1. Chapter 5 Class 12 Continuity and Differentiability
2. Serial order wise
3. Ex 5.8

Transcript

Ex 5.8, 1 Verify Rolleโs theorem for the function ๐ (๐ฅ) = ๐ฅ2 + 2๐ฅ โ 8, ๐ฅ โ [โ 4, 2]. ๐ (๐ฅ) = ๐ฅ2 + 2๐ฅ โ 8, ๐ฅ โ [โ 4, 2]. Rolleโs theorem is satisfied if Condition 1 ๐ ๐ฅ๏ทฏ=๐ฅ2 + 2๐ฅ โ 8 is continuous at โ4 , 2๏ทฏ Since ๐ ๐ฅ๏ทฏ=๐ฅ2 + 2๐ฅ โ 8 is a polynomial & Every polynomial function is continuous for all ๐ฅ โ๐ โ ๐ ๐ฅ๏ทฏ=๐ฅ2 + 2๐ฅ โ 8 is continuous at ๐ฅโ[โ 4, 2] Condition 2 ๐ ๐ฅ๏ทฏ=๐ฅ2 + 2๐ฅ โ 8 is differentiable at โ4 , 2๏ทฏ ๐(๐ฅ) =๐ฅ2 + 2๐ฅ โ 8 is a polynomial . & Every polynomial function is differentiable for all ๐ฅ โ๐ therefore ๐(๐ฅ) is differentiable at โ4 , 2๏ทฏ Condition 3 ๐ ๐ฅ๏ทฏ = ๐ฅ2 + 2๐ฅ โ 8 ๐ โ4๏ทฏ = โ4๏ทฏ๏ทฎ2๏ทฏ+2 โ4๏ทฏโ8 = 16 โ 8 โ 8 = 16 โ 16 = 0 & ๐(2) = 2๏ทฏ๏ทฎ2๏ทฏ+2 2๏ทฏโ8 = 4+4โ8 = 8โ8 = 0 Hence ๐ โ4๏ทฏ = ๐ 2๏ทฏ Now, ๐ ๐ฅ๏ทฏ = ๐ฅ2 + 2๐ฅ โ 8 ๐๏ทฎโฒ๏ทฏ ๐ฅ๏ทฏ = 2๐ฅ+2โ0 ๐๏ทฎโฒ๏ทฏ ๐ฅ๏ทฏ = 2๐ฅ+2 ๐๏ทฎโฒ๏ทฏ ๐ฅ๏ทฏ = 2๐+2 Since all three condition satisfied ๐๏ทฎโฒ๏ทฏ ๐๏ทฏ = 0 2๐+2 = 0 2c = โ 2 ๐ = โ 2๏ทฎ2๏ทฏ = โ1 Value of c = โ1 โ โ4 , 2๏ทฏ Thus, Rolleโs Theorem is satisfied.

Ex 5.8