Let `f(x)=(x-2)^2x^n, n in N ` Then f(x) has a minimum at

Options

A. `x=2 " for all " n in N `
B. x=2 ifn is odd
C. x=0 ifn is even
D. x=0 is if n is odd

Answer

A, C

Explanation

We have
`f(x)=(x-2)^2x^n`
`rArr f(x)=2(x-2)x^n+n(x-2)^2x^(n-1)`
`rArr f(x)=(x-2)x^(n-1){2x+n(x-2)}`
`f(x)={(n+2)x-2n}(x-2)x^(n-1)`
`therefore f(x)=0 rArr x=0 , rArr x=0 ,x=2,x=(2n)/(n+2)`
Now `f(0-h)={(n+2)h+2n}(2+h)(-1)^(n+1)h^(n-1)`
`f(0+h)={(n+2)h-2n}(h-2)h^(n-1)`
`rArr f(0+h)={(2n-(n+2)h}(2-h)h^(n-1)`
Clearly f(x)changes its sign from neagative to positive if n is even .Therefore f(x) attains minimum at x=0 if n is even .
If n is odd, then there is no changes in the sing of f(x) isn the neitghbourhood of x=0 ,So f(x)does not attains a maximum of minimum.
Similary we have
`f(2-h)=(-h)(2-h)^(n-1)(4-2h-nh)lt 0 " for all" n in N `
and `f(2+h)=h(2+h)=h(2+h)^(n-1)(4+2h+nh) gt 0 " for all " n in N `
Therefore f(x) changes its sign from negative to positive in ther neighbourhood of x=2.So x=2 is point of local minimum for all `n in N`