![Slide37.JPG](https://d1avenlh0i1xmr.cloudfront.net/fbfad03f-6159-4d8b-9829-3c9a46263183/slide37.jpg)
![Slide38.JPG](https://d1avenlh0i1xmr.cloudfront.net/ab378acf-e633-453d-b82f-97140c17fc40/slide38.jpg)
![Slide39.JPG](https://d1avenlh0i1xmr.cloudfront.net/7a438ddc-8635-40fd-ae2f-c502dc64dcab/slide39.jpg)
![Slide40.JPG](https://d1avenlh0i1xmr.cloudfront.net/4071c2d4-5f1e-4ea9-9e66-8918d4bc5380/slide40.jpg)
![Slide41.JPG](https://d1avenlh0i1xmr.cloudfront.net/f4a6e9da-c5cb-4a5f-812a-72050d277e42/slide41.jpg)
![Slide42.JPG](https://d1avenlh0i1xmr.cloudfront.net/04618260-2892-4eeb-bd53-905fe72a6a6e/slide42.jpg)
![Slide43.JPG](https://d1avenlh0i1xmr.cloudfront.net/6f53a8c1-10e3-4a38-a8cc-e75c17e43528/slide43.jpg)
![Slide44.JPG](https://d1avenlh0i1xmr.cloudfront.net/7201c920-dba7-44ac-b985-ad44f7e0858b/slide44.jpg)
Ex 6.3
Last updated at April 16, 2024 by Teachoo
Ex 6.3, 12 Find the maximum and minimum values of ๐ฅ + sin 2๐ฅ on [0, 2ฯ ] Let f(๐ฅ)=๐ฅ + sin 2๐ฅ Finding fโ(๐) ๐โ(๐ฅ)=๐(๐ฅ + sin 2๐ฅ)/๐๐ฅ =1+2 cosโก2๐ฅ Putting fโ(๐)=๐ 1 + 2 cos 2๐ฅ=0 2 cos 2๐ฅ=โ1 cos 2๐ฅ=(โ1)/2 cos 2๐ฅ=cosโกใ2๐/3ใ General solution for cos 2๐ฅ is 2๐ฅ=2๐๐ยฑ2๐/3 ๐ฅ=(2๐๐ ยฑ 2๐/3)/2 ๐ฅ= nฯ ยฑ ๐/3 Putting ๐=0 ๐ฅ=0(๐)ยฑ๐/3 =ยฑ๐/3 " " We know that ๐๐๐ 60ยฐ=1/2 And cos is negative in 2nd & 3rd quadrant ๐ = 180 โ 60 = 120 = 120 ร ๐/180 = 2๐/3 So, ๐ฅ=(โ๐)/3,๐/3 Since Given ๐ฅ โ[0 , 2๐] โด ๐ฅ=๐/3 only Putting ๐=1 ๐ฅ=(1)๐ยฑ๐/3 =๐ยฑ๐/3 =(3๐ + ๐)/3 , (3๐ โ ๐)/3 =4๐/3 , 2๐/3 Putting ๐=2 ๐ฅ=2(๐)ยฑ๐/3 ๐ฅ=2๐โ๐/3 & 2๐+๐/3 ๐ฅ=(6๐ โ ๐)/3 & (6๐ + ๐)/3 ๐ฅ=5๐/3 & 7๐/3 So, ๐ฅ=5๐/3 only Also, We are given interval ๐ฅ โ[0 , 2๐] Hence , calculating f(๐ฅ) at ๐ฅ=0 , ๐/3 , 2๐/3 , 4๐/3 , 5๐/3 , 2๐ Hence, f(๐ฅ) is Maximum at ๐ฅ=2๐ Maximum value of f(๐)=๐๐ & f(๐ฅ) is Minimum at ๐ฅ=0 Minimum value of f(๐)=๐