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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 EQ-Bank 15

Solve the following equation for \(x\):

\(2^{2x}=3(2^{x+1})-8\).   (3 marks)

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\(x=1\ \ \text{or} \ \ 2\)

Show Worked Solution
\(2^{2x}\) \(=3(2^{x+1})-8\)  
\(2^{2x}\) \(=3(2 \times 2^{x})-8\)  
\(0\) \(=2^{2x}-6\cdot2^{x}+8\)  

 
\(\text{Let}\ \ X=2^{x}\)

\(X^2-6X+8\) \(=0\)  
\((X-4)(X-2)\) \(=0\)  
\(X\) \(=4\ \ \text{or}\ \ 2\)  

 
\(2^{x}=4\ \ \Rightarrow\ \ x=2\)

\(2^{x}=2\ \ \Rightarrow\ \ x=1\)

Functions, 2ADV F1 EQ-Bank 16

  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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a.    `-1/3`

b.    `0.462`

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a.    `1/(root3(7+pi)) = (7+pi)^(-1/3)`

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

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)`

 

L&E, 2ADV E1 2011 HSC 1c

Solve   `2^(2x+1)=32`.    (2 marks) 

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`x=2`

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MARKER’S COMMENT: Many students correctly solved this by taking the logarithms of both sides.
`2^(2x+1)` `=32`
`2^(2x+1)` `=2^5`
`2x+1` `=5`
`:. x` `=2`