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Ex 6.3
Ex 6.3, 1 (ii)
Ex 6.3, 1 (iii) Important
Ex 6.3, 1 (iv)
Ex 6.3, 2 (i)
Ex 6.3, 2 (ii) Important
Ex 6.3, 2 (iii)
Ex 6.3, 2 (iv) Important
Ex 6.3, 2 (v) Important
Ex 6.3, 3 (i)
Ex 6.3, 3 (ii)
Ex 6.3, 3 (iii)
Ex 6.3, 3 (iv) Important
Ex 6.3, 3 (v)
Ex 6.3, 3 (vi)
Ex 6.3, 3 (vii) Important
Ex 6.3, 3 (viii)
Ex 6.3, 4 (i)
Ex 6.3, 4 (ii) Important
Ex 6.3, 4 (iii)
Ex 6.3, 5 (i)
Ex 6.3, 5 (ii)
Ex 6.3, 5 (iii) Important
Ex 6.3, 5 (iv)
Ex 6.3,6
Ex 6.3,7 Important
Ex 6.3,8
Ex 6.3,9 Important
Ex 6.3,10 You are here
Ex 6.3,11 Important
Ex 6.3,12 Important
Ex 6.3,13
Ex 6.3,14 Important
Ex 6.3,15 Important
Ex 6.3,16
Ex 6.3,17
Ex 6.3,18 Important
Ex 6.3,19 Important
Ex 6.3, 20 Important
Ex 6.3,21
Ex 6.3,22 Important
Ex 6.3,23 Important
Ex 6.3,24 Important
Ex 6.3,25 Important
Ex 6.3, 26 Important
Ex 6.3, 27 (MCQ)
Ex 6.3,28 (MCQ) Important
Ex 6.3,29 (MCQ)
Last updated at May 29, 2023 by Teachoo
Ex 6.3, 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]