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Calculus, 2ADV C4 2023 HSC 28

The curve  \(y=f(x)\)  is shown on the diagram. The equation of the tangent to the curve at point  \(T(-1,6)\)  is  \(y=x+7\). At a point \(R\), another tangent parallel to the tangent at \(T\) is drawn.
 

The gradient function of the curve is given by  \(\dfrac{dy}{dx}=3x^2-6x-8\).

Find the coordinates of \(R\).  (4 marks)

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\(R(3,-22)\)

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\(\text{Gradient at}\ R=1:\)

\(3x^2-6x-8\) \(=1\)  
\(3x^2-6x-9\) \(=0\)  
\(3(x^2-2x-3\) \(=0\)  
\(3(x+1)(x-3)\) \(=0\)  
♦ Mean mark 51%.

\(x\text{-coordinate of}\ R = 3\)

\(y\) \(=\int 3x^2-6x-8\ dx\)  
  \(=x^3-3x^2-8x+c\)  

 
\(\text{Graph passes through}\ (-1,6):\)

\(6\) \(=-1-3+8+c\)  
\(c\) \(=2\)  

 
\(y=x^3-3x^2-8x+2\)

\(\text{When}\ x=3:\)

\(y=27-27-24+2=-22\)

\(\therefore R(3,-22)\)

Calculus, 2ADV C3 2017 HSC 9 MC

The graph of  `y = f^{′}(x)`  is shown.
 

The curve  `y = f (x)`  has a maximum value of 12.

What is the equation of the curve  `y = f (x)`?

  1. `y = x^2-4x + 12`
  2. `y = 4 + 4x-x^2`
  3. `y = 8 + 4x-x^2`
  4. `y = x^2-4x + 16`
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`C`

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`text(Find the equation of)\ \ f^{′}(x):`

`m = -2,\ \ y text(-int) = 4`

`y = -2x + 4`

`f(x)` `= int -2x + 4\ dx`
  `= -x^2 + 4x + c`

 

`text(Maximum)\ \ f(x) = 12\ \ text(when)\ \ f^{′}(x) = 0:`

`-2x + 4` `= 0`
`x` `= 2`

 
`text(Substitute)\ \ x=2\ \ text(into)\ \ f(x):`

`:. 12` `= -2^2 + 4 ⋅ 2 + c`
`c` `= 8`

 

`:. f(x) = 8 + 4x-x^2`

`=>  C`

Calculus, 2ADV C4 2008 HSC 9c

A beam is supported at  `(-b, 0)`  and  `(b, 0)`  as shown in the diagram.
 

2008 9c

 
It is known that the shape formed by the beam has equation  `y = f(x)`, where  `f(x)`  satisfies

  `f^{″}(x)` `= k (b^2-x^2),\ \ \ \ \ `(`k` is a positive constant) 
and        `f^{′}(-b)` `= -f'(b)`.

 

  1. Show that  `f^{′}(x) = k (b^2x-(x^3)/3)`.   (2 marks)

  2. How far is the beam below the  `x`-axis at  `x = 0`?   (2 marks)

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  1. `text(Proof)\ text{(See Worked Solutions)}`
  2. `(5kb^4)/12\ text(units)`
Show Worked Solution
i.    `text(Show)\ \ f^{′}(x) = k (b^2x-x^3/3)`
`f^{″}(x)` `= k (b^2 – x^2)`
`f^{′}(x)` `= int k (b^2-x^2)\ dx`
  `= k int b^2-x^2\ dx`
  `= k (b^2x-x^3/3) + c`

 

`text(S)text(ince S.P. exists at)\ \ x = 0`

`=> f^{′}(x)` `= 0\ \ text(when)\ \  x = 0`
`0` `= k (b^2 * 0-0) + c`
`c` `= 0`

 

`:.\ f^{′}(x) = k (b^2x-x^3/3)\ \ \ text(… as required)`

 

ii.   `f(x)` `= int f^{′}(x)\ dx`
    `= k int b^2x-x^3/3\ dx`
    `= k ((b^2x^2)/2-x^4/12) + c`

 

`text(We know)\ \ f(x) = 0\ \ text(when)\ \ x = b`

`=> 0` `= k ( (b^2*b^2)/2-b^4/12) + c`
`c` `= -k ( (6b^4)/12-b^4/12)`
  `= -k ( (5b^4)/12 )`
  `= -(5kb^4)/12`

 

`:.\ text(When)\ \ x = 0, text(the beam is)\ \ (5kb^4)/12\ \ text(units)`

`text(below the)\ x text(-axis.)`

Calculus, 2ADV C4 2008 HSC 5a

The gradient of a curve is given by  `dy/dx = 1-6 sin 3x`. The curve passes through the point  `(0, 7)`.

What is the equation of the curve?   (3 marks)

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`y = x + 2 cos 3x + 5`

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`dy/dx` `= 1-6 sin 3x`
`y` `= int 1-6 sin 3x\ dx`
  `= x + 2 cos 3x + c`

 

`text(Passes through)\ (0,7):`

`=> 0 + 2 cos 0 + c` `= 7`
`2 + c` `= 7`
`c` `= 5`

 

`:.\ text(Equation is)\ \ \ y = x + 2 cos 3x + 5`

Calculus, 2ADV C4 2014 HSC 11f

The gradient function of a curve  `y = f(x)`  is given by  `f^{′}(x) = 4x-5`.  The curve passes through the point  `(2, 3)`.

Find the equation of the curve.  (2 marks)

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`f(x) = 2x^2-5x + 5`

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`f^{′}(x)` `= 4x-5`
`f(x)` `= int 4x-5\ dx`
  `= 2x^2-5x + C`

 

`text(Given)\ \ f(x)\ text(passes through)\ (2,3):`

`3` `= 2(2^2)-5(2) + C`
`3` `= 8-10 + C`
`C` `= 5`

 
`:.\ f(x) = 2x^2-5x + 5`

Calculus, 2ADV C4 2011 HSC 4c

The gradient of a curve is given by  `dy/dx = 6x-2`.  The curve passes through the point `(-1, 4)`. 

What is the equation of the curve?   (2 marks)

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 `y = 3x^2-2x-1`

Show Worked Solution

`dy/dx = 6x-2` 

`y` `= int 6x-2\ dx`
  `= 3x^2-2x + c`

 
`text{Since it passes through}\ (-1,4),`

`4` `= 3 (-1)^2-2(-1) + c`
`4` `= 3 + 2 + c`
`c` `= -1`

 
`:. y = 3x^2-2x-1`

Calculus, 2ADV C4 2013 HSC 16a

The derivative of a function `f(x)` is  `f^{′}(x) = 4x-3`.  The line  `y = 5x-7`  is tangent to the graph `f(x)`.

Find the function `f(x)`.   (3 marks) 

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`f(x) = 2x^2-3x + 1`

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`text(Solution 1)`

`f^{′}(x)` `=4x-3`
`f(x)` `= int 4x-3\ dx`
  `= 2x^2-3x + c`

 
`text(Intersection when)`

`2x^2-3x + c = 5x-7`

`2x^2-8x + (7 + c) = 0`

  
`text(S)text(ince)\ \ y=5x-7\ \ text(is a tangent)\ => Delta =0`

`b^2-4ac` `=0`
`(-8)^2-[4xx2xx(7+c)]` `= 0`
`64-56-8c` `=8`
`8c` `=8`
`c` `=1`

 
`:.f(x) = 2x^2-3x + 1`

 
`text(Solution 2)`

`f^{′}(x)=4x-3`

`y=5x-7\ \ text{(Gradient = 5)}`

`=>4x-3` `=5`
`x` `=2`

 
`f(x) = 2x^2-3x + 1`

`f(x)\ text{passes through (2, 3)}`

`f(2)` `=2xx 2^2-3(2)+c`
`3` `=8-6+c`
`c` `=1`

 
`:.f(x) = 2x^2-3x + 1`