Ex 6.5, 2 - Find max and min values (i) f(x) = |x + 2| - 1 - Ex 6.5

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  1. Chapter 6 Class 12 Application of Derivatives
  2. Serial order wise
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Ex 6.5,2 (Method 1) Find the maximum and minimum values, if any, of the following functions given by ๐‘–๏ทฏ ๐‘“ (๐‘ฅ)=|๐‘ฅ + 2| โ€“1 ๐‘“ (๐‘ฅ)=|๐‘ฅ + 2| โ€“1 Minimum value of ๐‘ฅ + 2๏ทฏ=0 Minimum value of f ๐‘ฅ๏ทฏ= minimum value of |๐‘ฅ + 2| โ€“1 =0โˆ’1 =โˆ’1 Hence minimum value of f ๐’™๏ทฏ=โˆ’๐Ÿ Ex 6.5,2 (Method 2) Find the maximum and minimum values, if any, of the following functions given by ๐‘–๏ทฏ ๐‘“ (๐‘ฅ)=|๐‘ฅ + 2| โ€“1 ๐‘“ (๐‘ฅ)=|๐‘ฅ + 2| โ€“1 f ๐‘ฅ๏ทฏ= ๐‘ฅ+2๏ทฏโˆ’1 wโ„Ž๐‘’๐‘Ÿ๐‘’ ๐‘ฅโ‰ฅโˆ’2๏ทฎโ€“ ๐‘ฅ+2๏ทฏโˆ’1 ๐‘คโ„Ž๐‘’๐‘Ÿ๐‘’ ๐‘ฅ<โˆ’2 ๏ทฏ๏ทฏ f ๐‘ฅ๏ทฏ= ๐‘ฅ+1 , ๐‘คโ„Ž๐‘’๐‘Ÿ๐‘’ ๐‘ฅโ‰ฅโˆ’2๏ทฎโˆ’๐‘ฅโˆ’3 , where ๐‘ฅ<โˆ’2๏ทฏ๏ทฏ We check sign of fโ€™ ๐‘ฅ๏ทฏ at ๐‘ฅ=โˆ’2 Thus, at x = โ€“2, fโ€™(x) changes sign from negative to positive โ‡’ ๐‘ฅ=โˆ’2 is point of minima of f ๐‘ฅ๏ทฏ Finding minimum value of f ๐‘ฅ๏ทฏ Putting ๐‘ฅ=โˆ’2 f ๐‘ฅ๏ทฏ =|๐‘ฅ + 2| โ€“1 f 2๏ทฏ=|โˆ’2+ 2| โ€“1 = 0 โ€“ 1 = โˆ’1 Minimum value of f ๐‘ฅ๏ทฏ is โ€“1 at ๐‘ฅ=โˆ’2 f ๐‘ฅ๏ทฏ has no maximum value Ex 6.5,2 (Method 1) Find the maximum and minimum values, if any, of the following functions given by (ii) ๐‘“(๐‘ฅ)= โ€“ | ๐‘ฅ +1|+3 f(๐‘ฅ)= โ€“ | ๐‘ฅ +1|+3 We know that ๐‘ฅ +1๏ทฏโ‰ฅ0 So, โˆ’ ๐‘ฅ +1๏ทฏโ‰ค0 Maximum value of g ๐‘ฅ๏ทฏ = maximum value of โ€“ | ๐‘ฅ +1|+3 = 0 + 3 = 3 Hence maximum value of f(x) is 3 Ex 6.5,2 (Method 2) Find the maximum and minimum values, if any, of the following functions given by (ii) ๐‘“(๐‘ฅ)= โ€“ |๐‘ฅ+1|+3 g(๐‘ฅ)= โ€“|๐‘ฅ+1|+3 g(๐‘ฅ)= โ€“ ๐‘ฅ+1๏ทฏ+3 ๐‘ฅโ‰ฅโˆ’1๏ทฎโ€“ โˆ’ ๐‘ฅ+1๏ทฏ๏ทฏ+3 ๐‘ฅ<โˆ’1๏ทฏ๏ทฏ g(๐‘ฅ)= โˆ’๐‘ฅ+2 ๐‘ฅโ‰ฅโˆ’1๏ทฎ๐‘ฅ+4 ๐‘ฅ<โˆ’1๏ทฏ๏ทฏ We check sign of fโ€™ ๐‘ฅ๏ทฏ at ๐‘ฅ=โˆ’1 Thus, at x = โ€“1, fโ€™(x) changes sign from positive to negative โ‡’ ๐‘ฅ=โˆ’1 is point of maxima of f ๐‘ฅ๏ทฏ Finding maximum value of f ๐‘ฅ๏ทฏ Putting ๐‘ฅ=โˆ’1 f ๐‘ฅ๏ทฏ=โˆ’ | ๐‘ฅ +1|+3 f ๐‘ฅ๏ทฏ=โˆ’ | ๐‘ฅ +1|+3 f โˆ’1๏ทฏ =โˆ’ |โˆ’1 +1|+3 =โˆ’ |0|+3 = 3 Hence Maximum value is 3 Ex 6.5,2 Find the maximum and minimum values, if any, of the following functions given by (iii) โ„Ž ๐‘ฅ๏ทฏ= sin โก(2๐‘ฅ)+ 5 โ„Ž ๐‘ฅ๏ทฏ= sin โก(2๐‘ฅ)+ 5 We know that โ€“1 โ‰ค sin ฮธ โ‰ค 1 โ€“1 โ‰ค sin 2๐‘ฅ โ‰ค 1 Adding 5 both sides โ€“1 + 5 โ‰ค sin2๐‘ฅ + 5 โ‰ค 1 + 5 4 โ‰ค sin 2๐‘ฅ + 5 โ‰ค 6 4 โ‰ค f ๐‘ฅ๏ทฏโ‰ค6 Hence Maximum value of f ๐’™๏ทฏ=๐Ÿ” & Minimum value of f ๐’™๏ทฏ=๐Ÿ’ Ex 6.5,2 Find the maximum and minimum values, if any, of the following functions given by (iv) ๐‘“ (๐‘ฅ)=|sinโก4๐‘ฅ+3| ๐‘“ (๐‘ฅ)=| sinโก4๐‘ฅ+3| We know that โ€“1 โ‰ค sin ฮธ โ‰ค 1 So, โ€“1 โ‰ค sin 4๐‘ฅ โ‰ค 1 Adding 3 both sides โ€“1 + 3 โ‰ค sin 4๐‘ฅ + 3 โ‰ค 1 + 3 2 โ‰ค sin 4๐‘ฅ +3 โ‰ค 4 Taking modulus |2| โ‰ค | sinโก4๐‘ฅ+3| โ‰ค |4| 2 โ‰ค | sinโก4๐‘ฅ+3| โ‰ค |4| 2 โ‰ค f ๐‘ฅ๏ทฏโ‰ค4 Hence Maximum value of f ๐‘ฅ๏ทฏ is 4 & Minimum value of f ๐‘ฅ๏ทฏ is 2 Hence Maximum value of f ๐’™๏ทฏ is 4 & Minimum value of f ๐’™๏ทฏ is 2 Ex 6.5,2 Find the maximum and minimum values, if any, of the following functions given by (v) โ„Ž ๐‘ฅ๏ทฏ=๐‘ฅ + 1 , ๐‘ฅ โˆˆ โˆ’1 , 1๏ทฏ Drawing graph of f ๐‘ฅ๏ทฏ=๐‘ฅ+1 โ„Ž ๐‘ฅ๏ทฏ have Maximum value of point closest to x = 1 & Minimum value of point closest to x = โ€“1 but its not possible to locate such points Thus the given function has neither the maximum value nor minimum value

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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.