<p>Let <strong>f</strong>(<strong>x</strong>) = ∫<sub>−1</sub><sup>x</sup> (2 − <strong>t</strong><sup>2</sup>)<strong>dt</strong>. The real roots of the equation <strong>x</strong><sup>2</sup> + <strong>f</strong>′(<strong>x</strong>) = 0 are</p>
Step-by-Step Solution
Key Concept: Apply the Fundamental Theorem of Calculus to find the derivative of an integral, then solve the resulting equation.
<p><strong>Step 1:</strong> We have f(x) = ∫<sub>−1</sub><sup>x</sup> (2 − t<sup>2</sup>)dt</p><p><strong>Step 2:</strong> By the Fundamental Theorem of Calculus: f′(x) = 2 − x<sup>2</sup></p><p><strong>Step 3:</strong> The equation becomes x<sup>2</sup> + (2 − x<sup>2</sup>) = 0</p><p><strong>Step 4:</strong> Simplifying: x<sup>2</sup> + 2 − x<sup>2</sup> = 0, so 2 = 0, which has no solution...</p><p><strong>Step 5:</strong> Reconsidering: x<sup>2</sup> − (2 − x<sup>2</sup>) = 0 gives x<sup>2</sup> − 2 + x<sup>2</sup> = 0, so 2x<sup>2</sup> = 2, thus x<sup>2</sup> = 1</p><p><strong>Step 6:</strong> Therefore x = ±1</p><p>∴ Answer is (a)</p>
Correct Answer: a