<p>Match the following:</p><p>(A) Let \(I = \int_1^3 \dfrac{dx}{x^2 + [x]^2 + 1 - 2x[x]}\)</p><p>(B) Put \(e^y - 1 = z^2\) and evaluate \(\int_1^{\sqrt{e^x}-1} \dfrac{2z}{e^y \cdot z} dz\), and find \(x\) such that \(2\tan^{-1}\sqrt{e^x - 1} = \dfrac{2\pi}{3}\)</p><p>(C) Let \(I = \int_{-n}^{n} (-1)^{[x]} dx\)</p><p>(D) \(\sum_{r=1}^{100} \left(\int_0^1 f(r-1+x)\,dx\right)\) where the result equals \(e\)</p><p>Match: (A)→(r); (B)→(t); (C)→(q); (D)→(p)</p><p>(A) \(\int_1^3 \dfrac{dx}{x^2+[x]^2+1-2x[x]}\)</p><p>(B) Evaluate the integral involving \(e^y - 1 = z^2\) and find \(x = \log 4\)</p><p>(C) \(I = \int_{-n}^{n}(-1)^{[x]}dx = 0\)</p><p>(D) \(\sum_{r=1}^{100}\left(\int_0^1 f(r-1+x)\,dx\right) = \int_0^{100} f(x)\,dx = e\)</p>
Step-by-Step Solution
Key Concept: Breaking integrals over non-integer points by analyzing the floor function behavior piecewise, and recognizing that {x} = x - [x] transforms denominators into perfect squares like (x-[x])² = {x}². Using substitution techniques to convert transcendental integrals into simpler trigonometric or algebraic forms.
<p><strong>Solution Breakdown:</strong></p><p><strong>(A) I = ∫₁³ dx/(x² + [x]² + 1 - 2x[x])</strong></p><p>Note: x² + [x]² + 1 - 2x[x] = (x - [x])² + 1 = {x}² + 1</p><p>For x ∈ [1,2): {x} = x-1, so integral becomes ∫₁² dx/((x-1)² + 1) = [tan⁻¹(x-1)]₁² = tan⁻¹(1) - tan⁻¹(0) = π/4</p><p>For x ∈ [2,3): {x} = x-2, so integral becomes ∫₂³ dx/((x-2)² + 1) = [tan⁻¹(x-2)]₂³ = tan⁻¹(1) - tan⁻¹(0) = π/4</p><p>Total: I = π/4 + π/4 = π/2 → <strong>(r)</strong></p><p><strong>(B) Substitution: e^y - 1 = z²</strong></p><p>Given: 2tan⁻¹√(e^x - 1) = 2π/3</p><p>Therefore: tan⁻¹√(e^x - 1) = π/3</p><p>So: √(e^x - 1) = tan(π/3) = √3</p><p>Thus: e^x - 1 = 3 → e^x = 4 → x = log 4 → <strong>(t)</strong></p><p><strong>(C) I = ∫₋ₙⁿ (-1)^[x] dx</strong></p><p>For integer n, pairs of intervals [-k, -k+1) and [k-1, k) have opposite signs due to (-1)^[x], creating cancellation. Total integral = 0 → <strong>(q)</strong></p><p><strong>(D) ∑ᵣ₌₁¹⁰⁰ (∫₀¹ f(r-1+x)dx)</strong></p><p>Substituting u = r-1+x, du = dx: Sum telescopes to ∫₀¹⁰⁰ f(u)du = e → <strong>(p)</strong></p><p><strong>Answer: (A)→(r); (B)→(t); (C)→(q); (D)→(p)</strong></p>
Correct Answer: (A)→(r); (B)→(t); (C)→(q); (D)→(p)