Area Under the Curve
Area Bounded by Curves
Grade 12
Question:
<p>Which of the following statement(s) is/are true for the function <i>f</i>(<i>x</i>) = (<i>x</i> – 1)<sup>2</sup>(<i>x</i> – 2) + 1 defined on [0, 2]?</p>
<p>(A) Range of <i>f</i> is <span>[23/27, 1]</span></p>
<p>(B) The coordinates of the turning point of the graph of <i>y</i> = <i>f</i>(<i>x</i>) occur at (1, 1) and <span>(5/3, 23/27)</span></p>
<p>(C) The value of <i>p</i> for which the equation <i>f</i>(<i>x</i>) = <i>p</i> has 3 distinct solutions lies in interval <span>(23/27, 1]</span></p>
<p>(D) The area enclosed by <i>y</i> = <i>f</i>(<i>x</i>), the lines <i>x</i> = 0 and <i>y</i> = 1 as <i>x</i> varies from 0 to 1 is <span>7/12</span></p>
Step-by-Step Solution
Key Concept: Analyze a cubic function by finding critical points, evaluating endpoints, and understanding when horizontal lines intersect the curve multiple times.
<p><strong>Solution:</strong> For <i>f</i>(<i>x</i>) = (<i>x</i> – 1)<sup>2</sup>(<i>x</i> – 2) + 1 on [0, 2]:</p><p><strong>(A)</strong> Find critical points: <i>f</i>'(<i>x</i>) = 0 gives turning points at <i>x</i> = 1 and <i>x</i> = 5/3. Evaluate: <i>f</i>(0) = 1, <i>f</i>(1) = 1, <i>f</i>(5/3) = 23/27, <i>f</i>(2) = 1. Range is [23/27, 1]. ✓</p><p><strong>(B)</strong> Turning points occur at (1, 1) and (5/3, 23/27). ✓</p><p><strong>(C)</strong> For 3 distinct solutions, the horizontal line <i>y</i> = <i>p</i> must intersect the curve at 3 points. This occurs when 23/27 < <i>p</i> ≤ 1. ✓</p><p><strong>(D)</strong> Area = ∫₀¹ [1 – ((<i>x</i> – 1)<sup>2</sup>(<i>x</i> – 2) + 1)] <i>dx</i> = ∫₀¹ [–(<i>x</i> – 1)<sup>2</sup>(<i>x</i> – 2)] <i>dx</i> = 7/12. ✓</p>
Correct Answer: A, B, C, D