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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 Dec. 16, 2024 by Teachoo

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

Transcript

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)
  1. Class 12
  2. Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

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

    CBSE Class 12 Sample Paper for 2021 Boards →
CBSE Class 12 Sample Paper for 2021 Boards →
Class 12
Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

About the Author

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