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

Transcript

Example 13 Find the intervals in which the function f given by f ï·ð¥ï·¯=ï·sinï·®ð¥ï·¯+ï·cosï·®ð¥ï·¯ , 0 â¤ ð¥ â¤ 2ð is strictly increasing or strictly decreasing. fï·ð¥ï·¯ = sin ð¥ + cos ð¥ 0 â¤ x â¤ 2Ï Step 1: Finding fâï·ð¥ï·¯ fï·ð¥ï·¯ = sin x + cos ð¥ fâï·ð¥ï·¯ = ï·ð ï·®ðð¥ï·¯ (sin ð¥ + cos ð¥) fâï·ð¥ï·¯ = ï·ðï·ï·sinï·®ð¥ï·¯ï·¯ï·®ðð¥ï·¯ + ï·ðï·ï·cosï·®ð¥ï·¯ï·¯ï·®ðð¥ï·¯ fâï·ð¥ï·¯ = cos ð¥ + ï·âð ððð¥ï·¯ fâï·ð¥ï·¯ = cosâ¡ð¥ â sinâ¡ð¥ Step 2: Putting fâï·ð¥ï·¯ = 0 cos ð¥ â sin ð¥ = 0 cos ð¥ = sin ð¥ â´ð¥=ï·ðï·®4ï·¯ ,ï·5ðï·®4ï·¯ ðð  0 â¤ ð¥ â¤ 2ð Step 3: Plotting points Points ð¥=ï·ðï·®4ï·¯ ,ï·5ðï·®4ï·¯ divide interval ï·0 , 2ðï·¯ into 3 disjoint intervals ï·0 , ï·ðï·®4ï·¯ï·¯, ï·ï·ðï·®4ï·¯,ï·5ðï·®4ï·¯ï·¯, ï·ï·5ðï·®4ï·¯ , 2ðï·¯ Step 4: Checking sign of fâï·ð¥ï·¯ = cos ð¥ â sin ð¥ Case 1 When ð¥ â ï·0 , ï·ðï·®4ï·¯ï·¯ as 0 â¤ ð¥ < ï·ðï·®4ï·¯ Thus, fâï·ð¥ï·¯ > 0 for ð¥ â ï·0 , ï·ðï·®4ï·¯ï·¯ Case 2 When ð¥ â ï·ï·ðï·®4ï·¯,ï·5ðï·®4ï·¯ï·¯ As ï·ðï·®4ï·¯ < x < ï·5ðï·®4ï·¯ Let us find value of fâ(x) at any value of ð¥ lies between ï·ðï·®4ï·¯,ï·5ðï·®4ï·¯ fâï·ð¥ï·¯ < 0 for ð¥ â ï·ï·ðï·®4ï·¯,ï·5ðï·®4ï·¯ï·¯ Case 3 When ð¥ â ï·ï·5ðï·®4ï·¯ , 2ðï·¯ as ï·5ðï·®4ï·¯ < ð¥ â¤ 2ð At ð¥ = 2Ï fâï·ð¥ï·¯ = cos ð¥ â sin ð¥ fâï·2ðï·¯ = cos 2ð â sin 2ð = cos ï·ð+ðï·¯ â 0 = â cos Ï â 0 = â ï·â1ï·¯ â0 = 1 > 0 So, fâï·ð¥ï·¯ > 0 at ð¥ = 2Ï Hence fâ(x) > 0 for ð¥ â ï·ï·5ðï·®4ï·¯ , 2ðï·¯ Thus, f is strictly increasing intervals ï·ð , ï·ðï·®ðï·¯ï·¯& ï·ï·ððï·®ðï·¯ , ððï·¯ f is strictly increasing intervals ï·ï·ðï·®ðï·¯ , ï·ððï·®ðï·¯ï·¯