Math AA SL

(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}}

.

Using Pascal’s triangle or the Binomial Theorem formula:

(x + 2)^{3} = x^{3} + (3)(x^{2})(2) + (3)(x)(2^{2}) + 2^{3} =

\boldsymbol{{\color{Purple} = x^{3} + 6x^{2} + 12x + 8}}

.

The sum of the integers is:

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

Therefore the sum of these integers is even.

.

(a)

The base and the number inside the logarithm are the same, so:

\boldsymbol{{\color{Purple} \log_{4}4 = 1}}

(b)

Using logarithmic laws we get that:

\log_{4}2 + \log_{4}32 = \log_{4}(2 \times 32) = \log_{4}64

Rewriting the expression and using logarithmic laws, we get:

\log_{4}64 = \log_{4}4^{3} = 3\log_{4}4, therefore:

\boldsymbol{{\color{Purple} \log_{4}2 + \log_{4}32 = 3}}

(c)

Using logarithmic laws we get that:

\log_{4}48 - \log_{4}3 = \log_{4}(\frac{48}{3}) = \log_{4}16

Rewriting the expression and using logarithmic laws, we get:

\log_{4}16 = \log_{4}4^{2} = 2\log_{4}4, therefore:

\boldsymbol{{\color{Purple} \log_{4}48 - \log_{4}3 = 2}}

.

(1 - 2p)^{4} = 1^{4} + 4(1)^{3}(-2p)^{1} + 6(1)^{2}(-2p)^{2} + 4(1)^{1}(-2p)^{3} + (-2p)^{4}

\boldsymbol{{\color{Purple} 16p^{4} - 32p^{3} + 24p^{2} - 8p + 1}}

.

(a)

Using logarithmic laws we can write that:

\ln40 = \ln(2 \times 20) = \ln2 + \ln20 =

\boldsymbol{{\color{Purple} = x + y}}

(b)

Using logarithmic laws we can write that:

\ln10 = \ln(\frac{20}{2}) = \ln20 - \ln2 =

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

(c)

Using logarithmic laws we can write that:

\ln80 = \ln(4 \times 20) = \ln(2^{2} \times 20) = \ln2^{2} + \ln20) = 2\ln2 + \ln20) =

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

.

To express the product of the sum and the difference of these integers we can write:

(a + b)(a - b)

Working further and simplifying, we get:

(a + b)(a - b) = a^{2} + ab - ab - b^{2} = a^{2} -  b^{2}

Therefore the product of the sum and the difference of these integers is equal to the difference of their squares.

.

(a)

The number of terms is one more than the expoinent in the expansion, so:

\boldsymbol{{\color{Purple} 7\: \textbf{terms}}}

(b)

Using the general term of the Binomial Theorem formula, for the x^{4} term we get:

^{6}C_{2} \times x^{6 - 2} \times 2^{2}

Using the formula for ^{n}C_{r}, we get that:

^{6}C_{2} = \frac{6!}{2!\times (6-2)!} = \frac{6 \times 5 \times 4!}{2!\times 4!} = \frac{6 \times 5}{2} = 15

So for the x^{4} term we get:

15 \times x^4 \times 4 = 60x^{4}

Therefore the coefficient of the x^{4} term is:

\boldsymbol{{\color{Purple} 60}}

.

Rearranging and using logarithmic laws, we get:

\ln x^{3} - \ln8 = 12

\ln \left ( \frac{x^{3}}{8} \right ) = 12

Changing the equation from logarithmic to exponential form, we get:

\frac{x^{3}}{8} = e^{12}

Solving for x gives us that:

x = \sqrt[3]{8e^{12}}, hence:

\boldsymbol{{\color{Purple} x = 2e^{4}}}

.

The sum of the squares of the integers is:

n^{2} + (n + 1)^{2} = n^{2} + n^{2} + 2n + 1 =

= 2n^{2} + 2n + 1 = 2(n^{2} + n) + 1

Therefore the sum of these integers is odd.

.

(a)

15 = \frac{6}{2} (u_{1} + (-5))

15 = 3u_{1} -15

\boldsymbol{{\color{Purple} u_{1} = 10}}

(b)

Finding the common difference:

-5 = 10 + (6-1)d

d = -3

Finding the tenth term:

u_{10} = 10 + (10 - 1)(-3)

\boldsymbol{{\color{Purple} u_{10} = -17}}

.

(a)

Using logarithmic laws, we get:

-2\ln\left ( \frac{1}{3} \right ) + 3\ln2 = \ln\left ( \frac{1}{3} \right )^{-2} + \ln2^{3} = \ln9 + \ln8, therefore:

\boldsymbol{{\color{Purple} -2\ln\left ( \frac{1}{3} \right ) + 3\ln2 = \ln72}}

(b)

Using our answer from part (a) we can get that:

\ln72= \ln(4x + 8)

72= 4x + 8

\boldsymbol{{\color{Purple} x = 16}}

.

Using the nth term of an arithmetic sequence formula we can write that:

u_{5} = u_{1} + (5-1)d

\log_{2}81 = 8 + 4d

Using logarithmic laws and simplifying we get that:

\log_{2}3^{4} = 8 + 4d

4\log_{2}3 = 8 + 4d

d = \log_{2}3 - 2

d = \log_{2}3 - 2\log_{2}2

d = \log_{2}3 - \log_{2}2^{2}

d = \log_{2}3 - \log_{2}4

\boldsymbol{{\color{Purple} d = \log_{2}\left ( \frac{3}{4} \right )}}

.

(a)

Finding the sum of the integers:

n + n + 1 + n + 2 + n + 3 = 4n + 6

Factorising we get 2(2n + 3), therefore the sum of these four integers is divisible by 2, so it is an even number.

(b)

Finding the sum of the squares:

n^{2} + (n + 1)^{2} + (n + 2)^{2} + (n + 3)^{2}

n^{2} + (n^{2} + 2n + 1) + (n^{2} + 4n + 4) + (n^{2} + 6n + 9)

4n^{2} + 12n + 14

4n^{2} + 12n + 12 + 2 = 4(n^{2} + 3n + 3) + 2

Since 4(n^{2} + 3n + 3) is divisible by 4, 4(n^{2} + 3n + 3) + 2 is not divisible by 4.

.

Sum of squares:

n^{2} + (n+2)^{2} = n^{2} + n^{2} + 4n + 4

Collecting terms and factorising 2:

2(n^{2} + 2n + 2)

Therefore the sum of the squares is divisible by two, hence it is an even number.

.

.(a) (i)

d = \frac{u_{5} - u_{3}}{2} = \frac{14-8}{2}

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

(a) (ii)

u_{1} = u_{3} - 2d = 8 - 2(3)

\boldsymbol{{\color{Purple} u_{1} = 2}}

(b)

Using the nth term of an arithmetic sequence formula:

35 = 2 + (n - 1)(3)

35 = 3n - 1

\boldsymbol{{\color{Purple} n = 12}}

.

Using the binomial expansion formula:

^{8}\textup{C}_{2} \times (2x)^{2} \times a^{6} = 224x^{2}

Calculating ^{8}\textup{C}_{2}:

\frac{8!}{2!(8-2)!} = \frac{8 \times 7 \times 6!}{2! \times 6!} = \frac{56}{2} = 28

So we get:

28 \times 4x^{2} \times a^{6} = 224x^{2}

a^{6} = 2

\boldsymbol{{\color{Purple} a = \pm \sqrt[6]{2}}}

.

(a)

Using exponential and logarithmic laws we get:

27^{\log_{3}5} = \left ( 3^{3} \right )^{\log_{3}5} =

= 3^{3\log_{3}5} = 3^{\log_{3}5^{3}} = 3^{\log_{3}125} =

\boldsymbol{{\color{Purple} = 125}}

(b)

Using logarithmic laws and rearranging the terms we get:

\log_{6}144 + \log_{6}6 - 2\log_{6}2 = \log_{6}144 - \log_{6}2^{2} + \log_{6}6 =

= \log_{6}144 - \log_{6}4 + 1 = \log_{6}\left ( \frac{144}{4} \right ) + 1 =

=\log_{6}36 + 1 = \log_{6}6^{2} + 1 = 2\log_{6}6 + 1 = 2 + 1 =

\boldsymbol{{\color{Purple} 3}}

.

Let’s represent the integers as n and n + 1

For the difference of the squares of the integers we get: 

(n + 1)^{2} - n^{2} = (n^{2} + 2n + 1) - n^{2} =

= n^{2}  + 2n + 1 - n^{2} = 2n + 1

For the sum of the integers we get:

n + (n + 1) = 2n + 1

Since 2n + 1 = 2n + 1, the difference of the squares of two consecutive positive integers is equal to the sum of the integers.

.

Using logarithmic laws we get:

\log_{2}(2x - 3) + \log_{2}(x + 5) = \log_{2}(3x - 3) + 2

\log_{2}((2x - 3)(x + 5)) = \log_{2}(3x - 3) + 2\log_{2}2

\log_{2}(2x^{2} + 7x - 15) = \log_{2}(3x - 3) + \log_{2}4

\log_{2}(2x^{2} + 7x - 15) = \log_{2}(4(3x - 3))

2x^{2} + 7x - 15 = 4(3x - 3)

Expanding, rearranging for zero and factorizing we get:

2x^{2} - 5x - 3 = 0

(2x + 1)(x - 3) = 0, therefore x = -\frac{1}{2} or x = 3

Checking the original equation we can see that if x = -\frac{1}{2}, then 2x - 3 < 0 and 3x - 3 < 0, hence:

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

.

Finding the common denominator and simplifying:

\frac{1}{\sqrt{x} - 1} - \frac{1}{\sqrt{x} + 1} = \frac{1}{\sqrt{x} - 1} \times \frac{\sqrt{x} + 1}{\sqrt{x} + 1} - \frac{1}{\sqrt{x} + 1} \times \frac{\sqrt{x} - 1}{\sqrt{x} - 1} =

= \frac{\sqrt{x} + 1}{(\sqrt{x} + 1)(\sqrt{x} - 1)} - \frac{\sqrt{x} - 1}{(\sqrt{x} + 1)(\sqrt{x} - 1)} = \frac{\sqrt{x} + 1}{x - 1} - \frac{\sqrt{x} - 1}{x - 1} =

= \frac{\sqrt{x} + 1 - (\sqrt{x} - 1)}{x - 1} = \frac{2}{x - 1}

.

Using exponent laws and rearranging we get that:

\left ( 2^{2} \right )^{x} = 3 - 2^{2} \times 2^{x}

\left ( 2^{x} \right )^{2} + 4 \times 2^{x} - 3 = 0

Using the substitution a = 2^{x}, we get:

a^{2} + 4a - 3 = 0

Using the quadratic formula to solve for a:

a = \frac{-4 \pm \sqrt{4^{2} - 4(1)(-3)}}{2(1)} = \frac{-4 \pm \sqrt{28}}{2} = \frac{-4 \pm 2\sqrt{7}}{2} = -2 \pm \sqrt{7}

Since 2^{x} > 0, we get that:

2^{x} = -2 + \sqrt{7}

Changing from an exponential to a logarithmic form:

\boldsymbol{{\color{Purple} x = \log_{2}(-2 + \sqrt{7})}}

.

Let’s represent the integers as n, n + 1, and n + 2

(a)

The sum of the integers is:

n + (n + 1) + (n + 2) = 3n + 3

Factorising, we get that the sum is 3(n + 1), therefore the sum of three consecutive integers is divisible by 3.

(b)

The sum of the squares of the integers is:

n^{2} + (n + 1)^{2} + (n + 2)^{2} =

= n^{2} + (n^{2} + 2n + 1) + (n^{2} + 4n + 4)

= 3n^{2} + 6n + 5

Factorising 3 from the first two terms we get:

3(n^{2} + 2n) + 5

Since 3(n^{2} + 2n) is divisible by 3 while 5 is not divisible by 3, the sum of the squares of three consecutive integers is never divisible by 3.

.

Let’s represent the integer as n.

Based on the information given in the question we can write:

\left ( n^{2} + \frac{1}{n} \right ) \times n  - 1= n^{3} + 1 - 1 = n^{3}

Since n is an even integer so divisible by 2, n^{3} is divisible by 8.

.

Using the nth term of a geometric sequence formula we get that:

u_{2} = 5r and u_{3} = 5r^{2}

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

v_{4} = 5 + (4-1)d = 5 + 3d and v_{16} = 5 + (16-1)d = 5 + 15d

Since u_{2} = v_{4} and u_{3} = v_{16}, we can write that:

5r = 5 + 3d and 5r^{2} = 5 + 15d

Multiplying the first equation by 5 and subtracting the result from the second equation gives us:

5r^{2} - 25r = -20

Rearranging for zero and dividing by 5 we get that:

r^{2} - 5r + 4 = 0

Solving for r we get that:

(r - 4)(r - 1) = 0, and since r \neq 1

\boldsymbol{{\color{Purple} r = 4}}

Substituting 4 for r into 5r = 5 + 3d we get:

5(4) = 5 + 3d and solving for d we get that:

\boldsymbol{{\color{Purple} d = 5}}

.

(a) (i)

d = u_{2} - u_{1} = u_{3} - u_{2}

k\textup{log}_{a}x - 2\textup{log}_{a}x =  \textup{log}_{a}x - k\textup{log}_{a}x

k - 2 = 1 - k, therefore:

\boldsymbol{{\color{Purple} k = \frac{3}{2}}}

(a) (ii)

d = u_{2} - u_{1} = \frac{3}{2}\textup{log}_{a}x - 2\textup{log}_{a}x = -\frac{1}{2}\textup{log}_{a}x

\boldsymbol{{\color{Purple} m = -\frac{1}{2}}}

(a) (iii)

Substituting into the sum of the first n terms of an arithmetic sequence formula:

4 = \frac{8}{2}\left ( 2 \times 2\textup{log}_{\sqrt{3}}\, x + (8-1)\left ( -\frac{1}{2}\textup{log}_{\sqrt{3}}\, x \right ) \right )

Simplifying and solving for x:

1 = 4\textup{log}_{\sqrt{3}}\, x - 3.5\textup{log}_{\sqrt{3}}\, x

\textup{log}_{\sqrt{3}}\, x = 2

x = (\sqrt{3})^{2}, so:

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

(b) (i)

r = \frac{u_{2}}{u_{1}} = \frac{u_{3}}{u_{2}}

\frac{k\textup{log}_{a}x}{2\textup{log}_{a}x} = \frac{\textup{log}_{a}x}{k\textup{log}_{a}x}

k^{2} = 2 \Rightarrow k = \pm \sqrt{2}

\boldsymbol{{\color{Purple} k = -\sqrt{2}}}

(b) (ii)

r = \frac{\textup{log}_{a}x}{-\sqrt{2}\textup{log}_{a}x} = -\frac{1}{\sqrt{2}}

Using the sum of an infinite geometric sequence formula:

\frac{5 \sqrt{2}}{1 + \sqrt{2}} = \frac{2\textup{log}_{a}x}{1 - \left ( -\frac{1}{\sqrt{2}} \right )}

Simplifying and multiplying the fraction on the right-hand side by \frac{\sqrt{2}}{\sqrt{2}}:

\frac{5 \sqrt{2}}{1 + \sqrt{2}} = \frac{2 \sqrt{2}\textup{log}_{a}x}{1 + \sqrt{2}}

\textup{log}_{a}x = \frac{5}{2}\Rightarrow x = a^{\frac{5}{2}}

\boldsymbol{{\color{Purple} a = x^{\frac{2}{5}}}}

.

Finding expressions for p and for q:

p = ^{5}\textrm{C}_{1} \times (2x)^{4} \times k = 80kx^{4}

q = ^{5}\textrm{C}_{2} \times (2x)^{3} \times k^{2} = \frac{5!}{2! \times 3!} \times (2x)^{3} \times k^{2} =80k^{2}x^{3}

q - p = 80k^{2} - 80k = 480

k^{2} - k - 6 = 0\Rightarrow (k - 3)(k + 2) = 0

\boldsymbol{{\color{Purple} k = 3}} or \boldsymbol{{\color{Purple} k = - 2}}

.

Let’s represent the integers as n, n + 1, n + 2, and n + 3

The mean of the integers can be written as:

\frac{n +(n + 1)+(n + 2)+(n + 3)}{4} = \frac{4n + 6}{4} = \frac{2n + 3}{2}

The median of the four integers is:

\frac{(n + 1)+(n + 2)}{2} = \frac{2n + 3}{2}

Since \frac{2n + 3}{2} = \frac{2n + 3}{2}, the mean of four consecutive positive integers is equal to the median of these integers.

.

(a)

Using the concept of the common difference:

d = u_{2} - u_{1} = u_{3} - u_{2} = u_{4} - u_{3}

\textup{log}_{b}m - \textup{log}_{b}32 = \textup{log}_{b}n - \textup{log}_{b}m = \textup{log}_{b}108 - \textup{log}_{b}n

Using the logarithmic law \textup{log}_{a}\frac{x}{y} = \textup{log}_{a}x - \textup{log}_{a}y:

\textup{log}_{b}\frac{m}{32} = \textup{log}_{b}\frac{n}{m} = \textup{log}_{b}\frac{108}{n}

Therefore \frac{m}{32} = \frac{n}{m} = \frac{108}{n} = r so 32\, , \,m\, , \,n and 108 are four consecutive terms of a geometric sequence.

(b)

Using the nth term of a geometric sequence formula:

u_{4} = u_{1} \times r^{3}

108 = 32 \times r^{3}\Rightarrow r = \frac{3}{2}

Based on what we found in part (a):

\frac{m}{32} = \frac{3}{2}\Rightarrow \boldsymbol{{\color{Purple} m = 48}}

\frac{108}{n} = \frac{3}{2}\Rightarrow \boldsymbol{{\color{Purple} n = 72}}

.

The information that there is exactly one point on the graph of g where x = 2 means that the equation g(x) = 2 has exactly one solution.

So we will start by writing that:

\log_{(p-2)}(x^{2} - 4x + 13) = 2

Converting from logarithmic to exponential form and working further we get that:

(p - 2)^{2} = x^{2} - 4x + 13

p^{2} - 4p + 4 = x^{2} - 4x + 13

x^{2} - 4x - p^{2} + 4p + 9 = 0

This equation still has exactly one solution, therefore its discriminant is equal to zero.

Hence we get that:

(-4)^{2} - (4)(1)(- p^{2} + 4p + 9) = 0

p^{2} - 4p - 5 = 0

(p + 1)(p - 5) = 0

p = -1 or p = 5

Since p>2, the final answer is:

\boldsymbol{{\color{Purple} p = 5}}

.

(a)

d = u_{2} - u_{1} = \ln(2x^{5}) - \ln(x^{5}) = \ln\left ( \frac{2x^{5}}{x^{5}} \right )

\boldsymbol{{\color{Purple} \textbf{Therefore} \, d = \ln2.}}

(b)

Using the nth term of an arithmetic sequence formula we can write that:

u_{6} = u_{1} + (n-1)d

10 = \ln(a^{5}) + (6-1)(\ln2)

10 = \ln(a^{5}) + 5\ln2

10 = \ln(a^{5}) + \ln2^{5}

10 = \ln(2^{5} \times a^{5})

Changing from logarithmic to exponential form:

2^{5} \times a^{5} = e^{10}

a^{5} = \frac{e^{10}}{2^{5}}

\boldsymbol{{\color{Purple} a= \frac{e^{2}}{2}}}

(c)

Let’s begin by finding the common ratio:

r = \frac{u_{2}}{u_{1}} = \frac{\frac{\ln(2x + 7)}{5}}{\ln(2x + 7)}

r = \frac{1}{5}

{\color{Purple} \textbf{Since} \: \boldsymbol{\left | r \right | < 1,} \: \textbf{the sum to infinity exists.}}

(d)

Using the sum of an infinite geometric sequence formula we get that:

S_{\infty} = \frac{\ln(2x + 7)}{1 - \frac{1}{5}} = \frac{\ln(2x + 7)}{\frac{4}{5}} = \frac{5\ln(2x + 7)}{4}

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

u_{11} = \ln(x^{5}) + (11-1)(\ln2) = \ln(x^{5}) + 10\ln2 = \ln(2^{10} \times x^{5})

The 11th term of the arithmetic sequence is four times the sum to infinity of the geometric sequence, so based on the two expressions that we found above we can write that:

4 \times \frac{5\ln(2x + 7)}{4} = \ln(2^{10} \times x^{5})

Working further we get that:

5\ln(2x + 7) = \ln(2^{10} \times x^{5})

\ln(2x + 7)^{5} = \ln(2^{10} \times x^{5})

(2x + 7)^{5} = 2^{10} \times x^{5}

2x + 7 = 2^{2} \times x

\boldsymbol{{\color{Purple} x = \frac{7}{2}}}

.