smc-1182-60-Inverse Trig functions

Calculus, SPEC1 2020 VCAA 6

Let `f(x) = arctan (3x - 6) + pi`.

Show that `f^{\prime}(x) = 3/(9x^2 - 36x + 37)`. (1 mark)
Hence, show that the graph of `f` has a point of inflection at `x = 2`. (2 marks)
Sketch the graph of `y = f(x)` on the axes provided below. Label any asymptotes with their equations and the point of inflection with its coordinates. (2 marks)

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`text(Proof)\ text{(See Worked Solutions)}`
`text(Proof)\ text{(See Worked Solutions)}`
`text(See Worked Solutions)`

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a.	`f^{\prime}(x)`	`= (d/(dx) (3x – 6))/(1 + (3x – 6)^2)`
		`= 3/(9x^2 – 36x + 37)`

b. `f^{\prime\prime}(x) = (3(18x – 36))/(9x^2 – 36x + 37)^2`

♦ Mean mark part (b) 42%.

`f^{\prime\prime}(x) = 0\ \ text(when)\ \ 18x – 36 = 0 \ => \ x = 2`

`text(If)\ \ x < 2, 18x – 36 < 0 \ => \ f^{\prime\prime}(x) < 0`

`text(If)\ \ x > 2, 18x – 36 > 0 \ => \ f^{\prime\prime}(x) > 0`

`text(S) text(ince)\ \ f^{\prime\prime}(x)\ \ text(changes sign about)\ \ x = 2,`

`text(a POI exists at)\ \ x = 2`

Calculus, SPEC1-NHT 2017 VCAA 3

Find the gradient of the curve with equation `x = sin (y/15)` when `x = 1/4`. Give your answer in the form `a sqrt b`, where `a, b \ in Z^+`. (3 marks)

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`m = 4 sqrt 15`

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`x`	`= sin (y/15)`
`sin^(-1)x`	`= y/15`
`y`	`= 15 sin^(-1)x`
`dy/dx`	`= 15/(sqrt(1-x^2))`

`text(When)\ \ x=1/4:`

`dy/dx`	`= 15/(sqrt(1-(1/4)^2))`
	`= 15/(sqrt(15/16))`
	`= 60/sqrt15`
	`=4sqrt15`

`:. m = 4 sqrt 15`

Calculus, SPEC2 2017 VCAA 6 MC

Given that `(dy)/(dx) = e^x\ text(arctan)(y)`, the value of `(d^2y)/(dx^2)` at the point `(0,1)` is

`1/2`
`(3pi)/8`
`−1/2`
`pi/4`
`−pi/8`

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`B`

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`(d^2y)/(dx^2)`	`= d/(dx)(e^x) xx text(arctan)(y) + d/(dx)(text(arctan)(y)) xx e^x`
	`= e^xtext(arctan)(y) + e^x (1/(1 + y^2))(dy)/(dx)`

♦ Mean mark 46%.

`(d^2y)/(dx^2)\|_(text{(0, 1)})`	`= e^0text(arctan)(1) + e^0(1/(1 + 1^2)) xxe^0text(arctan)(1)`
	`= 1 xx pi/4 + 1 xx 1/2 xx pi/4`
	`= (3pi)/8`

`=>B`