A) \[0\]
B) \[(-1)(n-1)!\]
C) \[n!-1\]
D) \[{{(-1)}^{n-1}}(n-1)!\]
E) \[{{(-1)}^{n}}(n-1)!\]
Correct Answer: B
Solution :
\[\because \] \[y=(1-x)(2-x)....(n-x)\] On taking log on both sides, we get \[log\text{ }y=log(1-x)+log(2-x)+...\]\[+\log (n-x)\] \[\frac{1}{y}\frac{dy}{dx}=\frac{1}{(1-x)}(-1)+\frac{1}{(2-x)}(-1)+....\] \[+\frac{1}{(n-x)}(-1)\] \[\frac{dy}{dx}=y\left[ \frac{(2-x)(3-x).....(n-x)(-1)+.....}{y} \right]\] \[{{\left( \frac{dy}{dx} \right)}_{x=1}}=1.2....(n-1)(-1)\] \[=(-1)(n-1)!\]
You need to login to perform this action.
You will be redirected in
3 sec
