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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

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

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

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

\(\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 P2 2023 SPEC1 8

A function \(f\) has the rule  \(f(x)=x\,e^{2x}\).

Use mathematical induction to prove that  \(f^{(n)}(x)=\big{(}2^{n}x+n\,2^{n-1}\big{)}e^{2x}\)  for  \(n \in \mathbb{Z}^{+}\), where  \(f^{(n)}(x)\)  represents the \(n\)th derivative of \(f(x)\). That is, \(f(x)\) has been differentiated \(n\) times.   (3 marks)

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\(\text{Proof (See Worked Solutions)}\)

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\(\text{If}\ \ n=1:\)

\(f(x)=x\,e^{2x}\ \ \Rightarrow \ \ f^{′}(x)=e^{2x} + 2x\,e^{2x} \)

\(f^{(1)}(x)=\big{(}2^{1}x + 1 \times 2^{1-1}\big{)}e^{2x} = e^{2x} + 2x\,e^{2x} \)

\(\therefore\ \text{True for}\ \ n=1.\)
 

\(\text{Assume true for}\ \ n=k \)

\(f^{(k)}(x)=\big{(}2^{k}x + k\,2^{k-1}\big{)}e^{2x}\)
 

\(\text{Prove true for}\ \ n=k+1 \)

\(\text{i.e.}\ f^{(k+1)}(x)=\big{(}2^{k+1}x + (k+1)2^{k}\big{)}e^{2x}\)

\(\text{LHS}\) \(= \dfrac{d}{dx} \big{(}f^{(k)}(x) \big{)} \)   
  \(= \dfrac{d}{dx} \Big{[}\big{(}2^{k}x + k\,2^{k-1}\big{)}e^{2x}\Big{]} \)  
  \(= 2^{k}e^{2x}+(2^{k}x+k\,2^{k-1}) \times 2 e^{2x} \)  
  \(=e^{2x} \big{(}2^{k}+2\cdot2^{k}x + 2k \cdot 2^{k-1}\big{)}\)  
  \(=e^{2x}\big{(}2^{k+1}+2^{k}+k \cdot 2^{k} \big{)} \)  
  \(=\big{(}2^{k+1}x + (k+1)2^{k}\big{)}e^{2x}\)  
  \(=\ \text{RHS}\)  

 
\(\Rightarrow \text{True for}\ \ n=k+1\)

\(\therefore\ \text{Since true for}\ \ n=1\text{, by PMI, true for}\ n \in \mathbb{Z}^{+} \)

Proof, EXT2 P2 EQ-Bank 6

Using mathematical induction and integration by parts, show

  `int_0^pi sin^nx\ dx=(n-1)/n int_0^pi sin^(n-2)x\ dx`  for  `n>=2.`   (4 marks)

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

`text{Prove}\ \ int_0^pi sin^nx\ dx=(n-1)/n int_0^pi sin^(n-2)x\ dx\ \ text{for}\ \ n>=2.`

`text{If}\ \ n=2:`

`text{LHS}` `=int_0^pi sin^(2)x\ dx`  
  `=1/2 int_0^pi (1-cos (2x))\ dx`  
  `=1/2 [x-1/2sin(2x)]_0^pi`  
  `=1/2[(pi-1/2sin(2pi))-0]`  
  `=pi/2`  

 

`text{RHS}` `=(2-1)/2 int_0^pi sin^0 x\ dx`  
  `=1/2int_0^pi 1\ dx`  
  `=1/2[x]_0^pi`  
  `=pi/2`  
  `=\ text{LHS}`  

 
`:.\ text{True for}\ \ n=2`
 

`text{Assume true for}\ \ n=k`

`text{i.e.}\ int_0^pi sin^kx\ dx=(k-1)/k int_0^pi sin^(k-2)x\ dx`
 

`text{Prove true for}\ \ n=k+1`

`text{i.e.}\ int_0^pi sin^(k+1)x\ dx=(k)/(k+1) int_0^pi sin^(k-1)x\ dx`

`text{Let}\ \ I=\ text{LHS} = int_0^pi sin^(k+1)x\ dx=int_0^pi sinx * sin^(k)x\ dx`

`text{Using IBP:}`

`u` `=sin^k x` `v^(′)=sinx`
`u^(′)` `=kcosx*sin^(k-1)x`     `v=-cosx`

 

`I` `=[uv]_0^pi-int_0^pi u^(′)v\ dx`  
  `=[-sin^kx*cosx]_0^pi-int_0^pi -kcosx*sin^(k-1)x*cosx\ dx`  
  `=[(-sin^k pi*cos pi)-(-sin^k 0*cos 0)+kint cos^2x*sin^(k-1)x\ dx`  
  `=0+kint (1-sin^2x)*sin^(k-1)x\ dx`  
  `=kint_0^pi sin^(k-1)x\ dx-kint_0^pi sin^(k+1)x\ dx`  
`I` `=kint_0^pi sin^(k-1)x\ dx-kI`  
`I+kI` `=kint_0^pi sin^(k-1)x\ dx`  
`I(k+1)` `=kint_0^pi sin^(k-1)x\ dx`  
`I` `=k/(k+1) int_0^pi sin^(k-1)x\ dx`  
  `=\ text{RHS}`  

 
`:.\ text{True for}\ \ n=k+1`

`:.\ text{Since true for} \ n=2,\ text{by PMI, true for integers} \ n>=2.`

Proof, EXT2* P2 2009 HSC 7a

  1. Use differentiation from first principles to show that  `d/(dx)(x)=1`.   (1 mark)

  2. Use mathematical induction and the product rule for differentiation to prove that  
     
        `d/(dx)(x^n)=nx^(n-1)`  for all positive integers `n`.   (2 marks)

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  1. `text{Proof (See Worked Solutions)}`
  2. `text{Proof (See Worked Solutions)}`
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i.   `text(Prove)\ \ d/(dx)(x)=1\ \ text(from first principles.)`

MARKER’S COMMENT: Students are reminded to use the number of marks allocated to a question as a guide to the work involved in answering it.
`d/(dx)(x)` `=lim_(h->0) (f(x+h)-f(x))/h`
  `=lim_(h->0)(x+h-x)/h`
  `=1\ \ \ text(… as required)`

 

ii.  `text(Prove)\ \ d/(dx)(x^n)=nx^(n-1)\ \ text(for integers)\ n>=1`

`text(If)\  n=1,`

`text(LHS)=d/(dx)(x^1)=1`

`text(RHS)=(1)x^0=1=text(LHS)`

`:.\ text(True for)\ n=1`

 
`text(Assume true for)\ n=k`

`text(i.e.)\ \ d/(dx)(x^k)=kx^(k-1)`

 `text(Prove true for)\ n=k+1`

IMPORTANT: Critical to read carefully and apply the product rule to prove the induction.

`text(i.e.)\ \ d/(dx)x^(k+1)=(k+1)x^k`

`text(LHS)` `=d/(dx)(x^(k+1))`
  `=d/(dx) (x*x^k)`
  `=d/(dx)(x) * x^k+x *d/(dx)(x^k)`
  `=(1xx x^k)+(x xxkx^(k-1))`
  `=x^k+kx^k`
  `=(k+1)x^k`
  `=\ text(RHS … as required)`

 
`=>\ text(True for)\ n=k+1`

`:.\ text(S)text(ince it is true for)\ n=1, text(by PMI, true for integral)\ n>=1`.