niks18 wrote:

\(m\) is the smallest positive integer such that for any integer \(n≥m\), the quantity \(n^3–7n^2+11n–5\) is positive. What is the value of \(m\)?

A. 4

B. 5

C. 8

D. 6

E. 11

OEGiven: \(n^3–7n^2+11n–5\)\(>0\), or

\(n^3-6n^2-n^2+5n+6n-5>0\), or

\(n^3-6n^2+5n-n^2+6n-5>0\)

\(n(n^2-6n+5)-1(n^2-6n+5)>0\). this implies \((n-1)(n^2-6n+5)>0\), this can be further factorized and we finally get

\((n-1)^2(n-5)>0\)---------(1).

Now \((n-1)^2\) will be always positive, hence for equation (1) to be positive \(n-5>0\)

or \(n>5\). Hence the smallest integer, satisfying the inequality is \(6\). so \(n≥6\) and as \(n≥m\). So \(m=6\)

Option

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I see from previous discussions that people are finding it hard to re-arrange the equation and factorize so presenting a solution -

One of the method that is very handy in factorizing a bit complex equation is to get one of the roots of the equation, as it will provide one of the factor of the expression. you can use the factor to divide the expression and get other factors. For e.g in this question

\(f(n)=n^3-7n^2+11n-5\), the root of the function will be a value where \(f(n)=0\). In GMAT, expression will be fairly simple, hence one root can be identified very easily. if you just ignore the \("n"\) in the expression and add it, you will realize that the summation is \(0\) i.e. one of the root of the expression \(1\).

so, \(f(1)=0\), this means \(n=1\) is one root of the equation.

so \(n-1=0\) or \((n-1)\) will be the factor of the equation. divide the equation by \((n-1)\), you will get quotient as \(n^2-6n+5\).

so our equation can be written as \((n-1)(n^2-6n+5)\). Now it becomes fairly simple to factorize

Not a big fan of solving through options as GMAT 700 level questions can hardly be solved through options and moreover it takes away the fun out of Mathematics