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Date May 2018 Marks available 1 Reference code 18M.2.sl.TZ2.6
Level SL only Paper 2 Time zone TZ2
Command term State Question number 6 Adapted from N/A

Question

Consider the curve y = 2x3 − 9x2 + 12x + 2, for −1 < x < 3

Sketch the curve for −1 < x < 3 and −2 < y < 12.

[4]
a.

A teacher asks her students to make some observations about the curve.

Three students responded.
Nadia said “The x-intercept of the curve is between −1 and zero”.
Rick said “The curve is decreasing when x < 1 ”.
Paula said “The gradient of the curve is less than zero between x = 1 and x = 2 ”.

State the name of the student who made an incorrect observation.

[1]
b.

Find the value of y when x = 1 .

[2]
c.

Find \(\frac{{{\text{dy}}}}{{{\text{dx}}}}\).

[3]
d.

Show that the stationary points of the curve are at x = 1 and x = 2.

[2]
e.

Given that y = 2x3 − 9x2 + 12x + 2 = k has three solutions, find the possible values of k.

[3]
f.

Markscheme

(A1)(A1)(A1)(A1)

Note: Award (A1) for correct window (condone a window which is slightly off) and axes labels. An indication of window is necessary. −1 to 3 on the x-axis and −2 to 12 on the y-axis and a graph in that window.
(A1) for correct shape (curve having cubic shape and must be smooth).
(A1) for both stationary points in the 1st quadrant with approximate correct position,
(A1) for intercepts (negative x-intercept and positive y intercept) with approximate correct position.

[4 marks]

a.

Rick     (A1)

Note: Award (A0) if extra names stated.

[1 mark]

b.

2(1)3 − 9(1)2 + 12(1) + 2     (M1)

Note: Award (M1) for correct substitution into equation.

= 7     (A1)(G2)

[2 marks]

 

 

 

 

 

c.

6x2 − 18x + 12     (A1)(A1)(A1)

Note: Award (A1) for each correct term. Award at most (A1)(A1)(A0) if extra terms seen.

[3 marks]

d.

6x2 − 18x + 12 = 0     (M1)

Note: Award (M1) for equating their derivative to 0. If the derivative is not explicitly equated to 0, but a subsequent solving of their correct equation is seen, award (M1).

6( − 1)(x − 2) = 0  (or equivalent)      (M1)

Note: Award (M1) for correct factorization. The final (M1) is awarded only if answers are clearly stated.

Award (M0)(M0) for substitution of 1 and of 2 in their derivative.

x = 1, x = 2 (AG)

[2 marks]

e.

6 < k < 7     (A1)(A1)(ft)(A1)

Note: Award (A1) for an inequality with 6, award (A1)(ft) for an inequality with 7 from their part (c) provided it is greater than 6, (A1) for their correct strict inequalities. Accept ]6, 7[ or (6, 7).

[3 marks]

f.

Examiners report

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

Syllabus sections

Topic 7 - Introduction to differential calculus » 7.5 » Values of x where the gradient of a curve is zero.
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