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L&E, 2ADV E1 EQ-Bank 2

Show  \(f(x)=\dfrac{1}{2}-\dfrac{1}{2^x+1}\)  is an odd function.  (3 marks)

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\(\text{Odd}\ \ \Rightarrow \  \ f(-x)=-f(x)\)

\(\begin{aligned}
f(x) & =\dfrac{1}{2}-\dfrac{1}{2^x+1} \\
f(-x) & =\dfrac{1}{2}-\dfrac{1}{2^{-x}+1} \times \dfrac{2^x}{2^x} \\
& =\dfrac{1}{2}-\dfrac{2^x}{1+2^x} \\
& =\dfrac{1}{2}-\dfrac{2^x+1-1}{2^x+1} \\
& =\dfrac{1}{2}-1+\dfrac{1}{2^x+1} \\
& =-\dfrac{1}{2}+\dfrac{1}{2^x+1} \\
& =-f(x)
\end{aligned}\)

 
\(\therefore f(x) \text { is odd.}\)

Show Worked Solution

\(\text{Odd}\ \ \Rightarrow \  \ f(-x)=-f(x)\)

\(\begin{aligned}
f(x) & =\dfrac{1}{2}-\dfrac{1}{2^x+1} \\
f(-x) & =\dfrac{1}{2}-\dfrac{1}{2^{-x}+1} \times \dfrac{2^x}{2^x} \\
& =\dfrac{1}{2}-\dfrac{2^x}{1+2^x} \\
& =\dfrac{1}{2}-\dfrac{2^x+1-1}{2^x+1} \\
& =\dfrac{1}{2}-1+\dfrac{1}{2^x+1} \\
& =-\dfrac{1}{2}+\dfrac{1}{2^x+1} \\
& =-f(x)
\end{aligned}\)

 
\(\therefore f(x) \text { is odd.}\)

L&E, 2ADV E1 2023 MET1 2

Solve  \(e^{2x}-12=4e^{x}\)  for  \(x\ \in\ R\).   (3 marks)

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\(x=\log_{e}6\)

Show Worked Solution
\(e^{2x}-12\) \(=4e^{x}\)
\(e^{2x}-4e^{x}-12\) \(=0\)

 
\(\text{Let}\ \ u=e^{x}:\)

\(u^2-4u-12\) \(=0\)
\((u-6)(u+2)\) \(=0\)
 

\(\Rightarrow u=6\ \ \ \text{or}\ -2\)

\(\therefore e^{x}\) \(=6\ \ \ \ \ \ \ \ \ \ \text{or}\ \ \ \ \ \ e^{x}=-2\ \text{(no solution)}\)  
\(x\) \(=\log_{e}6 \)  

Functions, 2ADV F1 SM-Bank 53

  1. If  `1/(root3(7+pi)) = (7+pi)^x`, find  `x`.  (1 mark)

  2. Calculate the value of  `1/(root3(7+pi))`  to 3 significant figures.   (1 mark)

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  1. `-1/3`
  2. `0.462`
Show Worked Solution

i.  `1/(root3(7+pi)) = (7+pi)^(-1/3)`

ii.   `1/(root3(7+pi))` `=0.4619…`
    `=0.462\ \ text{(to 3 sig. fig.)}`

L&E, 2ADV E1 2019 HSC 15a

Solve  `e^(2 ln x) = x + 6`   (2 marks)

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`x = 3 or -2`

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♦ Mean mark 47%.

`e^(2 ln x)` `= x + 6`
`ln e^(2 ln x)` `= ln (x + 6)`
`2 ln x` `= ln (x + 6)`
`ln x^2` `= ln (x + 6)`
`x^2` `= x + 6`
`x^2 – x – 6` `= 0`
`(x – 3) (x + 2)` `= 0`

 
`:. x = 3 \ \ (x>0)`

L&E, 2ADV E1 SM-Bank 11 MC

Solve the equation  `e^(4x) - 5e^(2x) + 4 = 0`  for `x`

  1. `x= 1, 4`
  2. `x= – 4, – 1`
  3. `x= 0, log_e 2`
  4. `x= – log_e 2, 0, log_e 2 `
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`C`

Show Worked Solution

`e^(4x) – 5e^(2x) + 4 = 0`

`text(Let)\ \ X=e^(2x)`

`X^2-5X+4` `=0`
`(X-4)(X-1)` `=0`
`X` `=4 or 1`

 

`:.e^(2x)` `=4` `e^(2x)` `=1`
`2x` `=log_e 4` `x` `=0`
`x` `=(2log_e 2)/2`    
  `=log_e 2`    

`=> C`

L&E, 2ADV E1 2004 HSC 6a

Solve the following equation for `x`:

`e^(2x) + 3e^x − 10 = 0`.  (2 marks)

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`x = ln 2`

Show Worked Solution
`e^(2x) + 3e^x − 10` `= 0`
`:. (e^x)^2 + 3e^x − 10` `= 0`

 
`text(Let)\ \ X = e^x,`

`:. X^2 + 3X – 10` `= 0`
`(X + 5)(X − 2)` `= 0`
`:. X =2 or -5`

 

`text(If)\ \ X` `=2`
`e^x` `=2`
`ln e^x` `=ln 2`
`x` `=ln 2`
`text(If)\ \ X` `=-5`
`e^x` `=-5\ \ \ text{(no solution)}`

 
 `:. x=ln 2`

L&E, 2ADV E1 2007 HSC 6a

Solve the following equation for `x`:

`2e^(2x) - e^x = 0`.  (2 marks)

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`x = ln\ 1/2`

Show Worked Solution

`text(Solution 1)`

`2e^(2x) – e^x = 0`

`text(Let)\ \ X = e^x`

`2X^2 – X` `= 0`
`X (2X – 1)` `= 0`

 
`X = 0 or 1/2`

 
`text(When)\ e^x = 0\  =>\ text(no solution)`

`text(When)\ e^x = 1/2`

`ln e^x` `= ln\ 1/2`
`:. x` `= ln\ 1/2`

 

`text(Solution 2)`

`2e^(2x)-e^x` `=0`
`2e^(2x)` `=e^x`
`ln 2e^(2x)` `=ln e^x`
`ln 2 +ln e^(2x)` `=x`
`ln 2 + 2x` `=x`
`x` `=-ln2`
  `=ln\ 1/2`

L&E, 2ADV E1 2015 HSC 8 MC

The diagram shows the graph of  `y = e^x (1 + x).`
 

How many solutions are there to the equation  `e^x (1 + x) = 1 - x^2`?

  1. `0`
  2. `1`
  3. `2`
  4. `3`
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`C`

Show Worked Solution

`text(The graphs intersect at 2 points)`

`:. e^x(1 + x) = 1 – x^2\ text(has 2 solutions)`

`=> C`

L&E, 2ADV E1 2010 HSC 4d

Let  `f(x)=1+e^x`.

Show that  `f(x)xxf(–x)=f(x)+f(–x)`.   (2 marks) 

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`text{Proof (See Worked Solutions).}`

 

Show Worked Solutions

`f(x)xxf(–x)`

`=(1+e^x)(1+e^-x)`

MARKER’S COMMENT: A common error in this question was not to realise that `e^xe^-x=e^0=1`.

`=1+e^-x+e^x+e^xe^-x`

`=e^x+e^-x+2`

 

`f(x)+f(–x)`

`=1+e^x+1+e^-x`

`=e^x+e^-x+2`

`=f(x)xxf(–x)\ \ …\ text(as required)`