Date | November 2019 | Marks available | 2 | Reference code | 19N.1.AHL.TZ0.H_10 |
Level | Additional Higher Level | Paper | Paper 1 (without calculator) | Time zone | Time zone 0 |
Command term | Show and Sketch | Question number | H_10 | Adapted from | N/A |
Question
Consider f(x)=2x−4x2−1, −1<x<1.
For the graph of y=f(x),
Find f′(x).
Show that, if f′(x)=0, then x=2−√3.
find the coordinates of the y-intercept.
show that there are no x-intercepts.
sketch the graph, showing clearly any asymptotic behaviour.
Show that 3x+1−1x−1=2x−4x2−1.
The area enclosed by the graph of y=f(x) and the line y=4 can be expressed as lnv. Find the value of v.
Markscheme
attempt to use quotient rule (or equivalent) (M1)
f′(x)=(x2−1)(2)−(2x−4)(2x)(x2−1)2 A1
=−2x2+8x−2(x2−1)2
[2 marks]
f′(x)=0
simplifying numerator (may be seen in part (i)) (M1)
⇒x2−4x+1=0 or equivalent quadratic equation A1
EITHER
use of quadratic formula
⇒x=4±√122 A1
OR
use of completing the square
(x−2)2=3 A1
THEN
x=2−√3 (since 2+√3 is outside the domain) AG
Note: Do not condone verification that x=2−√3⇒f′(x)=0.
Do not award the final A1 as follow through from part (i).
[3 marks]
(0, 4) A1
[1 mark]
2x−4=0⇒x=2 A1
outside the domain R1
[2 marks]
A1A1
award A1 for concave up curve over correct domain with one minimum point in the first quadrant
award A1 for approaching x=±1 asymptotically
[2 marks]
valid attempt to combine fractions (using common denominator) M1
3(x−1)−(x+1)(x+1)(x−1) A1
=3x−3−x−1x2−1
=2x−4x2−1 AG
[2 marks]
f(x)=4⇒2x−4=4x2−4 M1
(x=0 or) x=12 A1
area under the curve is ∫120f(x)dx M1
=∫1203x+1−1x−1dx
Note: Ignore absence of, or incorrect limits up to this point.
=[3ln|x+1|−ln|x−1|]120 A1
=3ln32−ln12(−0)
=ln274 A1
area is 2−∫120f(x)dx or ∫1204dx−∫120f(x)dx M1
=2−ln274
=ln4e227 A1
(⇒v=4e227)
[7 marks]