For how many integers n is n^4 + 6n < 6n^3 + n^2 ?

DhruvS

We plot the critical values on Number line

Critical values are the values of n for which the function result in a zero outcome

Function \(n*(n-6)*(n-1)*(n+1)\) will becomes 0 if
n = 0, or
(n-6) = 0 i.e. n = 6 or
(n-1) = 0 i.e. n = 1 or
(n-+1) = 0 i.e. n =-1 or

So we have 4 critical values i.e. {-1, 0, 1, 6} so we mark those values on Number Line and make loop between adjacent Critical values

Then we try with any value in a loop to see whether function will assume positive value or negative values and that's how we find favorable ranges.

GMATinsight Thank you for your reply. Please confirm my understanding below:
We plotted the values on the graph and now since n*(n-6)*(n-1)*(n+1)<0, i.e. the expression is less <0. We need to consider only the adjacent critical values and make two ranges==> -1<n<0 and 1<n<6. As there are no integers between -1<n<0, hence the only possible values are 2,3,4,5, according to 1<n<6.

Also can we not consider -1<n<6 to include all values ?

DhruvS

NO, we can NOT incluce values in range -1<n<6 simply because the function is not negative between 0 and 1
'
also because function is not negative at values n = 0 and n = 1

I hope it clear you doubt.

GMATinsight Understood. Thank You so much.

DhruvS

The tradition of thanking someone on GMAT CLUB is by pressing +1KUDOS button on the post that helped you...

Thanks for reminding.

Solution:

n^4 + 6n < 6n^3 + n^2

n^4 - 6n^3 - n^2 + 6n < 0

n^3(n - 6) - n(n - 6) < 0

(n^3 - n)(n - 6) < 0

n(n^2 - 1)(n - 6) < 0

n(n - 1)(n + 1)(n - 6) < 0

We see that on the left hand side of the last inequality, there are 4 factors.

If n is a negative integer less than -1, then all the 4 factors are negative. So the product of the 4 factors will be positive and will not be less than 0.

If n is -1, 0, 1 or 6, then one of the factors will be 0. So the product will be 0 and will not be less than 0.

If n is 2, 3, 4, or 5, the first 3 factors are positive while the last factor is negative. So the product will be negative and hence less than 0.

If n is a positive integer greater than 6, then all the 4 factors are positive. So the product will be positive and will not be less than 0.

Alternate Solution:

Let’s rewrite the given inequality as n^4 - n^2 < 6n^3 - 6n. Then:

n^2(n^2 - 1) < 6n(n^2 - 1)

n^2(n^2 - 1) - 6n(n^2 - 1) < 0

(n^2 - 1)(n^2 - 6n) < 0

In order to have the product of two expressions negative, one of the expressions must be negative and the other expression must be positive.

If n^2 - 1 is negative, then only n = 0 satisfies n^2 - 1 < 0. Notice that n = 0 does not satisfy n^4 + 6n < 6n^3 + n^2; thus it follows that n^2 - 1 must be positive.

Since n^2 - 1 > 0, |n| must be 2 or greater. Further, since n^2 - 1 is positive, n^2 - 6n must be negative; i.e. n^2 < 6n. Notice that this inequality is not satisfied for any negative integer and only satisfied for n = 1, 2, 3, 4 or 5. Combined with the earlier condition that |n| must be 2 or greater, we see that there are four integers satisfying the given inequality (namely 2, 3, 4 and 5).

Answer: C

Asked: For how many integers n is \(n^4 + 6n < 6n^3 + n^2\) ?

\(n^4 + 6n - 6n^3 - n^2 <0\)
\(n(n^3 - 6n^2 - n + 6) < 0\)
\(n(n-6)(n^2-1)<0\)
n(n-1)(n+1)(n-6) < 0

By WAVY LINE method: -
1<n<6
-1<n<0

n = {2,3,4,5}

IMO C

i.e. \((n^2-6n)*(n-1)*(n+1) < 0 \) ; could you please explain how you arrived at this from the previous step

sume3tss

CHeck the new steps in Blue

i.e. \(n^4 - n^2 < 6n^3 - 6n \)

i.e. \(n^2*(n-1)*(n+1) < 6n*(n-1)*(n+1) \)

i.e. \(n^2*(n-1)*(n+1) - 6n*(n-1)*(n+1) < 0\)

i.e. \((n-1)*(n+1)*(n^2 - 6n) < 0\)

i.e. \((n^2-6n)*(n-1)*(n+1) < 0 \)

i.e. \(n*(n-6)*(n-1)*(n+1) < 0 \)

This could be a very stupid thing to ask but bear with me please. is it okay to move 6n to the left side of inequality when we dont know whether n is positive or negative?