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Calculus, MET2 2024 VCAA 2 MC

A function  \(g: R \rightarrow R\)  has the derivative  \( { \displaystyle g^{\prime}(x)=x^3-x } \).

Given that  \(g(0)=5\), the value of \(g(2)\) is

  1. \(2\)
  2. \(3\)
  3. \(5\)
  4. \(7\)
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\(D\)

Show Worked Solution
\({ \displaystyle g^{\prime}(x)}\) \(=x^3-x\)
\(g(x)\) \(=\dfrac{x^4}{4}-\dfrac{x^2}{2}+c\)

 
\(\text{Given }g(0)=5,\ c=5\) 

\(g(x)=\dfrac{x^4}{4}-\dfrac{x^2}{2}+5\)

\(\therefore\ g(2)=\dfrac{2^4}{4}-\dfrac{2^2}{2}+5=7\)

\(\Rightarrow D\)

Calculus, MET2 2020 VCAA 3 MC

 Let  `f^(′)(x)=(2)/(sqrt(2x-3))`. 

If  `f(6)=4`, then
 

  1. `f(x)=2sqrt(2x-3)`
  2. `f(x)=sqrt(2x-3)-2`
  3. `f(x)=2sqrt(2x-3)-2`
  4. `f(x)=sqrt(2x-3)+2`
  5. `f(x)=sqrt(2x-3)`
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`=>C`

Show Worked Solution
`f^(′)(x)` `=2/(sqrt(2x-3))`  
`f(x)` `=2 int(2x-3)^{- 1/2}`  
  `=2*1/2*2(2x-3)^{1/2}+c`  
  `=2sqrt(2x-3)+c`  

 
`text(When)\ \ x=6, \ f(x)=4:`

`4=2sqrt(12-3) + c \ => \ c=-2`

`:. f(x) = 2sqrt(2x-3) – 2`

`=>C`

Calculus, MET1 2021 VCAA 2

Let  `f^{′}(x) = x^3 + x`.

Find  `f(x)`  given that  `f(1) = 2`.  (2 marks)

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`f(x) =1/4 x^4 + 1/2 x^2 +5/4`

Show Worked Solution
`f^{′}(x)` `= x^3 + x`  
`f(x)` `=int x^3 + x\ dx`  
  `=1/4 x^4 + 1/2 x^2 +c`  

 
`text(Given)\ f(1) = 2:`

`2` `= 1/4 + 1/2 + c`  
`c` `= 5/4`  

 
`:. f(x) =1/4 x^4 + 1/2 x^2 +5/4`

Calculus, MET1-NHT 2019 VCAA 2

Find  `f(x)`  given that  `f(1) = -7/4`  and  `f ^{\prime}(x) = 2x^2 - 1/4x^(-2/3)`.  (2 marks)

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`f(x) = 2/3x^3 – 3/4x^(1/3) – 5/3`

Show Worked Solution
`f(x)` `= int 2x^2 – 1/4x^(-2/3) dx`
  `= 2/3x^3 – 1/4 ⋅ 1/(1/3) x^(1/3) + c`
  `= 2/3x^3 – 3/4x^(1/3) + c`

 
`text(Given)\ \ f(1) = -7/4:`

`-7/4` `= 2/3 – 3/4 + c`
`c` `= -5/3`
`:. f(x)` `= 2/3x^3 – 3/4x^(1/3) – 5/3`

Calculus, MET2 2019 VCAA 5 MC

Let  `f prime(x) = 3x^2 - 2x`  such that  `f(4) = 0`.

The rule of  `f`  is

A.   `f(x) = x^3 - x^2`

B.   `f(x) = x^3 - x^2 + 48`

C.   `f(x) = x^3 - x^2 - 48`

D.   `f(x) = 6x - 2`

E.   `f(x) = 6x - 24`

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`C`

Show Worked Solution
`f(x)` `= int 3x^2 – 2x\ dx`
  `= x^3 – x^2 + c`

 

`text(When)\ \ x = 4, \ f(x) = 0`

`0 = 4^3 – 4^2 + c`

`c = -48`
 

`:. f(x) = x^3 – x^2 – 48`

`=>   C`

Calculus, MET2 2010 VCAA 6 MC

A function `g` with domain `R` has the following properties.

   ●    `g prime (x) = x^2 - 2x`

   ●    the graph of  `g(x)`  passes through the point  `(1, 0)`

`g (x)`  is equal to

A.   `2x - 2`

B.   `x^3/3 - x^2`

C.   `x^3/3 - x^2 + 2/3`

D.   `x^2 - 2x + 2`

E.   `3x^3 - x^2 - 1`

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`C`

Show Worked Solution
`g(x)` `= int (x^2 – 2x)\ dx`
  `= 1/3 x^3 – x^2 + c`

 

`text(Graph passes through)\ \ (1, 0),`

`0` `= 1/3 (1)^3 – (1)^2 + c`
`c` `= 2/3`
`:. g(x)` `= 1/3 x^3 – x^2 + 2/3`

`=>   C`

Calculus, MET2 2015 VCAA 16 MC

Let  `f(x) = ax^m`  and  `g(x) = bx^n`, where `a, b, m` and `n` are positive integers. The domain of  `f = text(domain of)\ g = R.`

If  `f prime (x)` is an antiderivative of  `g(x)`, then which one of the following must be true?

A.   `m/n` is an integer

B.   `n/m` is an integer

C.   `a/b` is an integer

D.   `b/a` is an integer

E.   `n - m = 2`

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`D`

Show Worked Solution
`text(Given)\ \ f prime(x)` `=int g(x)\ dx,`
`amx^(m – 1)` `= b/(n + 1) x^(n + 1)`
♦♦ Mean mark 22%.

 

`m – 1 = n + 1\ …\ (1)\ \ \ text{(equate powers)}`

`am = b/(n + 1)\ …\ (2)\ \ \ text{(equate coefficients)}`
 

`text(Solving simultaneous equations:)`

`m(n + 1) = b/a`

`text(S)text(ince)\ \ m, n + 1\ text(are both integers)\ \ text{(i.e. ∈}\ Z^+text{)},`

`=> m(n + 1) ∈ Z^+`

`:. b/a = m(n + 1) ∈ Z^+`

`=>   D`