If the integral I = ∫ (- (sin x)/(x) - ln xcos x)dx= f(x) + C (where, C is the constant of integration) and f(e) = - sin e, then the number of natural numbers less than [f((π)/(6))] is equal to (where, [·] is the

Correct answer: 0 Given Integral is I = - ∫ d(sin x ⋅ ln x) = - sin x ⋅ ln x + C ∴ f(x) = - sin x ⋅ ln x ⇒ f((π)/(6)) = - (1)/(2) ((π)/(6)) = (1)/(2) ((6)/(π)) ⇒ f((π)/(6)) ∈ (0, 1) ⇒ [f((π)/(6))] = 0

NTA Abhyas 2020

Exam: NTA Abhyas
Year: 2020
Subject: Mathematics
Chapter: Indefinite Integration

NTA Abhyas 2020 · Mathematics · Indefinite Integration

If the integral I = ∫ (- (sin x)/(x) - ln xcos x)dx= f(x) + C...

If the integral

I = ∫ \left(- \dfrac{sin x}{x} - ln xcos x\right)dx= f\left(x\right) + C

(where,

C

is the constant of integration)

and f\left(e\right) = - sin e,

then the number of natural numbers less than \left[f\left(\dfrac{\pi}{6}\right)\right]

is equal to (where, \left[\cdot\right]

is the greatest integer function)

Correct answer

Step-by-step solution

Given Integral is

I = - ∫ d\left(sin x ⋅ ln x\right)

= - sin x ⋅ ln x + C

∴ f\left(x\right) = - sin x ⋅ ln x

⇒ f\left(\dfrac{\pi}{6}\right) = - \dfrac{1}{2}\ln \left(\dfrac{\pi}{6}\right) = \dfrac{1}{2}\ln \left(\dfrac{6}{\pi}\right)

⇒ f\left(\dfrac{\pi}{6}\right) \in \left(0, 1\right) ⇒ \left[f\left(\dfrac{\pi}{6}\right)\right] = 0

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