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The function f(x)=17−9x+6x2−x3f(x)=17-9x+6x^2-x^3 is strictly increasing in the interval, is:

Solution

✅ Drop

Differentiate the function:

f′(x)=−9+12x−3x2f'(x)=-9+12x-3x^2

=−3(x2−4x+3)=-3(x^2-4x+3)

=−3(x−1)(x−3)=-3(x-1)(x-3)

For the function to be strictly increasing,

f′(x)>0f'(x)>0

So,

−3(x−1)(x−3)>0-3(x-1)(x-3)>0

which gives

1<x<31<x<3.

Therefore, the function is strictly increasing on (1,3)(1,3).

However, (1,3)(1,3) is not present in any of the given options. In fact, Option 3, (−∞,1]∪[3,∞)(-\infty,1]\cup[3,\infty), represents the region where the function is decreasing.

Hence, the question has no correct option and was dropped by NTA.

Wrong options only. Happens. Ask your crush.

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