Question
Let e be the base of natural logarithm and let $f:\{ 1,2,3,4\} \rightarrow \left\{ 1,e,e^{2},e^{3} \right\}$ and $g:\left\{ 1,e,e^{2},e^{3} \right\} \rightarrow \left\{ 1,\frac{1}{2},\frac{1}{3},\frac{1}{4} \right\}$ be two bijective functions such that $f$ is strictly decreasing and g is strictly increasing. If $\phi(x) = \left\lbrack f^{- 1}\left\{ {\text{ }g}^{- 1}\left( \frac{1}{2} \right) \right\} \right\rbrack^{x}$, then the area of the region
$R = \left\{ (x,y):x^{2} \leq y \leq \phi(x) \right.\ $, $0 \leq x \leq 1\}$ is:
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Solution
Sol. [Key Concept: Here $f^{- 1}(g^{- 1}(1/2))$ makes $\varphi(x{) = 2}^{x}$; the area is $\int_{0}^{1}{(2}^{x} - x^{2})\ dx$.]
$${f(x) = e^{x - 1} }{g(x) = \frac{1}{\log_{e}x + 1} }{f^{- 1}\left( {\text{ }g}^{- 1}(1/2) \right) = f^{- 1}(e) = 2 }{\phi(x) = 2^{x} }{\therefore\int_{0}^{1}\mspace{2mu}\left( 2^{x} - x^{2} \right)dx = \left( \frac{2^{x}}{ln2} - \frac{x^{3}}{3} \right)_{0}^{1} = \frac{1}{ln2} - \frac{1}{3} }{= \frac{3 - \mathcal{l}n2}{3\mathcal{l}n2}}$$
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