A) \[{{(-1)}^{n}}{{2}^{n-1}}(n-1)!\]
B) \[{{(-2)}^{n-1}}{{2}^{n}}(n-1)!\]
C) \[{{(-2)}^{n}}n!\]
D) \[{{(-1)}^{n-1}}{{2}^{n}}(n-1)!\]
E) \[{{2}^{n-1}}(n-1)!\]
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 \[\frac{1}{f(x)}f(x)=\frac{1}{(x-2)}+\frac{1}{(x-4)}\] \[+...+\frac{1}{(x-2n)}\] \[f(x)=(x-4)(x-6)...(x-2n)\] \[+(x-2)(x-6)....(x-2n)\] \[+.....+(x-2)(x-6)...(x-2(n-1))\] \[\therefore \] \[f(2)=(-2)(-4)....(2-2n)\] \[={{(-2)}^{n-1}}(1.2....(n-1))={{(-2)}^{n-1}}(n-1)!\]
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