Date | November 2017 | Marks available | 3 | Reference code | 17N.2.AHL.TZ0.H_10 |
Level | Additional Higher Level | Paper | Paper 2 | Time zone | Time zone 0 |
Command term | Sketch | Question number | H_10 | Adapted from | N/A |
Question
Consider the function f(x)=√xsinx, 0<x<πf(x)=√xsinx, 0<x<π.
Consider the region bounded by the curve y=f(x)y=f(x), the xx-axis and the lines x=π6, x=π3x=π6, x=π3.
Show that the xx-coordinate of the minimum point on the curve y=f(x)y=f(x) satisfies the equation tanx=2xtanx=2x.
Determine the values of xx for which f(x)f(x) is a decreasing function.
Sketch the graph of y=f(x)y=f(x) showing clearly the minimum point and any asymptotic behaviour.
Find the coordinates of the point on the graph of ff where the normal to the graph is parallel to the line y=−xy=−x.
This region is now rotated through 2π2π radians about the xx-axis. Find the volume of revolution.
Markscheme
attempt to use quotient rule or product rule M1
f′(x)=sinx(12x−12)−√xcosxsin2x (=12√xsinx−√xcosxsin2x) A1A1
Note: Award A1 for 12√xsinx or equivalent and A1 for −√xcosxsin2x or equivalent.
setting f′(x)=0 M1
sinx2√x−√xcosx=0
sinx2√x=√xcosx or equivalent A1
tanx=2x AG
[5 marks]
x=1.17
0<x⩽1.17 A1A1
Note: Award A1 for 0<x and A1 for x⩽1.17. Accept x<1.17.
[2 marks]
concave up curve over correct domain with one minimum point above the x-axis. A1
approaches x=0 asymptotically A1
approaches x=π asymptotically A1
Note: For the final A1 an asymptote must be seen, and π must be seen on the x-axis or in an equation.
[3 marks]
f′(x) (=sinx(12x−12)−√xcosxsin2x)=1 (A1)
attempt to solve for x (M1)
x=1.96 A1
y=f(1.96…)
=1.51 A1
[4 marks]
V=π∫π3π6xdxsin2x (M1)(A1)
Note: M1 is for an integral of the correct squared function (with or without limits and/or π).
=2.68 (=0.852π) A1
[3 marks]