Question

The value of $n$, for which the function $f\left(x\right)={(x^2-4)}^n(x^2-x+1),n\in N$ assumes a local minima at $x=2$, is

• A. an even number
• B. an odd number
• C. an irrational number
• D. cannot be determined

Answer 1

The correct option is A an even number

We have $f\left(x\right)={(x^2-4)}^n(x^2-x+1)$ assumes local minima at $x=2$

$\Rightarrow f\left(2\right)<f(2-h)$ and $f\left(2\right)<f(2+h)$, where $h>0$

Since $f\left(2\right)=0\Rightarrow f(2-h)>0$ and $f(2+h)>0,\forall h>0$

$\Rightarrow {(-h)}^n{(4-h)}^n[h^2-3h+1]>0$ and $\Rightarrow {\left(h\right)}^n{(4+h)}^n[h^2+5h+1]>0$

$\Rightarrow {(-h)}^n>0[\because (4-h)>0,h^2-3h+1>0,4+h>0,h^2+5h+1>0\forall h>0]$

$\Rightarrow n$ is an even number