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Graphs, SPEC1 2012 VCAA 10

Consider the functions with rules  `f(x) = arcsin (x/2) + 3/sqrt (25 x^2 - 1)`  and  `g(x) = arcsin (3x) - 3/sqrt (25x^2 - 1).`

    1. Find the maximal domain of  `f_1(x) = arcsin (x/2).`  (1 mark)
    2. Find the maximal domain of  `f_2(x) = 3/sqrt (25x^2 - 1).` (1 mark)
    3. Find the largest set of values of  `x in R`  for which  `f(x)`  is defined.  (1 mark)
  2. Given that  `h(x) = f(x) + g(x)`  and that  `theta = h(1/4)`, evaluate  `sin (theta).`

     

    Give your answer in the form  `(a sqrt b)/c, \ a, b, c in Z.`  (3 marks)

Show Answers Only
    1. `-2 <= x <= 2`
    2. `x < -1/5 uu x > 1/5`
    3. `-2 <= x < -1/5 uu 1/5 < x <= 2`
  2. `(5 sqrt 7)/16`
Show Worked Solution

a.i.   `text(Maximal Domain occurs when:)`

  `-1 <= x/2 <= 1`
  `{x: -2 <= x <= 2}`

 
a.ii.   `text(Maximal Domain occurs when:)`

♦♦ Mean mark part (a)(ii) 33%.

  `25x^2 – 1 > 0`
  `x^2 > 1/25`
  `{x: x < -1/5 uu x > 1/5}`

 

a.iii.  `text(Max domain for which)\ \ f(x)\ \ text(is defined:)`

♦♦ Mean mark part (a)(iii) 26%.

  `(-2 <= x <= 2) nn (x < -1/5 uu x > 1/5)`
  `{x: -2 <= x < -1/5 \ uu \ 1/5 < x <= 2}`

 

b.    `h(x)` `= sin^(-1) (x/2) + 3/sqrt(25x^2 – 1) + sin^(-1)(3x) – 3/sqrt(25x^2 – 1)`
    `= sin^(-1)(x/2) + sin^(-1)(3x)`

 
`text(When)\ \ x=1/4,\ \ h(x)=theta`

♦♦♦ Mean mark part (b) 20%.

`theta=sin^(-1) (1/8) + sin^(-1) (3/4)`

`text(Let)\ \ theta_1 = sin^(-1) (1/8),\ \ theta_2 = sin^(-1) (3/4)`

`text(Using)\ \ sin(theta_1 + theta_2)=sin theta_1 cos theta_2 + cos theta_1 sin theta_2:`

`sin(theta)` `= 1/8 * sqrt7/4 + sqrt63/8 * 3/4`
  `=sqrt7/32 + (9sqrt7)/32`
  `=(5 sqrt7)/16`

Calculus, SPEC2 2018 VCAA 1

Consider the function  `f: D -> R`, where  `f(x) = 2 text(arcsin)(x^2 - 1)`.

  1. Determine the maximal domain `D` and the range of  `f`.  (2 marks)
  2. Sketch the graph of  `y = f(x)`  on the axes below, labelling any endpoints and the `y`-intercept with their coordinates.  (3 marks)
     
    `qquad` 
     
  3. Find  `f prime(x)`  for  `x > 0`, expressing your answer in the form  `f prime(x) = A/sqrt(2 - x^2), \ A in R`.  (1 mark)
  4. Write down  `f prime(x)`  for  `x < 0`, expressing your answer in the form  `f prime(x) = B/sqrt(2 - x^2), \ B in R`.  (1 mark)
  5. The derivative  `f prime(x)`  can be expressed in the form  `f prime(x) = {g(x)}/sqrt(2 - x^2)` over its maximal domain.
  1. Find the maximal domain of  `f prime`.  (1 mark)
  2. Find  `g(x)`, expressing your answer as a piecewise (hybrid) function.  (1 mark)
  3. Sketch the graph of  `g` on the axes below.  (2 marks)

 

Show Answers Only
  1. `D [- sqrt 2 , sqrt 2]; qquad f(x) in [-pi, pi]`
  2. `text(See Worked Solutions)`
  3. `4/sqrt(2 – x^2)`
  4. `(-4)/sqrt(2 – x^2)`
    1. `x in (-sqrt2, 0) ∪ (0, sqrt 2)`
    2. `g(x) = {(-4text(,), x < 0), (\ \ \ 4text(,), x > 0):}`
    3. `text(See Worked Solutions)`
Show Worked Solution

a.   `-1 <= x^2 – 1 <= 1`

`=>\ 0 <= x^2<= 1`

`=>\ -sqrt2 <= x <=sqrt2`

`:.\ text(Domain:)\ [- sqrt 2 , sqrt 2]`

 

`sin^(-1)(x) in [-pi/2, pi/2]`

`=> 2 sin^(-1)(x) in [-2xx pi/2, 2 xx pi/2]`

`:.\ text(Range:)\ [-pi, pi]`

 

b.   

 

c.   `f(x) = 2 sin^(-1)(x^2 – 1)`

`f(x)` `=2 sin^(-1)(x^2 – 1)`

 
`:. f prime(x)= 4/sqrt(2 – x^2)\ \ \ (x > 0,\ \ text(by CAS))`

 

d.    `f prime(x)` `= (4x)/sqrt(x^2(2 – x^2))`
    `= (-4)/sqrt(2 – x^2)\ \ \ (x < 0)`

 

e.i.  `f prime(x)\ \ text(is defined when:)`

♦♦ Mean mark part (e)(i) 21%.

`2 – x^2 > 0\ \ and\ \ x!=0`

`:. x in (-sqrt2, 0) ∪ (0, sqrt 2)`

 

e.ii.  `g(x) = +- 4`

♦ Mean mark part (e)(ii) 49%, part (e)(iii) 48%.

`:.  g(x) = {(-4text(,), x < 0), (\ \ \ 4text(,), x > 0):}`

 

e.iii.   

Graphs, SPEC2 2015 VCAA 2 MC

The range of the function with rule  `f(x) = (2 - x)arcsin(x/2 - 1)`  is

A.   `[-pi,0]`

B.   `[-pi/2,pi/2]`

C.   `[-((2 - x)pi)/2,((2 - x)pi)/2]`

D.   `[0,4]`

E.   `[0,pi]`

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`A`

Show Worked Solution

`text(By CAS, create a graph of)\ \ f(x) = (2 – x)sin^(-1)(x/2 – 1).`

`=> A`

Algebra, SPEC2 2018 VCAA 2 MC

Consider the function  `f` with rule  `f(x) = 1/sqrt(sin^(-1)(cx + d))`, where  `c, d in R`  and  `c > 0`.

The domain of  `f` is

A.  `x > -d/c`

B.  `-d/c < x <= (1 - d)/c`

C.  `(-1 - d)/c <= x <= (1 - d)/c`

D. `x in R\ text(\) {-d/c}`

E.  `x in R` 

Show Answers Only

`B`

Show Worked Solution
`cx + d` `in [-1, 1]`
`cx` `in [-1-d, 1-d]`
`x` `in [(-1-d)/c, (1 – d)/c]\ \ …\ (1)`

 

`sin^(-1) (cx + d)` `> 0`
`cx + d` `>0`
`cx` `> -d`
`x` `> -d/c\ \ …\ (2)`

 

`(1) nn (2):`

`x in (-d/c, (1 – d)/c]`

 
`=>  B`