Question 70

The smallest integer $$n$$ such that $$n^3-11n^2+32n-28>0$$ is

Correct Answer: 8


We can see that at n = 2, $$n^3-11n^2+32n-28=0$$ i.e. (n-2) is a factor of $$n^3-11n^2+32n-28$$


We can further factorize n^2-9n+14 as (n-2)(n-7).


$$\Rightarrow$$ $$n^3-11n^2+32n-28>0$$

$$\Rightarrow$$ $$(n-2)^2(n-7)>0$$

Therefore, we can say that n-7>0

Hence, n$$_{min}$$ = 8

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