Let \( f: R \rightarrow R \) be a bijective function. Let \( g(x) \) be a continuous function su... VIDEO
Let \( f: R \rightarrow R \) be a bijective function. Let \( g(x) \) be a continuous function such that \( \int_{0}^{f(x)} g(t) d t=g(f(x))-1 \forall x \in R \)
Value of \( \int_{0}^{1} g(\{x\}) \cdot\left[\frac{1-x}{(1+x)^{3}}\right] d x \) is :
[Note : \( \{y\} \) denotes fractional part of \( y \).
(a) \( 1-\frac{e}{4} \)
(b) 1
(c) \( 1+\frac{e}{4} \)
(d) \( \frac{e}{4}-1 \)
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