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Calculus, EXT1 C2 2020 HSC 13c

Suppose  `f(x) = tan(cos^(-1)(x))`  and  `g(x) = (sqrt(1-x^2))/x`.

The graph of  `y = g(x)`  is given.
 

  1. Show that  `f^(′)(x) = g^(′)(x)`.  (4 marks)

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  2. Using part (i), or otherwise, show that  `f(x) = g(x)`.  (3 marks) 

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  1. `text(See Worked Solutions)`
  2. `text(See Worked Solutions)`
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i.   `f(x) = tan(cos^(-1)(x))`

♦ Mean mark (i) 50%.
`f^(′)(x)` `= -1/sqrt(1-x^2) · sec^2(cos^(-1)(x))`
  `= -1/sqrt(1-x^2) · 1/(cos^2(cos^(-1)(x)))`
  `= -1/(x^2sqrt(1-x^2))`

 
`g(x) = (1-x^2)^(1/2) · x^(-1)`

`g^(′)(x)` `= 1/2 · -2x(1-x^2)^(-1/2) · x^(-1)-(1-x^2)^(1/2) · x^(-2)`
  `= (-x)/(x sqrt(1-x^2))-sqrt(1-x^2)/(x^2)`
  `= (-x^2-sqrt(1-x^2) sqrt(1-x^2))/(x^2 sqrt(1-x^2))`
  `= (-x^2-(1-x^2))/(x^2sqrt(1-x^2))`
  `= -1/(x^2sqrt(1-x^2))`
  `=f^(′)(x)`

 

ii.   `f^(′)(x) = g^(′)(x)`

♦♦♦ Mean mark (ii) 15%.

`=> f(x) = g(x) + c`
 

`text(Find)\ c:`

`f(1)` `= tan(cos^(-1) 1)`
  `= tan 0`
  `= 0`

`g(1) = sqrt(1-1)/0 = 0`

`f(1) = g(1) + c`

`:. c = 0`

`:. f(x) = g(x)`

Filed Under: Harder Trig Calculus (Ext1), Inverse Functions Calculus (Ext1) Tagged With: Band 5, Band 6, smc-1037-10-Sin/Cos Differentiation, smc-1037-20-Tan Differentiation, smc-1038-60-Other

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