Q. 78

Mark (√) against A. k > 3

B. k ≥ 3

C. k < 3

D. k ≤ 3

Answer :

Given f(x) = kx3 -9x2+ 9x+3


f’(x) = 3kx2-18x+9


f’(x) = 3(kx2 – 6x + 3)>0


kx2 – 6x + 3 >0


For quadratic equation to be greater than 0. a>0 and D<0.


k>0 and (-6)2- 4(k)(3)<0


36 – 12k<0


12k>36


k>3


k>3.

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