Question:
If $ f(x)=(x-2)(x-4)(x-6)....(x-2n), $ then $ f'(2) $ is
If $ f(x)=(x-2)(x-4)(x-6)....(x-2n), $ then $ f'(2) $ is
Updated On: Jun 12, 2024
- $ {{(-1)}^{n}}{{2}^{n-1}}(n-1)! $
- $ {{(-2)}^{n-1}}{{2}^{n}}(n-1)! $
- $ {{(-2)}^{n}}n! $
- $ {{(-1)}^{n-1}}{{2}^{n}}(n-1)! $
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The Correct Option is B
Solution and Explanation
$ \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)! $
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)! $
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