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Date November 2018 Marks available 2 Reference code 18N.1.SL.TZ0.S_10
Level Standard Level Paper Paper 1 Time zone Time zone 0
Command term Find Question number S_10 Adapted from N/A

Question

Let f ( x ) = x 3 2 x 2 + a x + 6 . Part of the graph of f is shown in the following diagram.

The graph of f crosses the y -axis at the point P. The line L is tangent to the graph of f at P.

Find the coordinates of P.

[2]
a.

Find f ( x ) .

[2]
b.i.

Hence, find the equation of L in terms of a .

[4]
b.ii.

The graph of f has a local minimum at the point Q. The line L passes through Q.

Find the value of a .

[8]
c.

Markscheme

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

valid approach      (M1)

eg    f ( 0 ) ,   0 3 2 ( 0 ) 2 + a ( 0 ) + 6 ,   f ( 0 ) = 6 ,   ( 0 , y )

(0, 6)  (accept  x = 0 and  y = 6)     A1 N2

 

[2 marks]

a.

f = 3 x 2 4 x + a      A2 N2

 

[2 marks]

b.i.

valid approach      (M1)

eg    f ( 0 )

correct working      (A1)

eg    3 ( 0 ) 2 4 ( 0 ) + a ,  slope =  a ,   f ( 0 ) = a

attempt to substitute gradient and coordinates into linear equation      (M1)

eg    y 6 = a ( x 0 ) ,   y 0 = a ( x 6 ) ,   6 = a ( 0 ) + c L  = a x + 6

correct equation      A1 N3

eg   y = a x + 6 ,   y 6 = a x ,   y 6 = a ( x 0 )

 

[4 marks]

b.ii.

valid approach to find intersection      (M1)

eg    f ( x ) = L

correct equation      (A1)

eg    x 3 2 x 2 + a x + 6 = a x + 6

correct working      (A1)

eg    x 3 2 x 2 = 0 ,   x 2 ( x 2 ) = 0

x = 2 at Q      (A1)

 

valid approach to find minimum      (M1)

eg    f ( x ) = 0

correct equation      (A1)

eg    3 x 2 4 x + a = 0

substitution of their value of x at Q into their f ( x ) = 0 equation      (M1)

eg    3 ( 2 ) 2 4 ( 2 ) + a = 0 ,   12 8 + a = 0

a = −4     A1 N0

 

[8 marks]

c.

Examiners report

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

Syllabus sections

Topic 4—Statistics and probability » SL 4.1—Concepts, reliability and sampling techniques
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