/A]
c.
[N/A]
d.i.
[N/A]
d.ii.
[N/A]
e.

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 2—Functions » SL 2.5—Modelling functions
Topic 5—Calculus » SL 5.6—Stationary points, local max and min
Topic 2—Functions
Topic 5—Calculus

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