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Calculus, 2ADV C3 2025 HSC 12

Find the equation of the tangent to  \(y=5 x^3-\dfrac{2}{x^2}-9\)  at the point \((1,-6)\).   (3 marks)

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\(y=19 x-25\)

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\(y=5 x^3-2 x^{-2}-9\)

\(y^{\prime}=15 x^2+4 x^{-3} \)

\(\text{At} \ \  x=1:\)

\(y^{\prime}=15+4=19\)

\(\text{Equation of line} \ \  m=19 \ \ \text {through}\ \ (1,-6): \)

\(y+6\) \(=19(x-1)\)
\(y+6\) \(=19 x-19\)
\(y\) \(=19 x-25\)

Calculus, 2ADV C3 2024 MET2 16 MC

Suppose that a function \(f(x)\) and its derivative \(f^{\prime}(x)\) satisfy  \(f(4)=25\)  and  \(f^{\prime}(4)=15\).

Determine the gradient of the tangent line to the graph of  \( {\displaystyle y=\sqrt{f(x)} } \)  at  \( x=4 \).

  1. \(\sqrt{15}\)
  2. \(\dfrac{1}{10}\)
  3. \(\dfrac{15}{2}\)
  4. \(\dfrac{3}{2}\)
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\(D\)

Show Worked Solution
\(y\) \(=\sqrt{f(x)}={f(x)}^{\frac{1}{2}}\)
\(y^{\prime}\) \(=\dfrac{1}{2}\left({f(x)}^{-\frac{1}{2}}\right)\times f^{\prime}(x)=\dfrac{f^{\prime}(x)}{2\sqrt{f(x)}}\)

 
\(\text{Given}\ \ f(4)=25,\ f^{\prime}(4)=15\)

\(\therefore\ y^{\prime}=\dfrac{15}{2\sqrt{25}}=\dfrac{3}{2}\)

\(\Rightarrow D\)

♦♦ Mean mark 36%.

Calculus, 2ADV C3 2020 HSC 8 MC

The graph of  `y = f(x)`  is shown.
 

Which of the following inequalities is correct?

  1. `f^{″}(1) < 0 < f^{′}(1) < f(1)`
  2. `f^{″}(1) < 0 < f(1) < f^{′}(1)`
  3. `0 < f^{″}(1) < f^{′}(1) < f(1)`
  4. `0 < f^{″}(1) < f(1) < f^{′}(1)`
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`A`

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`y = f(x)\ text(is concave down at)\ \ x = 1`

Mean mark 52%.

`f^{″}(1) < 0`

`=>\ text(Eliminate C and D.)`
 
 

`text(T)text(angent)\ => y=mx+b`

COMMENT: Note the concavity causes any tangent to cut the positive y-axis. No y-axis scale is therefore required.

`text(At)\ \ x = 1, m_text(tang) = f^{′}(1)`

`text(At intersection:)`

`mx+b` `=f(1)`  
`f^{′}(1) xx 1 + b` `=f(1)`  
`:. f^{′}(1)` `<f(1),\ \ \  (b>0)`  

 
`:. f^{″}(1) < 0 < f^{′}(1) < f(1)`

`=>A`

Calculus, 2ADV C3 SM-Bank 7

The graph of  `f(x) = sqrt x (1 - x)`  for  `0<=x<=1`  is shown below.
 


 

  1. Calculate the area between the graph of  `f(x)` and the `x`-axis.  (2 marks)

  2. For `x` in the interval `(0, 1)`, show that the gradient of the tangent to the graph of  `f(x)`  is  `(1 - 3x)/(2 sqrt x)`.  (1 mark)

The edges of the right-angled triangle `ABC` are the line segments `AC` and `BC`, which are tangent to the graph of  `f(x)`, and the line segment `AB`, which is part of the horizontal axis, as shown below.

Let `theta` be the angle that `AC` makes with the positive direction of the horizontal axis.
 


 

  1. Find the equation of the line through `B` and `C` in the form  `y = mx + c`, for  `theta = 45^@`.  (3 marks)

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  1. `4/15\ text(units)^2`
  2. `text(Proof)\ \ text{(See Workes Solutions)}`
  3. `y = -x + 1`
Show Worked Solution
i.   `text(Area)` `= int_0^1 (sqrt x – x sqrt x)\ dx`
    `= int_0^1 (x^(1/2) – x^(3/2))\ dx`
    `= [2/3 x^(3/2) – 2/5 x^(5/2)]_0^1`
    `= (2/3 – 2/5) – (0 – 0)`
    `= 10/15 – 6/15`
    `= 4/15\ text(units)^2`

 

ii.   `f (x)` `= x^(1/2) – x^(3/2)`
  `f prime (x)` `= 1/2 x^(-1/2) – 3/2 x^(1/2)`
    `= 1/(2 sqrt x) – (3 sqrt x)/2`
    `= (1 – 3x)/(2 sqrt x)\ \ text(.. as required.)`

 

iii.  `m_(AC) = tan 45^@=1`

♦♦♦ Mean mark (Vic) part (iii) 20%.
MARKER’S COMMENT: Most successful answers introduced a pronumeral such as  `a=sqrtx`  to solve.

`=> m_(BC) = -1\ \ (m_text(BC) _|_ m_(AC))`

 
`text(At point of tangency of)\ BC,\  f prime(x) = -1`

`(1 – 3x)/(2 sqrt x)` `=-1`
`1-3x` `=-2sqrtx`
`3x-2sqrt x-1` `=0`

 
`text(Let)\ \ a=sqrtx,`

`3a^2-2a-1` `=0`
`(3a+1)(a-1)` `=0`
`a=1 or -1/3`   
`:. sqrt x` `=1` `or`   `sqrt x=- 1/3\ \ text{(no solution)}`
`x` `=1`    

 
`f(1)=sqrt1(1-1)=0\ \ =>B(1,0)`
 

`text(Equation of)\ \ BC, \ m=-1, text{through (1,0):}`

`y-0` `=-1(x-1)`
`y` `=-x+1`