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Proof, EXT2 P1 2024 HSC 14d

The following argument attempts to prove that  \(0=1\).

  1. We evaluate \(\displaystyle\int \frac{1}{x}\, d x\) using the method of integration by parts.
  \(\displaystyle \int \frac{1}{x}\,d x\) \(=\displaystyle \int \frac{1}{x} \times 1\, d x\)
    \(=\displaystyle\frac{1}{x} \times x-\int-\frac{1}{x^2} x\, d x\)
    \(=1+\displaystyle\int \frac{1}{x}\, d x\)
  1.  
    We may now subtract \(\displaystyle \int \frac{1}{x}\,d x\) from both sides to show that  \(0=1\).

Explain what is wrong with this argument.  (2 marks)

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\(\text {Consider the line in proof }\)

\(\displaystyle {\int \frac{1}{x}\, d x=1+\int \frac{1}{x}\, d x}\)

 \(\Rightarrow \text{ Each integral will have its own constant }\)

\(\Rightarrow \text{If constants are \(c_1\) and \(c_2, \abs{c_1-c_2}=1\) in this proof.}\)

\(\Rightarrow \text{Subtracting}\ \displaystyle \int \frac{1}{x}\,d x\ \text{from both sides invalidates the proof.}\)

Show Worked Solution

\(\text {Consider the line in proof }\)

\(\displaystyle {\int \frac{1}{x}\, d x=1+\int \frac{1}{x}\, d x}\)

♦♦ Mean mark 34%.

\(\Rightarrow \text{ Each integral will have its own constant }\)

\(\Rightarrow \text{If constants are \(c_1\) and \(c_2, \abs{c_1-c_2}=1\) in this proof.}\)

\(\Rightarrow \text{Subtracting}\ \displaystyle \int \frac{1}{x}\,d x\ \text{from both sides invalidates the proof.}\)

Proof, EXT2 P1 2024 HSC 14a

Prove that if \(a\) is any odd integer, then  \(a^2-1\)  is divisible by 8.   (2 marks) 

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\(\text {Prove if  \(a\) is odd, \(a^2-1\) is divisible by 8.}\)

\(\text {Let}\ \ a=2 n+1, \ \ n \in \mathbb{Z}\)

  \(a^2-1\) \(=(2 n+1)^2-1\)
    \(=4 n^2+4 n+1-1\)
    \(=4 n(n+1)\)

 
\(\text{If \(n\) is odd (\(n=2 k+1, k \in Z)\):}\)

\(a^2-1=4(2 k+1)(2 k+2)=8(2 k+1)(k+1) /8\)

\(\text{If \(n\) is even \((n=2 k)\):}\)

\(a^2-1=4(2 k)(2 k+1)=8 k(2 k+1) /8\)

\(\therefore \text{ If \(a\) is odd, \(a^2-1\) is divisible by 8}\)

Show Worked Solution

\(\text {Prove if  \(a\) is odd, \(a^2-1\) is divisible by 8.}\)

\(\text {Let}\ \ a=2 n+1, \ \ n \in \mathbb{Z}\)

  \(a^2-1\) \(=(2 n+1)^2-1\)
    \(=4 n^2+4 n+1-1\)
    \(=4 n(n+1)\)

 
\(\text{If \(n\) is odd (\(n=2 k+1, k \in Z)\):}\)

\(a^2-1=4(2 k+1)(2 k+2)=8(2 k+1)(k+1) /8\)

\(\text{If \(n\) is even \((n=2 k)\):}\)

\(a^2-1=4(2 k)(2 k+1)=8 k(2 k+1) /8\)

\(\therefore \text{ If \(a\) is odd, \(a^2-1\) is divisible by 8}\)

Proof, EXT2 P1 2022 HSC 16c

It is given that for positive numbers `x_(1),x_(2),x_(3),dots,x_(n)` with arithmetic mean `A`,
 

               `(x_(1)xxx_(2)xxx_(3)xx cdots xxx_(n))/(A^(n)) <= 1`    (Do NOT prove this.)

Suppose a rectangular prism has dimensions  `a,b,c`  and surface area `S`.

  1. Show that  `abc <= ((S)/(6))^((3)/(2))`.  (2 marks)

  2. Using part (i), show that when the rectangular prism with surface area `S` is a cube, it has maximum volume.  (2 marks)

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  1. `text{Proof (See Worked Solutions)}`
  2. `text{Proof (See Worked Solutions)}`
Show Worked Solution

i.    `text{Show}\ \ abc <= ((S)/(6))^((3)/(2))`

`S=2(ab+bc+ac)`

`A=(ab+bc+ac)/3`

`text{Using given relationship, where}\ x_1=ab, x_2=bc, …`


♦♦♦ Mean mark (i) 26%.
`(ab xx bc xx ca)/((ab+bc+ac)/3)^3` `<=1`  
`(ab xx bc xx ca)` `<=((ab+bc+ac)/3)^3`  
`(abc)^2` `<=((2(ab+bc+ac))/6)^3`  
`abc` `<=(S/6)^(3/2)`  

 

ii.   `text{If prism is a cube}\ \ a=b=c`

`=> V=a^3, \ \ S=6a^2`

`(S/6)^(3/2)=((6a^2)/6)^(3/2)=a^3`
 

`text{In the case of a cube:}`

`V=(S/6)^(3/2)`


♦♦♦ Mean mark (ii) 26%.

`text{Also,}\ \ V<=(S/6)^(3/2)\ \ \ text{(using part (i))}`

`:.\ text{For rectangular prisms with a given surface area, a cube}`

`text{has the maximum volume.}`

Proof, EXT2 P1 2022 HSC 2 MC

The following proof aims to establish that  – 4 = 0

`text{Let}` `a=-4`    
`=>` `a^2 = 16 \ text{and} ` `\ 4a + 4 = -12` `text{Line 1}`
`=>` `a^2 + 4a + 4 =` `4` `text{Line 2}`
`=>` `(a + 2)^2 =` `2^2` `text{Line 3}`
`=>` `a + 2 =` `2` `text{Line 4}`
`=>` `a =` `0`  

 
At which line is the implication incorrect?

  1. Line 1
  2. Line 2
  3. Line 3
  4. Line 4
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`D`

Show Worked Solution

`text{Consider Line 4:}`

`text{Given}\ \ (a + 2)^2=2^2\ \ =>\ \ a+2=+-2`

`=>D`

Proof, EXT2 P1 2021 HSC 15d

Prove that  `2^n + 3^n ≠ 5^n`  for all integers  `n ≥ 2`.  (2 marks)

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`text{See Worked Solution}`

Show Worked Solution

`text(Expanding)\ \ (2 + 3)^n:`

Mean mark 51%.
`5^n` `= (2 + 3)^n`
  `= 2^n +\ ^n C_1 *2^{n-1} *3 + \ … \ + \ ^n C_{n-1} *2 * 3^{n-1} + 3^n`

 

`text{S} text{ince} \ \ ^n C_1 * 2^{n-1} *3 + \ … \ + \ ^n C_{n-1} * 2 * 3^{n-1} > 0 \ \ text{for} \ n ≥ 2`
 

`=> 5^n > 2^n + 3^n \ \ text{for} \ n ≥ 2`

`:. 5^n ≠ 2^n + 3^n \ \ text{for} \ n ≥ 2`

Proof, EXT2 P1 2021 HSC 15b

For integers  `n ≥ 1`, the triangular numbers  `t_n`  are defined by  `t_n = (n(n + 1))/2`,  giving  `t_1 = 1, t_2 = 3, t_3 = 6, t_4 = 10`  and so on.

For integers  `n >= 1`,  the hexagonal numbers  `h_n`  are defined by  `h_n = 2n^2-n`,  giving  `h_1 = 1, h_2 = 6, h_3 = 15, h_4 = 28`  and so on.

  1. Show that the triangular numbers  `t_1, t_3 , t_5`, and so on, are also hexagonal numbers.  (2 marks)

  2. Show that the triangular numbers  `t_2, t_4 , t_6`, and so on, are not hexagonal numbers.  (1 mark)

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  1. `text(See Worked Solution)`
  2. `text(See Worked Solution)`
Show Worked Solution

i.   `t_n = (n(n + 1))/2`

♦ Mean mark part (i) 50%.

`text(Odd triangular numbers:)`

`→ \ text(Let)\ \ n = 2k-1\ \ text(for integers)\ \ k >= 1`

`t_(2k – 1)` `= ((2k-1)(2k-1 + 1))/2`
  `= (2k(2k-1))/2`
  `= 2k^2-k\ \ text{(definition of hexagonal numbers)}`

 
`:. t_1, t_3\ …\ text(are also hexagonal numbers.)`

 

ii.   `t_2, t_4, t_6, …`

♦ Mean mark part (ii) 1%!

`→\ text(Let)\ \ n = 2k\ \ text(for integers)\ \ k > = 1`

`t_(2k)` `= (2k(2k + 1))/2`
  `= 2k^2 + k`

 
`text(Find values for)\ \ k, n\ \ text(that satisfy:)`

`2k^2 + k` `= 2n^2 – n`
`2k^2-2n^2 + k + n` `= 0`
`2(k-n)(k + n) + (k + n)` `= 0`
`(k + n)(2k-2n + 1)` `= 0`

 
`k = -n =>\ text(no solution)\ (k, n >= 1)`

`2k = 2n-1 =>\ text(no solution)\ \ (2k\ \ text(is even,)\ \ 2n-1\ \ text{is odd)}`

`:. t_2, t_4, t_6, …\ text(are not hexagonal numbers.)`

Proof, EXT2 P1 2021 HSC 15a

For all non-negative real numbers `x` and `y, \ sqrt(xy) <= (x + y)/2`.  (Do NOT prove this.)

  1. Using this fact, show that for all non-negative real numbers `a`, `b` and `c`,
  2.     `sqrt(abc) <= (a^2 + b^2 + 2c)/4`.  (2 marks)

  3. Using part (i), or otherwise, show that for all non-negative real numbers `a`, `b` and `c`,  
  4.     `sqrt(abc) <= (a^2 + b^2 + c^2 + a + b + c)/6`.  (2 marks)

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  1. `text(See Worked Solution)`
  2. `text(See Worked Solution)`
Show Worked Solution

i.   `text(Show)\ \ sqrt(abc) <= (a^2 + b^2 + 2c)/4`

♦ Mean mark part (i) 50%.

`sqrt(abc) = sqrt((ab)c) <= (ab + c)/2\ …\ (1)`
 

`text(Let)\ \ x = a^2\ \ text(and)\ \ y = b^2:`

`sqrt(xy) = sqrt(a^2b^2) ` `<= (a^2 + b^2)/2`
`ab` `<= (a^2 + b^2)/2\ …\ (2)`

 
`text{Substitute (2) into (1):}`

`sqrt(abc) <= ((a^2 + b^2)/2 + c)/2`

`sqrt(abc) <= (a^2 + b^2 + 2c)/4`

 

ii.   `text(Show)\ sqrt(abc) <= (a^2 + b^2 + c^2 + a + b + c)/6`

♦♦♦ Mean mark part (ii) 26%.

`text{Similarly (to part a):}`

`text(If)\ \ x = b^2\ \ text(and)\ \ y = c^2`

  `sqrt(abc) <= (b^2 + c^2 + 2a)/4`

`text(If)\ \ x = c^2\ \ text(and)\ \ y = a^2`

  `sqrt(abc) <= (c^2 + a^2 + 2b)/4`

`:.3sqrt(abc)` `<= (a^2 + b^2 + 2c + b^2 + c^2 + 2a + c^2 + a^2 + 2b)/4`
`3sqrt(abc)` `<= (2(a^2 + b^2 + c^2 + a + b + c))/4`
`sqrt(abc)` `<= (a^2 + b^2 + c^2 + a + b + c)/6`

Proof, EXT2 P1 2018 HSC 15c

Let  `n`  be a positive integer and let  `x`  be a positive real number.

  1.  Show that  `x^n - 1 - n(x - 1) = (x - 1)(1 + x + x^2 + … + x^(n - 1) - n)`.  (1 mark)

  2.  Hence, show that  `x^n >= 1 + n(x - 1)`.  (2 marks)

  3.  Deduce that for positive real numbers `a` and `b`,
     
          `a^nb^(1-n)>=na + (1-n)b`  (2 marks)

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  1. `text(See Worked Solutions)`
  2. `text(See Worked Solutions)`
  3. `text(See Worked Solutions)`
Show Worked Solution
i.    `text(RHS)` `= (x – 1)underbrace{(1 + x + x^2 + … + x^(n – 1) – n)}_{text(GP where)\ \ a = 1,\ r = x}`
    `= (x – 1) ((1(x^n – 1))/(x – 1) – n)`
    `= x^n – 1 – n(x – 1)`
    `=\ text(LHS)`

 

ii.   `text(Let)\ P(x) = (x – 1)(1 + x + x^2 + … + x^(n – 1) – n)`

♦♦♦ Mean mark 11%.

`text(If)\ \ x = 1, P(x) = 0`
 

`text(If)\ \ 0 < x < 1, \ (x – 1) < 0, \ (1 + x + x^2 + … + x^(n – 1) – n) < 0`

`=> P(x) > 0`
 

`text(If)\ \ x > 1, \ (x – 1) > 0, \ (1 + x + … + x^(n – 1) – n) > 0`

`=> P(x) > 0`
 

`x^n – 1 – n(x – 1) >= 0`

`:. x^n >= 1 + n(x – 1)`

 

iii.   `x^n >= 1 + n(x – 1)\ \ text(for)\ \ x ∈ R^+`

♦♦ Mean mark 23%.

`text(S)text(ince)\ a,b ∈ R^+`

`(a/b)^n` `>= 1 + n(a/b – 1)`
`(a^n)/(b^n) xx b` `>= b + na – nb,\ \ \ \ (b > 0)`
`a^n b^(1 – n)` `>= na + (1 – n)b`

Proof, EXT2 P1 2015 HSC 15b

Suppose that  `x >= 0`  and  `n`  is a positive integer.

  1. Show that  `1 - x <= 1/(1 + x) <= 1.`  (2 marks) 

  2. Hence, or otherwise, show that 
     
          `1 - 1/(2n) <= n ln (1 + 1/n) <= 1.`  (2 marks)

  3. Hence, explain why
     
        `lim_(n -> oo) (1 + 1/n)^n = e.`  (1 mark)

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  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `text(Proof)\ \ text{(See Worked Solutions)}`
Show Worked Solution
i.    `1-x^2` `<=1\ \ \ text(for)\ x>=0`
  `(1-x)(1+x)` `<=1`
  `(1-x)` `<=1/(1+x)`

 

`text(S)text(ince)\ \ 1 + x >= 1\ \ text(when)\ \ x >= 0`

`=>1/(1 + x) <= 1`

`:. 1 – x <= 1/(1 + x) <= 1.`

♦♦ Mean mark 21%.
STRATEGY: The conversion of the middle term from a fraction into a logarithm should flag the need for integration of each term.
ii.    `int_0^(1/n) (1 – x)\ dx` `<= int_0^(1/n) (dx)/(1 + x) <= int_0^(1/n) 1\ dx`
  `[x – x^2/2]_0^(1/n)` `<= [ln (1 + x)]_0^(1/n) <= [x]_0^(1/n)`
  `1/n-1/(2n^2)` `<= ln (1 + 1/n) <= 1/n`
  `1 – 1/(2n) ` `<= n ln (1 + 1/n) <= 1,\ \ \ \ \ (n>=1)`

 

 

iii.    `lim_(n -> oo) (1 – 1/(2n))` `<= lim_(n -> oo){n ln (1 + 1/n)} <= lim_(n -> oo) (1)`
  `1` `<= lim_(n -> oo) {l ln (1 + 1/n)} <= 1`
♦♦ Mean mark 29%.
`:. lim_(n -> oo) (n ln (1 + 1/n))`  `=1`
`lim_(n -> oo) ln (1 + 1/n)^n` `=1`
`:.lim_(n -> oo) (1 + 1/n)^n` `=e`