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Calculus, EXT1 EQ-Bank 25

Graph the polynomial  \(P(x)=(x+1)^2(2-x)^3\)  on the grid below, clearly identifying all axis intercepts.   (3 marks)

 

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Show Worked Solution

\(P(x)=(x+1)^2(2-x)^3 \ \ \Rightarrow\ \ \text{zeros at} \ \ x=-1,2\)

\(\text{At} \ \ x=-1, m (\text{multiplicity})=2 \ \ \Rightarrow\ \ \text{curve is a tangent to} \ x \text{-axis}\)

\(\text{At} \ \ x=2, m=3 \ \ \Rightarrow\ \ \text{curve has horizontal POI.}\)

\(\text{At} \ \ x=0, P(x)=(1)^2(2)^3=8\)
 

Calculus, EXT1 EQ-Bank 28

\(P(x)\) is a polynomial where  \(P(\alpha)=0\)  and  \(P^{\prime}(\alpha)=0\).

  1. Show that \((x-\alpha)^2\) is a factor of \(P(x)\).   (2 marks)

  2. The curve  \(y=x^3+b x^2+c x+4\)  is tangent to the \(x\)-axis at  \(x=-1\). Find the values of \(b\) and \(c\).   (3 marks)

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a.    \(P(\alpha)=0\ \ \Rightarrow\ \ (x-a)\ \text{is a factor of \(P(x)\)}\)

\(P(x)=(x-\alpha) \cdot Q(x)\)
 

\(P^{\prime}(x)=Q(x)+(x-a) \cdot Q(x)\)

\(\text{Since}\ \ P^{\prime}(\alpha)=0:\)

\(Q(\alpha)=0 \ \ \Rightarrow\ \ (x-\alpha) \ \text{is a factor of} \ \ Q(\alpha)\)
 

\(\text{Let} \ \ Q(x)=(x-\alpha) \cdot R(x)\)

\(P(x)=(x-\alpha)^2 \cdot R(x)\)

\(\therefore \ (x-\alpha)^2 \ \text{is a factor of} \ P(x).\)
 

b.    \(b=6, c=9\)

Show Worked Solution

a.    \(P(\alpha)=0\ \ \Rightarrow\ \ (x-a)\ \text{is a factor of \(P(x)\)}\)

\(P(x)=(x-\alpha) \cdot Q(x)\)
 

\(P^{\prime}(x)=Q(x)+(x-a) \cdot Q(x)\)

\(\text{Since}\ \ P^{\prime}(\alpha)=0:\)

\(Q(\alpha)=0 \ \ \Rightarrow\ \ (x-\alpha) \ \text{is a factor of} \ \ Q(\alpha)\)
 

\(\text{Let} \ \ Q(x)=(x-\alpha) \cdot R(x)\)

\(P(x)=(x-\alpha)^2 \cdot R(x)\)

\(\therefore \ (x-\alpha)^2 \ \text{is a factor of} \ P(x).\)
 

b.    \(\text{Since the curve is tangent at} \ \ x=-1\)

\(x=-1 \ \ \text{is a double root}\)

\(P(x)=x^3+b x^2+c x+4\)

\(P(-1)=-1+b-c+4=0 \ \ \Rightarrow\ \ b-c=-3\ \ldots\ (1)\)
 

\(P^{\prime}(x)=3 x^2+26 x+c\)

\(P^{\prime}(-1)=3-2 b+c=0 \ \ \Rightarrow\ \ -2 b+c=-3\ \ldots\ (2)\)
 

\(\text{Add} \ (1)+(2):\)

\(-b=-6 \ \ \Rightarrow\ \ b=6\)

\(\text{Substitute \(\ b=6\ \) into (1):}\)

\(-6-c=-3 \ \ \Rightarrow\ \ c=9\).

\(\therefore b=6, c=9\)

Calculus, EXT1 EQ-Bank 26

Graph the polynomial  \(p(x)=(x-1)\left(x^2+3 x+1\right)\), clearly identifying all axis intercepts.   (3 marks)

 

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Show Worked Solution

\(p(x)=(x-1)\left(x^2+3 x+1\right)\)

\(\text{Zeros:} \ \ x=1, x=\dfrac{-3 \pm \sqrt{9-4 \cdot 1 \cdot 1}}{2}=\dfrac{-3 \pm \sqrt{5}}{2}\ \ (\approx-2.62,-0.38)\)

\(\text{At}\ \ x=1, m(\text{multiplicity})=1 \ \Rightarrow \ \text{curve crosses}\  x\text {-axis}\)

\(\text{At} \ \ x=\dfrac{-3 \pm \sqrt{5}}{2}, m=1 \ \Rightarrow \ \text{curve crosses} \ x\text {-axis}\)

\(p(0)=(-1)(1)=-1\)

Calculus, EXT1 EQ-Bank 12

Let  \(P(x)=x^3+a x^2+b x-4\)  where \(a\) and \(b\) are real numbers.

If  \(x=2\)  is a double root of  \(P(x)=0\), find the values of \(a\) and \(b\).   (3 marks)

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Show Worked Solution

\(P(x)=x^3+a x^2+b x-4\)

\(P^{\prime}(x)=3 x^2+2 a x+b\)

\(\text{Since} \ \ x=2\ \ \text{is a double root,}\)

\(P(2)=0:\)

\(8+4 a+2 b-4\) \(=0\)
\(2 a+b\) \(=-2\ \ldots\ (1)\)

 

\(P^{\prime}(2)=0:\)

\(12+4 a+b\) \(=0\)
\(4 a+b\) \(=-12\ \ldots\ (2)\)

 

\([(1) \times 2]-(2):\)

\(b=8\)
 

\(\text{Substitute} \ \ b=8 \ \ \text{into (1):}\)

\(2a+8\) \(=-2\)
\(a\) \(=-5\)

Functions, EXT1 F2 EQ-Bank 24

The polynomial  \(P(x)=2 x^3-m x^2+n x+27\)  has a double root and  \(P(-3)=P^{\prime}(-3)=0\). 

Find the values of \(m\) and \(n\) and hence find the other root of \(P(x)\).   (3 marks)

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\(m=-15, n=36\)

\(\gamma= -\dfrac{3}{2}\)

Show Worked Solution

\(P(x)=2 x^3-m x^2+n x+27\)

\(P^{′}(x)=6x^2-2m x+n\)

\(P(-3)\) \(=2 \times (-3)^{3}-(-3)^2m-3n+27\)  
\(0\) \(=-54-9m-3n+27\)  
\(9m+3n\) \(=-27\ \ …\ (1)\)  

 

\(P^{′}(-3)\) \(=6 \times (-3)^2+6m+n\)  
\(6m+n\) \(=-54\)  
\(18m+3n\) \(=-162\ …\ (2)\)  

 
   \( (2)-(1):\)

\(9m=-135\ \ \Rightarrow  m=-15\)

\(6 \times -15+n=-54\ \ \Rightarrow \ n=36\)
 

\(P(x)=2 x^3+15 x^2+36 x+27\)

\(\text{Using product of roots:}\)

\(\alpha \beta \gamma\) \(=\dfrac{-d}{a}\)  
\((-3)^{2} \gamma\) \(=\dfrac{-27}{2} \)  
\(\gamma\) \(=\dfrac{-27}{2 \times 9} = -\dfrac{3}{2}\)  

Functions, EXT1 F2 2023 HSC 14b

Consider the hyperbola  \(y=\dfrac{1}{x}\)  and the circle  \((x-c)^2+y^2=c^2\), where \(c\) is a constant.

  1. Show that the \(x\)-coordinates of any points of intersection of the hyperbola and circle are zeros of the polynomial  \(P(x)=x^4-2 c x^3+1\).   (1 mark)

  2. The graphs of  \(y=x^4-2 c x^3+1\)  for  \(c=0.8\)  and  \(c=1\) are shown.
     

  1. By considering the given graphs, or otherwise, find the exact value of  \(c>0\)  such that the hyperbola  \(y=\dfrac{1}{x}\)  and the circle  \((x-c)^2+y^2=c^2\)  intersect at only one point.   (3 marks)

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i.    \(\text{See Worked Solutions}\)

ii.   \(\sqrt[4]{\dfrac{16}{27}}\approx 0.877\)

Show Worked Solution

i.     \(y=\dfrac{1}{x}\ …\ (1) \)

\((x-c)^2+y^2=c^2\ …\ (2) \)

\(\text{Substitute (1) into (2):}\)

\((x-c)^2+\Big{(}\dfrac{1}{x}\Big{)}^2 \) \(=c^2\)  
\(x^2-2cx+c^2+\dfrac{1}{x^2}\) \(=c^2\)  
\(x^4-2cx^3+1\) \(=0\)  
Mean mark (i) 53%.

ii.    \(\text{Graphs show that for some value of}\ \ 0.8 \leq c \leq 1,\)

\(P(x)\ \text{has a minimum that touches the}\ x\text{-axis once.}\)

\(P(x)\) \(=x^4-2cx^3+1\)  
\(P^{′}(x)\) \(=4x^3-6cx^2\)  

 
\(\text{Find}\ x\ \text{when}\ P^{′}(x)=0: \)

\(4x^3-6cx^2\) \(=0\)  
\(2x^2(2x-3c)\) \(=0\)  
\(x\) \(=\dfrac{3c}{2}\ \ (x \neq 0)\)  

 
\(\text{Find}\ c\ \text{when}\ P(\frac{3c}{2})=0: \)

\(\Big{(} \dfrac{3c}{2} \Big{)}^4-2c\Big{(} \dfrac{3c}{2} \Big{)}^3+1 \) \(=0\)  
\(\dfrac{81c^4}{16}-\dfrac{54c^4}{8}+1\) \(=0\)  
\(\dfrac{(108-81)c^4}{16}\) \(=1\)  
\(\dfrac{27c^4}{16}\) \(=1\)  
\(c^4\) \(=\dfrac{16}{27}\)  
\(c\) \(=\sqrt[4]{\dfrac{16}{27}}\approx 0.877\)  
Mean mark (ii) 19%.

Functions, EXT1 F2 2022 HSC 13d

The monic polynomial, `P`, has degree 3 and roots `alpha, \beta, \gamma`.

It is given that

           `alpha^(2)+beta^(2)+gamma^(2)=85\ \ and`

           `P^(')(alpha)+P^(')(beta)+P^(')(gamma)=87.`

Find  `alpha beta+beta gamma+gamma alpha`.   (3 marks)

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`-2`

Show Worked Solution

`P(x)=x^3-(alpha+beta+gamma)x^2+(alphabeta+alphagamma+betagamma)x-alphabetagamma`

`P^(′)(x)=3x^2-2(alpha+beta+gamma)x+alphabeta+alphagamma+betagamma`

`P^(′)(alpha)+P^(′)(beta)+P^(′)(gamma)`

  `=3alpha^2-2(alpha+beta+gamma)alpha+alphabeta+alphagamma+betagamma`

`+ 3beta^2-2(alpha+beta+gamma)beta+alphabeta+alphagamma+betagamma`

`+ 3gamma^2-2(alpha+beta+gamma)gamma+alphabeta+alphagamma+betagamma`

  `=3(alpha^(2)+beta^(2)+gamma^(2))-2(alpha^(2)+beta^(2)+gamma^(2))`

`-2(alphabeta+alphagamma+alphabeta+betagamma+alphagamma+betagamma)+3(alphabeta+alphagamma+betagamma)`

  `=(alpha^(2)+beta^(2)+gamma^(2))-(alphabeta+alphagamma+betagamma)`

`text{Substituting in given values:}`

`85-(alphabeta+alphagamma+betagamma)` `=87`  
`:.alphabeta+alphagamma+betagamma` `=-2`  

♦♦ Mean mark 33%.

Functions, EXT1′ F2 2019 HSC 4 MC

The polynomial  `2x^3 + bx^2 + cx + d`  has roots `1` and `-3`, with one of them being a double root.

What is a possible value of `b`?

  1. `-10`
  2. `-5`
  3. `5`
  4. `10`
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`D`

Show Worked Solution

`f(x) = 2x^3 + bx^2 + cx + d`

`f^{′}(x) = 6x^2 + 2bx + c`

`text(Roots at 1 and −3:)`

`f(1) = 0`

`2 + b + c + d` `= 0`
`b + c + d` `= -2\ \ text{… (1)}`

 
`f(-3) = 0`

`-54 + 9b-3c + d` `= 0`
`9b-3c + d` `= 54\ \ text{… (2)}`

 
`(2)-(1)`

`8b-4c` `= 56`
`2b-c` `= 14\ \ text{… (3)}`

 
`text(If double root at 1:)`

`f^{′}(1) = 0`

`6 + 2b + c` `= 0`
`2b + c` `= -6\ \ text{… (4)}`

 
`(3) + (4)`

`4b= 8\ \ =>\ \ b=2`

 
`text(If double root at – 3:)`

`f^{′}(-3) = 0`

`54-6b + c` `= 0`
`-6b + c` `= -54\ \ text{… (5)}`

 
`(3) + (5)`

`-4b=-40\ \ =>\ \ b=10`

`=>   D`

Functions, EXT1′ F2 2018 HSC 11b

The polynomial  `p(x) = x^3 + ax^2 + b`  has a zero at `r` and a double zero at 4.

Find the values of `a`, `b` and `r`.   (3 marks)

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`a =− 6, b = 32, r = -2`

Show Worked Solution

`p(x) = x^3 + ax^2 + b`

`p^{′}(x) = 3x^2 + 2ax`
 

`text(Double root at 4:)`

`p^{′}(4)` `=0`
`3 xx 16 + 8a` `= 0`
`a` `= −6`

 

`p(4)` `=0`
`64 + 16a + b` `= 0`
`64-96 + b` `= 0`
`b` `= 32`

 
`text(Roots of)\ p(x)\ text(are)\ \ 4, 4, r:`

` 4 + 4 +r` `= -a/1 = 6`
`:. r` `= -2`

Functions, EXT1′ F2 2017 HSC 12d

Let `P(x)` be a polynomial.

  1. Given that `(x-alpha)^2` is a factor of `P(x)`, show that
  2. `qquad qquad P(alpha) = P^{′}(alpha) = 0`.   (2 marks)

  3. Given that the polynomial  `P(x) = x^4-3x^3 + x^2 + 4`  has a factor `(x-alpha)^2`, find the value of `alpha`.   (2 marks)

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i.    `text(Proof)\ \ text{(See Worked Solutions)}`

ii.   `2`

Show Worked Solution
i.   `P(x)` `= (x-alpha)^2 · Q(x)`
  `P^{′}(x)` `= 2 (x-alpha) · Q (x) + (x-alpha)^2 *Q^{′}(x)`
    `= (x-alpha) [2 Q (x) + (x-alpha) *Q^{′}(x)]`

 

`P (alpha)` `= 0 xx Q(x) = 0`
`P^{′}(alpha)` `= 0 [2Q (x) + 0 xx Q^{′}(x)] = 0`

 
`:. P(alpha) = P^{′}(alpha) = 0\ text(… as required.)`

 

ii.   `P(x)` `= x^4-3x^3 + x^2 + 4`
  `P^{′}(x)` `= 4x^3-9x^2 + 2x`
    `= x (4x^2-9x + 2)`
    `= x (4x-1) (x-2)`

 
`:. P^{′}(x) = 0\ \ text(when)\ \ x = 0, 1/4 or 2`

`=>\ text(Multiple roots may exist at)\ \ x=0, 1/4 or 2.`

`text(Test each root in)\ \ P(x):`

`P(0)` `= 0-0 + 0 + 4 = 4`
`P(1/4)` `= (1/4)^4-3(1/4)^3 + (1/4)^2 + 4= 4 \frac{5}{256}`
`P(2)` `= 16-3(8) + 4 + 4 = 0`

 
`(x-2)^2\ \text(is a factor of)\ P(x)`

`:. alpha = 2`

Functions, EXT1′ F2 2016 HSC 13d

Suppose  `p(x) = ax^3 + bx^2 + cx + d`  with `a, b, c` and `d` real, `a != 0.`

  1. Deduce that if  `b^2-3ac < 0`  then `p(x)` cuts the `x`-axis only once.   (2 marks)

  2. If  `b^2-3ac = 0`  and  `p(-b/(3a)) = 0`, what is the multiplicity of the root  `x = -b/(3a)?`   (2 marks)

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a.    `text(See Worked Solutions)`

b.    `3`

Show Worked Solution

a.    `p(x) = ax^3 + bx^2 + cx + d`

`p^{′}(x) = 3ax^2 + 2bx + c`

`=> p(x)\ text(will cut the)\ xtext(-axis once only if)\ \ Delta(p^{′}(x)) < 0`

`(2b)^2-4(3a)c` `< 0`
`4b^2-12ac` `< 0`
`b^2-3ac` `< 0`

 

b.    `p(-b/(3a)) = 0`

`p^{′}(−b/(3a))` `=3a(- b/(3a))^2 + 2b (- b/(3a))+c`
  `=- b^2/(3a)+c`
  `=0\ \ \ (text{given}\ \ b^2-3ac = 0)`

 
`:.\ text(Multiplicity at least 2.)`

♦♦ Mean mark 32%.

 
`p^{″}(x) = 6ax + 2b`

`p^{″}(−b/(3a))=6a(−b/(3a)) + 2b=0`

`:. text(Multiplicity of)\ \ x=− b/(3a)\ \ text(is 3.)`

Functions, EXT1′ F2 2016 HSC 2 MC

Which polynomial has a multiple root at  `x = 1?`

  1. `x^5-x^4-x^2 + 1`
  2. `x^5-x^4-x-1`
  3. `x^5-x^3-x^2 + 1`
  4. `x^5-x^3-x + 1`
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`=> C`

Show Worked Solution

`\text{By trial and error}`

`text(Consider option)\ C:`

`P(1)` `= 1-1-1 + 1 = 0`
`P^{′}(x)` `= 5x^4-3x^2-2x`
`P^{′}(1)` `= 5-3-2 = 0`

 
`:.\ text(Multiple root at)\ x = 1`

`=> C`

Functions, EXT1′ F2 2015 HSC 4 MC

The polynomial  `x^3 + x^2-5x + 3`  has a double root at  `x = alpha.`

What is the value of  `alpha ?`

  1. `-5/3`
  2. `-1`
  3. `1`
  4. `5/3`
Show Answers Only

`C`

Show Worked Solution

`P^{′}(x)= 3x^2 + 2x-5= (3x + 5) (x-1)`

 `:. text(Only possible double roots occur when)`

`x=1\ \ text(or)\ \ x=-5/3`
 

`P(1)=1+1-5+3=0`

`:. x = 1\ \ text(is double root.)`

`=>  C`

Functions, EXT1′ F2 2009 HSC 3c

Let  `P(x) = x^3 + ax^2 + bx + 5`, where `a` and `b` are real numbers.

Find the values of `a` and `b` given that `(x-1)^2` is a factor of `P(x).`   (3 marks)

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`a = 3,\ b = -9`

Show Worked Solution
`P(x)` `= x^3 + ax^2 + bx + 5`
`P(1)` `= 1 + a + b + 5=0`
 `:.b` `= -a-6\ \ \ \ …\ (1)`

 

`P^{′}(x)` `= 3x^2 + 2ax + b`
`P^{′}(1)` `= 3 + 2a + b=0`
 `2a+b` `= -3\ \ \ \ …\ (2)`

 
`text(Substitute)\ \ b=-a-6\ \ text{from (1) into (2):}`

`2a+(-a-6)=-3`

`:.a = 3,\ \ \ b = -3-6=-9`

Functions, EXT1′ F2 2013 HSC 15b

The polynomial  `P(x) = ax^4 + bx^3 + cx^2 + e`  has remainder `-3` when divided by  `x-1`. The polynomial has a double root at  `x = -1.`

  1. Show that  `4a + 2c = -9/2.`   (2 marks)

  2. Hence, or otherwise, find the slope of the tangent to the graph  `y = P(x)`  when  `x = 1.`   (1 mark)

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a.    `text(Proof)\ \ text{(See Worked Solutions)}`

b.    `-9`

Show Worked Solution
a.   `P(x)` `= ax^4 + bx^3 + cx^2 + e`
  `P^{′}(x)` `=4ax^3 + 3bx^2 + 2cx`

 
`P(1)=-3:`

`a+b+c+e=-3\ \ \ …\  (1)`

`P(-1)=0:`

`a-b+c+e=0\ \ \ …\ (2)`

`P^{′}(-1)=0`

`-4a+3b-2c=0\ \ \ …\ (3)`

 
`(1)-(2)`

`2b=-3\ \ =>\ \ b=-3/2`

 
`text(Substitute into)\ (3):`

`:.4a+2c=3b=-9/2\ \ \ \ text(… as required)`

 

b.    `P^{′}(1)` `= 4a + 3b + 2c`
  `= -9/2-9/2`
  `=-9`

Functions, EXT1′ F2 2014 HSC 14a

Let  `P(x) =x^5-10x^2 +15x-6`.

Show that  `x = 1`  is a root of `P(x)` of multiplicity three.   (2 marks)

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`text{See Worked Solutions}`

Show Worked Solution

`P(x) =x^5-10x^2 +15x-6`

`P(1) = 1-10 + 15-6 = 0`
 

`P^{′}(x)` `= 5x^4-20x + 15`
`P^{′}(1)` `= 5-20 + 15 = 0`
`P^{″}(x)` `= 20x^3-20`
`P^{″}(1)` `= 20-20 = 0`
`P^{‴}(x)` `= 60x^2`
`P^{‴}(1)` `= 60 ≠ 0`

 
`:.x = 1\ text(is a root of)\ P(x)\ text(of multiplicity 3.)`

Functions, EXT1 F2 2013 HSC 4 MC

Which diagram best represents the graph  `y = x (1-x)^3 (3-x)^2`?
 

2013 4 mc1

2013 4 mc2

2013 4 mc3

2013 4 mc4

Show Answers Only

`D`

Show Worked Solution

`y = x(1-x)^3 (3-x)^2`

`text(By elimination)`

`text(Consider when)\ \ x < 0:`

`y = text{(–ve)} xx text{(+ve)} xx text{(+ve)} < 0`

`:.\ text(Cannot be)\ A\ text(or)\ C`
 

`text(Consider the cubic factor)\ (1-x)^3:`

`text(The graph must have a stationary point at)\ x = 1`

`:.\ text(Cannot be)\ B`

`=>  D`