A) \[{{(-1)}^{n}}{{2}^{n-1}}.(n-1)!\]
B) \[{{(-2)}^{n-1}}(n-1)!\]
C) \[{{(-2)}^{n}}n!\]
D) \[{{(-1)}^{n-1}}{{2}^{n}}(n-1)!\]
Correct Answer: B
Solution :
\[\because \] \[f(x)=(x-2)(x-4)(x-6)....(x-2n)\] Taking log on both sides, we get \[\log f(x)=\log (x-2)+\log (x-4)\] \[+.....+\log (x-2n)\] On differentiating w.r.t.\[x,\]we get \[\frac{1}{f(x)}.f'(x)=\frac{1}{(x-2)}+\frac{1}{(x-4)}\] \[+....+\frac{1}{(x-2n)}\] \[f'(x)=(x-4)(x-6)....(x-2n)\] \[+(x-2)(x-6)....(x-2n)\] \[+...+(x-2)(x-4)...\{x-2(x-1)\}\] \[\therefore \]\[f'(2)=(-2)(-4).....(2-2n)\] \[={{(-2)}^{n-1}}[1.2....(n-1)]\] \[={{(-2)}^{n-1}}(n-1)!\]
You need to login to perform this action.
You will be redirected in
3 sec
