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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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\(y=19 x-25\)

Show Worked Solution

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

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

\(\text{At} \ \  x=1:\)

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

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

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

Calculus, 2ADV C3 SM-Bank 1 MC

The tangent to the graph of  `y = x^3 - ax^2 + 1`  at  `x = 1` passes through the origin.

The value of `a` is

  1. `1/2`
  2. `1`
  3. `3/2`
  4. `5/2`
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`B`

Show Worked Solution
`y` `= x^3 – ax^2 + 1`
`dy/dx` `= 3x^2 – 2ax`

 
`text{At} \ \ x = 1 \ => \  y = 2-a, \ dy/dx = 3-2a`
 

`text{T} text{angent passes through} \ (1,2-a) \ text{and} \ (0,0)`

`m_text{tang} = 2 – a`

`text{Equating gradients:}`

`3-2a` `= 2-a`
`:. a` `= 1`

 
`=> B`

Calculus, 2ADV C3 2021 HSC 31

By considering the equation of the tangent to  `y = x^2 - 1`  at the point  `(a, a^2 - 1)`, find the equations of the two tangents to  `y = x^2 - 1`  which pass through `(3, –8)`.  (4 marks)

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`y = 14x – 50`

`y = -2x – 2`

Show Worked Solution

`y = x^2 – 1`

♦♦ Mean mark 36%.

`(dy)/(dx) = 2x`

`text(At)\ \ x = a, (dy)/(dx) = 2a`
 

`text(Find equation of line)\ \ m = 2a, text(through)\ (a, a^2 – 1):`

`y – (a^2 – 1)` `= 2a(x – a)`
`y – a^2 + 1` `= 2ax – 2a^2`
`y` `= 2ax – a^2 – 1`

 
`text(If tangent passes through)\ (3, –8):`

`2a(3) – a^2 – 1` `= -8`
`6a – a^2 + 7` `= 0`
`a^2 – 6a – 7` `= 0`
`(a – 7)(a + 1)` `= 0`

 
`=> a = 7\ \ text(or)\ \ -1`
 

`:.\ text(Equation of tangents:)`

`y = 14x – 50`

`y = -2x – 2`

Calculus, 2ADV C3 2021 HSC 13

Find the exact gradient of the tangent to the curve  `y = x tan x`  at the point where  `x = pi/3`.  (3 marks)

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`sqrt3 + (4pi)/3`

Show Worked Solution

`y = x tan x`

`(dy)/(dx) = tan x + x sec^2 x`

`text(Find)\ \ m\ \ text(when)\ \ x = pi/3:`

`(dy)/(dx)` `= tan\ pi/3 + pi/3 · 1/(cos^2\ pi/3)`
  `= sqrt3 + pi/3 · 1/(1/4)`
  `= sqrt3 + (4pi)/3`

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`
Show Worked Solution

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 2018 HSC 12b

Find the equation of the tangent to the curve  `y = cos 2x`  at  `x = pi/6`.  (3 marks)

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`y = -sqrt 3 x + (sqrt 3 pi + 3)/6`

Show Worked Solution
`y` `= cos 2x`
`(dy)/(dx)` `= -2 sin 2x`

 
`text(When)\ \ x = pi/6:`

`y` `= cos  pi/3 = 1/2`
`(dy)/(dx)` `= -sin  pi/3 = -sqrt 3`

 
`text(Equation of tangent)\ \ m = -sqrt 3,\ text(through)\ (pi/6, 1/2):`

`y – y_1` `= m(x – x_1)`
`y – 1/2` `= -sqrt 3 (x – pi/6)`
`y` `= -sqrt 3 x + (sqrt 3 pi)/6 + 1/2`
`:. y` `= -sqrt 3 x + (sqrt 3 pi + 3)/6`

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`

Calculus, 2ADV C3 2016 HSC 11f

Find the gradient of the tangent to the curve  `y = tan x`  at the point where  `x = pi/8.`

Give your answer correct to 3 significant figures.  (2 marks)

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`1.17`

Show Worked Solution

`y = tan x = (sin x)/(cos x)`

`(dy)/(dx) = sec^2 x`

`text(When)\ \ x = pi/8,`

`:. (dy)/(dx)` `= sec^2(pi/8)`
  `= 1/(cos^2(pi/8))`
  `= 1.1715…`
  `= 1.17\ \ text{(to 3 sig fig)}`

Calculus, 2ADV C3 2007 HSC 2c

The point  `P (pi, 0)`  lies on the curve  `y = x sinx`. Find the equation of the tangent to the curve at  `P`.  (3 marks)

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`y = – pi x + pi^2`

Show Worked Solution

`y = x sin x`

`(dy)/(dx)` `= x xx d/(dx) (sin x) + d/(dx) x xx sin x`
  `= x  cos x + sin x`

 
`text(When)\ \ x = pi`

`(dy)/(dx)` `= pi xx cos pi + sin pi`
  `= pi (-1) + 0`
  `= – pi`

 
`text(Equation of line,)\ \ m = – pi,\ text(through)\ P(pi, 0):`

`y – y_1` `= m(x – x_1)`
`y – 0` `= – pi(x – pi)`
`:. y` `= – pi x + pi^2`

Calculus, 2ADV C3 2006 HSC 2c

Find the equation of the tangent to the curve  `y = cos 2x`  at the point whose `x`-coordinate is  `pi/6`.  (3 marks)

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`y = – sqrt 3 x + ((sqrt 3 pi)/6 + 1/2)`

Show Worked Solution

`y = cos 2x`

`dy/dx = -2 sin 2x`

`text(When)\ \ x = pi/6`

`y` `= cos (2 xx pi/6)`
  `= cos (pi/3)`
  `= 1/2`

 

`dy / dx` `= -2 sin (pi/3)`
  `= -2 xx sqrt 3 / 2`
  `= – sqrt 3`

 
`text(Equation of tangent,)\ \ m = – sqrt 3, text(through)\ \ (pi/6, 1/2):`

`y – y_1` `= m(x – x_1)`
`y – 1/2` `= – sqrt 3 ( x – pi/6)`
`y – 1/2` `= – sqrt 3 x + (sqrt 3 pi)/6`
`y` `= – sqrt 3 x + ((sqrt 3 pi)/6 + 1/2)`

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`

Show Worked Solution

`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 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`

Show Worked Solutions

`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.`