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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 C4 2021 HSC 24

The curve  `y = 3/(x - 1)`  intersects the line  `y = 3/2 x`  at the point (2, 3).

The region bounded by the curve  `y = 3/(x - 1)`, the line  `y = 3/2 x`, the  `x`-axis and the line  `x = 4`  is shaded in the diagram.
 

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

Show Answers Only

`3 + 3 log_e 3\ text(u)²`

Show Worked Solution
`text(Area)` `= int_0^2 3/2 x\ dx + int_2^4 3/(x – 1)\ dx`
  `= [3/4 x^2]_0^2 + 3[log_e(x – 1)]_2^4`
  `= 12/4 + 3[log_e 3 – log_e 1]`
  `= 3 + 3 log_e 3\ \ text(u²)`

Calculus, 2ADV C4 2019 HSC 12d

The diagram shows the graph of  `y = (3x)/(x^2 + 1)`.
 

 
The region enclosed by the graph, the `x`-axis and the line  `x = 3`  is shaded.

Calculate the exact value of the area of the shaded region.  (3 marks)

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`3/2 ln 10\ text(u²)`

Show Worked Solution
`text(Area)` `= int_0^3 (3x)/(x^2 + 1)\ dx`
  `= 3/2 int_0^3 (2x)/(x^2 + 1)\ dx`
  `= 3/2[ln(x^2 + 1)]_0^3`
  `= 3/2(ln 10 – ln 1)`
  `= 3/2 ln10\ \ text(u²)`

Calculus, 2ADV C4 2018 HSC 15b

The diagram shows the region bounded by the curve  `y = 1/(x + 3)`  and the lines  `x = 0`,  `x = 45`  and  `y = 0`. The region is divided into two parts of equal area by the line  `x = k`, where `k` is a positive integer. 
 

 
What is the value of the integer `k`, given that the two parts have equal areas?  (3 marks)

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`9`

Show Worked Solution
`text(Total Area)` `= int_0^45 1/(x + 3)`
  `= [ln (x + 3)]_0^45`
  `= ln 48 – ln 3`
  `= ln 16`

 

`=> int_0^k 1/(x + 3)` `= 1/2 xx ln 16`
`[ln (x + 3)]_0^k` `= ln 16^(1/2)`
`ln (k + 3) – ln 3` `= ln 4`
`ln ((k + 3)/3)` `= ln 4`
`(k + 3)/3` `= 4`
`:. k` `= 4 xx 3 – 3`
  `= 9`

Calculus, 2ADV C4 2015 HSC 10 MC

The diagram shows the area under the curve  `y = 2/x`  from  `x = 1`  to  `x = d`.

2012 2ua 10 mc

What value of `d` makes the shaded area equal to `2`?

  1. `e`
  2. `e + 1`
  3. `2e`
  4. `e^2`
Show Answers Only

`A`

Show Worked Solution

`int_1^d 2/x\ dx = 2`

`[2 log_e x]_1^d = 2`

`2 log_e d – 2 log_e 1` `= 2`
`2 log_e d` `= 2`
`log_e d` `= 1`
`:. d` `= e`

`=> A`

Calculus, 2ADV C4 2010 HSC 5c

The diagram shows the curve  `y=1/x`, for  `x>0`.

The area under the curve between  `x=a`  and  `x=1`  is  `A_1`. The area under the curve between  `x=1`  and  `x=b`  is  `A_2`.
 

2010 5c
 

The areas  `A_1`  and  `A_2`  are each equal to `1` square unit.

Find the values of  `a`  and  `b`.     (3 marks)

Show Answer Only

`a=1/e`

`b=e`

Show Worked Solutions
IMPORTANT: Note when `log_e a=1`, the definition of a log means that `e^1=a`. Many students failed to earn an easy 3rd mark by recognising this.
`int_a^1 1/x \ dx` `=1`
`[ln x]_a^1` `=1`
`ln1-lna` `=1`
`lna` `=-1`
`:.\ a` `=e^-1=1/e`

 

`int_1^b 1/x dx` `=1`
`[lnx]_1^b` `=1`
`lnb-ln1` `=1`
`ln b` `=1`
`:.b` `=e`