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Calculus, 2ADV C4 2019 HSC 13c

  1.  Differentiate `(ln x)^2`.   (2 marks)

    --- 4 WORK AREA LINES (style=lined) ---

  2.  Hence, or otherwise, find `int(ln x)/x\ dx`.   (1 mark)

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Show Answers Only

i.    `(2 ln x)/x`

ii.   `1/2 (ln x)^2 + C`

Show Worked Solution

i.    `y= (ln x)^2`

`(dy)/(dx)= 2 xx 1/x xx ln x= (2 ln x)/x`

♦ Mean mark (ii) 49%.

 

ii.   `int (ln x)/x\ dx=1/2 int (2 ln x)/x dx= 1/2 (ln x)^2 +C`

Filed Under: L&E Integration, L&E Integration, Log Calculus (Y12) Tagged With: Band 3, Band 5, smc-1203-30-Log (Indefinite), smc-1203-50-Diff then Integrate, smc-7187-30-Log (Indefinite), smc-7187-50-Diff then Integrate, smc-964-10-Differentiation, smc-964-50-Diff then integrate

Calculus, 2ADV C4 2008 HSC 3b

  1. Differentiate  `log_e (cos x)`  with respect to  `x`.   (2 marks)

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  2. Hence, or otherwise, evaluate  `int_0^(pi/4) tan x\ dx`.   (2 marks)

    --- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

a.    `-tan x`

b.    `-log_e (1/sqrt2)\ \ text(or)\ \ 0.35\ \ text{(2 d.p.)}`

Show Worked Solution
a.     `y` `= log_e (cos x)`
  `dy/dx` `= (-sin x)/(cos x)`
    `=-tan x`

 

b.     `int_0^(pi/4) tan x\ dx`
  `=-[log_e (cos x)]_0^(pi/4)`
  `=-[log_e(cos (pi/4))-log_e (cos 0)]`
  `=-[log_e (1/sqrt2)-log_e 1]`
  `=-[log_e (1/sqrt2)-0]`
  `=-log_e (1/sqrt2)`
  `= 0.346…\ = 0.35\ \ text{(2 d.p.)}`

Filed Under: Differentiation and Integration, Log Calculus, Log Calculus (Y12), Trig Integration, Trig Integration Tagged With: Band 3, Band 4, smc-1204-50-Diff then Integrate, smc-7188-50-Diff then Integrate, smc-964-10-Differentiation, smc-964-40-Trig overlap, smc-964-50-Diff then integrate

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