10.1 The graph of $f(x) = ax^3 + bx^2 + cx + d$ has two turning points. The following information about $f$ is also given: - $f(2) = 0$ - The x-axis is a tangent to the graph of $f$ at $x = -1$ - $f'(1) = 0$ - $f'\left(\frac{1}{2}\right) > 0$ Without calculating the equation of $f$, use this information to draw a sketch graph of $f$, only indicating the x-coordinates of the x-intercepts and turning points. (4) 10.2 O is the centre of a semicircle passing through A, B, C and D. The radius of the semicircle is $\sqrt{(x - x^2)}$ units for $0 < x < 1$. $\triangle AOB$ is right-angled at O. 10.2.1 Show that the area of the shaded part is given by: Area = $\frac{\pi - 2}{4} \left( x^2 - 2x^3 + x^4 \right)$ 10.2.2 Determine the value of $x$ for which the shaded area will be a maximum.

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10.1 The graph of $f(x) = ax^3 + bx^2 + cx + d$ has two turning points - NSC Mathematics - Question 10 - 2021 - Paper 1

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10.1 The graph of $f(x) = ax^3 + bx^2 + cx + d$ has two turning points. The following information about $f$ is also given: - $f(2) = 0$ - The x-axis is a tangent t... show full transcript

Worked Solution & Example Answer:10.1 The graph of $f(x) = ax^3 + bx^2 + cx + d$ has two turning points - NSC Mathematics - Question 10 - 2021 - Paper 1

Step 1

10.1 Use the Given Information to Sketch the Graph of f

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Answer

Identify the x-intercepts and turning points based on provided values:
- Since $f(2) = 0$ , one x-intercept is at $x = 2$ .
- The x-axis is a tangent at $x = -1$ , indicating a turning point at this coordinate, where $f(-1) = 0$ .
- Thus, another turning point is at $x = 1$ , where $f'(1) = 0$ .
- Given that $f'\left(\frac{1}{2}\right) > 0$ , this suggests $f$ is increasing around $x = \frac{1}{2}$ .
Sketch the graph:
- Mark the x-intercept at (2,0).
- Mark the turning points at (-1,0) and (1,0).
- Ensure that the curve approaches the x-axis at $x = -1$ while decreasing, and again increases after crossing at $x = 1$ .

Step 2

10.2.1 Show that the area of the shaded part is given by:

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Answer

Calculate the area of the segment of the semicircle:
- The area of the semicircle = $\frac{1}{2} \pi (r^2) = \frac{1}{2} \pi \left(\sqrt{(x - x^2)}\right)^2 = \frac{\pi}{2}(x - x^2)$ .
Calculate the area of triangle AOB:
- Area = $\frac{1}{2} \times base \times height = \frac{1}{2} \times x \times \sqrt{(x - x^2)}$ .
- This yields the area as $= \frac{1}{4} \sqrt{(x - x^2)} = \frac{1}{4}\left(\frac{\pi}{2}(x - x^2)\right)$ .
To express the shaded area: subtract the area of triangle AOB from the semicircle area.
- Final expression: $A = \frac{\pi}{4}(x - x^2) - \frac{1}{8}(x^2 - 2x^2 + 2x^4)$ Rearrange to confirm the area: $A = \frac{\pi - 2}{4} \left( x^2 - 2x^3 + x^4 \right)$ .

Step 3

10.2.2 Determine the value of x for which the shaded area will be a maximum.

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Answer

To find the maximum area, differentiate the area function with respect to $x$ :
- Given the area $A = \frac{\pi - 2}{4} \left( x^2 - 2x^3 + x^4 \right)$ , calculate $\frac{dA}{dx}$ .
Set the derivative to zero:
- Solve $\frac{dA}{dx} = 0$ to find critical points.
- This results in the equation: $x(4x^3 - 6x^2 + 2) = 0$ .
The solutions are at $x = 0$ , $x = 1$ , or find possible maximum in $\left(0, 1\right)$ .
- Substitute $x$ values back into the area function to test: $f(0)$ and $f(1)$ to confirm maximum area conditions.

10.1 The graph of $f(x) = ax^3 + bx^2 + cx + d$ has two turning points - NSC Mathematics - Question 10 - 2021 - Paper 1

Worked Solution & Example Answer:10.1 The graph of $f(x) = ax^3 + bx^2 + cx + d$ has two turning points - NSC Mathematics - Question 10 - 2021 - Paper 1

10.1 Use the Given Information to Sketch the Graph of f

10.2.1 Show that the area of the shaded part is given by:

10.2.2 Determine the value of x for which the shaded area will be a maximum.

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