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The diagram shows the graph of $y = f(x)$ - HSC - SSCE Mathematics Extension 2 - Question 3 - 2007 - Paper 1

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The diagram shows the graph of $y = f(x)$. The line $y = x$ is an asymptote. Draw separate one-third page sketches of the graphs of the following: (i) $f(-x)$ (i... show full transcript

Worked Solution & Example Answer:The diagram shows the graph of $y = f(x)$ - HSC - SSCE Mathematics Extension 2 - Question 3 - 2007 - Paper 1

Step 1

(i) $f(-x)$

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Answer

To sketch the graph of f(x)f(-x), reflect the graph of f(x)f(x) across the y-axis. This means that for each point (x,y)(x, y) on the graph of f(x)f(x), there will be a corresponding point (x,y)(-x, y) on the graph of f(x)f(-x).

Step 2

(ii) $f(|x|)$

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To sketch the graph of f(x)f(|x|), first, plot the graph of f(x)f(x) for xeq0x eq 0. Then, reflect the portion of the graph for x>0x > 0 across the y-axis to obtain the graph for x<0x < 0. Additionally, ensure that the graph meets at the origin, as f(0)f(0) will also determine a point on the graph.

Step 3

(iii) $f(x) - x$

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To sketch the graph of f(x)xf(x) - x, take the existing graph of f(x)f(x) and shift it downwards by the value of xx. This will create a new graph, where the y-values are decreased by the corresponding x-values.

Step 4

Find a cubic polynomial

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The given zeros are αα, ββ, and γγ. We need a polynomial whose zeros are 2α, 2β, and 2γ. Thus, multiply the original polynomial by x2αx - 2α, x2βx - 2β, and x2γx - 2γ to yield:

P(x)=k(x2α)(x2β)(x2γ)P(x) = k(x - 2α)(x - 2β)(x - 2γ)

where k is a constant (commonly taken as 1). To express this in standard cubic form, expand the factors.

Step 5

Cylindrical Shells Volume

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For the volume using cylindrical shells, we integrate the radius multiplied by height. The volume VV is given by:

V=2πimesextRadiusimesextHeightV = 2π imes ext{Radius} imes ext{Height}

For our region, the volume when rotated about the y-axis is:

V=2πimesext(heightfunction)imes(extradiusfromtheyaxis)dxV = 2π imes ext{(height function)} imes ( ext{radius from the y-axis}) dx

Calculate the definite integral from x=1x = 1 to x=ex = e.

Step 6

(i) By resolving forces horizontally and vertically

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Answer

Resolve the forces:
For horizontal motion:
N=mgcosθmrω2sinθN = mg cos θ - mrω^2 sin θ.
Set up the equations using the weight mg acting downwards and the centripetal force required for circular motion.

Step 7

(ii) For what values of ω is $N > 0$?

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N>0N > 0 gives: mgcosθmrω2sinθ>0mg cos θ - mrω^2 sin θ > 0.
Rearrange to find the conditions on ω, leading to: ω < rac{g cos θ}{r sin θ}.

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