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Ex 6.5

Ex 6.5, 1 (i)
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Ex 6.5, 1 (ii)

Ex 6.5, 1 (iii) Important

Ex 6.5, 1 (iv)

Ex 6.5, 2 (i)

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Ex 6.5, 4 (i)

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Ex 6.5, 27 (MCQ)

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Ex 6.5,29 (MCQ)

Last updated at Dec. 8, 2021 by Teachoo

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Ex 6.5, 1 (Method 1) Find the maximum and minimum values, if any, of the following functions given by (i) f (π₯) = (2π₯ β 1)^2 + 3f(π₯)=(2π₯β1)^2+3 Hence, Minimum value of (2π₯β1)^2 = 0 Minimum value of (2π₯β1^2 )+3 = 0 + 3 = 3 Square of number cant be negative It can be 0 or greater than 0 Also, there is no maximum value of π₯ β΄ There is no maximum value of f(x) Ex 6.5, 1 (Method 2) Find the maximum and minimum values, if any, of the following functions given by (i) f (π₯) = (2π₯ β 1)^2+3Finding fβ(x) f(π₯)=(2π₯β1)^2+3 fβ(π₯)= 2(2π₯β1) Putting fβ(π)=π 2(2π₯β1)=0 2π₯ β 1 = 0 2π₯ = 1 π₯ = 1/2 Thus, x = 1/2 is the minima Finding minimum value f(π₯)=(2π₯β1)^2+3 Putting π₯ = 1/2 f(1/2)=(2 Γ 1/2β1)^2+3= (1β1)^2+3= 3 β΄ Minimum value = 3 There is no maximum value Ex 6.5, 1 (Method 3) Find the maximum and minimum values, if any, of the following functions given by (i) π (π₯)= (2π₯ β 1)^2 + 3Double Derivative Test f(π₯)=(2π₯β1)^2+3 Finding fβ(π) fβ(π₯)=2(2π₯β1)^(2β1) = 2(2π₯β1) Putting fβ(π)=π 2(2π₯β1)=0 (2π₯β1)=0 2π₯ = 0 + 1 π₯ = 1/2 Finding fββ(π) fβ(π₯)=2(2π₯β1) fβ(π₯) = 4π₯ β 2 fββ(π₯)= 4 So, fββ (1/2) = 4 Since fββ (1/2) > 0 , π₯ = 1/2 is point of local minima Putting π₯ = 1/2 , we can calculate minimum value f(π₯) = (2π₯β1)^2+3 f(1/2)= (2(1/2)β1)^2+3= (1β1)^2+3= 3 Hence, Minimum value = 3 There is no Maximum value