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Ex 6.5

Ex 6.5, 1 (i)
Important

Ex 6.5, 1 (ii)

Ex 6.5, 1 (iii) Important

Ex 6.5, 1 (iv)

Ex 6.5, 2 (i)

Ex 6.5, 2 (ii) Important

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Ex 6.5, 2 (iv) Important

Ex 6.5, 2 (v) Important

Ex 6.5, 3 (i)

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Ex 6.5, 3 (iv) Important

Ex 6.5, 3 (v)

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Ex 6.5, 3 (vii) Important

Ex 6.5, 3 (viii)

Ex 6.5, 4 (i)

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Ex 6.5, 5 (i)

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Ex 6.5, 5 (iii) Important

Ex 6.5, 5 (iv)

Ex 6.5,6

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Ex 6.5,8

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Ex 6.5,10 You are here

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Ex 6.5, 27 (MCQ)

Ex 6.5,28 (MCQ) Important

Ex 6.5,29 (MCQ)

Last updated at April 15, 2021 by Teachoo

Ex 6.5, 10 Find the maximum value of 2π₯3 β 24π₯ + 107 in the interval [1, 3]. Find the maximum value of the same function in [β3, β1]. Let f(π₯)=2π₯^3β24π₯+107 Finding fβ(π₯) fβ(π₯)=π(2π₯^3 β 24π₯ + 107)/ππ₯ = 2 Γ 3π₯^2β24 = 6π₯^2β24 = 6 (π₯^2β4" " ) Putting f(π₯)=0 6 (π₯^2β4" " )=0 π₯^2β4" = 0 " π₯^2=4 π₯=Β±β4 π₯=Β±2 Thus, π₯=2 , β 2 Since x is in interval [1 , 3] π₯ = 2 is only Critical point Also, since given the interval π₯= β [1 , 3] We calculate f(x) at π₯= 1 , 2 & 3 Hence maximum value of f(π)=ππ at π = 3 in the interval [1 , 3] For the interval [βπ , βπ] π₯ = β2 is only Critical Point Also, since given the interval π₯= β [β3,β1] We calculate f(x) at π₯= β1 , β2 & β3 Hence maximum value of f(π)=πππ at π = β2 in the interval [β3,β1]