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If $\int_{0}^{\infty} \frac{\sin x}{x} d x = \frac{π}{2}$ then
Statement-1: $\int_{0}^{\infty} \frac{\sin a x \cos b x}{x} d x = π / 2 (a > b > 0)$
Statement-2: $lim_{x \to 0} \frac{\sin a x \cos b x}{x} = a$

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STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is a correct explanation for STATEMENT-1

STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is NOT a correct explanation for STATEMENT-1

STATEMENT-1 is True, STATEMENT-2 is False

STATEMENT-1 is False, STATEMENT-2 is True

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detailed solution

Correct option is B

sin⁡axcos⁡bx=12[sin⁡(a+b)x+sin⁡(a−b)x]∫0∞ sin⁡axcos⁡bxxdx=12∫0∞ sin⁡(a+b)xxdx+∫0∞ sin⁡(a−b)xxdx=12∫0∞ sin⁡ttdt+∫0∞ sin⁡uudu=π/2

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If ∫0∞ sin⁡xxdx=π2 thenStatement-1: ∫0∞ sin⁡axcos⁡bxxdx=π/2(a>b>0)Statement-2: limx→0 sin⁡axcos⁡bxx=a

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free study material

If $\int_{0}^{\infty} \frac{\sin x}{x} d x = \frac{π}{2}$ then
Statement-1: $\int_{0}^{\infty} \frac{\sin a x \cos b x}{x} d x = π / 2 (a > b > 0)$
Statement-2: $lim_{x \to 0} \frac{\sin a x \cos b x}{x} = a$