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

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Ex 6.2,6 Find the intervals in which the following functions are strictly increasing or decreasing: (a) 𝑥2 + 2𝑥 – 5 f (𝑥) = 𝑥2 + 2𝑥 – 5 Step 1: Calculating f’ (𝑥) f’ (𝑥) = 2𝑥 + 2 f’ ( 𝑥) = 2 (𝑥 + 1) Step 2: Putting f’ (𝑥) = 0 2 (𝑥 + 1) = 0 (𝑥 + 1) = 0 𝑥 = –1 Step 3: Plotting point on real line Point 𝑥 = – 1 divide the real line into 2 disjoint interval i.e. − −1 & −1 , uc1 Step 4: Hence f is decreasing for − –1 f is increasing for –1 ,uc1 Ex 6.2,6 Find the intervals in which the following functions are strictly increasing or decreasing: (b) 10 – 6𝑥 – 2𝑥2 Step 1: Calculating f’ (𝑥) f 𝑥 = 10 – 6𝑥 – 2𝑥2 f’ 𝑥 = –6 – 4𝑥 f’ 𝑥 = 2( –3 – 2𝑥) Step 2: Putting f’ 𝑥 = 0 2−3 −2𝑥=0 –3 – 2𝑥 = 0 –2𝑥 = 3 𝑥 = −32 Step 3: Plotting point on real line Point 𝑥 = −32 divide the real line into 2 disjoint interval i.e. − −32 & −32 , uc1 Step 4: Hence f is decreasing for 𝒙 > −𝟑𝟐 f is increasing for 𝒙 < −𝟑𝟐 Ex 6.2,6 Find the intervals in which the following functions are strictly increasing or decreasing: (c) –2𝑥3 – 9𝑥2 – 12𝑥 + 1 f 𝑥 = –2𝑥3 – 9𝑥2 – 12𝑥 + 1 Step 1: Calculating f’𝑥 f’ 𝑥 = –6𝑥2 –18𝑥 – 12 + 0 f’ 𝑥 = –6𝑥2+3𝑥+2 f’ 𝑥 = –6𝑥2+2𝑥+𝑥+2 f’ 𝑥 = –6𝑥𝑥+2+1(𝑥+2) f’ 𝑥 = –6𝑥+1( 𝑥+2) f’ 𝑥 = –6𝑥+1 𝑥+2 Step 2: Putting f’ 𝑥 = 0 f’ 𝑥 = 0 – 6 𝑥+1 𝑥+2 = 0 𝑥+1 𝑥+2 = 0 So, 𝑥 = –1 , –2 Step 3: Plotting value of 𝑥 on real line Point 𝑥 = –2 & –1 divide the real line into 3 disjoint intervals i.e. − −2 −2 , −1 & −1 , uc1 Step 4: Hence, f is increasing for –2 < 𝒙 < –1 & decreasing for 𝒙 < –2 & 𝒙 > –1 Ex 6.2,6 Find the intervals in which the following functions are strictly increasing or decreasing: (d) 6 – 9𝑥 – 𝑥2 f𝑥 = 6 – 9𝑥 – 𝑥2 Step 1: Calculating f’ 𝑥 f’ 𝑥 = –9 – 2𝑥 f’ 𝑥 = –2𝑥+9 Step 2: Putting f’𝑥 = 0 –2𝑥+9 = 0 2𝑥 + 9 = 0 2𝑥 = – 9 𝑥 = −92 Step 3: Plotting point 𝑥 = −92 on real line Point 𝑥 = −92 divide the real line into disjoint interval i.e. − −9 2 & −92 , uc1 Step 4: Hence f is Strictly Increasing for 𝒙 < −𝟗𝟐 & Strictly Decreasing for 𝒙 > −𝟗𝟐 Ex 6.2,6 Find the intervals in which the following functions are strictly increasing or decreasing: (e) 𝑥 + 13 𝑥 – 33 f𝑥 = 𝑥+13 𝑥−33 Step 1: Calculating f’𝑥 f𝑥 = 𝑥+13 𝑥−33 f’𝑥= [𝑥+13]′𝑥−33 +𝑥−33′𝑥+13 f’𝑥=3𝑥+12𝑥−33 + 3𝑥−32𝑥+13 f’𝑥=3𝑥+12𝑥−32 𝑥−3+ 𝑥+1 f’𝑥=3𝑥+12𝑥−32 2𝑥−2 f’𝑥= 6𝑥+12 𝑥−32 𝑥−1 Step 2: Putting f’𝑥=0 6𝑥+12 𝑥−32 𝑥−1 = 0 Hence, 𝑥 = –1 , 1 , & 3 Step 3: Plotting values of x on real line. The point x = –1 , 1 & 3 divide the line segment into 4 disjoint intervals i.e. − −1 −1 , 1 1 , 3 & 3 , uc1 Note that:- f’𝑥 = 6 𝑥+12 𝑥−32 𝑥−1 Step 4: Plotting values of x on real line. Hence, f is strictly increasing for 1 < 𝒙 < 3 & 𝒙 > 3 f is strictly decreasing for 𝒙 < –1 & −𝟏<𝒙< 1

Chapter 6 Class 12 Application of Derivatives

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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.