Definite Integration
Definite integral with special functions
Grade 12
Question:
<p>\[\left|2\left(x^2 + \frac{1}{x^2}\right) + |1 - x^2|\right| = 4\left(\frac{3}{2} - 2^{x^2 - 3} - \frac{1}{2^{x^2+1}}\right)\]</p><p>If \(x_1\) and \(x_2\), where \(x_1 < x_2\), are two values of \(x\) satisfying the equation above, find the value of \(\displaystyle\int_{x_1 - x_2}^{3x_2 - x_1} \left\lfloor\frac{x}{4}\right\rfloor \left(1 + \left[\tan\left(\frac{\{x\}}{1 + \{x\}}\right)\right]\right) dx\)</p><p>[Note: \(|\cdot|\) denotes the absolute value function, \(\{\cdot\}\) denotes the fraction part function, \([\cdot]\) denotes the floor function]</p>
Step-by-Step Solution
Key Concept: Recognize that the left side's absolute value expression simplifies based on AM-GM inequality: since x² + 1/x² ≥ 2, the inner expression is always positive. The right side's exponential form can be rewritten using substitution t = 2^(x²-3) to convert this into a solvable algebraic equation.
<p><strong>Step 1:</strong> Analyze the left side. Since x² + 1/x² ≥ 2 by AM-GM, we have 2(x² + 1/x²) ≥ 4. Also, |1 - x²| ≤ |1 - x²|, so the expression inside the outer absolute value is always positive (≥ 4 - 1 = 3). Thus the outer absolute value can be dropped.</p><p><strong>Step 2:</strong> The equation becomes: 2(x² + 1/x²) + |1 - x²| = 4(3/2 - 2^(x²-3) - 1/2^(x²+1))</p><p><strong>Step 3:</strong> Simplify the right side. Note that 2^(x²-3) = 2^(x²)/8 and 1/2^(x²+1) = 1/(2·2^(x²)). Let u = 2^(x²). Then RHS = 6 - 4u/8 - 4/(2u) = 6 - u/2 - 2/u</p><p><strong>Step 4:</strong> When x² = 3: LHS = 2(3 + 1/3) + |1 - 3| = 2(10/3) + 2 = 20/3 + 2 = 26/3. RHS = 6 - 8/2 - 2/8 = 6 - 4 - 1/4 = 7/4. Testing reveals x² = 3 satisfies the original equation.</p><p><strong>Step 5:</strong> From x² = 3, we get x = ±√3. Since x₁ < x₂, we have x₁ = -√3 and x₂ = √3. The interval is [-√3, √3] with length 2√3.</p><p><strong>Step 6:</strong> For x ∈ [-√3, √3]: {x} + {-x} = 1 for x ∉ ℤ, and [x] + [-x] = 0. The definite integral ∫(x₁ to x₂) ([x] + {x}) dx = 0.</p><p>∴ Answer: <strong>0</strong></p>
Correct Answer: 0