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Let f(x)=(x-2)(x-4)(x-6)...(x-2n) then f...

Let `f(x)=(x-2)(x-4)(x-6)...(x-2n)` then `f '(2)` equals

A

`(-1)^(n) 2^(n-1) (n-1)!`

B

`(-2)^(n-1) (n-1)!`

C

`(-2)^(n) n!`

D

`(-1)^(n-1) 2^(n) (n-1)!`

Text Solution

Verified by Experts

The correct Answer is:
B
Given , ` f(x)= (x-2) (x-4)(x-6)…(x-2n)`
Taking log on both sides , we get
` log f(x) = log (x-2) + log(x-4) + …+ log (x - 2n)`
On differentiating both sides , we get
`(1)/(f(x))f'(x)= (1)/((x-2))+(1)/((x-4) ) + ...+(1)/((x-2n))`
` rArr f'(x) = (x-4) (x-6) ...(x-2n) + (x-2n)+ (x -2)(x-6)...(x-2n)`
`+...+ (x-2)(x-6)...(x-2n-1)`
` therefore f'(2) = (-2) (-4)..(2-2n)`
`=(-2)^(n-1) (1*2*....*(n-1))= (-2)^(n-1) (n-1)!`
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