The diagram shows a sketch of $y = f'(x)$, the derivative function of $y = f(x)$. The curve $y = f(x)$ has a horizontal asymptote $y = 1$. (i) Identify and classify the turning points of the curve $y = f(x)$. (ii) Sketch the curve $y = f(x)$ given $f(0) = 0 /2$ and $y = f(x)$ is continuous. On your diagram, clearly identify and label any important features. (b) A cylindrical hole of radius $r$ is bored through a sphere of radius $R$. The hole is perpendicular to the $xy$ plane and its axis passes through the origin $O$, which is the centre of the sphere. The resulting solid is denoted by $S$. The cross-section of $S$ shown in the diagram is distance $h$ from the $xy$ plane. (i) Show that the area of the cross-section shown above is $ ext{A} = ext{π}(R^2 - h^2 - r^2)$. (ii) Determine the volume of $S$, and express your answer in terms of $b$ alone, where $2b$ is the length of $h$. (c) Use differentiation to show that $ anigg(rac{1}{2}x + 1igg) + anigg(rac{1}{2} rac{1}{2x+1}igg)$ is constant for $2x + 1 > 0$. What is the exact value of the constant?

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The diagram shows a sketch of $y = f'(x)$, the derivative function of $y = f(x)$ - HSC - SSCE Mathematics Extension 2 - Question 4 - 2001 - Paper 1

Question 4

The diagram shows a sketch of $y = f'(x)$, the derivative function of $y = f(x)$. The curve $y = f(x)$ has a horizontal asymptote $y = 1$. (i) Identify and class... show full transcript

Worked Solution & Example Answer:The diagram shows a sketch of $y = f'(x)$, the derivative function of $y = f(x)$ - HSC - SSCE Mathematics Extension 2 - Question 4 - 2001 - Paper 1

Step 1

Identify and classify the turning points of the curve $y = f(x)$.

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Answer

To identify the turning points of the curve $y = f(x)$ , we look for the values of $x$ where the derivative $f'(x)$ changes sign. From the graph, we observe that there are two critical points where $f'(x) = 0$ : one approximately at $x = 1$ and another at $x = 5$ .

At $x = 1$ , the derivative transitions from positive to negative, indicating a local maximum. At $x = 5$ , the derivative transitions from negative to positive, indicating a local minimum. Thus, we classify $x = 1$ as a local maximum and $x = 5$ as a local minimum.

Step 2

Sketch the curve $y = f(x)$ given $f(0) = 0 / 2$ and $y = f(x)$ is continuous.

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Answer

To sketch the curve $y = f(x)$ , we start by plotting the points. Given that $f(0) = 0 / 2 = 0$ , we place a point at (0,0).

From our earlier analysis of the derivative graph, we know:

As $x$ approaches $- rac{2}{3}$ , $y = f(x)$ approaches 1, due to the horizontal asymptote.
The local maximum at $x=1$ is where $f(1)$ may reach a point above $y=1$ .

Using these features, our sketch will reflect smooth transitions between these points, making sure to approach the horizontal asymptote on both ends and displaying the correct local behaviors at identified critical points.

Step 3

Show that the area of the cross-section shown above is $ ext{A} = ext{π}(R^2 - h^2 - r^2)$.

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Answer

To derive the area of the cross-section, we can analyze the geometry involved. The area of the circular section can be expressed as follows:

The area of the entire circle from the sphere is given by: $A_{sphere} = ext{π}R^2$
The hole's area is determined by its radius $r$ : $A_{hole} = ext{π}r^2$
We cut through a height $h$ , which allows us to account for the spherical cap's section from the sphere.

Therefore, the area of the cross-section $A$ can be calculated as: $A = ext{Area of sphere} - ext{Area of hole}$ $A(h) = ext{π}(R^2 - h^2 - r^2)$

Step 4

Determine the volume of $S$, and express your answer in terms of $b$ alone, where $2b$ is the length of $h$.

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Answer

The volume of the solid $S$ can be obtained by integrating the area of the cross-section along the height $h$ . Given that $h = 2b$ , we express the limits of integration accordingly:

$V = ext{π} imes ext{integral from 0 to 2b of }(R^2 - h^2 - r^2) ext{ dh}$

Upon integration, we find:

$V = ext{π} igg[R^2h - rac{h^3}{3} - r^2h igg]_{0}^{2b}$

Carrying out the evaluation gives us a volume expressed in terms of $b$ .

Step 5

Use differentiation to show that $ anigg(rac{1}{2}x + 1igg) + anigg(rac{1}{2} rac{1}{2x+1}igg)$ is constant for $2x + 1 > 0$.

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Answer

Utilizing differentiation, we denote $f(x) = anigg( rac{1}{2}x + 1igg) + anigg( rac{1}{2} rac{1}{2x+1}igg)$ . First, we compute the derivative of $f(x)$ : $f'(x) = rac{1}{ ext{cos}^2( rac{1}{2}x + 1) imes rac{1}{2}} + rac{1}{ ext{cos}^2( rac{1}{2(2x + 1)}) imes rac{1}{4}} imes rac{1}{2}$ By examining the function $f(x)$ , if the derivative is equal to zero, then it indicates that $f(x)$ is a constant on that interval. The exact constant can be evaluated specifically at points such as $x = 0$ .

The diagram shows a sketch of $y = f'(x)$, the derivative function of $y = f(x)$ - HSC - SSCE Mathematics Extension 2 - Question 4 - 2001 - Paper 1

Worked Solution & Example Answer:The diagram shows a sketch of $y = f'(x)$, the derivative function of $y = f(x)$ - HSC - SSCE Mathematics Extension 2 - Question 4 - 2001 - Paper 1

Identify and classify the turning points of the curve $y = f(x)$.

Sketch the curve $y = f(x)$ given $f(0) = 0 / 2$ and $y = f(x)$ is continuous.

Show that the area of the cross-section shown above is $ ext{A} = ext{π}(R^2 - h^2 - r^2)$.

Determine the volume of $S$, and express your answer in terms of $b$ alone, where $2b$ is the length of $h$.

Use differentiation to show that $ anigg( rac{1}{2}x + 1igg) + anigg( rac{1}{2} rac{1}{2x+1}igg)$ is constant for $2x + 1 > 0$.

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Use differentiation to show that $ anigg(rac{1}{2}x + 1igg) + anigg(rac{1}{2} rac{1}{2x+1}igg)$ is constant for $2x + 1 > 0$.