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


Ex 6.3, 12 Find the maximum and minimum values of ๐‘ฅ + sin 2๐‘ฅ on [0, 2ฯ€ ] Let f(๐‘ฅ)=๐‘ฅ + sin 2๐‘ฅ Finding fโ€™(๐’™) ๐‘“โ€™(๐‘ฅ)=๐‘‘(๐‘ฅ + sin 2๐‘ฅ)/๐‘‘๐‘ฅ =1+2 cosโก2๐‘ฅ Putting fโ€™(๐’™)=๐ŸŽ 1 + 2 cos 2๐‘ฅ=0 2 cos 2๐‘ฅ=โˆ’1 cos 2๐‘ฅ=(โˆ’1)/2 cos 2๐‘ฅ=cosโกใ€–2๐œ‹/3ใ€— General solution for cos 2๐‘ฅ is 2๐‘ฅ=2๐‘›๐œ‹ยฑ2๐œ‹/3 ๐‘ฅ=(2๐‘›๐œ‹ ยฑ 2๐œ‹/3)/2 ๐‘ฅ= nฯ€ ยฑ ๐œ‹/3 Putting ๐‘›=0 ๐‘ฅ=0(๐œ‹)ยฑ๐œ‹/3 =ยฑ๐œ‹/3 " " We know that ๐‘๐‘œ๐‘  60ยฐ=1/2 And cos is negative in 2nd & 3rd quadrant ๐œƒ = 180 โˆ’ 60 = 120 = 120 ร— ๐œ‹/180 = 2๐œ‹/3 So, ๐‘ฅ=(โˆ’๐œ‹)/3,๐œ‹/3 Since Given ๐‘ฅ โˆˆ[0 , 2๐œ‹] โˆด ๐‘ฅ=๐œ‹/3 only Putting ๐‘›=1 ๐‘ฅ=(1)๐œ‹ยฑ๐œ‹/3 =๐œ‹ยฑ๐œ‹/3 =(3๐œ‹ + ๐œ‹)/3 , (3๐œ‹ โˆ’ ๐œ‹)/3 =4๐œ‹/3 , 2๐œ‹/3 Putting ๐‘›=2 ๐‘ฅ=2(๐œ‹)ยฑ๐œ‹/3 ๐‘ฅ=2๐œ‹โˆ’๐œ‹/3 & 2๐œ‹+๐œ‹/3 ๐‘ฅ=(6๐œ‹ โˆ’ ๐œ‹)/3 & (6๐œ‹ + ๐œ‹)/3 ๐‘ฅ=5๐œ‹/3 & 7๐œ‹/3 So, ๐‘ฅ=5๐œ‹/3 only Also, We are given interval ๐‘ฅ โˆˆ[0 , 2๐œ‹] Hence , calculating f(๐‘ฅ) at ๐‘ฅ=0 , ๐œ‹/3 , 2๐œ‹/3 , 4๐œ‹/3 , 5๐œ‹/3 , 2๐œ‹ Hence, f(๐‘ฅ) is Maximum at ๐‘ฅ=2๐œ‹ Maximum value of f(๐’™)=๐Ÿ๐… & f(๐‘ฅ) is Minimum at ๐‘ฅ=0 Minimum value of f(๐’™)=๐ŸŽ

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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 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.