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Date May 2017 Marks available 1 Reference code 17M.2.SL.TZ2.T_6
Level Standard Level Paper Paper 2 Time zone Time zone 2
Command term Write down Question number T_6 Adapted from N/A

Question

Consider the function f ( x ) = x 4 + a x 2 + 5 , where a is a constant. Part of the graph of y = f ( x ) is shown below.

M17/5/MATSD/SP2/ENG/TZ2/06

It is known that at the point where x = 2 the tangent to the graph of y = f ( x ) is horizontal.

There are two other points on the graph of y = f ( x ) at which the tangent is horizontal.

Write down the y -intercept of the graph.

[1]
a.

Find f ( x ) .

[2]
b.

Show that a = 8 .

[2]
c.i.

Find f ( 2 ) .

[2]
c.ii.

Write down the x -coordinates of these two points;

[2]
d.i.

Write down the intervals where the gradient of the graph of y = f ( x ) is positive.

[2]
d.ii.

Write down the range of f ( x ) .

[2]
e.

Write down the number of possible solutions to the equation f ( x ) = 5 .

[1]
f.

The equation f ( x ) = m , where m R , has four solutions. Find the possible values of m .

[2]
g.

Markscheme

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

5     (A1)

 

Note:     Accept an answer of ( 0 ,   5 ) .

 

[1 mark]

a.

( f ( x ) = ) 4 x 3 + 2 a x     (A1)(A1)

 

Note:     Award (A1) for 4 x 3 and (A1) for + 2 a x . Award at most (A1)(A0) if extra terms are seen.

 

[2 marks]

b.

4 × 2 3 + 2 a × 2 = 0     (M1)(M1)

 

Note:     Award (M1) for substitution of x = 2 into their derivative, (M1) for equating their derivative, written in terms of a , to 0 leading to a correct answer (note, the 8 does not need to be seen).

 

a = 8     (AG)

[2 marks]

c.i.

( f ( 2 ) = ) 2 4 + 8 × 2 2 + 5     (M1)

 

Note:     Award (M1) for correct substitution of x = 2 and  a = 8 into the formula of the function.

 

21     (A1)(G2)

[2 marks]

c.ii.

( x = )   2 ,   ( x = )  0     (A1)(A1)

 

Note:     Award (A1) for each correct solution. Award at most (A0)(A1)(ft) if answers are given as ( 2   , 21 ) and ( 0 ,   5 ) or ( 2 ,   0 ) and ( 0 ,   0 ) .

 

[2 marks]

d.i.

x < 2 ,   0 < x < 2     (A1)(ft)(A1)(ft)

 

Note:     Award (A1)(ft) for x < 2 , follow through from part (d)(i) provided their value is negative.

Award (A1)(ft) for 0 < x < 2 , follow through only from their 0 from part (d)(i); 2 must be the upper limit.

Accept interval notation.

 

[2 marks]

d.ii.

y 21     (A1)(ft)(A1)

 

Notes:     Award (A1)(ft) for 21 seen in an interval or an inequality, (A1) for “ y ”.

Accept interval notation.

Accept < y 21 or f ( x ) 21 .

Follow through from their answer to part (c)(ii). Award at most (A1)(ft)(A0) if x is seen instead of y . Do not award the second (A1) if a (finite) lower limit is seen.

 

[2 marks]

e.

3 (solutions)     (A1)

[1 mark]

f.

5 < m < 21 or equivalent     (A1)(ft)(A1)

 

Note:     Award (A1)(ft) for 5 and 21 seen in an interval or an inequality, (A1) for correct strict inequalities. Follow through from their answers to parts (a) and (c)(ii).

Accept interval notation.

 

[2 marks]

g.

Examiners report

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

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
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Topic 2—Functions » SL 2.4—Key features of graphs, intersections using technology
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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