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

Transcript

Ex 6.2,6 Find the intervals in which the following functions are strictly increasing or decreasing: (a) ๐‘ฅ2 + 2๐‘ฅ โ€“ 5 f (๐‘ฅ) = ๐‘ฅ2 + 2๐‘ฅ โ€“ 5 Step 1: Calculating fโ€™ (๐‘ฅ) fโ€™ (๐‘ฅ) = 2๐‘ฅ + 2 fโ€™ ( ๐‘ฅ) = 2 (๐‘ฅ + 1) Step 2: Putting fโ€™ (๐‘ฅ) = 0 2 (๐‘ฅ + 1) = 0 (๐‘ฅ + 1) = 0 ๐‘ฅ = โ€“1 Step 3: Plotting point on real line Point ๐‘ฅ = โ€“ 1 divide the real line into 2 disjoint interval i.e. (โˆ’โˆž , โˆ’1) & (โˆ’1 , โˆž) Step 4: Hence f is strictly decreasing for (โˆ’โˆž ," โ€“1" ) f is strictly increasing for ( "โ€“1" ,โˆž) Ex 6.2,6 Find the intervals in which the following functions are strictly increasing or decreasing: (b) 10 โ€“ 6๐‘ฅ โ€“ 2๐‘ฅ2 Step 1: Calculating fโ€™ (๐‘ฅ) f (๐‘ฅ) = 10 โ€“ 6๐‘ฅ โ€“ 2๐‘ฅ2 fโ€™ (๐‘ฅ) = โ€“6 โ€“ 4๐‘ฅ fโ€™ (๐‘ฅ) = 2( โ€“3 โ€“ 2๐‘ฅ) Step 2: Putting fโ€™ (๐‘ฅ) = 0 2(โˆ’3 โˆ’2๐‘ฅ)=0 โ€“3 โ€“ 2๐‘ฅ = 0 โ€“2๐‘ฅ = 3 ๐‘ฅ = (โˆ’3)/2 Step 3: Plotting point on real line Point ๐‘ฅ = (โˆ’3)/2 divide the real line into 2 disjoint interval i.e. (โˆ’โˆž , (โˆ’3)/2) & ((โˆ’3)/2 , โˆž) Step 4: Hence f is strictly decreasing for ๐’™ > (โˆ’๐Ÿ‘)/๐Ÿ f is strictly increasing for ๐’™ < (โˆ’๐Ÿ‘)/๐Ÿ Ex 6.2,6 Find the intervals in which the following functions are strictly increasing or decreasing: (c) โ€“2๐‘ฅ3 โ€“ 9๐‘ฅ2 โ€“ 12๐‘ฅ + 1 f (๐‘ฅ) = โ€“2๐‘ฅ3 โ€“ 9๐‘ฅ2 โ€“ 12๐‘ฅ + 1 Step 1: Calculating fโ€™(๐‘ฅ) fโ€™ (๐‘ฅ) = โ€“6๐‘ฅ2 โ€“18๐‘ฅ โ€“ 12 + 0 fโ€™ (๐‘ฅ) = โ€“6(๐‘ฅ2+3๐‘ฅ+2) fโ€™ (๐‘ฅ) = โ€“6(๐‘ฅ2+2๐‘ฅ+๐‘ฅ+2) fโ€™ (๐‘ฅ) = โ€“6(๐‘ฅ(๐‘ฅ+2)+1(๐‘ฅ+2)) fโ€™ (๐‘ฅ) = โ€“6((๐‘ฅ+1)( ๐‘ฅ+2)) fโ€™ (๐‘ฅ) = โ€“6(๐‘ฅ+1) (๐‘ฅ+2) Step 2: Putting fโ€™ (๐‘ฅ) = 0 fโ€™ (๐‘ฅ) = 0 โ€“ 6 (๐‘ฅ+1) (๐‘ฅ+2) = 0 (๐‘ฅ+1) (๐‘ฅ+2) = 0 So, ๐‘ฅ = โ€“1 , โ€“2 Step 3: Plotting value of ๐‘ฅ on real line Point ๐‘ฅ = โ€“2 & โ€“1 divide the real line into 3 disjoint intervals i.e. (โˆ’โˆž , โˆ’2) (โˆ’2 , โˆ’1) & (โˆ’1 , โˆž) Step 4: Hence, f is strictly increasing for โ€“2 < ๐’™ < โ€“1 & strictly decreasing for ๐’™ < โ€“2 & ๐’™ > โ€“1 Ex 6.2,6 Find the intervals in which the following functions are strictly increasing or decreasing: (d) 6 โ€“ 9๐‘ฅ โ€“ ๐‘ฅ2 f(๐‘ฅ) = 6 โ€“ 9๐‘ฅ โ€“ ๐‘ฅ2 Step 1: Calculating fโ€™ (๐‘ฅ) fโ€™ (๐‘ฅ) = โ€“9 โ€“ 2๐‘ฅ fโ€™ (๐‘ฅ) = โ€“(2๐‘ฅ+9) Step 2: Putting fโ€™(๐‘ฅ) = 0 โ€“(2๐‘ฅ+9) = 0 2๐‘ฅ + 9 = 0 2๐‘ฅ = โ€“ 9 ๐‘ฅ = (โˆ’9)/2 Step 3: Plotting point ๐‘ฅ = (โˆ’9)/2 on real line Point ๐‘ฅ = (โˆ’9)/2 divide the real line into disjoint interval i.e. (โˆ’โˆž , (โˆ’9)/( 2)) & ((โˆ’9)/2 , โˆž) Step 4: Hence f is Strictly Increasing for ๐’™ < (โˆ’๐Ÿ—)/๐Ÿ & Strictly Decreasing for ๐’™ > (โˆ’๐Ÿ—)/๐Ÿ Ex 6.2,6 Find the intervals in which the following functions are strictly increasing or decreasing: (e) (๐‘ฅ + 1)^3 (๐‘ฅ โ€“ 3)^3 f(๐‘ฅ) = (๐‘ฅ+1)3 (๐‘ฅโˆ’3)3 Step 1: Calculating fโ€™(๐‘ฅ) f(๐‘ฅ) = (๐‘ฅ+1)3 (๐‘ฅโˆ’3)3 Using product rule in (๐‘ฅ+1)3 (๐‘ฅโˆ’3)3 As (๐‘ข๐‘ฃ)^โ€ฒ=๐‘ข^โ€ฒ ๐‘ฃ+๐‘ฃ^โ€ฒ ๐‘ข fโ€™(๐‘ฅ)= ใ€–[(๐‘ฅ+1)^3]ใ€—^โ€ฒ (๐‘ฅโˆ’3)^3 +[(๐‘ฅโˆ’3)3]^โ€ฒ (๐‘ฅ+1)^3 fโ€™(๐‘ฅ)=3(๐‘ฅ+1)2(๐‘ฅโˆ’3)3 + 3(๐‘ฅโˆ’3)2(๐‘ฅ+1)3 fโ€™(๐‘ฅ)=3(๐‘ฅ+1)2(๐‘ฅโˆ’3)2 ((๐‘ฅโˆ’3)+ (๐‘ฅ+1)) fโ€™(๐‘ฅ)=3(๐‘ฅ+1)2(๐‘ฅโˆ’3)2 (2๐‘ฅโˆ’2) fโ€™(๐‘ฅ)= 6(๐‘ฅ+1)2 (๐‘ฅโˆ’3)2 (๐‘ฅโˆ’1) Step 2: Putting fโ€™(๐‘ฅ)=0 6(๐‘ฅ+1)2 (๐‘ฅโˆ’3)2 (๐‘ฅโˆ’1) = 0 Hence, ๐‘ฅ = โ€“1 , 1 , & 3 Step 3: Plotting values of x on real line. The point x = โ€“1 , 1 & 3 divide the line segment into 4 disjoint intervals i.e. (โˆ’โˆž , โˆ’1) (โˆ’1 , 1) (1 , 3) & (3 , โˆž)Note that:- fโ€™(๐‘ฅ) = 6 (๐‘ฅ+1)^2 (๐‘ฅโˆ’3)^2 (๐‘ฅโˆ’1) Step 4: Plotting values of x on real line. Hence, f is strictly increasing for 1 < ๐‘ฅ < 3 & ๐‘ฅ > 3 i.e. (1, 3) and (3, โˆž) f is strictly decreasing for ๐‘ฅ < โ€“1 & โˆ’1<๐‘ฅ< 1 i.e. (โ€“ โˆž, โ€“1) and (โ€“ 1, 1)

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.