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Last updated at Jan. 7, 2020 by Teachoo

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Ex 6.5, 2 Find the maximum and minimum values, if any, of the following functions given by (i) π (π₯)=|π₯ + 2| β1 π (π₯)=|π₯ + 2| β1 Minimum value of |π₯ + 2|=0 Minimum value of f(π₯)= minimum value of |π₯ + 2| β1 =0β1 =β1 Hence minimum value of f(π)=βπ And there is no maximum value of f(x) Ex 6.5, 2 Find the maximum and minimum values, if any, of the following functions given by (ii) π(π₯)= β | π₯ +1|+3 f(π₯)= β | π₯ +1|+3 We know that | π₯ +1|β₯0 So, β| π₯ +1|β€0 Maximum value of g(π₯) = maximum value of β | π₯ +1|+3 = 0 + 3 = 3 Hence maximum value of f(x) is 3 And there is no minimum value of f(x) Ex 6.5, 2 Find the maximum and minimum values, if any, of the following functions given by (iii) β(π₯)= sin β‘(2π₯)+ 5 β(π₯)= sin β‘(2π₯)+ 5 We know that β1 β€ sin ΞΈ β€ 1 β1 β€ sin 2π₯ β€ 1 Adding 5 both sides β1 + 5 β€ sin2π₯ + 5 β€ 1 + 5 4 β€ sin 2π₯ + 5 β€ 6 4 β€ f(π₯)β€6 Hence Maximum value of f(π)=π & Minimum value of f(π)=π Ex 6.5, 2 Find the maximum and minimum values, if any, of the following functions given by (iv) π (π₯)=|sinβ‘4π₯+3| π (π₯)=| sinβ‘4π₯+3| We know that β1 β€ sin ΞΈ β€ 1 So, β1 β€ sin 4π₯ β€ 1 Adding 3 both sides β1 + 3 β€ sin 4π₯ + 3 β€ 1 + 3 2 β€ sin 4π₯ +3 β€ 4 Taking modulus |2| β€ | sinβ‘4π₯+3| β€ |4| 2 β€ | sinβ‘4π₯+3| β€ |4| 2 β€ f(π₯)β€4 Hence Maximum value of f(π) is 4 & Minimum value of f(π) is 2 Ex 6.5, 2 Find the maximum and minimum values, if any, of the following functions given by (v) β(π₯)=π₯ + 1 , π₯ β (β1 , 1) Drawing graph of f(π₯)=π₯+1 β(π₯) have Maximum value of point closest to x = 1 & Minimum value of point closest to x = β1 but its not possible to locate such points Thus the given function has neither the maximum value nor minimum value

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Chapter 6 Class 12 Application of Derivatives

Serial order wise

About the Author

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.