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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)`
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`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)
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  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`