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

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo