Activities
P.o.t.W. #5
■ GDC is allowed ■ for SL and HL students A slice of bread is modelled by a two-dimensional shape consisting of a rectangle with the graph of \(y = 1 - {x^2},\; - 1 \le x \le 1 on top of the rectangle,...
Activities
P.o.t.W. #4
■ No GDC ■ for HL students A quarter circle is inscribed in a square with sides of one unit and AB is a radius of the quarter circle as shown in the figure. At point B a line is drawn tangent to the...
Activities
P.o.t.W. #3
■ No GDC ■ for SL and HL students total marks: 12 Consider the graph of \(g\left( x \right) = x\ln x shown at right. (a) Write down the domain of \(g. [1 mark] (b) Determine the range of \(g. ...
Activities
P.o.t.W. #2
■ GDC is allowed ■ for SL and HL students The rules for a lottery are as follows. Each lottery ticket costs $10. When a person purchases a lottery ticket they select five different numbers from 1 to...
Activities
P.o.t.W. #1
■ GDC is allowed ■ SL: parts (a) & (b) HL: all parts (a) Determine the exact area bounded by the graph of \(f\left( x \right) = \sqrt {4 - {x^2}} and the x-axis. (b) Consider...
Activities
P.o.t.W. #1 Solution
(a) The graph of \(f\left( x \right) = \sqrt {4 - {x^2}} is a semicircle with radius 2 units. Hence, the area bounded by the graph of \(f and the x-axis is \(\frac{1}{2}\pi{r^2} = \frac{1}{2}\pi{2^2}...
Activities
P.o.t.W. #14 Solution
(a) Let the total number of coins in the well at time \(t = n be S. Then, \(1 + 2 + 3 + \ldots + \left( {n - 2} \right) + \left( {n - 1} \right) + n = S (1) This can be re-written, such that...
Activities
P.o.t.W. #20 Solution
In order for the lengths of \(a, \(b and 1 to form a triangle it must be true that \(a + b > 1. If it’s true that \(a + b > 1, then what condition must \(a and \(b satisfy so that \(a, \(b and 1 are...
Activities
P.o.t.W. #29 Solution
The perpendicular bisector of the chord of a circle must be a diameter of the circle. \textrm{FG} is a perpendicular bisector of the chord that measures 13 units. Hence, {\mathrm{FG}} is a diameter of...
Activities
P.o.t.W. #27 Solution
Since \({\textrm{AB}} = 1 then \({\textrm{BC}} = 1; so, the radius of the larger circle is 2 . This makes \({\textrm{CQ}} = 2 and triangle CQB a 30-60-90 triangle where angle BCQ \( = 60^\circ = \dfrac{\pi}{3} radians....