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Question 20 - CBSE Class 12 Sample Paper for 2022 Boards (MCQ Based - for Term 1) - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

Last updated at May 29, 2023 by

The least value of the function 𝑓(π‘₯) = 2 π‘π‘œπ‘  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

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Question 20 The least value of the function 𝑓(π‘₯) = 2 π‘π‘œπ‘  π‘₯ + π‘₯ in the closed interval ["0," πœ‹/2] is: (a) 2 (b) πœ‹/2 + √3 (c) πœ‹/2 (d) The least value does not exist Let f(π‘₯)="2 π‘π‘œπ‘  π‘₯ + π‘₯" Finding f’(𝒙) 𝑓’(π‘₯)=𝑑(2 cos⁑π‘₯ + π‘₯)/𝑑π‘₯ =βˆ’2 sin⁑π‘₯+1 Putting f’(𝒙)=𝟎 βˆ’2 sin⁑π‘₯+1=0 2 sin π‘₯=1 sin π‘₯=1/2 ∴ x = 𝝅/πŸ” Since we are given closed interval ["0," 𝝅/𝟐] Let’s check value of f(x) at 0, 𝝅/πŸ” and 𝝅/𝟐 f(0) = 2 cos 0 + 0 = 2 Γ— 1 + 0 = 2 f(𝝅/πŸ”) = 2 cos 𝝅/πŸ” + 𝝅/πŸ” = 2 Γ— √3/2 + 𝝅/πŸ” = √3 + 𝝅/πŸ” f(𝝅/𝟐) = 2 cos 𝝅/𝟐 + 𝝅/𝟐 = 2 Γ— 0 + 𝝅/𝟐 = 𝝅/𝟐 Since value of f(x) is lowest at x = 𝝅/𝟐 So, the correct answer is (C)

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.