For x (0, 5π/2) , define f(x) = _0^x t t dt . Then f has: - Mathematics MCQ FIITJEE AIEEE 2011 (Set Q) 2026 | ExamTest.live

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For x \in (0, 5 π /2), define f (x) = \int_{0}^{x} t sin t d t . Then f has:

Solution & Step-by-step Explanation

Using Leibniz Rule: f^{'} (x) = x sin x .For critical points, f^{'} (x) = 0 . Since x > 0, sin x = 0 ⟹ x = π, 2 π .Second derivative: f^{''} (x) = sin x + x cos x .At x = π : f^{''} (π) = sin π + π cos π = 0 + π (- 1) = - π < 0 . (Local Maximum)At x = 2 π : f^{''} (2 π) = sin 2 π + 2 π cos 2 π = 0 + 2 π (1) = 2 π > 0 . (Local Minimum)

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For x \in (0, 5 π /2), define f (x) = \int_{0}^{x} t sin t d t . Then f has:

local minimum at π and 2 π

local minimum at π and local maximum at 2 π

local maximum at π and local minimum at 2 π

local maximum at π and 2 π

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