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Proof, EXT2 P1 2025 HSC 14d

Positive real numbers \(a, b, c\) and \(d\) are chosen such that  \(\dfrac{1}{a}, \dfrac{1}{b}, \dfrac{1}{c}\)  and  \(\dfrac{1}{d}\)  are consecutive terms in an arithmetic sequence with common difference \(k\), where  \(k \in \mathbb{R} , k>0\).

Show that  \(b+c<a+d\).   (3 marks)

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\(\text{AP:} \ \dfrac{1}{a}, \dfrac{1}{b}, \dfrac{1}{c}, \dfrac{1}{d} \ \ \text{where common difference}=k\ \ (k>0)\)

\(\text{Show} \ \ b+c<a+d\)

\(\dfrac{1}{a}<\dfrac{1}{a}+k(k>0)\)

\(\dfrac{1}{a}<\dfrac{1}{b}\)

\(\text{Similarly for}\ \ \dfrac{1}{b}<\dfrac{1}{c}\  \ldots\  \text{such that}\)

\(\dfrac{1}{a}<\dfrac{1}{b}<\dfrac{1}{c}<\dfrac{1}{d} \ \ \Rightarrow\ \ a>b>c>d\ \ldots\ (1)\)
 

\(\dfrac{1}{b}=\dfrac{1}{a}+k \ \ \Rightarrow \ \ k=\dfrac{1}{b}-\dfrac{1}{a}=\dfrac{a-b}{a b}\ \ \Rightarrow \ \ a-b=kab\)

\(\text {Similarly:}\)

\(\dfrac{1}{d}=\dfrac{1}{c}+k \ \ \Rightarrow \ \ k=\dfrac{c-d}{cd}\ \ \Rightarrow\ \ c-d=kcd\)
 

\(\text{Using \(\ a>b>c>d\ \)  (see (1) above):}\)

\(a-b>c-d\)

\(b+c<a+d\)

Show Worked Solution

\(\text{AP:} \ \dfrac{1}{a}, \dfrac{1}{b}, \dfrac{1}{c}, \dfrac{1}{d} \ \ \text{where common difference}=k\ \ (k>0)\)

\(\text{Show} \ \ b+c<a+d\)

\(\dfrac{1}{a}<\dfrac{1}{a}+k\ \ (k>0)\)

\(\dfrac{1}{a}<\dfrac{1}{b}\)

\(\text{Similarly for}\ \ \dfrac{1}{b}<\dfrac{1}{c}\  \ldots\  \text{such that}\)

\(\dfrac{1}{a}<\dfrac{1}{b}<\dfrac{1}{c}<\dfrac{1}{d} \ \ \Rightarrow\ \ a>b>c>d\ \ldots\ (1)\)
 

\(\dfrac{1}{b}=\dfrac{1}{a}+k \ \ \Rightarrow \ \ k=\dfrac{1}{b}-\dfrac{1}{a}=\dfrac{a-b}{a b}\ \ \Rightarrow \ \ a-b=kab\)

\(\text {Similarly:}\)

\(\dfrac{1}{d}=\dfrac{1}{c}+k \ \ \Rightarrow \ \ k=\dfrac{c-d}{cd}\ \ \Rightarrow\ \ c-d=kcd\)
 

\(\text{Using \(\ a>b>c>d\ \)  (see (1) above):}\)

\(a-b>c-d\)

\(b+c<a+d\)

Proof, EXT2 P1 2024 HSC 13d

It is known that for all positive real numbers \(x, y\)

\(x+y \geq 2 \sqrt{x y} .\)     (Do NOT prove this.)

Show that if \(a, b, c\) are positive real numbers with  \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\)  then  \(a \sqrt{b c}+b \sqrt{a c}+c \sqrt{a b} \leq a b c\).   (3 marks)

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\(\text{If }\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1, \ \ \ a, b, c \in \mathbb{R} ^{+}\)

\(\text {Prove} \ \ a \sqrt{b c}+b \sqrt{a c}+c \sqrt{a b} \leqslant a b c\)

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\) \(=1\)  
\(\dfrac{b c+a c+a b}{a b c}\) \(=1\)  
\(ab+bc+ac\) \(=abc\ …\ (1)\)  

 
\(x+y \geqslant 2 \sqrt{x y}, \quad x, y \in \mathbb{R}^{+} \text{(given)}\)

\(a+b \geqslant 2 \sqrt{a b} \ \Rightarrow \ \sqrt{a b} \leqslant \dfrac{1}{2}(a+b)\)

\(\text{Similarly,}\)

\(\sqrt{bc} \leqslant \dfrac{1}{2}(b+c) \ \text{and}\ \ \sqrt{a c} \leqslant \dfrac{1}{2}(a+c)\)

  \(a \sqrt{b c}+b \sqrt{a c}+c \sqrt{a b}\) \(\leqslant a \times \dfrac{1}{2}(b+c)+b \times \dfrac{1}{2}(a+c)+c \times \dfrac{1}{2}(a+b)\)
    \(\leqslant \dfrac{1}{2} a b+\dfrac{1}{2} a c+\dfrac{1}{2} a b+\dfrac{1}{2} b c+\dfrac{1}{2} a c+\dfrac{1}{2} b c\)
    \(\leqslant a b+b c+a c\)
    \(\leqslant a b c\ \  \text{(see (1) above)}\)

Show Worked Solution

\(\text{If }\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1, \ \ \ a, b, c \in \mathbb{R} ^{+}\)

\(\text {Prove} \ \ a \sqrt{b c}+b \sqrt{a c}+c \sqrt{a b} \leqslant a b c\)

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\) \(=1\)  
\(\dfrac{b c+a c+a b}{a b c}\) \(=1\)  
\(ab+bc+ac\) \(=abc\ …\ (1)\)  

 
\(x+y \geqslant 2 \sqrt{x y}, \quad x, y \in \mathbb{R}^{+} \text{(given)}\)

\(a+b \geqslant 2 \sqrt{a b} \ \Rightarrow \ \sqrt{a b} \leqslant \dfrac{1}{2}(a+b)\)

\(\text{Similarly,}\)

\(\sqrt{bc} \leqslant \dfrac{1}{2}(b+c) \ \text{and}\ \ \sqrt{a c} \leqslant \dfrac{1}{2}(a+c)\)

  \(a \sqrt{b c}+b \sqrt{a c}+c \sqrt{a b}\) \(\leqslant a \times \dfrac{1}{2}(b+c)+b \times \dfrac{1}{2}(a+c)+c \times \dfrac{1}{2}(a+b)\)
    \(\leqslant \dfrac{1}{2} a b+\dfrac{1}{2} a c+\dfrac{1}{2} a b+\dfrac{1}{2} b c+\dfrac{1}{2} a c+\dfrac{1}{2} b c\)
    \(\leqslant a b+b c+a c\)
    \(\leqslant a b c\ \  \text{(see (1) above)}\)

Proof, EXT2 P1 2021 HSC 5 MC

Which of the following statements is FALSE?

  1. `∀ a, b ∈ RR,`                                             `a < b \ => \ a^3 < b^3`
  2. `∀ a, b ∈ RR,`                                             `a < b \ => e^{-a} > e^{-b}`
  3. `∀ a, b ∈ (0, + ∞),`                               `a < b \ => \ text{ln} \ a < text{ln} \ b`
  4. `∀ a, b ∈ RR, text{with} \ a,b ≠ 0,`                    `a < b \ => \ 1/a > 1/b`
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`D`

Show Worked Solution

`text{By contradiction:}`

♦ Mean mark 50%.

`text{Consider} \ D`

`text{Let} \ \ a = -1 \ \ text{and} \ \ b = 1,`

`a < b \ -> \ -1 < 1 \ \ text{(TRUE)}`

`1/a > 1/b \ -> \ -1 > 1 \ \ text{(FALSE)}`
 

`=>\ D \ text{is false}`

Proof, EXT2 P1 2012 HSC 16c

Let  `n`  be an integer where  `n > 1`. Integers from  `1`  to  `n`  inclusive are selected randomly one by one with repetition being possible. Let  `P(k)`  be the probability that exactly  `k`  different integers are selected before one of them is selected for the second time, where  `1 ≤ k ≤ n`.

  1. Explain why  `P(k) = ((n − 1)!k)/(n^k(n − k)!)`.   (2 marks)

  2. Suppose  `P(k) ≥ P(k − 1)`. Show that  `k^2- k- n ≤ 0`.   (2 marks)

  3. Show that if  `sqrt(n + 1/4) > k − 1/2`  then the integers  `n`  and  `k`  satisfy  `sqrtn > k − 1/2`.   (2 marks)

  4. Hence show that if  `4n + 1`  is not a perfect square, then  `P(k)`  is greatest when  `k`  is the closest integer to  `sqrtn`. 

     

    You may use part (iii) and also that  `k^2 − k − n >0`  if  `P(k)< P(k − 1)`.   (2 marks)

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

i.   `text(Let)\ P(1)=text{(After 1st number chosen, the second draw matches)}`

`P(1) =n/n xx 1/n=1/n`

`text(Let)\ P(2)=text{(After 1st two numbers chosen, the third draw matches)}`

`P(2) = n/n xx (n − 1)/n xx 2/n`

♦♦♦ Mean mark part (i) 7%.

`P(3)=n/n xx (n-1)/n xx (n-2)/n xx 3/n`

`vdots`

 
`=>text{On the}\ (k+1)text(th draw)`

`P(k)` `= n/n xx (n − 1)/n xx (n − 2)/n xx …\ xx (n − k+1)/n xx k/n`
  `=((n-1)!)/n^k xx k/(1 xx 2 xx … xx (n-k))`
  `=((n-1)!\ k)/(n^k (n-k)!)`

♦ Mean mark part (ii) 38%.

 

ii. `P(k)` `≥ P(k − 1)`
  `((n − 1)!\ k)/(n^k(n − k)!)` `≥ ((n − 1)!  (k − 1))/(n^(k − 1)(n − k + 1)!)`
  `k(n − k + 1)` `≥ n(k − 1)`
  `kn − k^2 + k` `≥ nk − n`
  `:.k^2 − k − n` `≤ 0`

 

♦♦♦ Mean mark part (iii) 10%.
iii.   `sqrt(n + 1/4)` `> k − 1/2`
  `n + 1/4` `> (k − 1/2)^2`
  `n + 1/4` `> k^2 − k + 1/4`
  `n` `>k^2 − k`
  `n` `> k(k − 1)`

 

`text(S)text(ince)\ n\ text(and)\ k\ text(are integers such that)\ 1 ≤ k ≤ n.`

`=>n\ text(is at least the next integer after)\ k.`

`=>(k −1)\ text(is the integer before)\ k.`

`:.n` `>k^2 − k +1/4`
`n`  `>(k-1/2)^2`
`:.sqrt n` `>k-1/2`

 

iv.  `text(Find the largest integer)\ \ k\ \ text(for which)\ \ P(k) ≥ P(k − 1)`

`P(k) ≥ P(k − 1)\ \ text(when)\ \ k^2 − k − n ≤ 0\ \ \ text{(part (ii))}`

 

`text(Solving)\ \ k^2 − k − n ≤ 0`

♦♦♦ Mean mark part (iii) 1%.

`(1 – sqrt(4n + 1))/2 ≤ k ≤ (1 + sqrt(4n+1))/2`

`text(Given)\ \ n>0,\ \ (1 – sqrt(4n + 1))/2<0`

`:.1 ≤ k ≤ (1 + sqrt(4n+1))/2`

`text(S)text(ince)\ \ 4n+1\ \ text(is not a perfect square and)\ \ k\ \ text(is an integer)`

`k<` ` (1 + sqrt(1 + 4n))/2`
`k<` ` 1/2 + sqrt(n + 1/4)`
`k-1/2<` `sqrt(n + 1/4)`
`k-1/2<` `sqrt n\ \ \ \ \ text{(from part (iii))}`
`k<` `1/2+sqrt n`

 

`:. P(k)\ text(is greatest when)\ \ k\ \ text(is the closest integer to)\ n.`

Proof, EXT2 P1 2012 HSC 15a

  1. Prove that  `sqrt(ab) ≤ (a + b)/2`, where  `a ≥ 0`  and  `b ≥ 0`.   (1 mark)

  2. If  `1 ≤ x ≤ y`,  show that  `x(y − x + 1) ≥ y`.   (2 marks)

  3. Let  `n`  and  `j`  be positive integers with  `1 ≤ j ≤ n`.
     
    Prove that  `sqrtn ≤ sqrt(j(n − j + 1)) ≤ (n + 1)/2.`  (2 marks)

  4. For integers  `n ≥ 1`, prove that
     
        `(sqrtn)^n ≤ n! ≤ ((n + 1)/2)^n`.   (1 mark)

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  1. `text{Proof (See Worked Solutions)}`
  2. `text{Proof (See Worked Solutions)}`
  3. `text{Proof (See Worked Solutions)}`
  4. `text{Proof (See Worked Solutions)}`
Show Worked Solution
i. `(sqrta − sqrtb)^2` `≥ 0`
  `a − 2sqrt(ab) + b` `≥ 0`
  `a + b` `≥ 2sqrt(ab)`
  `sqrt(ab)` `≤ (a + b)/2`

 

ii.  `text(Solution 1)`

♦♦ Mean mark part (ii) 28%.

`text(S)text(ince)\ \ 1 ≤ x ≤ y`

`y-x` `>=0`
`y(x-1)-x(x-1)` `>=0,\ \ \ \ (x-1>=0)`
`xy-x^2+x-y` `>=0`
`:.xy-x^2+x` `>=y`

 

`text(Solution 2)`

`x( y − x + 1)` `= xy − x^2 + x`
  `= -y + xy − x^2 + x + y`
  `= y(x − 1) − x(x − 1)+ y`
  `= (x − 1)( y − x) + y`

 

`text(S)text(ince)\ \ x − 1 ≥ 0\ \ text(and)\ \  y − x ≥ 0`

`=>(x − 1)( y − x) + y` `>=y`
`:.x(y − x + 1)` `>=y`

 

iii.  `text(Let)\ c = n − j + 1, \ c > 0\ text(as)\ n ≥ j.`

♦ Mean mark part (iii) 39%.

 

`=> sqrt(j(n − j + 1))\ \ text(can be written as)\ \ sqrt(jc).`

`sqrt(jc)` `≤ (j + c)/2\ \ \ \ text{(part (i))}`
`sqrt(j(n − j + 1))` `≤ (j + n-j+1)/2`
`=>sqrt(j(n − j + 1))` `≤ (n+1)/2`
`j(n − j +1)` `≥ n\ \ \ \ text{(part (ii))}`
`=>sqrt(j(n − j + 1))` `≥ sqrt n`

 

`:.sqrtn ≤ sqrt(j(n − j + 1)) ≤ (n + 1)/2`

 

iv. `text{From (iii)}\ n ≤ j(n− j +1) ≤ (n + 1)/2`

♦♦♦ Mean mark part (iv) just 2%!

`text(Let)\ j\ text(take on the values from 1 to)\ n.`

`j = 1:` `sqrtn ≤ sqrt(1(n)) ≤ (n + 1)/2`
`j = 2:` `sqrtn ≤ sqrt(2(n−1)) ≤ (n + 1)/2`
  `vdots`
`j = n:` `sqrtn ≤ sqrt(n(1)) ≤ (n + 1)/2`

 

`text{Multiply the corresponding parts of each line}`

`(sqrtn)^n ≤ sqrt(n! xx n!) ≤ ((n + 1)/2)^n`

`(sqrtn)^n ≤ n! ≤ ((n + 1)/2)^n`

Proof, EXT2 P1 2013 HSC 16a

  1. Find the minimum value of  `P(x) = 2x^3 - 15x^2 + 24x + 16`, for  `x >= 0.`  (2 marks)

  2. Hence, or otherwise, show that for  `x >= 0`,
     
        `(x + 1) (x^2 + (x + 4)^2) >= 25x^2.`  (1 mark)

  3. Hence, or otherwise, show that for  `m >= 0`  and  `n >= 0`,
     
        `(m + n)^2 + (m + n + 4)^2 >= (100mn)/(m + n + 1).`  (2 marks)

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  1. `0`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `text(Proof)\ \ text{(See Worked Solutions)}`
Show Worked Solution
♦ Mean mark 43%.

STRATEGY: The best responses carefully considered the domain restrictions, notably `P(0).`
i. `P(x)` `= 2x^3 – 15x^2 + 24x + 16,\ \ \ x >= 0`
  `P prime(x)` `= 6x^2 – 30x +24`
    `= 6 (x^2 – 5x + 4)`
    `= 6 (x – 1) (x – 4)`
  `P″(x)` ` = 12x – 30`

 

`text(MAX or MIN when)\ \ P prime (x)=0`

`text(i.e. when)\ \ x=1  or  4`

`P″(1) = -6 < 0\ \ \ \ P″(4) = 18 > 0`

`:.text(Minimum turning point of)\ \ P(x)\ \ text(at)\ \ x = 4,`

`P(4) = 128 – 240 + 96 + 16 = 0`

 
`text(Checking limits:)`

`P(0) = 16`

`text(As)\ \ x->oo,\ \ y->oo`

`:.text(Minimum value of)\ \ P(x)\ \ text(is)\ \ 0\ \ text(when)\ \ x = 4.`

 

♦ Mean mark 47%.

ii.  `text(LHS)` `= (x + 1) (x^2 + (x + 4)^2)`
  `= (x + 1) (2x^2 + 8x + 16)`
  `= 2x^3 + 8x^2 + 16x + 2x^2 + 8x + 16`
  `= 2x^3 + 10x^2 + 24x + 16`
  `= (2x^3 – 15x^2 + 24x + 16) + 25x^2`
  `= P(x) + 25x^2`

 

`text(The minimum value of)\ \ P(x)\ \ text(for)\ \ x>= 0\ \ text(is)\ \ 0,`

`=>P(x) + 25x^2` `>= 25x^2`
`:.(x + 1) (x^2 + (x + 4)^2)` `>= 25x^2,\ \ \ text(for)\ x >= 0`

 

♦♦ Mean mark 19%.

STRATEGY: The best responses carefully considered the domain restrictions, notably `P(0).`
iii.   `(m + n)^2 + (m + n + 4)^2`
  `= (m + n)^2 + (m + n)^2 + 8(m + n) + 16`
  `= 2 (m + n)^2 + 8 (m + n) + 16`

 

`text(Let)\ \ x = m + n`

`(m + n)^2 + (m + n + 4)^2 = 2x^2 + 8x + 16`

`text{Using part (ii)}`

`(x + 1) (2x^2 + 8x + 16)` `>= 25x^2`
`2x^2 + 8x + 16` `>= (25x^2)/(x + 1),\ \ \ x >= 0`
`(m + n)^2 + (m + n + 4)^2` `>= (25 (m + n)^2)/(m + n + 1)`

 

`text(Consider)\ \ (m – n)^2`

`text(S)text(ince)\ \ (m – n)^2` `>= 0`
`m^2 + n^2` `>= 2mn`
`(m+n)^2-2mn` `>= 2mn`
`(m + n)^2` `>= 4mn`
`:.(m + n)^2 + (m + n + 4)^2` `>= (100mn)/(m + n + 1)`

Proof, EXT2 P1 2014 HSC 15a

Three positive real numbers `a`, `b` and `c` are such that  `a + b + c = 1`  and  `a ≤ b ≤ c`.

By considering the expansion of  `(a + b + c)^2`, or otherwise, show that 

`qquad 5a^2 + 3b^2 +c^2 ≤ 1`.  (2 marks)

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`text{Proof (See Worked Solutions)}`

 

Show Worked Solution

`a + b+ c = 1,\ \ a ≤ b ≤ c`

♦♦ Mean mark 13%.

`(a + b+ c)^2 = 1`

`a^2 + b^2 + c^2 + 2ab+ 2bc + 2ac = 1`

`text(S)text(ince)\ a ≤ b\ \ =>a^2 ≤ ab`

`text(S)text(ince)\ b ≤ c\ \ =>b^2 ≤ bc`

`text(S)text(ince)\ a ≤ c\ \ =>a^2 ≤ ac`
 

`=> a^2 + b^2 + c^2 + 2a^2 + 2b^2 + 2a^2` `≤ 1`
`:.5a^2 + 3b^2 + c^2` `≤ 1\ \ \ text(… as required)`