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Functions, SPEC2 2023 VCAA 2 MC

The graph of  \(y=\dfrac{x^3}{a x^2+b x+c}\)  has asymptotes given by  \(y=2 x+1\)  and  \(x=1\). The values of \(a, b\) and \(c\) are, respectively

  1. \(2,-4,2\)
  2. \( \dfrac{1}{2},-\dfrac{1}{4},-\dfrac{1}{4} \)
  3. \( \dfrac{1}{2}, \dfrac{1}{4},-\dfrac{3}{4} \)
  4. \( \dfrac{1}{2},-\dfrac{1}{4},-\dfrac{3}{4} \)
  5. \(2,-4,-8\)
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\(B\)

Show Worked Solution

\(\text{Using partial fractions (by calc):}\)

\(\text{Remainder}\ =  \dfrac{x}{a}-\dfrac{b}{a^2} =  2x+1\)

\(a=\dfrac{1}{2},\ \ b=-\dfrac{1}{4} \)

\(x=1\ \ \text{is a solution of}\ \ ax^2+bx+c=0\)

\(c=-\dfrac{1}{4} \)

\(\Rightarrow B\)

Calculus, SPEC2 2021 VCAA 1

Let  `f(x) = ((2x - 3)(x + 5))/((x - 1)(x + 2))`.

  1. Express  `f(x)`  in the form  `A + (Bx + C)/((x - 1)(x + 2))`,  where `A`, `B` an `C` are real constants.  (1 mark)
  2. State the equation of the asymptotes of the graph of `f`.  (2 marks)
  3. Sketch the graph of `f` on the set of axes below. Label the asymptotes with their equations, and label the maximum turning point and the point of inflection with their coordinates, correct to two decimal places. Label the intercepts with the coordinate axes.  (3 marks)
     

     
  4. Let  `g_k(x) = ((2x - 3)(x + 5))/((x - k)(x + 2))`, where `k` is a real constant.
  5.  i. For what values of `k` will the graph of `g_k`, have two asymptotes?  (2 marks)
  6. ii. Given that the graph of `g_k` has more than two asymptotes, for what values of `k` will the graph of `g_k` have no stationary points?  (2 marks)
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  1. `f(x) = 2 + (5x – 11)/((x – 1)(x + 2))`
  2. `text(Horizontal asymptote:)\ y = 2`

  3.  
  4.  i. `k = 3/2 => g_k(x) = 2 + 6/(x + 2)`
  5. ii. `k , -5\ \ text(or)\ \ k > 3/2`
Show Worked Solution

a.   `text{By CAS  (prop Frac}\ f(x)):`

`f(x) = 2 + (5x – 11)/((x – 1)(x + 2))`

 

b.   `text(Vertical asymptotes:)\ x = 1, x = –2`

`text(As)\ \ x -> ∞, y -> 2`

`text(Horizontal asymptote:)\ y = 2`

♦ Mean mark part (c) 48%.

 

c.   

 

d.i.   `text(Two asymptotes only when:)`

♦♦ Mean mark part (d)(i) 24%.

`k = -2 \ => \ g_k(x) = 2 – (23 + x)/((x + 2)^2)`

`k = -5 \ => \ g_k(x) = 2 – 7/(x + 2)`

`k = 3/2 \ => \ g_k(x) = 2 + 6/(x + 2)`

 

d.ii.   `text(By CAS, solve)\ \ d/(dx)(g_k(x)) = 0\ \ text(for)\ \ x:`

♦♦♦ Mean mark part (d)(ii) 16%.

`x = (-4k + 15 ± sqrt(-21(2k^2 + 7k – 15)))/(2k + 3)`
 

`text(No solutions occur when:)`

`k = -3/2\ \ text(or)`

`2k^2 + 7k – 15 < 0`

`=> k < -5\ \ text(or)\ \ k > 3/2`

Algebra, SPEC2 2020 VCAA 7 MC

For non-zero constants `a` and `b`, where  `b < 0`, the expression  `1/(ax(x^2 + b))`  in partial fraction form with linear denominators, where  `A, B`  and  `C`  are real constants, is

  1. `A/(ax) + (Bx + C)/(x^2 + b)`
  2. `A/(ax) + B/(x + sqrtb) + C/(x - sqrtb)`
  3. `A/x + B/(ax + sqrt|b|) + C/(ax - sqrt|b|)`
  4. `A/x + B/(x + sqrt|b|) + C/(x - sqrt|b|)`
  5. `A/(ax) + B/((x + sqrtb)^2) + C/(x + sqrtb)`
Show Answers Only

`D`

Show Worked Solution

♦♦♦ Mean mark 26%.
MARKER’S COMMENT: Option A results from not considering `b<0`.
`1/(ax(x^2 + b))` `= 1/a (A_1/x + B_1/(x + sqrt|b|) + C_1/(x – sqrt|b|))`
  `= A/x + B/(x + sqrt|b|) + C/(x – sqrt|b|)`

 
`=>D`

 

Algebra, SPEC1 VCE SM-Bank 7

Given that  `(16x - 43)/((x - 3)^2 (x + 2))`  can be written as

 
`(16x - 43)/((x - 3)^2 (x + 2)) = a/(x - 3)^2 + b/(x - 3) + c/(x + 2)`,

  
where  `a, b` and `c in RR`, find  `a, b and c.`  (3 marks)

    1.  
Show Answers Only

`a = 1, b = 3, c = -3`

 

Show Worked Solution

`(16x – 43)/((x – 3)^2 (x + 2)) = a/(x – 3)^2 + b/(x – 3) + c/(x + 2)`

`16x – 43 = a (x + 2) + b (x – 3) (x + 2) + c (x – 3)^2`
 

`text(When)\ \ x = 3,\ \ 5a =5\ \ =>a=1`

`text(When)\ \ x=-2,\ \ 25c=-75\ \ =>c=-3`

`text(When)\ \ x=0`

`-43` `= 2(1) – 6b + (-3)(-3)^2`
`6b` `= 18`
`b` `=3`

 
`:.a=1, b=3, c=-3`

Algebra, SPEC1 VCE SM-Bank 4

Find  `A, B`  and  `C in RR`, such that

`1/(x(x^2 + 2)) = A/x + (Bx + C)/(x^2 + 2).`   (2 marks)

Show Answers Only

`A = 1/2,\ \ \ B = -1/2,\ \ \ C = 0`

Show Worked Solution
`1/(x(x^2 + 2))` `= A/x + (Bx + C)/(x^2 + 2)`
`1` `= A (x^2 + 2) + (Bx + C) x`
`1` `= (A+B)x^2 + Cx + 2A`

 

`2A = 1,\ \ =>A=1/2`

`C=0`

`A + B = 0,\ \ =>B=-1/2`

`:.A = 1/2,\ \ \ \ B = -1/2,\ \ \ \ C = 0`

Algebra, SPEC1 VCE SM-Bank 1

Find numbers  `A, B` and `C in RR`, such that

`(x^2 + 8x + 11)/((x -3)(x^2 + 2)) = A/(x - 3) + (Bx + C)/(x^2 + 2).`  (2 marks)

Show Answers Only

`A = 4,\ B = -3,\ C = -1`

Show Worked Solution

`(x^2 + 8x + 11)/((x -3)(x^2 + 2)) = A/(x – 3) + (Bx + C)/(x^2 + 2)`

`x^2 + 8x + 11` `=A(x^2 + 2)+(Bx+C)(x-3)`
  `=(A+B)x^2+(C-3B)x+(2A-3C)`

 

`A+B=1\ \ \ => B=1-A`

`C-3B=8\ \ \ =>C=11-3A`

`2A-3C=11 \ \ \ =>2A-33+9A=11\ \ \ =>A=4`

`B=1-4=-3`

`C=11-12=-1`

 `:.A=4, B=-3, C=-1`

Algebra, SPEC2-NHT 2017 VCAA 5 MC

Given that `A, B, C` and `D` are non-zero rational numbers, the expression  `(3x + 1)/(x(x - 2)^2)`  can be represented in partial fraction form as

A.   `A/x + B/((x - 2))`

B.   `A/x + B/(x - 2)^2`

C.   `A/x + B/((x - 2)) + C/(x - 2)^2`

D.   `A/x + B/x^2 + C/((x - 2))`

E.   `A/x + (Bx)/((x - 2)) + (Cx + D)/(x - 2)^2`

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`C`

Show Worked Solution

`(3x + 1)/(x(x – 2)^2) = A/x + B/(x – 2) + C/(x – 2)^2`

 
`=>   C`

Algebra, SPEC2 2018 VCAA 3 MC

Which one of the following, where `A, B, C` and `D` are non-zero real numbers, is the partial fraction form for the expression

      `(2x^2 + 3x + 1)/{(2x + 1)^3 (x^2 - 1)}?`

A.  `A/(2x + 1) + B/(x - 1) + C/(x + 1)`

B.  `A/(2x + 1) + B/(2x - 1)^2 + C/(2x + 1)^3 + (Dx)/(x^2 - 1)`

C. `A/(2x + 1) + (Bx + C)/(x^2 - 1)`

D.  `A/(2x + 1) + B/(2x + 1)^2 + C/(x - 1)`

E.  `A/(2x + 1) + (Bx + C)/(2x + 1)^2 + D/(x - 1)` 

Show Answers Only

`D`

Show Worked Solution

`(2x^2 + 3x + 1)/{(2x + 1)^3 (x^2 – 1)}`

♦ Mean mark 46%.

`= ((2x^2 + 2x) + (x + 1))/((2x + 1)^3 (x – 1) (x + 1))`

`= (2x(x + 1) + (x + 1))/((2x + 1)^3 (x – 1) (x + 1))`

`= ((2x + 1)(x + 1))/((2x + 1)^3 (x – 1) (x + 1))`

`= 1/((2x + 1)^2(x – 1))`

`= A/(2x + 1) + B/(2x + 1)^2 + C/(x – 1)`

 
`=>  D`