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Figure 3 shows the shaded region R which is bounded by the curve $y = -2x^2 + 4x$ and the line $y = \frac{3}{2}$ - Edexcel - A-Level Maths Pure - Question 1 - 2005 - Paper 2

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Question 1

Figure-3-shows-the-shaded-region-R-which-is-bounded-by-the-curve-$y-=--2x^2-+-4x$-and-the-line-$y-=-\frac{3}{2}$-Edexcel-A-Level Maths Pure-Question 1-2005-Paper 2.png

Figure 3 shows the shaded region R which is bounded by the curve $y = -2x^2 + 4x$ and the line $y = \frac{3}{2}$. The points A and B are the points of intersection o... show full transcript

Worked Solution & Example Answer:Figure 3 shows the shaded region R which is bounded by the curve $y = -2x^2 + 4x$ and the line $y = \frac{3}{2}$ - Edexcel - A-Level Maths Pure - Question 1 - 2005 - Paper 2

Step 1

(a) the x-coordinates of the points A and B

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Answer

To find the x-coordinates of the points A and B, we need to solve for the points of intersection between the curve and the line.

  1. Set the equations equal to each other: −2x2+4x=32-2x^2 + 4x = \frac{3}{2}

  2. Rearrange this equation to set it to zero: −2x2+4x−32=0-2x^2 + 4x - \frac{3}{2} = 0 Multiply through by 2 to eliminate the fraction: −4x2+8x−3=0-4x^2 + 8x - 3 = 0

  3. Simplifying gives: 4x2−8x+3=04x^2 - 8x + 3 = 0

  4. Apply the quadratic formula where a=4a = 4, b=−8b = -8, and c=3c = 3: x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} x=8±(−8)2−4⋅4⋅32⋅4x = \frac{8 \pm \sqrt{(-8)^2 - 4 \cdot 4 \cdot 3}}{2 \cdot 4} x=8±64−488x = \frac{8 \pm \sqrt{64 - 48}}{8} x=8±168x = \frac{8 \pm \sqrt{16}}{8} x=8±48x = \frac{8 \pm 4}{8}

  5. Therefore, we get: x=128=32x = \frac{12}{8} = \frac{3}{2} and x=48=12x = \frac{4}{8} = \frac{1}{2}.

  6. Thus, the x-coordinates of A and B are x=12x = \frac{1}{2} and x=32x = \frac{3}{2}.

Step 2

(b) the exact area of R

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Answer

To find the area of region R, we will integrate the difference between the top curve and the bottom line.

  1. The area A is given by: A=∫1232(curve−line) dxA = \int_{\frac{1}{2}}^{\frac{3}{2}} (\text{curve} - \text{line}) \, dx. Here, the curve is y=−2x2+4xy = -2x^2 + 4x and the line is y=32y = \frac{3}{2}.

  2. Thus, we have: A=∫1232(−2x2+4x−32) dx. A = \int_{\frac{1}{2}}^{\frac{3}{2}} ( -2x^2 + 4x - \frac{3}{2}) \, dx.

  3. Calculate the integral: =∫1232(−2x2+4x−32) dx= \int_{\frac{1}{2}}^{\frac{3}{2}} ( -2x^2 + 4x - \frac{3}{2}) \, dx =[−23x3+2x2−32x]1232= \left[ -\frac{2}{3}x^3 + 2x^2 - \frac{3}{2}x \right]_{\frac{1}{2}}^{\frac{3}{2}}

  4. Evaluate at the bounds: Evaluating at x=32x = \frac{3}{2}: =−23(32)3+2(32)2−32(32)= -\frac{2}{3}(\frac{3}{2})^3 + 2(\frac{3}{2})^2 - \frac{3}{2}(\frac{3}{2}) =−23⋅278+2⋅94−94= -\frac{2}{3} \cdot \frac{27}{8} + 2 \cdot \frac{9}{4} - \frac{9}{4} =−94+92−94= -\frac{9}{4} + \frac{9}{2} - \frac{9}{4} =92−184=18−94=94= \frac{9}{2} - \frac{18}{4} = \frac{18 - 9}{4} = \frac{9}{4}.

  5. Next, evaluating at x=12x = \frac{1}{2}: =−23(12)3+2(12)2−32(12)= -\frac{2}{3}(\frac{1}{2})^3 + 2(\frac{1}{2})^2 - \frac{3}{2}(\frac{1}{2}) =−23⋅18+2⋅14−34= -\frac{2}{3} \cdot \frac{1}{8} + 2 \cdot \frac{1}{4} - \frac{3}{4} =−112+12−34= -\frac{1}{12} + \frac{1}{2} - \frac{3}{4} =−112+612−912=−1+9−612=−412=−13 = -\frac{1}{12} + \frac{6}{12} - \frac{9}{12} = -\frac{1 + 9 - 6}{12} = -\frac{4}{12} = -\frac{1}{3}.

  6. Therefore, the area is: A=(94−(−13))=94+13=2712+412=3112. A = \left( \frac{9}{4} - \left(-\frac{1}{3}\right) \right) = \frac{9}{4} + \frac{1}{3} = \frac{27}{12} + \frac{4}{12} = \frac{31}{12}.

  7. The final result for the area of region R is given by: A=116,A = \frac{11}{6}, which is indeed equivalent to 1561 \frac{5}{6}.

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