Question

If $f(x)=(x-2)(x-4)(x-6)....(x-2n),$ then $f^{\prime }(2)$ is

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

Answer 1

Answer: (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
$\frac{1}{f(x)}f(x)=\frac{1}{(x-2)}+\frac{1}{(x-4)}$ $+...+\frac{1}{(x-2n)}$ $f(x)=(x-4)(x-6)...(x-2n)$ $+(x-2)(x-6)....(x-2n)$ $+.....+(x-2)(x-6)...(x-2(n-1))$
$\therefore$ $f(2)=(-2)(-4)....(2-2n)$
$={(-2)}^{n-1}(\mathrm{1.2....}(n-1))={(-2)}^{n-1}(n-1)!$