Question

If $f(x)=(x-2)(x-4)(x-6)....(x-2n),$ then $f'(2)$ is

• A. ${{(-1)}^{n}}{{2}^{n-1}}(n-1)!$
• B. ${{(-2)}^{n-1}}{{2}^{n}}(n-1)!$
• C. ${{(-2)}^{n}}n!$
• D. ${{(-1)}^{n-1}}{{2}^{n}}(n-1)!$

Answer 1

The Correct Option is B

$\because$ $f(x)=(x-2)(x-4)(x-6)....(x-2n)$ Taking log on both sides, we get $\log f(x)=\log (x-2)+\log (x-4)$ $+....+\log (x-2n)$
aOn differentiating w.r.t. $x,$ we get
$\frac{1}{f(x)}f(x)=\frac{1}{(x-2)}+\frac{1}{(x-4)}$ $+...+\frac{1}{(x-2n)}$ $f(x)=(x-4)(x-6)...(x-2n)$ $+(x-2)(x-6)....(x-2n)$ $+.....+(x-2)(x-6)...(x-2(n-1))$
$\therefore$ $f(2)=(-2)(-4)....(2-2n)$
$={{(-2)}^{n-1}}(1.2....(n-1))={{(-2)}^{n-1}}(n-1)!$