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CAT Previous Year Paper | CAT Polynomials and Inequalities Questions | Question 14

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Question 14 : The smallest integer n such that n3 - 11n2 + 32n - 28 > 0 is [TITA]


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Explanatory Answer

Method of solving this CAT Question from Polynomials and Inequalities

We have to find the smallest integer n such that n3 - 11n2 + 32n - 28 > 0
For this we can assume some values for n, such that n = 10 , 9 , 8 ,.. so on and can find the smallest integer
n3 - 11n2 + 32n - 28 > 0
Let us now assume that n = 10
1000 – 1100 + 320 - 28 > 0
192 > 0 Hence this n = 10 works out

Similarly n = 9 , 8 also works so we can try out with n = 8 and if it works we can try for lower numbers otherwise we can try with 9
n3 - 11n2 + 32n - 28 > 0
512 – 704 + 256 – 28 > 0
36 > 0
Since n = 8 works lets check for n = 7
343 – 539 + 224 – 28 = 0
Hence the smallest integer was found to be 8

The other way of thinking about it can be by substituting it with smaller numbers and checking if something can factorize it
So \\frac{P(x)}{x-a}) where remainder = 0
Let's substitute n = 1 with n3 - 11n2 + 32n - 28
1 – 11 + 32 – 28 = -6 this doesn’t work and now we can substitute n = 2 and check it out
8 – 44 – 64 – 28 = 0
So this number n3 - 11n2 + 32n - 28 is a multiple of n-2

CAT 2018 Question Paper Quants Slot 2 Polynomials and Inequalities

(n-2)(n2 – 9n + 14)
(n – 2)(n - 2)(n - 7) > 0
This will be greater than 0 when n = 7 and when n is between 2 and 7 it will be negative
When n = 1 ,
1 - 11 + 32 - 28 = -6 so this doesn’t work
Hence the smallest integer which will work is n = 8

The question is "The smallest integer n such that n3 - 11n2 + 32n - 28 > 0 is [TITA]"

Hence, the answer is 8

 

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