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Calculus, MET2 2020 VCAA 4

The graph of the function  `f(x)=2xe^((1-x^(2)))`, where  `0 <= x <= 3`, is shown below.
 

  1. Find the slope of the tangent to `f` at  `x=1`.   (1 mark)

  2. Find the obtuse angle that the tangent to `f` at  `x = 1`  makes with the positive direction of the horizontal axis. Give your answer correct to the nearest degree.   (1 mark)

  3. Find the slope of the tangent to `f` at a point  `x =p`. Give your answer in terms of  `p`.   (1 mark)

  4.  i. Find the value of `p` for which the tangent to `f` at  `x=1` and the tangent to `f` at  `x=p`  are perpendicular to each other. Give your answer correct to three decimal places.   (2 marks)

  5. ii. Hence, find the coordinates of the point where the tangents to the graph of `f` at  `x=1`  and  `x=p`  intersect when they are perpendicular. Give your answer correct to two decimal places.   (3 marks)

Two line segments connect the points `(0, f(0))`  and  `(3, f(3))`  to a single point  `Q(n, f(n))`, where  `1 < n < 3`, as shown in the graph below.
 
         
 

  1.   i. The first line segment connects the point `(0, f(0))` and the point `Q(n, f(n))`, where `1 < n < 3`.
  2.      Find the equation of this line segment in terms of  `n`.   (1 mark)

  3.  ii. The second line segment connects the point `Q(n, f(n))` and the point  `(3, f(3))`, where  `1 < n < 3`.
  4.      Find the equation of this line segment in terms of `n`.   (1 mark)

  5. iii. Find the value of `n`, where  `1 < n < 3`, if there are equal areas between the function `f` and each line segment.
  6.      Give your answer correct to three decimal places.   (3 marks)

Show Answers Only
  1. `-2`
  2. `117^@`
  3. `2(1-2p^(2))e^(1-p^(2))” or “(2e-4p^(2)e)e^(-p^(2))`
  4.  i. `0.655`
  5. ii. `(0.80, 2.39)`
  6.   i. `y_1=2e^((1-n^2))x`
  7.  ii. `y_2=(2n e^((1-n^2))-6e^(-8))/(n-3) (x-3) + 6e^(-8)`
  8. iii. `n= 1.088`
Show Worked Solution

a.   `f(x)=2xe^((1-x^(2)))`

`f^{′}(1)=-2`

♦ Mean mark part (b) 37%.

 

b.   `text{Solve:}\ tan theta =-2\ \ text{for}\ \ theta in (pi/2, pi)`

`theta = 117^@`
 

c.   `text{Slope of tangent}\ = f^{′}(p)`

`f^{′}(p)=2(1-2p^(2))e^(1-p^(2))\ \ text{or}\ \ (2e-4p^(2)e)e^(-p^(2))`
 

d.i.   `text{If tangents are perpendicular:}`

`f^{′}(p) xx-2 =-1\ \ =>\ \ f^^{′}(p)=1/2`

`text{Solve}\ \ 2(1-2p^(2))e^(1-p^(2))=1/2\ \ text{for}\ p:`

`p=0.655\ \ text{(to 3 d.p.)}`
 

d.ii.  `text{Equation of tangent at}\ \ x=1: \ y=4-2x`

♦ Mean mark part (d)(ii) 41%.

`text{Equation of tangent at}\ \ x=p: \ y= x/2 + 1.991…`

`text{Solve}\ \ 4-2x = x/2 + 1.991…\ \ text{for}\ x:`

`=> x = 0.8035…`

`=> y=4-2(0.8035…) = 2.392…`

`:.\ text{T}text{angents intersect at (0.80, 2.39)}`

♦ Mean mark part (e)(i) 44%.

 
e.i.  `Q (n, 2n e^(1-n^2))`

`m_(OQ) = (2n e^((1-n^2)) – 0)/(n-0) = 2e^((1-n^2))`

`:.\ text{Equation of segment:}\ \ y_1=2e^((1-n^2))x`

♦♦ Mean mark part (e)(ii) 28%.
 

e.ii.  `P(3, f(3)) = (3, 6e^(-8))`

`m_(PQ) = (2n e^((1-n^2))-6e^(-8))/(n-3)`

`text{Equation of line segment:}`

`y_2-6e^(-8)` `=(2n e^((1-n^2))-6e^(-8))/(n-3) (x-3)`  
`y_2` `=(2n e^((1-n^2))-6e^(-8))/(n-3) (x-3) + 6e^(-8)`  
♦♦ Mean mark part (e)(iii) 28%.

 

e.iii.  `text{Find}\ n\ text{where shaded areas are equal.}`

`text{Solve}\ int_(0)^(n)(f(x)-y_(1))\ dx=int_(n)^(3)(y_(2)-f(x))\ dx\ \ text{for} n:`

`=> n= 1.088\ \ text{(to 3 d.p.)}`

Calculus, MET2 2021 VCAA 3

Let  `q(x) = log_e (x^2-1)-log_e (1-x)`.

  1. State the maximal domain and the range of `q`.   (2 marks)

  2.  i. Find the equation of the tangent to the graph of `q` when  `x =-2`.   (1 mark)

  3. ii. Find the equation of the line that is perpendicular to the graph of `q` when  `x =-2`  and passes through the point  (-2, 0).   (1 mark)

Let  `p(x) = e^{-2x}-2e^-x + 1.`

  1. Explain why `p` is not a one-to one function.   (1 mark)

  2. Find the gradient of the tangent to the graph of `p` at  `x = a`.   (1 mark)

The diagram below shows parts of the graph of `p` and the line  `y = x + 2`.
 
 
                     
 
The line  `y = x + 2`  and the tangent to the graph of `p` at  `x = a`  intersect with an acute angle of `theta` between them.

  1. Find the value(s) of `a` for which  `theta = 60^@`. Give your answer(s) correct to two decimal places.   (3 marks)

  2. Find the `x`-coordinate of the point of intersection between the line  `y = x + 2` and the graph of `p`, and hence find the area bounded by  `y = x + 2`, the graph of `p` and the `x`-axis, both correct to three decimal places.   (3 marks)

Show Answers Only
  1. `text{Domain:} \ x ∈ (-∞, -1)`
    `text{Range:} \ y ∈ R`
  2. i.  `-x-2`
    ii. `x + 2`
  3. `text{Not a one-to-one function as it fails the horizontal line test.}`
  4. `-2e^{-2a} + 2e^{-a}`
  5. `-0.67`
  6. `1.038\ text(u)^2`
Show Worked Solution

a.    `text(Method 1)`

`x^2-1 > 0 \ \ => \ \ x > 1 \ \ ∪ \ \ x < -1`

`1-x > 0 \ \ => \ \ x < 1`
 
`:. \ x ∈ (-∞, -1)`
 

`text(Method 2)`

`text{Sketch graph by CAS}`

`text{Asymptote at} \ x = -1`

`text{Domain:} \ x ∈ (-∞, -1)`

`text{Range:} \ y ∈ R`
 
 

b.     i.   `text{By CAS (tanLine} \ (q (x), x, -2)):`

`y = -x-2`
 

ii.   `text{By CAS (normal} \ (q (x), x, -2)):`

`y = x + 2`
  

c.    `text{Sketch graph by CAS.}`

`p(x) \ text{is not a one-to-one function as it fails the horizontal line test}`

`text{(i.e. it is a many-to-one function)}`
 

d.   `p^{′}(x) = -2e^{-2x} + 2e^-x`

`p^{prime}(a) = -2e^{-2a} + 2e^{-a}`
 
 

e. 

 
`text{Case 1}`

`text{By CAS, solve:}`

`2e^{-a} -2e^{-2a} =-tan (15^@) \ \ text{for}\ a:`

`a = -0.11`
 

`text{Case 2}`

`text{Case 1}`

`text{By CAS, solve:}`

`2e^{-a}-2e^{-2a} = -tan 75^@\ \ text{for}\ a:`

`a = -0.67`
 

f.     `text{At intersection,}`

`x + 2 = e^{-2x} -2e^{-x} + 1`

`x = -0.750`
 

`text{Area}` `= int_{-2}^{-0.750} x + 2\ dx + int_{-0.750}^0 e^{-2x}-2e^{-x} + 1\ dx`
  `= 1.038 \ text(u)^2`

Calculus, MET2 2011 VCAA 17 MC

The normal to the curve with equation  `y = x^(3/2) + x`  at the point  `(4,12)`  is parallel to the straight line with equation

A.   `4x = y`

B.   `4y + x = 7`

C.   `y = x/4 + 1`

D.   `x - 4y = −5`

E.   `4y + 4x = 20` 

Show Answers Only

`=> B`

Show Worked Solution

`text(Solution 1:)`

♦ Mean mark 48%.
`y` `= x^(3/2) + x`
`dy/dx` `=3/2 x^(1/2)+1`

 

`text(At)\ \ x=4,`

`m=3/2 sqrt4 +1 = 4`

`m_text(norm) = – 1/4`

`text(Consider the gradients of each option:)`

`m_text(option B) = -1/4`

`=>B`

 

`text(Solution 2:)`

`text(Equation of normal at)\ \ x = 4:`

`y = −1/4x + 13qquad[text(CAS: normal Line) (x^(3/2) + x, x, 4)]`

`m = −1/4`

`=> B`

Calculus, MET1 2006 VCAA 8

A normal to the graph of  `y = sqrt x`  has equation  `y = -4x + a`, where `a` is a real constant. Find the value of `a.`   (4 marks)

Show Answers Only

`a = 18`

Show Worked Solution

`text(Normal) = text(line) _|_ text(to tangent)`

♦ Mean mark 40%.

`m_{text(norm)}=-4\ \ => \ m_tan=1/4`

`y=sqrt x\ \ => \ dy/dx=1/2 x^(-1/2)`

`text(Find)\ x\ text(when:)`

`1/(2 sqrt x)` `=1/4`
`sqrt x` `=2`
`x` `=4`

 

`:.\ text(Point of tangency is)\ \ (4, 2)`
 

`text(Find equation of normal:)`

`y-y_1` `= m (x-x_1)`
`y-2` `= -4(x-4)`
`:. y` `= -4x + 18`
`:. a` `= 18`

Calculus, MET1 2007 VCAA 9

The graph of  `f: R -> R`,  `f(x) = e^(x/2) + 1`  is shown. The normal to the graph of `f` where it crosses the `y`-axis is also shown.
 

MET1 2007 VCAA Q9
 

  1. Find the equation of the normal to the graph of `f` where it crosses the `y`-axis.   (2 marks)

  2. Find the exact area of the shaded region.   (3 marks)

Show Answers Only
  1. `y = -2x + 2`
  2. `(2sqrte-2)\ text(u)²`
Show Worked Solution

a.   `text(Normal is)\ ⊥\ text(to tangent)`

MARKER’S COMMENT: A common error was made in calculating the point of tangency. Be careful!

`text(Point of tangency:)\ (0,2)`

`text(Gradient of normal:)`

`f^{prime}(x)` `= 1/2 e^(x/2)`
`f^{prime}(0)` `= 1/2`

  
`:. m_text(norm) = -2`
  

`text(Equation of normal:)`

`y-y_1` `= m(x-x_1)`
`y-2` `=-2(x-0)`
`y` `=-2x+2`

 

♦ Mean mark 44%.
b.    `:.\ text(Area)` `= int_0^1 (e^(x/2) + 1-(-2x +2))\ dx`
    `= int_0^1 (e^(x/2) + 2x-1)\ dx`
    `= [2e^(x/2) + x^2-x]_0^1`
    `= (2e^(1/2) + 1^2-1)-(2e^0)`
    `= (2sqrte-2)\ text(u)²`

Calculus, MET2 2012 VCAA 9 MC

The normal to the graph of  `y = sqrt (b - x^2)`  has a gradient of 3 when  `x = 1.`

The value of `b` is

A.   `-10/9`

B.   `10/9`

C.   `4`

D.   `10`

E.   `11`

Show Answers Only

`D`

Show Worked Solution
`y` `=sqrt (b – x^2)`
`dy/dx` `=(-2x)/(2 sqrt(b-x^2))`
   

`text(When)\ \ x=1,`

`dy/dx` `=(-1)/sqrt(b-1)`
   

`text(S)text(ince)\ \ m_(norm) xx m_(tan) = -1,`

`dy/dx=- 1/3`

 

`(-1)/sqrt(b-1)` `=- 1/3`
`sqrt(b-1)` `=3`
`b-1` `=9`
`b` `=10`

 
`=>   D`