SmarterEd

Aussie Maths & Science Teachers: Save your time with SmarterEd

Algebra, MET2 2023 VCAA 16 MC

Let \(f(x)=e^{x-1}\).

Given that the product function \(f(x)\times g(x)=e^{(x-1)^2}\), the rule for the function \(g\) is

  1. \(g(x)=e^{x-1}\)
  2. \(g(x)=e^{(x-2)(x-1)}\)
  3. \(g(x)=e^{(x+2)(x-1)}\)
  4. \(g(x)=e^{x(x-2)}\)
  5. \(g(x)=e^{x(x-3)}\)
Show Answers Only

\(B\)

Show Worked Solution
\(f(x)\times g(x)\) \(=e^{(x-1)^2}\)
\(e^{x-1}\times g(x)\) \(=e^{(x-1)^2}\)
\( g(x)\) \(=\dfrac{e^{(x-1)^2}}{e^{(x-1)}}\)
  \(=e^{(x-1)^2}\times e^{(x-1)^-1}\)
  \(=e^{x^2-3x+2}\)
  \(=e^{(x-2)(x-1)}\)

 
\(\Rightarrow B\)

Algebra, MET1 2023 VCAA 2

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

Show Answers Only

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

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

 
\(\text{Let}\ \ u=e^{x}:\)

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

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

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

Algebra, MET1 2015 VCAA 7b

Solve  `3e^t = 5 + 8e^(−t)`  for `t`.  (3 marks)

Show Answers Only

`log_e(8/3)`

Show Worked Solution

`3e^t – 5 – 8e^(−t) = 0`

♦ Part (b) mean mark 44%.
MARKER’S COMMENT: Many students could not create a quadratic, and others who did made mistakes by trying to solve using the quadratic formula rather than simple factorising.

`text(Multiply both sides by)\ e^t:`

`3e^(2t) – 5e^t – 8 = 0`

`text(Let)\ \ y = e^t`

`log_2((6 – x)/(4 – x))` `= 2`
`3y^2 – 5y – 8` `= 0`
`(3y – 8)(y + 1)` `= 0`
`y` `=8/3\ quadquadquadquadquad text(or)\ \ \ \ ` `y` `=-1\ \ text{(No solution)}`
`e^t` `= 8/3`  
`:. t` `= log_e(8/3)\ \ \ `    

Algebra, MET2 2017 VCAA 8 MC

If  `y = a^(b - 4x) + 2`, where  `a > 0`, then `x` is equal to

  1. `1/4(b - log_a(y - 2))`
  2. `1/4(b - log_a(y + 2))`
  3. `b - log_a(1/4(y + 2))`
  4. `b/4 - log_a(y - 2)`
  5. `1/4(b + 2 - log_a(y))`
Show Answers Only

`A`

Show Worked Solution

`text(Solving for)\ x:`

`y – 2` `= a^(b – 4x)`
`b – 4x` `= log_a(y – 2)`
`4x` `= b – log_a(y – 2)`
`:. x` `= 1/4(b – log_a(y – 2))`

 
`=> A`

Algebra, MET2 2008 VCAA 12 MC

Let  `f: R -> R,\ f(x) = e^x + e^(–x).`

For all  `u in R,\ f(2u)`  is equal to

  1. `f(u) + f(-u)`
  2. `2 f(u)`
  3. `(f(u))^2 - 2`
  4. `(f(u))^2`
  5. `(f(u))^2 + 2`
Show Answers Only

`C`

Show Worked Solution

`text(Solution 1)`

♦ Mean mark 44%.

`text(Define)\ \ f(x) = e^x + e^-x`

`text(Enter each functional equation)`

`[text(i.e.)\ \ f(2u) = (f(u))^2 – 2]`

`text(until CAS output is “true”)`

`=>   C`

 

`text(Solution 2)`

`f(2u)` `=e^(2u) + e^(-2u)`
`(f(u))^2` `=(e^u + e^(-u))^2`
  `=e^(2u) + 2 + e^(-2u)`
   

`:. f(2u) = (f(u))^2-2`

`=>C`

Algebra, MET1 SM-Bank 9

Solve the following equation for `x`:

`2e^(2x) - e^x = 0`.  (2 marks)

Show Answers Only

`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`

Algebra, MET1 2011 VCAA 2b

Solve the equation  `4^x - 15 × 2^x = 16`  for `x.`  (3 marks)

Show Answers Only

`x = 4`

Show Worked Solution
`4^x – 15 xx 2^x – 16` `= 0`
`2^(2x) – 15 xx 2^x – 16` `= 0`

 

♦♦ Mean mark 33%.
MARKER’S COMMENT: “Poorly answered”. Many students incorrectly stated that if  `y=2^x`, then `2y=4^x`.

`text(Let)\ \ y = 2^x`

`y^2 – 15y – 16` `= 0`
`(y – 16) (y + 1)` `= 0`
`y` `= 16` `\ \ \ or\ \ \ ` `y` `= – 1`
`2^x` `= 16`   `2^x` `= – 1`
`:. x` `= 4`   `text(No solution)`