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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.}\)

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

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

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

♦♦ 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 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`

Proof, EXT2 P1 2006 HSC 8a

Suppose  `0 <= t <= 1/sqrt 2.`

  1. Show that  `0 <= (2t^2)/(1 - t^2) <= 4t^2.`   (2 marks)

  2. Hence show that  `0 <= 1/(1 + t) + 1/(1 - t) - 2 <= 4t^2.`   (1 mark)

  3. By integrating the expressions in the inequality in part (ii) with respect to  `t`  from  `t = 0`  to  `t = x\ \  text{(where}\ \ 0 <= x <= 1/sqrt2\ \ text{)}`, show that
     
        `0 <= log_e ((1 + x)/(1 - x)) - 2x <= (4x^3)/3.`  (2 marks)

  4. Hence show that for  `0 <= x <= 1/sqrt 2`
     
        `1 <= ((1 + x)/(1 - x)) e^(-2x) <= e^((4x^3)/3).`  (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)}`
  4. `text(Proof)\ \ text{(See Worked Solutions)}`
Show Worked Solution
i.    `0` `<= t <= 1/sqrt 2`
  `0` `<= t^2 <= 1/2`
  `0` `>= -t^2 >= -1/2`
  `1` `>= 1 – t^2 >= 1/2`
  `1` `<= 1/(1 – t^2) <= 2`
  `2t^2` `<= (2t^2)/(1 – t^2) <= 4t^2,\ \ \ \ (2t^2>0)`
  `:. 0` `<= (2t^2)/(1 – t^2) <= 4t^2`

 

ii.   `1/(1 + t) + 1/(1 – t) – 2`

`=((1-t)+(1+t)-2(1-t^2))/(1-t^2)`

`=(2t^2)/(1-t^2)`

 

`text{Substituting into part (i)}`

`:. 0 <= 1/(1 + t) + 1/(1 – t) – 2 <= 4t^2`

 

iii.   `int_0^x 0\ dt` `<= int_0^x (1/(1 + t) + 1/(1 – t) – 2)\ dt <= int_0^x 4t^2\ dt`
  `0` `<= [log_e (1 + t) – log_e (1 – t) – 2t]_0^x <= [(4t^3)/3]_0^x`
  `0` `<= [log_e (1 + x) – log_e (1 – x) – 2x]<= [(4x^3)/3]`
  `0` `<= log_e ((1 + x)/(1 – x)) – 2x <= (4x^3)/3`

 

iv.   `text(S)text(ince)\ \ e^a>e^b\ \ text(when)\ \ a>b` 

`e^0` `<= e^([ln ((1 + x)/(1 – x)) – 2x]) <= e^((4x^3)/3)`
`1` `<= e^(ln ((1 + x)/(1 – x))) xx e^(-2x) <= e^((4x^3)/3)`
`1` `<= ((1 + x)/(1 – x)) e^(-2x) <= e^((4x^3)/3)`

 

Proof, EXT2 P1 2007 HSC 7a

  1. Show that  `sin x < x`  for  `x > 0.`   (2 marks)

  2. Let  `f(x) = sin x - x + x^3/6`.

     

    Show that the graph of  `y = f(x)`  is concave up for  `x > 0.`   (2 marks)

  3. By considering the first two derivatives of  `f(x)`,show that  `sin x > x - x^3/6`  for  `x > 0.`   (2 marks)

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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.    `text(Let)\ \ g(x)` `=sin x-x`
  `g′(x)` `=cosx-1<=1\ \ \ text(for all)\ x>0`
MARKER’S COMMENT: A geometric proof using arc length and a right angled triangle caused problems as few students could deal with the case  `x>pi`.

 

`=>g(x)\ \ text(is a decreasing function)`

`text(When)\ \ x=0,\ \ g(0)=0`

 

`text(Considering)\ \ g(x)\ \ text(when)\ \ x>0,`

`g(x)` `<0`
`sinx -x` `<0`
`sin x` `<x\ \ \ text(for all)\ x>0`

 

ii. `f(x)` `=sin x – x + x^3/6`
  `f prime (x)` `=cos x – 1 + x^2/2`
  `f ″ (x)` `=x – sin x`
  `:.\ f″ (x)` `> 0\ \ \ \ text{(using part (i))}`

 

`:. f(x)\ \ text(is concave up for)\ \ x > 0.`

 

iii.  `f″(x)>0\ \ \ \ text{(part (ii))}`

`=>f′(x)\ \ text(is an increasing function)`

`text(When)\ \ x=0,\ \ f′(0)=0`

`=>f′(x)>0\ \ \ text(for)\ \ x>0`

 

`:. f(x)\ \ text(has a positive gradient that steepens)`

`text(for)\ \ x>0, and f(0)=0`
 

`f(x)` `>0`
`sin x – x + x^3/6` `>0`

 
`:.sin x > x – x^3/6\ \ \ text(for)\ \ x>0`

Proof, EXT2 P1 2014 HSC 16b

Suppose `n` is a positive integer. 

  1. Show that 
     
        `-x^(2n) ≤ 1/(1 + x^2) − (1 − x^2 + x^4 − x^6 + … + (-1)^(n − 1) x^(2n − 2)) ≤ x^(2n)`.  (3 marks)

  2. Use integration to deduce that
     
    `-1/(2n + 1) ≤ pi/4 − (1 − 1/3 + 1/5 − … + (-1)^(n − 1) 1/(2n − 1)) ≤ 1/(2n + 1)`.  (2 marks)

  3. Explain why   `pi/4 = 1 − 1/3 + 1/5 − 1/7 + …`.  (1 mark)

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

i.   `text(Let)\ \ S_n=1 − x^2 + x^4 − x^6 + … + (-1)^(n − 1) x^(2n − 2)`

`=>text(GP where)\ \ a=1, r=-x^2`

♦♦ Mean mark 12%.
STRATEGY: Applying the GP formula and simplifying the middle term is worth 2 full marks in this question.
`:.S_n` `=(1(1-(-x^2)^n))/(1-(-x^2))`
  `= (1 − (-x^2)^n)/(1 + x^2)`

 

`:.1/(1 + x^2) − (1 − x^2 + x^4 − x^6 + … + (-1)^(n − 1) x^(2n − 2))​`

`=1/(1 + x^2)-(1 − (-x^2)^n)/(1 + x^2)`

`=((-x^2)^n)/(1 + x^2)`

`=((-1)^nx^(2n))/(1 + x^2)`

 

`text(S)text(ince)\ \ (1 + x^2)>=1,\ \ \ -x^(2n) ≤ ((-1)^nx^(2n))/(1 + x^2) ≤ x^(2n)`

`:.\ text(We can conclude)`

`-x^(2n) ≤ 1/(1 + x^2) − (1 − x^2 + x^4 − x^6 + … + (-1)^(n − 1) x^(2n − 2))\ ≤ x^(2n)` 

 

ii.   `text{Integrating part (i) between 0 and 1}`

♦ Mean mark 43%.
`int_0^1 -x^(2n)\ dx` `=(-1)/(2n + 1)[x^(2n + 1)]_0^1`
  `=(-1)/(2n + 1)`
`int_0^1 1/(1 + x^2)` `=[tan^(-1) x]_0^1`
  `=pi/4`
`int_0^1 (1 − x^2 + x^4 − x^6 + … + (-1)^(n − 1) x^(2n − 2))\ dx` 
`=[x − (x^3)/3 + (x^5)/5 −… + ((-1)^(n − 1)x^(2n − 1))/(2n − 1)]_0^1`
`=1 − 1/3 + 1/5 − … + ((-1)^(n − 1))/(2n − 1)`
`int_0^1 x^(2n)\ dx` `=1/(2n + 1)[x^(2n + 1)]_0^1`
  `=1/(2n + 1)`

 
`:.\ text(We can conclude)`

`(-1)/(2n + 1) ≤ pi/4 − (1 − 1/3 + 1/5 − … + (-1)^(n − 1) 1/(2n − 1)) ≤ 1/(2n + 1)`

 

iii.  `(-1)/(2n + 1) ≤ pi/4 − (1 − 1/3 + 1/5 − … + ((-1)^(n − 1))/(2n − 1)) ≤ 1/(2n + 1)`

`text(As)\ n → ∞,`

`=>(-1)/(2n + 1) → 0^-\ \ text(and)\ \ 1/(2n + 1) → 0^+`

`=> pi/4 − (1 − 1/3 + 1/5 − … + 1/(2n − 1)) → 0`

♦♦ Mean mark 28%.

`:. pi/4 = 1 -1/3 + 1/5 − 1/7 + …`