1. Chapter 6 Class 12 Application of Derivatives
2. Serial order wise

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Example 12 Find intervals in which the function given by f (x) = sin 3x, x, â ï·0, ï·ðï·®2ï·¯ï·¯ is (a) increasing (b) decreasing. fï·ð¥ï·¯ = sin 3ð¥ where ð¥ â ï·0 ,ï·ðï·®2ï·¯ï·¯ Step 1 :- Finding fâ(x) fï·ð¥ï·¯ = sin 3ð¥ fâï·ð¥ï·¯ = ï·ðï·ï·sinï·®3ð¥ï·¯ï·¯ï·®ðð¥ï·¯ fâï·ð¥ï·¯ = cos 3ð¥ . ï·ðï·3ð¥ï·¯ï·®ðð¥ï·¯= cos 3ð¥. 3 = 3. cos 3ð¥ Step 2: Putting fâï·ð¥ï·¯ = 0 3 cos 3ð¥ = 0 cos 3ð¥ = 0 We know that cos Î¸ = 0 When Î¸ = ï·ðï·®2ï·¯ & ï·3ðï·®2ï·¯ â 3ð¥ = ï·ðï·®2ï·¯ & 3ð¥ = ï·3ðï·®2ï·¯ ð¥ = ï·ðï·®2 Ã3ï·¯ & ð¥ = ï·3ðï·®2 Ã 3ï·¯ ð¥ = ï·ðï·®6ï·¯ & ð¥ = ï·ðï·®2ï·¯ Since ð¥ = ï·ðï·®6ï·¯ â ï·0 ,ï·ðï·®2ï·¯ï·¯ & ð¥ = ï·ðï·®2ï·¯ â ï·0,ï·ðï·®2ï·¯ï·¯ both values of ð¥ are valid Step 3: Plotting point Since ð¥ â ï·0 ,ï·ðï·®2ï·¯ ï·¯ we start number line from 0 & end at ï·ðï·®2ï·¯ Point ð¥ = ï·ðï·®6ï·¯ divide the interval ï·0 ,ï·ðï·®2ï·¯ï·¯ into two disjoint intervals ï·0 ,ï·ðï·®6ï·¯ï·¯ and ï·ï·ðï·®6ï·¯, ï·ðï·®2ï·¯ï·¯ Step 4: Checking sign of fâï·ð¥ï·¯ fâï·ð¥ï·¯ = 3. cos 3ð¥ Case 1 In ð¥ â ï·0 ,ï·ðï·®6ï·¯ï·¯ 0<ð¥<ï·ðï·®6ï·¯ 3Ã0<3ð¥<ï·3ðï·®6ï·¯ 0<3ð¥<ï·ðï·®2ï·¯ So when ð¥ â ï·0 ,ï·ðï·®6ï·¯ï·¯, then 3ð¥ â ï·0 , ï·ðï·®2ï·¯ï·¯ And we know that cos ð>0 for ð â ï·0 , ï·ðï·®2ï·¯ï·¯ cos 3x >0 for 3x â ï·0 , ï·ðï·®2ï·¯ï·¯ cos 3x >0 for x â ï·0 , ï·ðï·®6ï·¯ï·¯ 3 cos 3x >0 for x â ï·0 , ï·ðï·®6ï·¯ï·¯ ðâ²(ð¥)>0 for x â ï·0 , ï·ðï·®6ï·¯ï·¯ Since fâ(x) â¥ 0 for ð¥ â ï·0 , ï·ðï·®6ï·¯ï·¯ Thus, f(x) is increasing for ð¥ â ï·0 , ï·ðï·®6ï·¯ï·¯ Case 2 Since ð¥ â ï·ï·ðï·®6ï·¯, ï·ðï·®2ï·¯ï·¯ ï·ðï·®6ï·¯<ð¥<ï·ðï·®2ï·¯ 3Ã ï·ðï·®6ï·¯<3ð¥<ï·3ðï·®2ï·¯ ï·ðï·®2ï·¯<3ð¥<ï·3ðï·®2ï·¯ So when ð¥ âï·ï·ðï·®6ï·¯ , ï·ðï·®2ï·¯ï·¯, then 3ð¥ â ï·ï·ðï·®2ï·¯ , ï·3ðï·®2ï·¯ï·¯ We know that, cos ð<0 for ð â ï·ï·ðï·®2ï·¯ , ï·3ðï·®2ï·¯ï·¯ cos 3ð¥<0 for 3ð¥ â ï·ï·ðï·®2ï·¯ , ï·3ðï·®2ï·¯ï·¯ cos 3ð¥<0 for ð¥ â ï·ï·ðï·®6ï·¯ , ï·ðï·®2ï·¯ï·¯ 3 cos 3ð¥<0 for ð¥ â ï·ï·ðï·®6ï·¯ , ï·ðï·®2ï·¯ï·¯ fâ(x) <0 for ð¥ â ï·ï·ðï·®6ï·¯ , ï·ðï·®2ï·¯ï·¯ Since fâ(x) â¤ 0 for ð¥ â ï·ï·ðï·®6ï·¯,ï·ðï·®2ï·¯ï·¯ Thus, f(x) is decreasing for ð¥ â ï·ï·ðï·®6ï·¯,ï·ðï·®2ï·¯ï·¯ Thus, f(x) is increasing for ð â ï·ð , ï·ðï·®ðï·¯ï·¯ & f(x) is strictly decreasing for ð â ï·ï·ðï·®ðï·¯ , ï·ðï·®ðï·¯ï·¯