CAT 2018 Slot 2 - QUANT Question 4

Question 4

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


Correct Answer: 8

Solution

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

$$\dfrac{n^3-11n^2+32n-28}{n-2}=n^2-9n+14$$

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

$$n^3-11n^2+32n-28=(n-2)^2(n-7)$$

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

$$\Rightarrow$$ $$(n-2)^2(n-7)>0$$

Therefore, we can say that n-7>0

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


View Video Solution


Comments

Register with

OR

Boost your Prep!

Download App