Date | May 2010 | Marks available | 8 | Reference code | 10M.2.hl.TZ1.10 |
Level | HL only | Paper | 2 | Time zone | TZ1 |
Command term | Determine and Find | Question number | 10 | Adapted from | N/A |
Question
The diagram below shows the graphs of y=|32x−3|, y=3 and a quadratic function, that all intersect in the same two points.
Given that the minimum value of the quadratic function is −3, find an expression for the area of the shaded region in the form ∫t0(ax2+bx+c)dx, where the constants a, b, c and t are to be determined. (Note: The integral does not need to be evaluated.)
Markscheme
|32x−3|=0 when x = 2 (A1)
the equation of the parabola is y=p(x−2)2−3 (M1)
through (0, 3)⇒3=4p−3⇒p=32 (M1)
the equation of the parabola is y=32(x−2)2−3 (=32x2−6x+3) A1
area =2∫20(3−32x)−(32x2−6x+3)dx M1M1A1
Note: Award M1 for recognizing symmetry to obtain 2∫20,
M1 for the difference,
A1 for getting all parts correct.
=∫20(−3x2+9x)dx A1
[8 marks]
Examiners report
This was a difficult question and, although many students obtained partial marks, there were few completely correct solutions.