If the function f(x)=x3+3(a−7)x2+3a2−9x−1 has positive points of extremum then 

If the function f(x)=x3+3(a7)x2+3a29x1 has positive points of extremum then 

  1. A

    a(,3)3,

  2. B

    a(,3)(3,29/7]

  3. C

    (,7)

  4. D

    (,29/7)

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    Solution:

    We have,

    f(x)=x3+3(a7)x2+3a29x1.f(x)=3x2+6x(a7)+3a29

    f the function f (x) has a positive point of minimum, then f' (x) = 0 must have positive roots

     x=0 is less than the roots of f(x)=0

     f(0)>0 O and Discriminant  0

     4(a7)24a290 and 3a29>0

     7a290 and a29>0 a297 and (a<3 or a>3)a(,3)(3,29/7]

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