Activities
P.o.t.W. #23 Solution
■ GDC allowed ■ (a) (i) The period of the tangent function is \(\pi. Hence, any two angles \(\theta and \(\theta + \pi will have the same tangent value; i.e. \(\tan \left( \theta \right) = \tan...
Activities
P.o.t.W. #2 Solution
(a) \({}_{12}{C_5} = 792; There are 792 ways that five different #s can be chosen from the numbers 1 to 12. (b) (i) The number of ways that three winning numbers can be chosen from the five winning...
Activities
P.o.t.W. #3 Solution
(a) domain: \(x > 0 (b) find the coordinates of the minimum point on the graph of \(g \(g'\left( x \right) = x \cdot \frac{1}{x} + \ln x = 1 + \ln x = 0\;\;\; \Rightarrow \;\;\;\ln x = - 1\;\;\; \Rightarrow...
Activities
P.o.t.W. #36 Solution
Let {u_1} be the first term and d be the common difference of the arithmetic series.2500d = 2500\;\;\;\; \Rightarrow \;\;\;\;d = 1 Then the sum of the first fifty terms, {S_{50}}, is \displaystyle{S_{50}}=\frac{{50}}{2}\left[{2{u_1}...
Activities
P.o.t.W. #33 Solution
area of region =\displaystyle\int_0^1{x\arctan x\,{{d}}x} Apply integration by parts to find \int{x\arctan x\,{{d}}x} Let u = \arctan x, then du=\displaystyle\frac{1}{{{x^2} + 1}}\differentialD x\displaystyle;...
Activities
P.o.t.W. #5 Solution
total area of shape \( = 2 + 2\int_0^1 {\left( {1 - {x^2}} \right)dx = 2 + 2\left[ {x - \frac{1}{3}{x^3}} \right]} _0^1 = 2 + 2\left[ {\left( {1 - \frac{1}{3}} \right) - 0} \right] = 2 + \frac{4}{3} =...
Activities
P.o.t.W. #34 Solution
h\left( x \right) = ax + b h\left( {h\left( x \right)} \right) = a\left( {ax + b} \right) + b = {a^2}x + ab + b h\left( {h\left( {h\left( x \right)} \right)} \right) = a\left( {{a^2}x + ab + b} \right)...
Activities
P.o.t.W. #10 Solution
Method 1 double angle identity for cosine: \(\cos 2\alpha = 2{\cos ^2}\alpha - 1 substituting \(\theta for \(2\alpha gives: \(\cos \theta = 2{\cos ^2}\left( {\frac{\theta }{2}} \right) - 1\; from the...
Activities
P.o.t.W. #26 Solution
■ No GDC ■ (a) (i) \(\displaystyle\int_0^a {{x^2}{\textrm{d}}x} = \left. {\frac{1}{3}{x^3}} \right]_0^a = \frac{1}{3}{a^3} - \frac{1}{3}{0^3} = \frac{1}{3}{a^3} Q.E.D. (ii) \(\displaystyle\frac{{\textrm{d}}}{{{\textrm{d}}a}}\left(...
Activities
P.o.t.W. #24 Solution
■ GDC allowed ■ . . perimeter: \(a + b + c = 90\;\;\; \Rightarrow \;\;\;a + b = 90 - c Pythagorean theorem: \({a^2} + {b^2} = {c^2} similar triangles: \(\dfrac{c}{b} = \dfrac{a}{{11.25}}\;\;\; \Rightarrow...