smc-1089-50-Other

Calculus, 2ADV C3 SM-Bank 2

Given \(y=x e^{-3 x}\), prove that

\(\dfrac{d^2 y}{d x^2}+6 \dfrac{d y}{d x}+9 y=0\) (3 marks)

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\(\text{Proof (See worked solution}\)

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\(y=x e^{-3 x}\)

\(\dfrac{d y}{d x}=e^{-3 x}-3 x e^{-3 x}\)

\(\dfrac{d^2 y}{d x^2}\)	\(=-3 e^{-3 x}-3 e^{-3 x}+3 \cdot 3 x e^{-3 x}\)
	\(=-6 e^{-3 x}+9 x e^{-3 x}\)

\(\text {Substituting into equation: }\)

\(\dfrac{d^2 y}{d x^2}+6 \dfrac{d y}{d x}+9 y\)

\(=-6 e^{-3 x}+9 x e^{-3 x}+6\left(e^{-3 x}-3 x e^{-3 x}\right)+9 x e^{-3 x}\)

\(=-6 e^{-3 x}+9 x e^{-3 x}+6 e^{-3 x}-18 x e^{-3 x}+9 x e^{-3 x}\)

\(=0\)

Calculus, 2ADV C3 2021 HSC 16

For what values of `x` is `f(x) = x^2 - 2x^3` increasing? (3 marks)

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`x ∈ (0, 1/3)`

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`f(x)`	`= x^2 – 2x^3`
`f′(x)`	`= 2x – 6x^2`

`f(x)\ \ text(is increasing when)\ \ f′(x) > 0`

`2x – 6x^2 > 0`

`2x(1 – 3x) > 0`

`x ∈ (0, 1/3)`

Calculus, 2ADV C3 2014 HSC 14a

Find the coordinates of the stationary point on the graph `y = e^x − ex`, and determine its nature. (3 marks)

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`(1,0)\ =>text(MINIMUM)`

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`y`	`= e^x – ex`
`dy/dx`	`= e^x – e`
`(d^2 y)/(dx^2)`	`= e^x`

`text(S.P. when)\ \ dy/dx = 0`

`e^x – e`	`= 0`
`e^x`	`= e^1`
`x`	`= 1`

`text(At)\ \ x = 1`

`y`	`= e^1 – e = 0`
`(d^2 y)/(dx^2)`	`= e > 0\ \ => text(MIN)`

`:.\ text(MINIMUM S.P. at)\ (1,0)`

Calculus, 2ADV C3 2010 HSC 8d

Let `f(x) = x^3-3x^2 + kx + 8`, where `k` is a constant.

Find the values of `k` for which `f(x)` is an increasing function. (2 marks)

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`k>3`

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`f(x)`	`= x^3-3x^2 + kx + 8`
`f^{′}(x)`	`= 3x^2-6x + k`

`f(x)\ text(is increasing when)\ \ f^{′}(x) > 0`

`=> 3x^2-6x + k > 0`

♦♦ Mean mark 28%.
MARKER’S COMMENT: The arithmetic required to solve `36-12k<0` proved the undoing of many students.

`f^{′}(x)\ text(is always positive)`

`=> f^{′}(x)\ text(is a positive definite.)`

`text(i.e. when)\ \ a > 0\ text(and)\ Delta < 0`

`a=3>0`

`Delta = b^2-4ac`

`:. (-6)^2-(4 xx 3 xx k)`	`<0`
`36-12k`	`<0`
`12k`	`>36`
`k`	`>3`

`:.\ f(x)\ text(is increasing when)\ \ k > 3.`

Calculus, 2ADV C3 2013 HSC 12a

The cubic `y = ax^3 + bx^2 + cx + d` has a point of inflection at `x = p`.

Show that `p= - b/(3a)`. (2 marks)

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

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`text(Show)\ \ p= – b/(3a)`

`y`	`=ax^3 + bx^2 + cx + d`
`y prime`	`=3ax^2 + 2bx + c`
`y″`	`=6ax + 2b`

`text(Given P.I. occurs when)\ \ x = p`

`=> y″=0\ \ text(when)\ \ x=p`

`:.\ 6ap + 2b`	`=0`
`6ap`	`=-2b`
`p`	`= -(2b)/(6a)`
	`=-b/(3a)\ \ \ text(… as required)`