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Question

f(x)=6x5 has a maximum at x = 0. 


  1. True

  2. False


Solution

The correct option is B

False


If  f(x) has derivative upto nth order and f(c)=f(c)..fn1(c)=0, then  

A)    n is even, fn(c)<0=>x=c is a point of maximum.

B)     n is even, fn(c)>0=>x=c is a point of minimum.

C)    n is odd, fn(c)<0=>f(x) is decreasing about x=c

D)    n is odd, fn(c)>0=>f(x) is increasing about x=c

So, we will differentiate until we get a non-zero value at x=0

f(x)=6x5

f(x)=30x4

f(0)=0

f(x)=120x3

f(0)=0

f(x)=f3(x)=360x2

f3(0)=0

f(x)=f4(x)=720x

f4(x)=0

f(x)=f5(x)=720

f5(0)=720

which is non –zero.

So, we get n = 5, which is odd.

n is odd,n is odd, fn(c)>0=>f(x) is increasing about x=c

That is f(x) is increasing at x=0. It does not have a maximum or minimum at x=0

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