The least value of the function π(π₯) = 2 πππ π₯ + π₯ in the closed interval [0, π/2] is:
(a) 2 (b) π/2 + √3
(c) π/2 (d) The least value does not exist
CBSE Class 12 Sample Paper for 2022 Boards (MCQ Based - for Term 1)
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CBSE Class 12 Sample Paper for 2022 Boards (MCQ Based - for Term 1)
Last updated at April 16, 2024 by Teachoo
Question 20 The least value of the function π(π₯) = 2 πππ π₯ + π₯ in the closed interval ["0," π/2] is: (a) 2 (b) π/2 + β3 (c) π/2 (d) The least value does not exist Let f(π₯)="2 πππ π₯ + π₯" Finding fβ(π) πβ(π₯)=π(2 cosβ‘π₯ + π₯)/ππ₯ =β2 sinβ‘π₯+1 Putting fβ(π)=π β2 sinβ‘π₯+1=0 2 sin π₯=1 sin π₯=1/2 β΄ x = π /π Since we are given closed interval ["0," π /π] Letβs check value of f(x) at 0, π /π and π /π f(0) = 2 cos 0 + 0 = 2 Γ 1 + 0 = 2 f(π /π) = 2 cos π /π + π /π = 2 Γ β3/2 + π /π = β3 + π /π f(π /π) = 2 cos π /π + π /π = 2 Γ 0 + π /π = π /π Since value of f(x) is lowest at x = π /π So, the correct answer is (C)