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Question 36 (iii) (Choice 2) - CBSE Class 12 Sample Paper for 2023 Boards - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

Last updated at Dec. 13, 2024 by

Find the points of local maximum/local minimum, if any, in the interval (0, 12) as well as the points of absolute maximum/absolute minimum in the interval [0, 12]. Also, find the corresponding local maximum/local minimum and the absolute maximum/absolute minimum values of the function.

 

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Question 36 (iii) - Choice 2 Find the points of local maximum/local minimum, if any, in the interval (0, 12) as well as the points of absolute maximum/absolute minimum in the interval [0, 12]. Also, find the corresponding local maximum/local minimum and the absolute maximum/absolute minimum values of the function.Now, our function 𝑓(𝑥) = −0.1𝑥^2 + 𝑚𝑥 + 98 Putting m = 1.2 𝑓(𝑥) = −0.1𝑥^2 + 1.2𝑥 + 98 𝑓(𝑥) = 〖−𝑥〗^2/10+12𝑥/10+98 𝑓(𝑥) = 𝟏/𝟏𝟎(−𝒙^𝟐+𝟏𝟐𝒙+𝟗𝟖𝟎) Finding f’(𝒙) 𝑓(𝑥) = 𝟏/𝟏𝟎(−𝒙^𝟐+𝟏𝟐𝒙+𝟗𝟖𝟎) Differentiating wwr.t x 𝑓’ (𝑥) =1/10(−2𝑥+12) Putting f’(𝒙) = 0 𝟏/𝟏𝟎 (−𝟐𝒙+𝟏𝟐)=𝟎 −2𝑥 + 12 = 0 −2𝑥 = −12 𝑥 = (−12)/(−2) 𝒙 =𝟔 Finding f’(𝒙) 𝑓(𝑥) = 𝟏/𝟏𝟎(−𝒙^𝟐+𝟏𝟐𝒙+𝟗𝟖𝟎) Differentiating wwr.t x 𝑓’ (𝑥) =1/10(−2𝑥+12) Putting f’(𝒙) = 0 𝟏/𝟏𝟎 (−𝟐𝒙+𝟏𝟐)=𝟎 −2𝑥 + 12 = 0 −2𝑥 = −12 𝑥 = (−12)/(−2) 𝒙 =𝟔 Finding f’’(𝒙) 𝑓’(𝑥) = 1/10(−2𝑥+12) Differentiating wr.t x 𝑓’^′ (𝑥)=1/10 × −2 𝒇’^′ (𝒙)= −20 Since f’’(𝑥) < 0 ∴ x = 6 is the local maxima Now, finding absolute minimum and maximum values Finding absolute minimum and maximum Since we are given interval [𝟎 , 𝟏𝟐] Hence, calculating f(𝑥) at 𝑥 = 0, 6, 12Hence, Absolute minimum value is 98 at 𝒙=𝟎, 𝟏𝟐 Absolute maximum value is 102.2 at 𝒙=𝟔

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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