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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)

Ex 6.5, 3 (ii)

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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, 4 (iii)

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 6.5,10

Ex 6.5,11 Important 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, 11 It is given that at π₯ = 1, the function π₯4 β 62π₯2 + ππ₯+ 9 attains its maximum value, on the interval [0, 2]. Find the value of a.We have f(π₯)=π₯4 β 62π₯2 + ππ₯+ 9 Finding fβ(π) fβ(π₯)=π(π₯^4β 62π₯^2 + ππ₯ + 9)/ππ₯ = γ4π₯γ^3β62 Γ2π₯+π = γ4π₯γ^3β124π₯+π Given that at π₯=1, f(π₯)=π₯^4β62π₯^2+ππ₯+9 attain its Maximum Value i.e. f(π₯) maximum at π₯=1 β΄ πβ(π₯)=0 at π₯=1 Now, fβ(1)=0 γ4π₯γ^3β124π₯+π = 0 4(1)^3β124(1)+a=0 4 β 124 + a = 0 β120 + a = 0 a = 120 Hence, a = 120