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If y = (1 + (2)/(x)) (1 + (2)/(x))(1 + (...

If `y = (1 + (2)/(x)) (1 + (2)/(x))(1 + (3)/(x))...(1 + (n)/(x))`
` x ne 0 , "then " (dy)/(dx) ` when x = - 1 is

A

`n!`

B

`(n-1)!`

C

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

D

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

Text Solution

Verified by Experts

The correct Answer is:
C
Given ,`y = (1 + (1)/(x)) (1 + (2)/(x)) (1 +(3)/(x))...(1+(n)/(x))`
On taking log both sides , we get
`log y = log (1 + (1)/(x)) + log(1 + (2)/(x)) = … + log (1 +(n)/(x))`
On differentiating both sides w.r.t.x, we get
`(1)/(y)(dy)/(d) =(1)/( (1 + (1)/(x))) (-(1)/(x^(2))) + (1)/((1 + (2)/(x))) (-(2)/(x^(2))) + ...+ (1)/((1 + (n)/(x))) (-(n)/(x^(2)))`
`(dy)/(dx) = - ((1 + (2)/(3))(1 + (3)/(x))...(1 +(n)/(x)) 2 (1 + (1)/(x)) (1 +(3)/(x)) (1 +(3)/(x)) ...(1 + (n)/(x)))/(x^(2))`
`therefore |((dy)/(dx))|_(x = - 1) = ((-1)(-1)(-2)(-3)...(1-n))/(1) - 0 -0`
`= (-1)^(n) (1) (2) (3) ...(n-1) = (-1)^(n) (n-1)!`
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