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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:}\)

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

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

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

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

   \(y=e\, \ln e=e\)

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

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

   \(e=\dfrac{1}{2e} \times e^2+c\)

   \(c=\dfrac{1}{2} e\)

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 C3 2005 HSC 5c

Find the coordinates of the point  `P`  on the curve  `y = 2e^x + 3x`  at which the tangent to the curve is parallel to the line  `y = 5x - 3`.  (3 marks)

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`(0, 2)`

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`text(Gradient of)\ \ y = 5x − 3\ text(is)\ 5.`

`y` `= 2e^x + 3x`
`(dy)/(dx)` `= 2e^x + 3`

 
`text(Find)\ \ x\ \ text(when)\ \ (dy)/(dx) = 5`

`5` `= 2e^x + 3`
`2e^x` `= 2`
`e^x` `= 1`
`x` `= 0`

 
`text(When)\ \ x = 0`

`y` `= 2e^0 + (3 xx 0)`
  `= 2`

 
`:.P\ \ text{has coordinates (0, 2)}`

Calculus, 2ADV C3 2014 HSC 15c

The line  `y = mx`  is a tangent to the curve  `y = e^(2x)`  at a point  `P`. 

  1. Sketch the line and the curve on one diagram.   (1 mark)

  2. Find the coordinates of  `P`.     (3 marks)

  3. Find the value of  `m`.   (1 mark)

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  1.   
  2. `P(1/2 ln (m/2), m/2)`
  3. `2e`
Show Worked Solution
i. 

 

ii. `y` `= e^(2x)`
  `dy/dx` `= 2e^(2x)`

 
`text(Gradient of)\ \ y = mx\ \ text(is)\ \ m`

♦ Mean mark 40%
COMMENT: Given `y= e^(ln(m/2))`, it follows `y=m/2`. Make sure you understand the arithmetic behind this (NB. Simply take the `ln` of both sides).

`text(Gradients equal when)`

`2e^(2x)` `= m`
`e^(2x)` `= m/2`
`ln e^(2x)` `= ln (m/2)`
`2x` `= ln (m/2)`
`x` `= 1/2 ln (m/2)`

 
`text(When)\ \ x = 1/2 ln (m/2)`

`y` `= e^(2 xx 1/2 ln (m/2))`
  `= e^(ln(m/2))`
  `= m/2`

 
`:.\ P (1/2 ln (m/2), m/2)`

 

iii.   `y=mx\ \ text(passes through)\ \ (0,0)\ text(and)\ (1/2 ln (m/2), m/2)`

♦♦ Mean mark 30%.

`text(Equating gradients:)`

`(m/2 – 0)/(1/2 ln (m/2) – 0)`  `=m`
`m/2` `=m xx 1/2 ln(m/2)`
`ln (m/2)` `= 1`
`m/2` `= e^1`
`m` `= 2e`