Math AA SL

The sum of the integers is:

k + (k + 2) = 2k + 2 = 2(k + 1)

Therefore the sum of these integers is even.

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(a)

Using the Pythagorean identity, \sin^{2}\theta + \cos^{2}\theta = 1 we can write that:

\sin^{2}x + \left (\frac{2}{3}  \right )^{2} = 1

\sin^{2}x = 1 - \frac{4}{9} = \frac{5}{9}

\boldsymbol{{{\color{Purple} \sin x = \frac{\sqrt{5}}{3} }}}

 (b)

Using the identity \tan \theta = \frac{\sin \theta}{\cos \theta} we can write that:

\tan x = \frac{\frac{\sqrt{5}}{3}}{\frac{2}{3}} = \frac{\sqrt{5}}{3} \times \frac{3}{2}

\boldsymbol{{{\color{Purple} \tan x = \frac{\sqrt{5}}{2} }}}

(c)

Using the identity \sin 2\theta = 2\sin\theta\cos\theta we can write that:

\sin 2x = 2\left ( \frac{\sqrt{5}}{3} \right )\left ( \frac{2}{3} \right )

\boldsymbol{{{\color{Purple} \sin 2x = \frac{4\sqrt{5}}{9} }}}

(d)

Using the identity (the other two identities also work here) \cos 2\theta = \cos^{2}\theta - \sin^{2}\theta we can write that:

\cos 2x = \left ( \frac{2}{3} \right )^{2} - \left ( \frac{\sqrt{5}}{3} \right )^{2} = \frac{4}{9} - \frac{5}{9}

\boldsymbol{{{\color{Purple} \cos 2x = -\frac{1}{9} }}}

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(a)

Reading from the graph we can see, that the maximum speed of the object is:

\boldsymbol{{\color{Purple} 5\:\textbf{m\,s}^{-1} }}

(b)

Acceleration is represented by the gradient of the velocity-time graph in the diagram.

(b) (i)

At t = 6 seconds the graph is horizontal, therefore:

\boldsymbol{{\color{Purple} \textbf{Acceleration} = 0}}

(b) (ii)

Using the idea that \textup{gradient} = \frac{\textup{rise}}{\textup{run}}, we get that:

\textup{gradient} = \frac{-3}{1} = -3, hence:

\boldsymbol{{\color{Purple} \textbf{Acceleration} = -3\: \textbf{m\,s}^{-2}}}

(c)

Distance travelled is represented by the area under the graph.

Using triangles and squares to find the area under the graph between t = 0 and t = 6, we get that:

\textup{Distance travelled} = \frac{3 \times 2}{2} + 2 \times 2 + \frac{2 \times 3}{2} + 1 \times 5 = 15

Therefore:

\boldsymbol{{\color{Purple} \textbf{Distance travelled} = 15\:\textbf{m}}}

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(a)

f(4) = 2(4)^{2} - 1

\boldsymbol{{\color{Purple} f(4) = 31}}

(b)

(g\circ f)(x) = g(f(x)) = (2x^{2} - 1) + 4

\boldsymbol{{\color{Purple} (g\circ f)(x) = 2x^{2} + 3}}

(c)

Using our answer from part (b) we get that:

2x^{2} + 3 = 21

x^{2} = 9

\boldsymbol{{\color{Purple} x = \pm 3}}

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(a) (i)

The median is represented by the vertical line inside the box, therefore:

\boldsymbol{{\color{Purple} \textbf{Median} = 15 }}

(a) (ii)

The lower quartile (Q_{1}) is represented by the lower end of the box, therefore:

\boldsymbol{{\color{Purple} Q_{1} = 14 }}

(b)

Using the interquartile range formula, we can write that:

\textup{IQR} = Q_{3} - Q_{1} = 17 - 14

\boldsymbol{{\color{Purple} \textbf{IQR} = 3 }}

(c)

To test for outliers at the higher end of the data set, we need to find:

Q_{3} + 1.5 \times \textup{IQR} = 17 + 1.5 \times 3 = 21.5

\boldsymbol{{\color{Purple} 21 < 21.5,\,  \textbf{so}\:  21 \: \textbf{is not an outlier.} }}

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(a)

Using the cosine rule we can write that:

\textup{PQ}^{2} = 8^{2} + 13^{2} - 2(8)(13)\cos55^{\circ} \approx 113.696

\boldsymbol{{\color{Purple} \textbf{PQ} \approx 10.7\: \textbf{m}}}

(b)

Using the triangle area formula we can write that:

A_{\textup{PQR}} = \frac{1}{2}(8)(13)\sin55^{\circ}

\boldsymbol{{\color{Purple} A_{\textbf{PQR}} \approx 42.6\: \textbf{m}^{2}}}

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(a)

d = 5 - 2

{\boldsymbol{\color{Purple} d = 3}}

(b)

Using the nth term of an arithmetic sequence formula, we get that:

u_{11} = 2 + (11-1)(3) = 2 + 30

\boldsymbol{{\color{Purple} u_{11} = 32}}

(c)

Using the sum of n terms of an arithmetic sequence formula (the second version in the Formula Booklet), we get that:

S_{11} = \frac{11}{2}\left ( 2 + 32 \right ) = 11 \times 17

\boldsymbol{{\color{Purple} S_{11} = 187}}

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(a)

Rewriting the expression for y using negative exponents, we get that:

y = 5x^{2} + x - 4 + 2x^{-1} - 7x^{-3}

Using the derivative of x^{n} rule, we get that:

\frac{dy}{dx} = 2 \times 5x^{2-1} + x^{1-1} + (-1) \times 2x^{-1 - 1} - (-3) \times 7x^{-3-1}

\boldsymbol{{\color{Purple} \frac{\boldsymbol{dy}}{\boldsymbol{dx}} = 10x + 1 - 2x^{-2} + 21x^{-4}(=10x + 1 -\frac{2}{x^{2}} + \frac{21}{x^{4}})}}

(b)

To find the gradient of the tangent to the graph of f at x = -1, we need to calculate f'(-1).

f'(-1) = 10(-1) + 1 -\frac{2}{(-1)^{2}} + \frac{21}{(-1)^{4}} = -10 + 1 - 2 + 21 = 10

To find the y-coordinate of the point where x = -1, we need to calculate f(-1).

f(-1) = 5(-1)^{2} + (-1) - 4 + \frac{2}{(-1)} - \frac{7}{(-1)^{3}} = 5 - 1 - 4 - 2 + 7 = 5

So using the equation of a straight line formula y - y_{1} = m(x - x_{1}), we get that:

y - 5 = 10(x - (-1))

\boldsymbol{{\color{Purple} y - 5 = 10(x + 1)}}

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(a) (i)

The vertical asymptote is at the value that is not part of the domain of f, so the equation of the vertical asymptote is:

\boldsymbol{{\color{Purple} x = -2}}

(a) (ii)

The horizontal asymptote is at the quotient of the x-coefficients, so we get:

y = \frac{2}{1}

\boldsymbol{{\color{Purple} y = 2}}

(b) (i)

To find the x-intercept, we can substitute zero for f(x) and get that:

\frac{2x - 1}{x + 2} = 0

2x - 1 = 0

x = \frac{1}{2}

Therefore the coordinates of the x-intercept are:

\boldsymbol{{\color{Purple} \left ( \frac{1}{2},0 \right )}}

(b) (ii)

To find the y-intercept, we can substitute zero for x and get that:

f(0) = \frac{2(0) - 1}{(0) + 2} = -\frac{1}{2}

Therefore the coordinates of the y-intercept are:

\boldsymbol{{\color{Purple} \left ( 0,-\frac{1}{2} \right )}}

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(a)

a = \textup{P}((X\cup Y)')

\boldsymbol{{\color{Purple} a = 0.2}}

(b)

\textup{P}(X\cup  Y) = 1 - \textup{P}((X\cup  Y)') = 1 - 0.2 = 0.8

Using the combined events formula, we can write that:

\textup{P}(X\cup  Y) = \textup{P}(X) + \textup{P}(Y) - \textup{P}(X\cap Y)

0.8 = 0.6 + 0.3 - \textup{P}(X\cap Y)

\textup{P}(X\cap Y) = 0.1\Rightarrow c = 0.1

b = \textup{P}(X) - \textup{P}(X\cap Y) = 0.6 - 0.1 = 0.5

d = \textup{P}(Y) - \textup{P}(X\cap Y) = 0.3 - 0.1 = 0.2

Therefore:

\boldsymbol{{\color{Purple} b = 0.5\, ,\, c = 0.1\, ,\, d = 0.2}}

(c) (i)

\textup{P}(X') = 1 - \textup{P}(X)  = 1 - 0.6

\boldsymbol{{\color{Purple} \textbf{P}(X') = 0.4}}

(c) (ii)

\textup{P}((X\cap Y)') = a + b + d = 0.2 + 0.5 + 0.2

\boldsymbol{{\color{Purple} \textbf{P}((X\cap Y)') = 0.9 }}

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