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Calculus, 2ADV C3 2025 HSC 24

The graphs of  \(y=e\, \ln x\)  and  \(y=a x^2+c\)  are shown. The line  \(y=x\)  is a tangent to both graphs at their point of intersection.
 

Find the values of \(a\) and \(c\).   (4 marks)

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\(a=\dfrac{1}{2e}, \ c=\dfrac{1}{2} e\)

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\begin{array}{ll}
y=e\, \ln x\ \ldots \ (1) & y=a x^2+c\ \ldots \ (2) \\
y^{\prime}=\dfrac{e}{x} & y^{\prime}=2 a x
\end{array}

\(\text{At point of tangency, gradient}=1.\)

\(\text{Find \(x\) such that:}\)

   \(\dfrac{e}{x}=1 \ \Rightarrow \ x=e\)

\(\text{At} \ \ x=e, \ y^{\prime}=2 a x=1:\)

   \(2ae=1 \ \Rightarrow \ a=\dfrac{1}{2e}\)

♦ Mean mark 41%.

\(\text{Find point of tangency using (1):}\)

   \(y=e\, \ln e=e\)

\(\text{Point of tangency at} \ (e,e).\)
 

\(\text{Since} \ (e, e) \text{ lies on (2):}\)

   \(e=\dfrac{1}{2e} \times e^2+c\)

   \(c=\dfrac{1}{2} e\)

Calculus, 2ADV C3 2024 HSC 29

Consider the curve  \(y=ax^2+bx+c\), where  \(a \neq 0\).

At a particular point, the tangent and normal to the curve are given by  \(t(x)=2x+3\)  and  \(n(x)=-\dfrac{1}{2}x-2\)  respectively.

The curve has a minimum turning point at  \(x=-4\).

Find the values of \(a, b\) and \(c\).   (4 marks)

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\(a=\dfrac{1}{2}, b=4, c=5\)

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\(y\) \(=ax^{2}+bx+c\)  
\(y^{′}\) \(=2ax+b\)  
♦♦ Mean mark 35%.

\(\text{SP’s when}\ \ y^{′}=0\ \text{and}\ \ x=-4:\)

\(-8a+b=0\ …\ (1)\)

\(\text{Find point where}\ t(x), n(x)\ \text{and curve intersect:}\)

\(2x+3\) \(=-\dfrac{1}{2}x-2\)  
\(4x+6\) \(=-x-4\)  
\(5x\) \(=-10\)  
\(x\) \(=-2\)  

 
\(\Rightarrow\ \text{Intersection at}\ (-2,-1)\)
 

\(\text{At}\ \ x=-2,\ m_{\text{tang}} = -4a+b\)

\(-4a+b=2\ …\ (2) \)

   \(\text{Subtract}\ (2)-(1):\)

\(4a=2\ \ \Rightarrow\ \ a=\dfrac{1}{2}\)

   \(\text{Substitute}\ \ a=\dfrac{1}{2}\ \ \text{into (1):}\)

\(\Rightarrow\ \ b=4\)

\(\text{Since}\ (-2,-1)\ \text{lies on}\ y: \)

\(-1=\dfrac{1}{2}(-2)^{2}+4(-2)+c\)

\(\Rightarrow c=5\)

\(\therefore a=\dfrac{1}{2}, b=4, c=5\)

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`

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`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`