Check sibling questions

Ex 5.8, 6 - Examine the applicability of Mean Value Theorem

Ex 5.8, 6 - Chapter 5 Class 12 Continuity and Differentiability - Part 2
Ex 5.8, 6 - Chapter 5 Class 12 Continuity and Differentiability - Part 3 Ex 5.8, 6 - Chapter 5 Class 12 Continuity and Differentiability - Part 4

Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


Transcript

Question 6 Examine the applicability of Mean Value Theorem in the function (𝑖) 𝑓 (𝑥) = [𝑥] 𝑓𝑜𝑟 𝑥 ∈ [5, 9]Greatest Integer less than equal to 𝑥 𝑓 (𝑥) =[𝑥] is not continuous & differentiable ⇒ Condition of Mean Value Theorem is not satisfied. Therefore, Mean Value Theorem is not applicable . Question 6 Examine the applicability of Mean Value Theorem in the function (𝑖𝑖) 𝑓 (𝑥) = [𝑥] 𝑓𝑜𝑟 𝑥 ∈ [−2, 2]Greatest Integer less than equal to 𝑥 𝑓 (𝑥) =[𝑥] is not continuous & differentiable ⇒ Condition of Mean Value Theorem is not satisfied. Therefore, Mean Value Theorem is not applicable . Question 2 Examine the applicability of Mean Value Theorem in the function (𝑖𝑖𝑖) 𝑓 (𝑥) = 𝑥2 – 1 𝑓𝑜𝑟 𝑥 ∈ [1, 2]𝑓 (𝑥) = 𝑥2 – 1 𝑓𝑜𝑟 𝑥 ∈ [1 , 2] Condition 1 𝑓(𝑥) = 𝑥2 – 1 𝑓(𝑥) is a polynomial & Every polynomial function is continuous ⇒ 𝑓(𝑥) is continuous at 𝑥∈[1, 2]Conditions of Mean value theorem 𝑓(𝑥) is continuous at (𝑎 , 𝑏) 𝑓(𝑥) is derivable at (𝑎 , 𝑏) If both conditions satisfied, then there exist some c in (𝑎 , 𝑏) such that 𝑓′(𝑐) = (𝑓(𝑏) − 𝑓(𝑎))/(𝑏 − 𝑎) 𝑓(𝑏)" = " 𝑓(2) = (2)^2−1 = 4−1 = 3 By Mean Value Theorem 𝑓^′ (𝑐) = (𝑓(𝑏) − 𝑓(𝑎))/(𝑏 − 𝑎) 2𝑐 = (3 − (0))/(2 − 1) 2𝑐 = 3/1 𝑐 = 3/2 𝒄 = 𝟑/𝟐 lies between 1 & 2 i.e., 𝑐 = 3/2∈ [1, 2] Hence Mean Value Theorem is satisfied. Question 6 Examine the applicability of Mean Value Theorem in the function (𝑖) 𝑓 (𝑥) = [𝑥] 𝑓𝑜𝑟 𝑥 ∈ [5, 9] Greatest Integer less than equal to 𝑥 𝑓 𝑥﷯ = 𝑥﷯ is not continuous & differentiable ⇒ Condition of Mean Value Theorem is not satisfied. Therefore, Mean Value Theorem is not applicable . Question 6 Examine the applicability of Mean Value Theorem in the function (𝑖𝑖) 𝑓 (𝑥) = [𝑥] 𝑓𝑜𝑟 𝑥 ∈ [−2, 2] Greatest Integer less than equal to 𝑥 𝑓 𝑥﷯ = 𝑥﷯ is not continuous & differentiable ⇒ Condition of Mean Value Theorem is not satisfied. Therefore, Mean Value Theorem is not applicable . Question 2 Examine the applicability of Mean Value Theorem in the function (𝑖𝑖𝑖) 𝑓 (𝑥) = 𝑥2 – 1 𝑓𝑜𝑟 𝑥 ∈ [1, 2] 𝑓 (𝑥) = 𝑥2 – 1 𝑓𝑜𝑟 𝑥 ∈ [1 , 2] Condition 1 𝑓 𝑥﷯ = 𝑥2 – 1 𝑓(𝑥) is a polynomial & Every polynomial function is continuous ⇒ 𝑓 𝑥﷯ is continuous at 𝑥∈[1, 2] Condition 2 𝑓 𝑥﷯=𝑥2 – 1 𝑓(𝑥) is a polynomial & Every polynomial function is differentiable ⇒ 𝑓 𝑥﷯ is differentiable at 𝑥∈ 1, 2﷯ Condition 3 𝑓 𝑥﷯ = 𝑥2 – 1 𝑓﷮′﷯ 𝑥﷯ = 2𝑥 –0 𝑓﷮′﷯ 𝑐﷯ = 2𝑐 𝑓 𝑥﷯ = 𝑥2 – 1 𝑓 𝑎﷯ = 𝑓 1﷯ = 1﷯﷮2﷯−1 = 1 – 1 = 0 𝑓 𝑏﷯ = 𝑓 2﷯ = 2﷯﷮2﷯−1 = 4−1 = 3 By Mean Value Theorem 𝑓﷮′﷯ 𝑐﷯ = 𝑓 𝑏﷯ − 𝑓 𝑎﷯﷮𝑏 − 𝑎﷯ 2𝑐 = 3 − 0﷯﷮2 − 1﷯ 2𝑐 = 3﷮1﷯ 𝑐 = 3﷮2﷯ 𝒄 = 𝟑﷮𝟐﷯ lies between 1 & 2 i.e., 𝑐 = 3﷮2﷯∈ [1, 2] Hence Mean Value Theorem is satisfied.

Ask a doubt
Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

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.