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Ex 6.5
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
Ex 6.5, 2 (iii)
Ex 6.5, 2 (iv) Important
Ex 6.5, 2 (v) Important
Ex 6.5, 3 (i)
Ex 6.5, 3 (ii)
Ex 6.5, 3 (iii)
Ex 6.5, 3 (iv) Important
Ex 6.5, 3 (v)
Ex 6.5, 3 (vi)
Ex 6.5, 3 (vii) Important
Ex 6.5, 3 (viii)
Ex 6.5, 4 (i)
Ex 6.5, 4 (ii) Important
Ex 6.5, 4 (iii)
Ex 6.5, 5 (i)
Ex 6.5, 5 (ii)
Ex 6.5, 5 (iii) Important
Ex 6.5, 5 (iv)
Ex 6.5,6
Ex 6.5,7 Important
Ex 6.5,8
Ex 6.5,9 Important
Ex 6.5,10
Ex 6.5,11 Important
Ex 6.5,12 Important
Ex 6.5,13
Ex 6.5,14 Important
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Ex 6.5,16 You are here
Ex 6.5,17
Ex 6.5,18 Important
Ex 6.5,19 Important
Ex 6.5, 20 Important
Ex 6.5,21
Ex 6.5,22 Important
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Ex 6.5,24 Important
Ex 6.5,25 Important
Ex 6.5, 26 Important
Ex 6.5, 27 (MCQ)
Ex 6.5,28 (MCQ) Important
Ex 6.5,29 (MCQ)
Last updated at March 16, 2023 by Teachoo
Ex 6.5, 16 Find two positive numbers whose sum is 16 and the sum of whose cubes is minimum.Let first number be π₯ Now, First number + second number =16 π₯ + second number = 16 second number = 16 β π₯ Now, Sum of Cubes = (ππππ π‘ ππ’ππππ )^3+(π πππππ ππ’ππππ )^3 Let S(π₯) = π₯3 + (16βπ₯)^3 We Need to Find Minimum Value of s(π₯) Finding Sβ(π₯) Sβ(π₯)= π(π₯^3+ (16 β π₯)^3 )/ππ₯ = 3π₯2 + 3(16βπ₯)^2. (0β1) = 3π₯2 + 3(16βπ₯)^2 (β1) = 3π₯2 β 3((16)^2+(π₯)^2β2(16)(π₯)) = 3π₯2 β 3(256+π₯^2β32π₯) = 3π₯2 β 3(256)β3π₯^2+3(32)π₯ = β3(256β32π₯) Putting Sβ(π₯)=0 β3(256β32π₯)=0 256 β 32π₯ = 0 32π₯ = 256 π₯ = 256/32 π₯ = 8 Finding Sββ(π₯) Sβ(π₯)=β3(256β32π₯) Sββ(π₯)=π(β3(256 β 32π₯))/ππ₯ = β3 π(256 β 32π₯)/ππ₯ = β3 [0β32] = 96 > 0 Since Sββ(π₯)>0 for π₯ = 8 π₯ = 8 is point of local minima & S(π₯) is minimum at π₯ = 8 Hence, 1st number = x = 8 & 2nd number = 16 β x = 16 β 8 = 8