# Ex 6.2,9 - Chapter 6 Class 12 Application of Derivatives

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

Ex 6.2,9 Prove that = 4 sin (2 + cos ) is an increasing function of in 0, 2 . = 4 sin 2 + cos We need to prove that function is increasing function of in 0 , 2 i.e. we need to prove that > 0 for 0 , 2 We have = 4 sin 2 + cos = 4 sin 2 + cos = 4 sin 2 + cos = 4 sin 2 + cos 1 = 4 sin 2 + cos 2 + cos 4 sin 2 + cos 2 1 = 4 cos 2 + cos 0 sin 4 sin 2 + cos 2 1 = 8 cos + 4 2 + 2 2 + cos 2 1 = 8 cos + 4 1 2 + cos 2 1 = 8 cos + 4 2 + cos 2 1 = 8 cos + 4 2 + cos 2 2 + cos 2 = 8 cos + 4 4 + 2 + 4 2 + cos 2 = 8 cos + 4 4 2 4 2 + cos 2 = 8 cos 4 cos + 4 4 2 2 + cos 2 = 4 cos + 0 2 2 + cos 2 = 4 cos 2 2 + cos 2 For the function to be increasing = cos 4 cos 2 + cos 2 > 0 Since denominator is a square, We need to show that cos 4 cos > 0 for 0 , 2 Now, 0 cos 1 Multiplying by 1 1 cos 0 Adding 4 both sides 1 + 4 cos +4 < 0 + 4 3 4 cos2 < 4 Thus, (4 cos2 ) is positive And cos is also positive Hence cos (4 cos2 ) is also positive cos (4 cos2 ) > 0 for 0 , 2 Hence = 4 sin 2 + is an increasing function for ,

Chapter 6 Class 12 Application of Derivatives

Serial order wise

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.