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Algebra, MET2-NHT 2019 VCAA 7 MC

If  `m = int_1^3 (2)/(x)\ dx`, then the value of  `e^m`  is

  1.  `log_e (9)`
  2.  `–9`
  3.  `(1)/(9)`
  4.  `9`
  5.  `–(1)/(9)`
Show Answers Only

`D`

Show Worked Solution
`int_1^3 (2)/(x)\ dx` `= [2 log_e x]_1^3`
`m` `= 2 log_e 3 – 2 log_e 1`
`m` `= 2 log_e 3`
`e^(2 log_e 3)` `= e^(log_e 9)`
  `= 9`

 
`=>D`

Calculus, MET1-NHT 2019 VCAA 3

  1. Evaluate  `int_2^7 1/(x + sqrt 3)\ dx`  and  `int_2^7 1/(x - sqrt 3)\ dx`.  (2 marks)
  2. Show that  `1/2 (1/(x - sqrt 3) + 1/(x + sqrt 3)) = x/(x^2 - 3)`.  (1 mark)
  3. Use your answers to part a. and part b. to evaluate  `int_2^7 x/(x^2 - 3)\ dx`  in the form  `1/a log_e(b)`, where  `a` and `b` are positive integers.  (1 mark)
Show Answers Only
  1. `log_e((7 – sqrt 3)/(2 – sqrt 3))`
  2. `text(Proof)\ text{(See Worked Solutions)}`
  3. `1/2 log_e 46`
Show Worked Solution
a.   `int_2^7 1/(x + sqrt 3)\ dx` `= [log_e (x + sqrt 3)]_2^7`
    `= log_e(7 + sqrt 3) – log_e (2 + sqrt 3)`
    `= log_e ((7 + sqrt 3)/(2 + sqrt 3))`

 

`int_2^7 1/(x – sqrt 3)\ dx` `= [log_e (x – sqrt 3)]_2^7`
  `= log_e (7 – sqrt 3) – log_e (2 – sqrt 3)`
  `= log_e ((7 – sqrt 3)/(2 – sqrt 3))`

 

b.   `1/2(1/(x – sqrt 3) + 1/(x + sqrt 3))` `= 1/2 ((x + sqrt 3 + x – sqrt 3)/((x – sqrt 3)(x + sqrt 3)))`
    `= 1/2 ((2x)/(x^2 – 3))`
    `= x/(x^2 – 3)\ \ text(… as required)`

 

c.   `int_2^7 x/(x^2 – 3)` `= 1/2 int_2^7 1/(x – sqrt 3) + 1/(x + sqrt 3)\ dx`
    `= 1/2[log_e ((7 – sqrt 3)/(2 – sqrt 3)) + log_e ((7 + sqrt 3)/(2 + sqrt 3))]`
    `= 1/2 log_e (((7 – sqrt 3)(7 + sqrt 3))/((2 – sqrt 3)(2 + sqrt 3)))`
    `= 1/2 log_e ((49 – 3)/(4 – 3))`
    `= 1/2 log_e 46`

Calculus, MET1 2017 VCAA 2

Let  `y = x log_e(3x).`

  1. Find  `(dy)/(dx)`.  (2 marks)
  2. Hence, calculate  `int_1^2 (log_e(3x) + 1) dx`. Express your answer in the form  `log_e(a)`, where `a` is a positive integer.  (2 marks)
Show Answers Only
  1. `(dy)/(dx) = log_e (3x) + 1`
  2. `log_e(12)`
Show Worked Solution

a.  `text(Using Product Rule:)`

`(hg)′` `= h′ g + h g′`
`:. (dy)/(dx)` `= 1 xx log_e (3x) + x (3/(3x))`
  `= log_e (3x) + 1`

 

b.  `text(Using Integration by Recognition:)`

MARKER’S COMMENT: Too many students ignored the hence instruction and were not able to form an integral.

`int (log_e(3x) + 1) dx = x log_e (3x)`

`:. int_1^2 (log_e (3x) + 1) dx`

`= [x log_e (3x)]_1^2`

`= (2 log_e (6)) – (log_e(3))`

`= log_e(6^2) – log_e(3)`

`= log_e (36/3)`

`= log_e(12)`

Calculus, MET2 2007 VCAA 9 MC

Let  `k = int_-2^-1 1/x\ dx`, then `e^k` is equal to

  1. `log_e(2)`
  2. `1`
  3. `2`
  4. `e`
  5. `1/2`
Show Answers Only

`E`

Show Worked Solution

`text(The integral can be seen as the negative)`

`text(equivalent of the area under)\ \ y=1/x`

`text(between)\ \ x=1 and x=2.`

`k` `= – [log_e x]_1^2`
  `=-(log_e(2)-log_e(1))`
  `= – log_e (2)`
  `= log_e (1/2)`
`:. e^k` `= e^(log_e(1/2))`
  `= 1/2`

 
`=>   E`

Calculus, MET2 2010 VCAA 22 MC

Let  `f` be a differentiable function defined for  `x > 2`  such that

`int_3^(ab + 2) f (x)\ dx = int_3^(a + 2) f(x)\ dx + int_3^(b + 2) f(x)\ dx`  where  `a > 1  and  b > 1.`

The rule for  `f (x)` is

  1. `sqrt (x - 2)`
  2. `log_e (x - 2)`
  3. `sqrt (2x - 4)`
  4. `log_e (2x - 2)`
  5. `1/(x - 2)`
Show Answers Only

`E`

Show Worked Solution

`text(Solution 1)`

`text(Consider option)\ E:`

`int_3^(ab + 2) (1/(x-2))\ dx` `= int_3^(a + 2) (1/(x-2))\ dx + int_3^(b + 2) (1/(x-2))\ dx`
`[log_e (x-2)]_3^(ab+2)` `=[log_e (x-2)]_3^(a+2) + [log_e (x-2)]_3^(b+2)`
`log_e(ab)-log_e1` `=(log_e a-log_e1) + (log_e b-log_e1)`
`log_e(ab)` `=log_e a + log_e b`

 

`text(Solution 2)`

♦♦♦ Mean mark 29%.

`text(Define each specific function)`

`(text{i.e.}\ \ f(x) = 1/(x – 2))`

`text(Enter functional equation until)`

`text(CAS output is “true”.)`

`=>   E`