IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

the.sniper

More than can be correct:

Let f(x)=(x-1)^m(x-2)ⁿ, x E R ,then each critical point of f(x) is either local maximum or local minimum when :

a) m=2,n=3

b) m=2,n=4

c) m=3,n=4

d) m=4,n=2

~ The Sniper

raulrag009

Differentiate

m(x-2)+n(x-1)=0

x= (2m+n)/(m+n)

check by puttin value fom options

hsbhatt

m+n should be even is the requirement you are looking for.

So (b) and (d) are correct

the.sniper

sir the answer u have posted is right but may i have the solution.... getting a bit confused

hsbhatt

Hey sorry, I thought you already have the solution.

I have to correct my answer a bit. It should be both m and n should be even.

With an expression like (x-1)^m(x-2)ⁿ, the first derivative vanishes at 3 points, x=1, x=2 and a third point.

We can dispose of the third point easily as you can see that f"(x) is non-zero at that point.

So the critical points to worry about are x=1 and x=2.

You can easily see that f^k(x) will go to zero at both points as long as k<min(m,n) as (x-1) and (x-2) appear in every term.

Each order of derivative reduces the degree of (x-1) and (x-2) by 1. So there is going to be a term that goes like (x-1)^m-k (x-2)ⁿ.

Let's assume that m<n. So, this term becomes (x-2)ⁿ and every other term has (x-1) as a factor. Hence f^m(x) becomes non-zero at x =1 and is zero for all f^k(x) k<m. So from the sufficiency condition for x=1 to be an extremum, m has to be even. (I mean the sufficency condition that if f^k(a) =0, for k<t, and f^t(a) is non zero, then a is an extremum point if t is even and not if t is odd)

Similarly at fⁿ(x), we have non-zero derivative at x =2 and so for extremum condition, n has to be even.

If m and n are not even, the extrema will not occur at these points.

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