Let fx and gx be continous, positive functions such that f−x=gx−1 , fx=gxg−x and ∫−2020fxdx=2024 ,then ∫−2020fxgxdx is

I=∫−2020fxgxdxUsing King’s rule, we have I=∫−2020f−xg−xdx=∫−2020f−x.fxgxdxAddition the two, we get: 2I=∫−2020fx.1+f−xgxdx=∫−2020fx.1+f−xgxdx=∫−2020fxdx=2020 ⇒I=1012

Let fx and gx be continous, positive functions such that f−x=gx−1 , fx=gxg−x and ∫−2020fxdx=2024 ,then ∫−2020fxgxdx is

I=∫−2020fxgxdxUsing King’s rule, we have I=∫−2020f−xg−xdx=∫−2020f−x.fxgxdxAddition the two, we get: 2I=∫−2020fx.1+f−xgxdx=∫−2020fx.1+f−xgxdx=∫−2020fxdx=2020 ⇒I=1012

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Let $f (x)$ and $g (x)$ be continous, positive functions such that $f (- x) = g (x) - 1$ , $f (x) = \frac{g (x)}{g (- x)}$ and $\int_{- 20}^{20} f (x) d x = 2024$ ,then $\int_{- 20}^{20} \frac{f (x)}{g (x)} d x$ is

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1012

506

2024

2012

answer is A.

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Detailed Solution

$I = \int_{- 20}^{20} \frac{f (x)}{g (x)} d x$
Using King’s rule, we have
$I = \int_{- 20}^{20} \frac{f (- x)}{g (- x)} d x = \int_{- 20}^{20} \frac{f (- x) . f (x)}{g (x)} d x$
Addition the two, we get:
$2 I = \int_{- 20}^{20} \frac{f (x) . (1 + f (- x))}{g (x)} d x = \int_{- 20}^{20} \frac{f (x) . (1 + f (- x))}{g (x)} d x = \int_{- 20}^{20} f (x) d x = 2020$

$\Rightarrow I = 1012$