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Date November 2018 Marks available 4 Reference code 18N.1.SL.TZ0.S_10
Level Standard Level Paper Paper 1 Time zone Time zone 0
Command term Find and Hence Question number S_10 Adapted from N/A

Question

Let f(x)=x32x2+ax+6f(x)=x32x2+ax+6. Part of the graph of ff is shown in the following diagram.

The graph of ff crosses the yy-axis at the point P. The line L is tangent to the graph of ff 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),  032(0)2+a(0)+6,  f(0)=6,  (0,y)

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

 

[2 marks]

a.

f=3x24x+a     A2 N2

 

[2 marks]

b.i.

valid approach      (M1)

eg   f(0)

correct working      (A1)

eg   3(0)24(0)+a,  slope = a,  f(0)=a

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

eg   y6=a(x0),  y0=a(x6),  6=a(0)+cL =ax+6

correct equation      A1 N3

eg  y=ax+6,  y6=ax,  y6=a(x0)

 

[4 marks]

b.ii.

valid approach to find intersection      (M1)

eg   f(x)=L

correct equation      (A1)

eg   x32x2+ax+6=ax+6

correct working      (A1)

eg   x32x2=0,  x2(x2)=0

x=2 at Q      (A1)

 

valid approach to find minimum      (M1)

eg   f(x)=0

correct equation      (A1)

eg   3x24x+a=0

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

eg   3(2)24(2)+a=0,  128+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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Topic 4—Statistics and probability
Topic 5—Calculus

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