1. Chapter 5 Class 12 Continuity and Differentiability
2. Serial order wise

Transcript

Ex 5.8, 4 Verify Mean Value Theorem, if ๐ (๐ฅ) = ๐ฅ2 โ 4๐ฅ โ 3 in the interval [๐, ๐], where ๐= 1 ๐๐๐ ๐= 4 ๐ (๐ฅ) = ๐ฅ2 โ 4๐ฅ โ 3 ๐ฅโ ๐, ๐๏ทฏ where a = 1 & b = 4 Mean Value Theorem satisfied if Condition 1 ๐ ๐ฅ๏ทฏ is continuous ๐ ๐ฅ๏ทฏ=๐ฅ2 โ 4๐ฅ โ 3 ๐ ๐ฅ๏ทฏ is a polynomial & Every polynomial function is continuous โ ๐ ๐ฅ๏ทฏ is continuous at ๐ฅโ[1, 4] Condition 2 If ๐ ๐ฅ๏ทฏ is differentiable ๐ ๐ฅ๏ทฏ = ๐ฅ2 โ 4๐ฅ โ 3 ๐ ๐ฅ๏ทฏ is a polynomial & Every polynomial function is differentiable โ ๐ ๐ฅ๏ทฏ is differentiable at ๐ฅโ 1, 4๏ทฏ Condition 3 ๐ ๐ฅ๏ทฏ = ๐ฅ2 โ 4๐ฅ โ 3 ๐๏ทฎโฒ๏ทฏ ๐ฅ๏ทฏ = 2๐ฅโ4 ๐๏ทฎโฒ๏ทฏ ๐๏ทฏ = 2๐โ4 ๐(๐) = ๐(1) = 1๏ทฏ๏ทฎ2๏ทฏโ4 1๏ทฏโ3 = 1 โ 4 โ 3 = โ6 ๐ ๐๏ทฏ = ๐ 4๏ทฏ = 4๏ทฏ๏ทฎ2๏ทฏโ4 4๏ทฏโ3 = 16 โ 16 โ 3 = โ 3 By Mean Value Theorem ๐๏ทฎโฒ๏ทฏ ๐๏ทฏ = ๐ ๐๏ทฏ โ ๐ ๐๏ทฏ๏ทฎ๐ โ ๐๏ทฏ ๐๏ทฎโฒ๏ทฏ ๐๏ทฏ = โ3 โ โ6๏ทฏ๏ทฎ4 โ 1๏ทฏ ๐๏ทฎโฒ๏ทฏ ๐๏ทฏ = โ3 + 6๏ทฎ3๏ทฏ ๐๏ทฎโฒ๏ทฏ ๐๏ทฏ = 3๏ทฎ3๏ทฏ ๐๏ทฎโฒ๏ทฏ ๐๏ทฏ = 1 2c โ 4 = 1 2c = 1 + 4 2c = 5 c = 5๏ทฎ2๏ทฏ Value of c = 5๏ทฎ2๏ทฏ which is lies between (1, 4) c = ๐๏ทฎ๐๏ทฏโ ๐, ๐๏ทฏ Hence Mean Value Theorem satisfied