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

Transcript

Ex 6.2,7 Show that ð¦ = log(1 + ð¥) â ï·2ð¥ï·®2 + ð¥ï·¯ , ð¥ > â 1 , is an increasing function of ð¥ throughout its domain. ð¦ = log ï·1+ð¥ï·¯ â ï·2ð¥ï·®2 + ð¥ï·¯ , ð¥ > â1 We need to show that y is strictly increasing function for ð¥ > â1 i.e. we need to show that ï·ðð¦ ï·®ðð¥ï·¯ > 0 for ð¥ > â1 Step 1: Finding ï·ðð¦ï·®ðð¥ï·¯ ð¦ = log ï·1+ð¥ï·¯ â ï·2ð¥ ï·®2 + ð¥ï·¯ ï·ðð¦ ï·®ðð¥ï·¯ = ï·ðï·ï·logï·®ï·1 + ð¥ï·¯ â ï·2ð¥ï·®2 + ð¥ï·¯ï·¯ï·¯ï·®ðð¥ï·¯ ï·ðð¦ ï·®ðð¥ï·¯ = ï·ðï·ðððï·1+ð¥ï·¯ï·¯ï·®ðð¥ï·¯ â ï·ðï·®ðð¥ï·¯ï·ï·2ð¥ï·®2+ð¥ï·¯ï·¯ ï·ðð¦ ï·®ðð¥ï·¯ = ï·1ï·®1 + ð¥ï·¯ . ï·1+ð¥ï·¯â â ï·ï·ï·ï·2ð¥ï·¯ï·®â²ï·¯ï·2 + ð¥ï·¯ âï· ï·2 + ð¥ï·¯ï·®â²ï·¯2ð¥ï·®ï·2 + ð¥ï·¯2ï·¯ï·¯ ï·ðð¦ ï·®ðð¥ï·¯ = ï·1ï·®1 + ð¥ï·¯ . ï·0+1ï·¯ â ï·ï·2ï·2 + ð¥ï·¯ â ï·0 + 1ï·¯2ð¥ï·®ï·2 + ð¥ï·¯2ï·¯ï·¯ ï·ðð¦ ï·®ðð¥ï·¯ = ï·1ï·®1 + ð¥ï·¯ âï·ï·4 + 2ð¥ â 2ð¥ï·®ï·2 + ð¥ï·¯2ï·¯ï·¯ ï·ðð¦ ï·®ðð¥ï·¯ = ï·1ï·®1 + ð¥ï·¯ â ï·ï·4ï·®ï·2 + ð¥ï·¯2ï·¯ï·¯ ï·ðð¦ ï·®ðð¥ï·¯ = ï·ï·2 + ð¥ï·¯2 â 4ï·1 + ð¥ï·¯ï·®ï·1 + ð¥ï·¯ï·2 + ð¥ï·¯2ï·¯ ï·ðð¦ ï·®ðð¥ï·¯ = ï·ï·2ï·¯2 + ï·ð¥ï·¯2 + 2ï·2ï·¯ï·ð¥ï·¯ â 4 â 4ð¥ï·®ï·1 + ð¥ï·¯ï·2 + ð¥ï·¯2ï·¯ ï·ðð¦ ï·®ðð¥ï·¯ = ï·4 + ð¥2 + 4ð¥ â 4 â 4ð¥ï·®ï·1 + ð¥ï·¯ï·2 + ð¥ï·¯2ï·¯ ï·ðð¦ ï·®ðð¥ï·¯ = ï·ð¥2ï·®ï·1 + ð¥ï·¯ï·2 + ð¥ï·¯2ï·¯ ï·ðð¦ ï·®ðð¥ï·¯ = ï·ï·ï·ð¥ï·®2 + ð¥ï·¯ï·¯ï·®2ï·¯ ï·1ï·®1 + ð¥ï·¯ Now, ï·ðð¦ ï·®ðð¥ï·¯ = ï·ï·ï·ð¥ï·®2 + ð¥ï·¯ï·¯ï·®2ï·¯ ï·1ï·®1 + ð¥ï·¯ Now, finding value where ï·ðð¦ï·®ðð¥ï·¯ > 0 ï·ðð¦ï·®ðð¥ï·¯ > 0 ï·ï·ï·ð¥ï·®2+ð¥ï·¯ï·¯ï·®2ï·¯.ï·ï·1ï·®1+ð¥ï·¯ï·¯ > 0 ï·ï·1ï·®1+ð¥ ï·¯ï·¯>0 This is possible only when 1 + ð¥ > 0 i.e. ð¥ > â1 So, ï·ððï·®ððï·¯ > 0 for ð > â1 Hence proved