f(x) = 3 k x^2 - 18 x + 9 > 0

for wat values of x, the given function is increasing ??

how to find ??

plz explain in detail.....

Dear Mohit,

for increasing func.    dy/dx > 0

6kx - 18 > 0

x > 3/k

For a func to be always positive a > 0 and D < 0   ( for quadratic func. )

so,  3k > 0 and 324 - 4*3k*9 < 0

k > 3

So,  x > 3/k  (where k > 3 ) for the given func. to be always increasing.

Please feel free to ask your queries here. We are all IITians and here to help you in your IIT JEE preparation.

All the best.

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