If 4a+2b+c=0, then the equation 3ax2+2bx+c=0 has at least one real root lying in the interval 

If 4a+2b+c=0, then the equation 3ax2+2bx+c=0 has at least one real root lying in the interval 

  1. A

    (0, 1)

  2. B

    (1, 2)

  3. C

    (0, 2)

  4. D

    none of these

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    Solution:

    Let f(x)=ax3+bx2+cx.Then, f(0)=0 and 

    f(2)=2(4a+2b+c)=0. Therefore, 0 and 2 are roots of f(x). Therefore, f(x)=0 i.e. 3ax2+2bx+c=0 has a root between 0 and 2.

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