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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 2020 HSC 29

The diagram shows the graph of  `y = c ln x, \ c > 0`.
 

  1. Show that the equation of the tangent to  `y = c ln x`  at  `x = p`, where  `p > 0`, is
     
    `\ \ \ \ \ y = c/p x - c + c ln p`.  (2 marks)

  2. Find the value of `c` such that the tangent from part (a) has a gradient of 1 and passes through the origin.  (2 marks)

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  1. `text(See Worked Solutions)`
  2. `c = e`
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a.   `y = c ln x`

`(dy)/(dx) = c/x`

`text(At)\ x = p,`

`m_text(tang) = c/p`

`text(T)text(angent passes through)\ (p, c ln p)`

 
`:.\ text(Equation of tangent)`

`y – c ln p` `= c/p (x – p)`
`y` `= c/p x – c + c ln p`

 

b.   `text(If)\ m_text(tang) = 1,`

♦ Mean mark part (b) 40%.
`c/p` `= 1`
`c` `= p`

 
`text(If tangent passes through)\ (0, 0)`

`0` `= −c + c ln c`
`0` `= c(ln c – 1)`

 
`ln c = 1\ \ (c > 0)`

`:. c = e`

Calculus, 2ADV C3 2005 HSC 2d

Find the equation of the tangent to  `y = log_ex`  at the point  `(e, 1)`.  (2 marks)

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`y = x/e`

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`y = log_ex`

`dy/dx = 1/x`

`text(At)\ \ (e, 1),`

`m = 1/e`
 

`text(Equation of tangent,)\ \ m = 1/e,\ text(through)\ (e, 1)`

`y – y_1` `= m(x – x_1)`
`y – 1`  `= 1/e(x -e)`
`y – 1`  `= x/e – 1`
`y`  `= x/e`

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`
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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`

Calculus, 2ADV C3 2010 HSC 2c

Find the gradient of the tangent to the curve  `y=ln (3x)`  at the point where  `x=2`.     (2 marks) 

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`1/2`

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`y=ln\ (3x)`

CAUTION: Read the question carefully! MANY wasted valuable exam time finding the equation of the tangent here.

`dy/dx=3/(3x)=1/x`
 

`text(At)\ \ x=2,`   

`dy/dx=1/2`

`:.\ text(The gradient at)\ \ x=2\ \ text(is)\ \ 1/2.`