CBSE Class 12 Sample Paper for 2022 Boards (MCQ Based - for Term 1)

Class 12
Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

## (a) 2Β  Β  Β  Β (b) Ο/2 + β3 (c) Ο/2Β  Β  (d) The least value does not exist

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Question 20 The least value of the function π(π₯) = 2 πππ  π₯ + π₯ in the closed interval ["0," π/2] is: (a) 2 (b) π/2 + β3 (c) π/2 (d) The least value does not exist Let f(π₯)="2 πππ  π₯ + π₯" Finding fβ(π) πβ(π₯)=π(2 cosβ‘π₯ + π₯)/ππ₯ =β2 sinβ‘π₯+1 Putting fβ(π)=π β2 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] Letβs 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 β3/2 + π/π = β3 + 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)