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Calculus, 2ADV C3 2025 MET2 16 MC

Consider the function  \(h(x)=a\, \log _e(b x)\), where \(a, b \in R\).

Given that its derivative \(h^{\prime}(x)\) has range \((0, \infty)\), which of the following must be true?

  1. \( a>0\)  only
  2. \( a>0\)  and  \(b<0\)
  3. \(a>0\)  and  \(b>0\)
  4. \(a b>0\)
Show Answers Only

\(D\)

Show Worked Solution

\(h(x)=a\, \log _e(b x)\)

\(h^{\prime}(x)=a\, \times \dfrac{b}{bx}=\dfrac{a}{x}\)

\(\text{Consider} \ \ h(x)=a\, \log _e(b x) \ \Rightarrow \ b x>0\)
 

\(\text{Case 1:} \ \ b>0 \ \Rightarrow \ x>0 \ (\text{since} \ \ b x>0 )\)

\(\dfrac{a}{x} \ \ \text{is only positive when}\ \  a>0\)
 

\(\text {Case 2:} \ \ b<0 \ \Rightarrow \ x<0 \ \ (\text{since} \ \ b x>0)\)

\(\dfrac{a}{x} \ \ \text{is only positive when} \ \  a<0\)

\(\text{In both cases,} \ ab>0\)

\(\Rightarrow D\)

Calculus, 2ADV C3 2025 HSC 10 MC

The graph of  \(y=f(x)\), with all its stationary points, is shown.
 

How many stationary points does the graph of  \(y=f\left(e^x\right)\)  have?

  1. 0
  2. 1
  3. 2
  4. 3
Show Answers Only

\(C\)

Show Worked Solution

\(y=f(e^{x})\ \ \Rightarrow\ \ y^{\prime}=e^{x} \times f(e^{x}) \)

\(\text{Find number of \(x\) values where}\ \ y^{\prime}=0.\)

\(\text{Since}\ e^{x} \in (0, \infty)\ \text{for all}\ x: \)

\(\text{Stationary points of \(f(e^x)\) = 2 (SP’s of \(f(x)\) for}\ x \in (0, \infty)).\)

\(\Rightarrow C\)

♦♦♦ Mean mark 29%.

Calculus, 2ADV C3 2025 HSC 9 MC

The diagram shows the graph of  \(y=f^{\prime}(x)\).
 

Given  \(f(1)=6\), which interval includes the best estimate for \(f(1.1)\) ?

  1. \([6.2,6.4)\)
  2. \([6.0,6.2)\)
  3. \([5.8,6.0)\)
  4. \([5.6,5.8)\)
Show Answers Only

\(B\)

Show Worked Solution

\(\text{Gradient of \(f(x)\)  at  \(x=1\)  is 2 (see graph).}\)

\(\text{Gradient of \(f(x)\)  at  \(x=1.1\)  is slightly below 2 (see graph).}\)

\(\text{As \(x\) increases 0.1 (from 1.0 to 1.1), \(y\) will increase less than 0.2 units.}\)

\(\therefore f(1.1) \in [6.0,6.2)\)

\(\Rightarrow B\)

♦♦♦ Mean mark 27%.

Calculus, 2ADV C3 EQ-Bank 5 MC

When  \(x=-2\)  on the curve  \(y=f(x)\), the following is true:

\(\dfrac{dy}{dx}<0\)  and  \(\dfrac{d^2 y}{d x^2}>0\)

At  \(x=-2, f(x)\) is

  1. Increasing and concave up
  2. Decreasing and concave up
  3. Increasing and concave down
  4. Decreasing and concave down
Show Answers Only

\(B\)

Show Worked Solution

\(\text{At}\ \ x=-2: \)

\(\dfrac{dy}{dx}<0 \ \Rightarrow \ \text{Decreasing}\)

\(\dfrac{d^2 y}{d x^2}>0 \ \Rightarrow \  \text{Concave up}\)

\(\Rightarrow B\)

Calculus, 2ADV C3 EQ-Bank 28

Given  \(y=x e^{-3 x}\), prove that

\(\dfrac{d^2 y}{d x^2}+6 \dfrac{d y}{d x}+9 y=0\)   (3 marks)

Show Answers Only

\(\text{Proof (See worked solution}\)

Show Worked Solution

\(y=x e^{-3 x}\)

\(\dfrac{d y}{d x}=e^{-3 x}-3 x e^{-3 x}\)

\(\dfrac{d^2 y}{d x^2}\) \(=-3 e^{-3 x}-3 e^{-3 x}+3 \cdot 3 x e^{-3 x}\)
  \(=-6 e^{-3 x}+9 x e^{-3 x}\)

 
\(\text {Substituting into equation: }\)

\(\dfrac{d^2 y}{d x^2}+6 \dfrac{d y}{d x}+9 y\)

\(=-6 e^{-3 x}+9 x e^{-3 x}+6\left(e^{-3 x}-3 x e^{-3 x}\right)+9 x e^{-3 x}\)

\(=-6 e^{-3 x}+9 x e^{-3 x}+6 e^{-3 x}-18 x e^{-3 x}+9 x e^{-3 x}\)

\(=0\)

Calculus, 2ADV C3 2024 MET2 3*

A function is defined as  \(f(x)=\dfrac{x-h}{(x+1)(x-4)}\)

Determine the range of \(h\) where the graph of \(f(x)\) have no turning points.   (3 marks)

Show Answers Only

\(-1 \leq h \leq 4\)

Show Worked Solution

\(f(x)=\dfrac{x-h}{(x+1)(x-4)}\)

\(\text{Let}\ \ u=x-h\ \ \Rightarrow\ \ u^{′} = 1\)

 \(v=x^2-3x-4\ \ \Rightarrow\ \ v^{′} = 2x-3\)

\(f^{′}(x)=\dfrac{(x^2-3x-4)-(2x-3)(x-h)}{(x+1)^2(x-4)^2}\)

\(\text{Solve}\ \ f^{′}(x)=0:\)

\(x^2-3x-4-2x^2+2xh+3x-3h\) \(=0\)  
\(-x^2+2xh-(4+3h)\) \(=0\)  
\(x^2-2xh+(4+3h)\) \(=0\)  
\(x\) \(= \dfrac{2h \pm \sqrt{4h^2-4(4+3h)}}{2}\)  
  \(=h \pm \sqrt{h^2-3h-4}\)  

 
\(\text{No turning points occur if}\ \ h^2-3h-4=(h-4)(h+1)<0\)

\(-1 \lt h \lt 4\)

\(\text{If}\ \ h=-1\ \ \text{or}\ \ 4, f(x)\ \text{is linear (no TPs)}\)

\(\therefore -1 \leq h \leq 4\)

Calculus, 2ADV C3 2024 13 MC

The function  \(f(x)=\dfrac{x}{2}+\dfrac{2}{x}\)  undergoes the following sequence of transformations to become \(g(x)\):

  1. dilation by a factor of 3 from the \(y\)-axis
  2. translation by 1 unit in the negative direction of the \(y\)-axis.

The function \(g\) has a local minimum at the point with the coordinates

  1. \((6,1)\)
  2. \(\left(\dfrac{2}{3}, 1\right)\)
  3. \((2,5)\)
  4. \(\left(2,-\dfrac{1}{3}\right)\)
Show Answers Only

\(A\)

Show Worked Solution

\(\text{Dilate by a factor of 3 from the}\ y\text{-axis:}\)

\(f(x) \rightarrow f_1(x)=\dfrac{\frac{x}{3}}{2}+\dfrac{2}{\frac{x}{3}}=\dfrac{x}{6}+\dfrac{6}{x}\)

\(\text{Translate 1 unit down:}\)

\(f_1(x) \rightarrow g(x)=\dfrac{x}{6}+\dfrac{6}{x}-1\)

\(g'(x)=\dfrac{1}{6}-\dfrac{6}{x^{2}}\)

\(\text{Max/min when}\ \ \dfrac{1}{6}-\dfrac{6}{x^{2}}=0\)

\(\Rightarrow A\)

♦ Mean mark 45%.
 

Calculus, 2ADV C4 2024 HSC 22

The graph of the function  \(f(x) = \ln(1 + x^{2})\)  is shown.
 

  1. Prove that \(f(x)\) is concave up for  \(-1 < x < 1\).   (3 marks)
  2. A table of function values, correct to 4 decimal places, for some \(x\) values is provided.

\begin{array} {|c|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & 0 & 0.25 & 0.5 & 0.75 & 1 \\
\hline
\rule{0pt}{2.5ex} \ln(1+x^2) \rule[-1ex]{0pt}{0pt} & \ \ \ \ 0\ \ \ \  & 0.0606 & 0.2231 & 0.4463 & 0.6931 \\
\hline
\end{array}

  1. Using the function values provided and the trapezoidal rule, estimate the shaded area in the diagram.   (2 marks)
  2. Is the answer to part (b) an overestimate or underestimate? Give a reason for your answer.   (1 mark)
Show Answers Only

a.    \(f(x)= \ln(1+x^2)\)

\(f^{\prime}(x)=\dfrac{2x}{1+x^2}\)

\(f^{\prime\prime}(x)\) \(=\dfrac{2(1+x^2)-2x(2x)}{(1+x^2)^2}\)
  \(=\dfrac{2+2x^2-4x^2}{(1+x^2)^2}\)
  \(=\dfrac{2(1-x^2)}{(1+x^2)^2}\)

 
\(\text{Consider domain}\ x \in(-1,1)\ \ \Rightarrow\ \ 1-x^2>0 \)

\((1+x^2)^2 \gt 0\ \ \text{for all}\ x\)

\(\Rightarrow \ f^{\prime\prime}(x) \gt 0\ \text{for}\ x \in(-1,1) \)

\(\therefore f(x)\ \text{is concave up for}\ x \in(-1,1) \)
 

b.    \(\text{Total shaded area}\ \approx 0.5383\ \text{(4 d.p.)}\)

c.    \(\text{Trapezoidal rule assumes straight lines join points in the table.}\)

\(\text{Since the graph is concave up in the given domain, the trapezoidal}\)

\(\text{rule will overestimate the area.}\)

Show Worked Solution

a.   \(f(x)= \ln(1+x^2)\)

\(f^{′}(x)=\dfrac{2x}{1+x^2}\)

\(f^{″}(x)\) \(=\dfrac{2(1+x^2)-2x(2x)}{(1+x^2)^2}\)  
  \(=\dfrac{2+2x^2-4x^2}{(1+x^2)^2}\)  
  \(=\dfrac{2(1-x^2)}{(1+x^2)^2}\)  

 
\(\text{In domain}\ x \in(-1,1)\ \ \Rightarrow\ \ 1-x^2>0 \)

\((1+x^2)^2 \gt 0\ \ \text{for all}\ x\)

\(\Rightarrow \ f^{″}(x) \gt 0\ \text{for}\ x \in(-1,1) \)

\(\therefore f(x)\ \text{is concave up for}\ x \in(-1,1) \)

♦ Mean mark (a) 47%.

b.   \(\text{Total shaded area}\)

\(\approx 2 \times \dfrac{h}{2}[ y_0 + 2(y_1+y_2+y_3) + y_4] \)

\(\approx 2 \times \dfrac{0.25}{2}[ 0 + 2(0.0606+0.2231+0.4463) + 0.6931] \)

\(\approx 0.538275 \)

\(\approx 0.5383\ \text{(4 d.p.)}\)

♦ Mean mark (b) 50%.

c.   \(\text{Trapezoidal rule assumes straight lines join points in the table.}\)

\(\text{Since the graph is concave up in the given domain, the trapezoidal}\)

\(\text{rule will overestimate the area.}\)

♦ Mean mark (c) 49%.

Calculus, 2ADV C3 2024 HSC 11

The graph of the function \(g(x)\) is shown.
 

Using the graph, complete the table with the words positive, zero or negative as appropriate.   (3 marks)

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \textit{\(x\)-value} \rule[-1ex]{0pt}{0pt} & \textit{First derivative of \(g(x)\) at \(x\)} \rule[-1ex]{0pt}{0pt} & \textit{Second derivative of \(g(x)\) at \(x\)} \\
\hline
\rule{0pt}{2.5ex} \text{\(x=-3\)} \rule[-1ex]{0pt}{0pt} & \text{ } \rule[-1ex]{0pt}{0pt} & \text{ } \\
\hline
\rule{0pt}{2.5ex} \text{\(x=1\)} \rule[-1ex]{0pt}{0pt} & \text{ } \rule[-1ex]{0pt}{0pt} & \text{ } \\
\hline
\rule{0pt}{2.5ex} \text{\(x=5\)} \rule[-1ex]{0pt}{0pt} & \text{ } \rule[-1ex]{0pt}{0pt} & \text{ } \\
\hline
\end{array}

Show Answers Only

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \textit{\(x\)-value} \rule[-1ex]{0pt}{0pt} & \textit{First derivative of \(g(x)\) at \(x\)} \rule[-1ex]{0pt}{0pt} & \textit{Second derivative of \(g(x)\) at \(x\)} \\
\hline
\rule{0pt}{2.5ex} \text{\(x=-3\)} \rule[-1ex]{0pt}{0pt} & \text{positive} \rule[-1ex]{0pt}{0pt} & \text{negative} \\
\hline
\rule{0pt}{2.5ex} \text{\(x=1\)} \rule[-1ex]{0pt}{0pt} & \text{zero} \rule[-1ex]{0pt}{0pt} & \text{zero} \\
\hline
\rule{0pt}{2.5ex} \text{\(x=5\)} \rule[-1ex]{0pt}{0pt} & \text{positive} \rule[-1ex]{0pt}{0pt} & \text{positive} \\
\hline
\end{array}

Show Worked Solution

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \textit{\(x\)-value} \rule[-1ex]{0pt}{0pt} & \textit{First derivative of \(g(x)\) at \(x\)} \rule[-1ex]{0pt}{0pt} & \textit{Second derivative of \(g(x)\) at \(x\)} \\
\hline
\rule{0pt}{2.5ex} \text{\(x=-3\)} \rule[-1ex]{0pt}{0pt} & \text{positive} \rule[-1ex]{0pt}{0pt} & \text{negative} \\
\hline
\rule{0pt}{2.5ex} \text{\(x=1\)} \rule[-1ex]{0pt}{0pt} & \text{zero} \rule[-1ex]{0pt}{0pt} & \text{zero} \\
\hline
\rule{0pt}{2.5ex} \text{\(x=5\)} \rule[-1ex]{0pt}{0pt} & \text{positive} \rule[-1ex]{0pt}{0pt} & \text{positive} \\
\hline
\end{array}

Calculus, 2ADV C3 EQ-Bank 11

The graph of \(f(x)\) is drawn below. Draw a sketch of \(f^{\prime}(x)\) on the same graph.   (2 marks)
 

 

Show Answers Only

Show Worked Solution

\(\text{Plot x-intercepts at the x-values of turning points.}\)

\(\text{Determine if gradient of}\ f(x)\ \text{is positive/negative to draw shape.}\)

 

Calculus, 2ADV C3 2023 HSC 6 MC

The following table gives the signs of the first and second derivatives of a function  \(y=f(x)\)  for different values of \(x\).
 

\(x\) \(-2\ \ \) \(0\) \(2\)
\( f^{′}(x) \) \(+\) \(0\) \(+\)
\(f^{″}(x)\) \(-\) \(0\) \(+\)

 
Which of the following is a possible sketch of  \(y=f(x)\)?
 

Show Answers Only

\(C\)

Show Worked Solution

\(\text{By elimination:}\)

\(\text{Gradient is positive at}\ x=-2\ \text{and}\ 2\ \ (\text{Eliminate}\ B\ \text{and}\ D)\)

\(\text{SP and POI at}\ x=0\ \ (\text{Eliminate}\ A)\)

\(\Rightarrow C\)

Calculus, 2ADV C3 2021 HSC 7 MC

The diagram shows part of  `y = f(x)`  which has a local minimum at  `x = –2`  and a local maximum at  `x = 3`.
 

Which of the following shows the correct relationship between  `f^(″)(–2), \ f(0)`  and  `f^(′)(3)`?

  1. `f(0) < f^(′)(3) < f^(″)(–2)`
  2. `f(0) < f^(″)(–2) < f^(′)(3)`
  3. `f^(″)(–2) < f^(′)(3) < f(0)`
  4. `f^(″)(–2) < f(0) < f^(′)(3)`
Show Answers Only

`A`

Show Worked Solution

Mean mark 52%.
`f^(″)(–2)` `> 0\ \ \ (text(concave up at)\ \  x = –2)`
`f(0)` `< 0\ \ \ (text(see graph))`
`f^(′)(3)` `= 0\ \ \ (text(S.P.))`

 
`:. f(0) < f^(′)(3) < f^(″)(–2)`

`=> A`

Calculus, 2ADV C3 2020 HSC 10 MC

The graph shows two functions  `y = f(x)`  and  `y = g(x)`.

Define  `h(x) = f(g(x))`.

How many stationary points does  `y = h(x)`  have for  `1 <= x <= 5`?

  1. 0
  2. 1
  3. 2
  4. 3
Show Answers Only

`D`

Show Worked Solution

`h(x) = f(g(x))`

♦♦♦ Mean mark 11%.

`h^{′}(x) = g^{′}(x) xx f^{′}(g(x))`
  

`text(S.P.’s occur when)\ \ g^{′}(x) = 0\ \ text(or)\ \ f^{′}(g(x)) = 0`

`g^{′}(x) = 0\ \ text(when)\ \ x =3\ (text(from graph))`

`f^{′}(x) = 0\ \ text(when)\ \ x ~~ 1  \ \ text{(i.e.}\ xtext{-value is just under 1)}`
 

`text(Find values of)\ x\ text(when)\ g(x) ~~ 1:`

`text(By inspection, there are 2 values where)`

`g(x) ~~ 1, \ x ∈ [1, 5]`

`:.\ text(There are 3 S.P.’s for)\ y = h(x), \ x ∈ [1, 5]`

`=> D`

Calculus, 2ADV C3 2020 HSC 5 MC

Which of the following could represent the graph of  `y = −x^2 + bx + 1`, where  `b > 0`?
 

A. B.
C. D.
Show Answers Only

`C`

Show Worked Solution

`y = −x^2 + bx + 1`

`(dy)/(dx) = −2x + b`

`text(S.P. occurs when)\ \ (dy)/(dx) = 0:`

`−2x + b` `= 0`
`x` `= b/2`

 
`text(S)text(ince)\ b > 0, text(SP occurs when)\ x > 0`

`=>C`

Calculus, 2ADV C3 2018 HSC 14c

Let  `f(x) = x^3 + kx^2 + 3x-5`, where `k` is a constant.

Find the values of `k` for which `f(x)` has NO stationary points.   (3 marks)

Show Answers Only

`-3 < k < 3`

Show Worked Solution

`f(x) = x^3 + kx^2 + 3x-5`

♦ Mean mark 49%.

`f^{′}(x) = 3x^2 + 2kx + 3`
 

`text(No S.P.’s exist if)\ f^{′}(x)\ text(has no roots,)`

`Delta` `< 0`
`b^2-4ac` `< 0`
`(2k)^2-4 xx 3 xx 3` `< 0`
`4k^2-36` `< 0`
`k^2-9` `< 0`
`(k-3) (k + 3)` `< 0`

 

`:. -3 < k < 3`

Calculus, 2ADV C3 2018 HSC 9 MC

The diagram shows the graph of  `f^{′}(x)`, the derivative of a function.
 

For what value of `x` does the graph of the function  `f(x)`  have a point of inflection?

  1. `x = a`
  2. `x = b`
  3. `x = c`
  4. `x = d`
Show Answers Only

`B`

Show Worked Solution

`text(P.I. will occur when the graph of)\ \ f^{′}(x)`

♦ Mean mark 41%.

`text(has a turning point).`

`:. x = b`

`=>  B`

Calculus, 2ADV C3 2017 HSC 4 MC

The function  `f(x)` is defined for  `a <= x <= b.`

On this interval,  `f ^{′}(x) > 0 and f^{″}(x) < 0.`

Which graph best represents  `y = f(x)`?
 

(A)   (B)  
(C)   (D)  
Show Answers Only

`A`

Show Worked Solution

`text(Interpreting)\ \ f^{′}(x) > 0,`

`=>\ text(gradient is always positive)`

`text(Interpreting)\ \ f^{″}(x) < 0,`

`=>\ text(Curve is concave down)`

`=>  A`

Calculus, 2ADV C3 2014 HSC 14e

The diagram shows the graph of a function `f(x)`. 

The graph has a horizontal point of inflection at `A`, a point of inflection at `B` and a maximum turning point at `C`.

  
On the diagram above, sketch the graph of the derivative `f^{′}(x)`.   (3 marks)

Show Answers Only

 

Show Worked Solution

2UA HSC 2014 14ei

Calculus, 2ADV C3 2014 HSC 14a

Find the coordinates of the stationary point on the graph  `y = e^x − ex`, and determine its nature.   (3 marks)

Show Answers Only

`(1,0)\ =>text(MINIMUM)`

Show Worked Solution
`y` `= e^x – ex`
`dy/dx` `= e^x – e`
`(d^2 y)/(dx^2)` `= e^x`

 
`text(S.P. when)\ \ dy/dx = 0`

`e^x – e` `= 0`
`e^x` `= e^1`
`x` `= 1`

 
`text(At)\ \ x = 1`

`y` `= e^1 – e = 0`
`(d^2 y)/(dx^2)` `= e > 0\ \  => text(MIN)`

 
`:.\ text(MINIMUM S.P. at)\ (1,0)`

Calculus, 2ADV C3 2009 HSC 8a

Geometry and Calculus, 2UA 2009 HSC 8a

The diagram shows the graph of a function  `y = f(x)`. 

  1. For which values of  `x`  is the derivative,  `f^{′}(x)`, negative?    (1 mark)

  2. What happens to  `f^{′}(x)`  for large values of  `x`?    (1 mark)

  3. Sketch the graph  `y = f^{′}(x)`.     (2 marks)

Show Answers Only

a.    `f^{′}(x) < 0\ text(when)`

 

`-1 < x < 3`

b.    `text(As)\ x -> oo`

 

`f^{′}(x) -> 0`

 c.
    Geometry and Calculus, 2UA 2009 HSC 8a Answer

Show Worked Solution

a.    `f^{′}(x) < 0\ text(when)`

`-1 < x < 3`

♦♦ Exact data not available.

 

b.   `text(As)\ x -> oo,`

`f^{′}(x) -> 0`

 

♦♦ Exact data not available.
MARKER’S COMMENT: Poorly drawn graphs with axes not labelled and inaccurate scales were common.

c.

  Geometry and Calculus, 2UA 2009 HSC 8a Answer

Calculus, 2ADV C3 2010 HSC 9b

Let  `y=f(x)`  be a function defined for  `0 <= x <= 6`, with  `f(0)=0`. 

The diagram shows the graph of the derivative of  `f`,  `y = f^{prime}(x)`. 

The shaded region `A_1` has area 4 square units. The shaded region `A_2` has area 4 square units. 

  1. For which values of  `x` is  `f(x)` increasing?   (1 mark)

  2. What is the maximum value of  `f(x)`?   (1 mark)

  3. Find the value of  `f(6)`.   (1 mark)

  4. Draw a graph of  `y =f(x)`  for  `0 <= x <= 6`.   (2 marks)

Show Answers Only

a.    `f(x)\ text(is increasing when)\ 0 <= x < 2`

b.    `text(MAX value of)\ f(x) = 4`

c.    `-6`

d.    
   

Show Worked Solution
a.     `f(x)\ text(is increasing when)\ \ f^{prime}(x) > 0`
  `text(From the graph)`
  `f(x)\ text(is increasing when)\ 0 <= x < 2`

 

b.     `f^{prime}(x) = 0\ \ text(when)\ \ x=2`
  `:.\ text(MAX at)\ \ x = 2`
  `int_0^2 f^{prime}(x)\ dx = 4\ \ \ (text(given since)\ A_1 = 4 text{)}`
  `text(We also know)`
`int_0^2\ f^{prime}(x)\ dx` `= [f(x)]_0^2`
  `= f(2)-f(0)`
  `= f(2)\ \ \ \ text{(since}\ f(0) = 0 text{)}`

`=> f(2) = 4` 

♦♦♦ Parts (b) and (c) proved particularly difficult for students with mean marks of 12% and 11% respectively.

 
`:.\ text(MAX value of)\ \ f(x) = 4`
 

c.     `int_0^4 f^{prime}(x)` `= A_1-A_2`
    `=0`

`text(We also know)`

`int_0^4 f^{prime}(x)\ dx` `= int_2^4 f^{prime}(x)\ dx + int_0^2 f^{prime}(x)\ dx`
  `=[f(x)]_2^4 + 4`
  `= f(4)-f(2) + 4\ \ \ (text(note)\ f(2)=4)`
  `=f(4)`

`=> f(4) = 0`

 
`text(Gradient)=-3\  text(from)\ \ x = 4\ \ text(to)\ \ x = 6`

`:.\ f(6)` `=-3 (6- 4)`
  `=-6`

  

♦♦ Mean mark (d) 28%
EXAM TIP: Clearly identify THE EXTREMES when given a defined domain. In this case, the origin is obvious graphically, and the other extreme at `x=6`, is CLEARLY LABELLED! 
d.     2UA HSC 2010 9bi

Calculus, 2ADV C3 2010 HSC 8d

Let  `f(x) = x^3-3x^2 + kx + 8`, where `k` is a constant.

Find the values of `k` for which `f(x)` is an increasing function.   (2 marks)

Show Answers Only

`k>3`

Show Worked Solution
`f(x)` `= x^3-3x^2 + kx + 8`
`f^{′}(x)` `= 3x^2-6x + k`

  
`f(x)\ text(is increasing when)\ \ f^{′}(x) > 0`

`=> 3x^2-6x + k > 0`

♦♦ Mean mark 28%.
MARKER’S COMMENT: The arithmetic required to solve `36-12k<0`  proved the undoing of many students.

 

`f^{′}(x)\ text(is always positive)`

`=> f^{′}(x)\ text(is a positive definite.)`

`text(i.e. when)\ \ a > 0\ text(and)\ Delta < 0`
 

`a=3>0`

`Delta = b^2-4ac`

`(-6)^2-(4 xx 3 xx k)` `<0`
`36-12k` `<0`
`12k` `>36`
`k` `>3`

 

`:.\ f(x)\ text(is increasing when)\ \ k > 3.`

Calculus, 2ADV C3 2011 HSC 9c

The graph  `y = f(x)`  in the diagram has a stationary point when  `x = 1`, a point of inflection when  `x = 3`, and a horizontal asymptote  `y = –2`.
 

 Geometry and Calculus, 2UA 2011 HSC 9c
 

Sketch the graph  `y = f^{′}(x)` , clearly indicating its features at  `x = 1`  and at  `x = 3`, and the shape of the graph as  `x -> oo`.   (3 marks)

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Geometry and Calculus, 2UA 2011 HSC 9c Answer

Show Worked Solution
 
♦ Mean mark 43%
IMPORTANT: Examiners regularly ask questions that require the graphing of an `f ^{′}(x)` given the `f(x)` graph and vice-versa. KNOW IT!

Geometry and Calculus, 2UA 2011 HSC 9c Answer

Calculus, 2ADV C3 2012 HSC 4 MC

The diagram shows the graph  `f(x)`.
 

2012 4 mc
 

Which of the following statements is true? 

  1. `f^{′}(a)>0\ \ text(and)\ \ f^{″}(a)<0`  
  2. `f^{′}(a)>0\ text(and)\ \ f^{″}(a)>0`  
  3. `f^{′}(a)<0\ \ text(and)\ \ f^{″}(a)<0`  
  4.  `f^{′}(a)<0\ \ text(and)\ \ f^{″}(a)>0`  
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`A`

Show Worked Solution

`text(At)\ \ x=a,`

`f^{′}(a) > 0\ \ \ :.\ text(Cannot be)\ C\ text(or)\ D`

`f^{″}(a) < 0\ \ text(because)\ \ f(x)\ text(is concave down at)\ x=a`

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

`=>  A`

Calculus, 2ADV C3 2013 HSC 12a

The cubic  `y = ax^3 + bx^2 + cx + d`  has a point of inflection at  `x = p`. 

Show that  `p= - b/(3a)`.   (2 marks)

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

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`text(Show)\ \ p= – b/(3a)`

`y` `=ax^3 + bx^2 + cx + d`
`y prime` `=3ax^2 + 2bx + c`
`y″` `=6ax + 2b`

 
`text(Given P.I. occurs when)\ \ x = p`

`=> y″=0\ \ text(when)\ \ x=p`

`:.\ 6ap + 2b` `=0`
`6ap` `=-2b`
`p` `= -(2b)/(6a)`
  `=-b/(3a)\ \ \ text(… as required)`

Calculus, 2ADV C3 2013 HSC 8 MC

The diagram shows points  `A`,  `B`,  `C`  and  `D`  on the graph  `y = f(x)`.
 

2013 8 mc

 
 At which point is  `f^{′}(x) > 0`  and  ` f^{″}(x)= 0`? 

  1. `A`  
  2. `B`  
  3. `C`  
  4. `D`  
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`B`

Show Worked Solution
♦ Mean mark 48%

`text(At)\ A,\ \ \ f^{′}(x) <0`

`text(At)\ C,\ \ \ f^{′}(x) =0`

`:.\ text(It cannot be)\ A\ text(or)\ C.`

`text(At)\ D,\ \ \ f^{″}(x) >0\ \ \ text{(concave up)}`

`text(At)\ B,\ \ \ f^{″}(x) =0\ \ \ text{(concavity changes)}`

`=> B`