SmarterEd

Aussie Maths & Science Teachers: Save your time with SmarterEd

Calculus, 2ADV C4 2022 HSC 28

The graph of the circle  `x^2+y^2=2`  is shown.

The interval connecting the origin, `O`, and the point `(1,1)` makes an angle `theta` with the positive `x`-axis.
 

  1. By considering the value of `theta`, find the exact area of the shaded region, as shown on the diagram.  (2 marks)

Part of the hyperbola  `y=(a)/(b-x)-1`  which passes through the points `(0,0)` and `(1,1)` is drawn with the circle  `x^2+y^2=2`  as shown.
 

  1. Show that  `a=b=2`.  (2 marks)

  2. Using parts (a) and (b), find the exact area of the region bounded by the hyperbola, the positive `x`-axis and the circle as shown on the diagram.   (3 marks)

Show Answers Only
  1. `(pi-2)/4\ text{u}^2`
  2. `text{Proof (See Worked Solutions)}`
  3. `(8ln2+pi-6)/4\ text{u}^2`
Show Worked Solution

a.   `tan theta=1\ \ =>\ \ theta = pi/4`

`text{Using Pythagoras,}`

`r=sqrt(1^2+1^2)=sqrt2`

`text{Shaded Area}` `=A_text{sector}-A_Delta`  
  `=(pi/4)/(2pi) xx pi r^2-1/2 xx b xx h`  
  `=1/8xxpixx(sqrt2)^2-1/2xx1xx1`  
  `=pi/4-1/2`  
  `=(pi-2)/4\ text{u}^2`  

 


Mean mark (a) 54%.

b.   `text{Show}\ \ a=b=2`

`y=(a)/(b-x)-1\ \ text{passes through}\ \ (0,0):`

`0` `=a/(b-0)-1`  
`a/b` `=1`  
`a` `=b`  

 
`y=(a)/(b-x)-1\ \ text{passes through}\ \ (1,1):`

`1` `=a/(b-1)-1`  
`a/(b-1)` `=2`  
`a` `=2(b-1)`  
`a` `=2b-2`  
`b` `=2b-2\ \ text{(using}\ a=b)`  
`b` `=2`  

 
`:.a=b=2`
 


♦ Mean mark (b) 47%.
c.    `int_0^1 2/(2-x)-1\ dx` `=int_0^1 -2 xx (-1)/(2-x)-1\ dx`
    `=[-2ln|2-x|-x]_0^1`
    `=[(-2ln1-1)-(-2ln2-0)]`
    `=-1+2ln2`

 

`:.\ text{Total Area}` `=2ln2-1 + (pi-2)/4`  
  `=(8ln2-4+pi-2)/4`  
  `=(8ln2+pi-6)/4\ text{u}^2`  

♦♦ Mean mark (c) 36%.

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)

Show Answers Only
  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 C4 2005 HSC 8b

2005 8b

The shaded region in the diagram is bounded by the circle of radius 2 centred at the origin, the parabola  `y = x^2– 3x + 2`, and the `x`-axis.

By considering the difference of two areas, find the area of the shaded region.  (3 marks)

Show Answers Only

`(pi − 5/6)\ \ text(u²)`

Show Worked Solution

`text(Shaded Area = Area in the quarter circle less)`

`text(the area below the parabola between)\ x= 0 and 1.`

`text(Area of)\ 1/4\ text(circle)` `= 1/4 xx pir^2`
  `= 1/4 xx pi xx 2^2`
  `= pi\ \ \ text(u²)`

 

`text(Area below the parabola between)\ x= 0 and 1`

`=int_0^1y\ dx`

`= int_0^1x^2 − 3x + 2\ dx`

`= [x^3/3 − 3/2x^2 + 2x]_0^1`

`= [(1/3 − 3/2 + 2) − 0]`

`= 5/6`

 

`:.\ text(Shaded Area) = (pi − 5/6)\ \ \ text(u²)`

Calculus, 2ADV C4 2008 HSC 10a


 

In the diagram, the shaded region is bounded by  `y = log_e (x – 2)`, the  `x`-axis and the line  `x = 7`.

Find the exact value of the area of the shaded region.   (5 marks)

Show Answers Only

`5 log_e 5 – 4\ \ \ text(u²)`

Show Worked Solution

`text(Shaded Area)\ text{(} A_1 text{)}` `=\ text(Rectangle) – A_2`
`text(Area of Rectangle)\ ` `= 7 xx log_e 5`

 

`text(Finding the Area of)\ \ A_2`

`y` `= log_e (x – 2)` 
`x – 2` `= e^y`
`x` `= e^y + 2`
`:. A_2` `= int_0^(log_e 5) x\ dy`
  `= int_0^(log_e 5) e^y + 2\ dy`
  `= [e^y + 2y]_0^(log_e 5)`
  `= [(e^(log_e 5) + 2 log_e 5) – (e^0 + 0)]`
  `= (5 + 2 log_e 5) – 1`
  `= 4 + 2 log_e 5`
   
`:.\ A_1` `= 7 log_e 5 – (4 + 2 log_e 5)`
  `= 5 log_e 5 – 4\ \ \ text(u²)`

Calculus, 2ADV C3 2009 HSC 10

`text(Let)\ \ f(x) = x - (x^2)/2 + (x^3)/3`

  1. Show that the graph of  `y = f(x)`  has no turning points.   (2 marks)

  2. Find the point of inflection of  `y = f(x)`.     (1 mark)

  3. i. Show that `1 - x + x^2 - 1/(1 + x) = (x^3)/(1 + x)`  for  `x != -1`.   (1 mark)

     

    ii. Let  `g(x) = ln (1 + x)`.

     

        Use the result in part c.i. to show that  `f prime (x) >= g prime (x)`  for all  `x >= 0`.   (2 marks)

  1. Sketch the graphs of  `y = f(x)`  and  `y = g(x)`  for  `x >= 0`.    (2 marks)

  2. Show that  `d/(dx) [(1 + x) ln (1 + x) - (1 + x)] = ln (1 + x)`.   (2 marks)

  3. Find the area enclosed by the graphs of  `y = f(x)`  and  `y = g(x)`, and the straight line  `x = 1`.   (2 marks)

Show Answers Only
  1. `text{Proof  (See Worked Solutions)}`
  2. `(1/2, 5/12)`
  3. i. `text{Proof  (See Worked Solutions)}`

     

    ii. `text{Proof  (See Worked Solutions)}` 

  4.  
        Geometry and Calculus, 2UA 2009 HSC 10 Answer
  5. `text{Proof  (See Worked Solutions)}`
  6. `1 5/12\ – 2ln2\ \ text(u²)`
Show Worked Solution
a.    `f(x) = x\ – (x^2)/2 + (x^3)/3`
♦♦ Mean mark 28% for all of Q10 (note that data for each question part is not available).
 

`text(Turning points when)\ f prime (x) = 0`

`f prime (x) = 1\ – x + x^2`

`x^2\ – x + 1 = 0`

`text(S)text(ince)\ \ Delta` `= b^2\ – 4ac`
  `= (–1)^2\ – 4 xx 1 xx 1`
  `= -3 < 0 => text(No solution)`

 
`:.\ f(x)\ text(has no turning points)`

 

b.    `text(P.I. when)\ f″(x) = 0`
`f″(x)` `= -1 + 2x = 0`
`2x` `= 1`
`x` `= 1/2`

`text(Check for change in concavity)`

`f″(1/4)` `= -1/2 < 0`
`f″(3/4)` `= 1/2 > 0`

`=>\ text(Change in concavity)`

`:.\ text(P.I. at)\ \ x = 1/2`

 

`f(1/2)` `= 1/2\ – ((1/2)^2)/2 + ((1/2)^3)/3`
  `= 1/2\ – 1/8 + 1/24`
  `= 5/12`

`:.\ text(Point of Inflection at)\ (1/2, 5/12)`

 

c.i.    `text(Show)\ 1\ – x + x^2\ – 1/(1 + x) = (x^3)/(1 + x),\ \ \ x != -1` 
`text(LHS)` `= (1+x)/(1+x)\ – (x(1+x))/(1+x) + (x^2(1+x))/((1+x))\ – 1/(1+x)`
  `= (1 + x\ – x\ – x^2 + x^2 + x^3\ – 1)/(1+x)`
  `= (x^3)/(1+x)\ \ \ text(… as required)`

 

c.ii.   `text(Let)\ g(x) = ln(1+x)`
  `g prime (x) = 1/(1 + x)`
`f prime (x)\ – g prime (x)` `= 1\ – x + x^2\ – 1/(1+x)`
  `= (x^3)/(1 + x)\ \ text{(using part (i))}`

`text(S)text(ince)\ (x^3)/(1 + x) >= 0\ text(for)\ x >= 0`

`f prime (x)\ – g prime (x) >= 0`
`f prime (x) >= g prime (x)\ text(for)\ x >= 0`
MARKER’S COMMENT: When 2 graphs are drawn on the same set of axes, you must label them. 
 

d. 

Geometry and Calculus, 2UA 2009 HSC 10 Answer
e.    `text(Show)\ d/(dx) [(1 + x) ln (1 + x)\ – (1 + x)] = ln (1 + x)`
  `text(Using)\ d/(dx) uv=uv′+vu′`
`text(LHS)` `= (1+x) xx 1/(1 + x) + ln(1+x)xx1 +  – 1`
  `= 1+ ln(1+x)\ – 1`
  `= ln(1+x)`
  `=\ text(RHS    … as required)`

 

f.    `text(Area)` `= int_0^1 f(x)\ – g(x)\ dx`
    `= int_0^1 (x\ – (x^2)/2 + (x^3)/3\ – ln(x+1))\ dx`
    `= [x^2/2\ – x^3/6 + (x^4)/12\ – (1 + x) ln (1+x) + (1+x)]_0^1`
    `text{(using part (e) above)}`
    `= [(1/2 – 1/6 + 1/12 – (2)ln2 + 2) – (ln1 + 1)]`
    `= 5/12\ – 2ln2 + 2\ – 1`
    `= 1 5/12\ – 2 ln 2\ \ text(u²)`