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  1. Chapter 7 Class 12 Integrals
  2. Serial order wise

Transcript

Ex 7.1, 22 If ๐‘‘/๐‘‘๐‘ฅ f(x) = 4x3 โˆ’ 3/๐‘ฅ4 such that f(2) = 0, then f(x) is x4 + 1/๐‘ฅ3 โˆ’ 129/8 (B) x3 + 1/๐‘ฅ4 + 129/8 (C) x4 + 1/๐‘ฅ3 + 129/8 (D) x3 + 1/๐‘ฅ4 โˆ’ 129/8 Given ๐‘‘/๐‘‘๐‘ฅ f(x) = 4x3 โˆ’ 3/๐‘ฅ4 Integrating both sides โˆซ1โ–’ใ€–๐‘‘/๐‘‘๐‘ฅ ๐‘“(๐‘ฅ) ใ€—=โˆซ1โ–’(4๐‘ฅ^3โˆ’ 3/๐‘ฅ^4 )๐‘‘๐‘ฅ โˆซ1โ–’๐‘‘/๐‘‘๐‘ฅ ๐‘“(๐‘ฅ)=4โˆซ1โ–’ใ€–๐‘ฅ^3 ๐‘‘๐‘ฅใ€—โˆ’3โˆซ1โ–’ใ€–1/๐‘ฅ^4 ๐‘‘๐‘ฅใ€— ๐‘“(๐‘ฅ)=4โˆซ1โ–’ใ€–๐‘ฅ^3 ๐‘‘๐‘ฅใ€—โˆ’3โˆซ1โ–’ใ€–๐‘ฅ^(โˆ’4) ๐‘‘๐‘ฅใ€— ๐‘“(๐‘ฅ)=4 ๐‘ฅ^(3 + 1)/(3 + 1)โˆ’3 ๐‘ฅ^(โˆ’4 + 1)/(โˆ’4 + 1)+๐ถ ๐‘“(๐‘ฅ)=4 ๐‘ฅ^4/4 โˆ’ 3 ๐‘ฅ^(โˆ’3)/(โˆ’3)+๐ถ ๐‘“(๐‘ฅ)=๐‘ฅ^4+๐‘ฅ^(โˆ’3)+๐ถ ๐‘“(๐‘ฅ)=๐‘ฅ^4+ 1/๐‘ฅ^3 +๐ถ Given ๐‘“(2)=0 Putting ๐‘ฅ=2 in (1) ๐‘“(2)=(2)^4+ 1/(2)^3 +๐ถ 0=16+ 1/8 +๐ถ 0= (128 + 1)/8 +๐ถ 0= 129/8 +๐ถ ๐ถ=(โˆ’129)/8 Putting ๐ถ=(โˆ’129)/8 in (1) ๐‘“(๐‘ฅ)=๐‘ฅ^4+ 1/๐‘ฅ^3 +๐ถ โ‡’ ๐’‡(๐’™)=๐’™^๐Ÿ’+ ๐Ÿ/๐’™^๐Ÿ‘ โˆ’๐Ÿ๐Ÿ๐Ÿ—/๐Ÿ– โˆด Option (A) is correct.

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.