If fx-n-110x-n20-n- then f20f-20-

Given : nbsp; f(x)=∏_n=1^10(x+n)^(20+n) Taking ln on both sides, we get ln f(x)=∑_n=1^10(20+n)ln(x+n) Differentiate w.r.t. x f^'(x)f(x)=∑_n=1^10(20+n)1(x+n) P ...

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Standard XII

Mathematics

Property 2

If fx=∏n=110x...

Question

If $f (x) = 10 \prod n = 1 (x + n)^{(20 + n)},$ then $\frac{f (20)}{f^{'} (20)} =$

\frac{1}{100}

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\frac{1}{10}

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1

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Solution

The correct option is B $\frac{1}{10}$
Given :
$f (x) = 10 \prod n = 1 (x + n)^{(20 + n)}$
Taking $ln$ on both sides, we get
$ln f (x) = 10 \sum n = 1 (20 + n) ln (x + n)$
Differentiate w.r.t. $x$
$\frac{f^{'} (x)}{f (x)} = 10 \sum n = 1 (20 + n) \frac{1}{(x + n)}$
Put $x = 20,$ we have
$\frac{f^{'} (20)}{f (20)} = 10 \sum n = 1 (20 + n) \frac{1}{(20 + n)}$
$\Rightarrow \frac{f^{'} (20)}{f (20)} = 10 \sum n = 1 1 = 10$
$∴ \frac{f (20)}{f^{'} (20)} = \frac{1}{10}$

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If f(x)=10∏n=1(x+n)(20+n), then f(20)f′(20)=

The correct option is B 110Given : f(x)=10∏n=1(x+n)(20+n) Taking ln on both sides, we get lnf(x)=10∑n=1(20+n)ln(x+n) Differentiate w.r.t. x f′(x)f(x)=10∑n=1(20+n)1(x+n) Put x=20, we have f′(20)f(20)=10∑n=1(20+n)1(20+n) ⇒f′(20)f(20)=10∑n=11=10 ∴f(20)f′(20)=110

If $f (x) = 10 \prod n = 1 (x + n)^{(20 + n)},$ then $\frac{f (20)}{f^{'} (20)} =$