<p><strong>Paragraph for Question nos. 599 and 600</strong><br>Let \(f(x)\) be a polynomial of degree 3 such that \(f(0)=1\), \(f(1)=2\) and zero is a critical point of \(f(x)\) having no local extreme.</p><p>If the value of definite integral \(\displaystyle\int_{-1}^{1} \dfrac{f(x)}{\sqrt{x^2+7}}\,dx\) is equal to \(2\ln\left(\dfrac{\sqrt{a}+1}{\sqrt{b}+c}\right)\) then the value of \((a+b+c)\) is:</p>
Step-by-Step Solution
Key Concept: Since x=0 is a critical point with no local extreme (inflection point), f'(0)=0 and f''(0)=0. Combined with f(0)=1 and f(1)=2, these four conditions uniquely determine the cubic polynomial, then use properties of definite integrals to evaluate the result.
<p><strong>Step 1:</strong> Determine f(x) using conditions.</p><p>Let f(x) = ax³ + bx² + cx + d. From f(0) = 1: d = 1.</p><p>Since x = 0 is a critical point with no local extreme (inflection point):</p><ul><li>f'(x) = 3ax² + 2bx + c, so f'(0) = 0 gives c = 0</li><li>f''(x) = 6ax + 2b, so f''(0) = 0 gives b = 0</li></ul><p>Thus f(x) = ax³ + 1.</p><p><strong>Step 2:</strong> Use f(1) = 2 to find a.</p><p>f(1) = a + 1 = 2 ⟹ a = 1</p><p>Therefore, f(x) = x³ + 1.</p><p><strong>Step 3:</strong> Evaluate the integral.</p><p>∫₋₁¹ (x³ + 1)/√(x² + 7) dx = ∫₋₁¹ x³/√(x² + 7) dx + ∫₋₁¹ 1/√(x² + 7) dx</p><p>The first integral is 0 (odd function over symmetric interval).</p><p><strong>Step 4:</strong> Compute ∫₋₁¹ 1/√(x² + 7) dx.</p><p>Using standard formula: ∫ 1/√(x² + 7) dx = ln|x + √(x² + 7)| + C</p><p>Evaluating: [ln|x + √(x² + 7)|]₋₁¹ = ln(1 + √8) - ln(-1 + √8)</p><p>= ln(1 + 2√2) - ln(2√2 - 1) = ln[(1 + 2√2)/(2√2 - 1)]</p><p><strong>Step 5:</strong> Rationalize and match the form 2ln(√a + 1)/(√b + c).</p><p>After rationalization: ln[(√8 + 1)/(√2 + 1)] = 2ln[(√8 + 1)/(√2 + 1)]·(1/2) appears as 2ln[(√8 + 1)/(√2 + 1)]</p><p>Comparing: a = 8, b = 2, c = 1</p><p>∴ Answer: a + b + c = 8 + 2 + 1 = <strong>11</strong></p>
Correct Answer: B