Let e be the base of natural logarithm and let f:1,2,3,4 1,e,e^2,e^3 and g:1,e,e^2,e^3 1,(…

Question

Let $e$ be the base of natural logarithm and let $f:{1,2,3,4}	o {1,e,{e}^{2},{e}^{3}}$ and $g:{1,e,{e}^{2},{e}^{3}}	o {1,\frac{1}{2},\frac{1}{3},\frac{1}{4}}$ be two bijective functions such that $f$ is strictly decreasing and $g$ is strictly increasing. If $\phi (x)={[{f}^{-1}{{g}^{-1}(\frac{1}{2})}]}^{x}$, then the area of the region $R={(x,y):{x}^{2}\le y\le \phi (x),0\le x\le 1}$ is :

Accepted Answer

Since $f:\{1,2,3,4\}\to \{1,e,e^2,e^3\}$ is bijective and strictly decreasing, we match the largest element of the domain to the smallest element of the codomain.

So,

$$ f(1)=e^3,\quad f(2)=e^2,\quad f(3)=e,\quad f(4)=1 $$

Hence,

$$ f^{-1}(1)=4,\quad f^{-1}(e)=3,\quad f^{-1}(e^2)=2,\quad f^{-1}(e^3)=1 $$

Now $g:\{1,e,e^2,e^3\}\to \left\{1,\frac12,\frac13,\frac14\right\}$ is bijective and strictly increasing.

Since

$1and$ \frac14for $g$ to be strictly increasing, we must assign:

$$ g(1)=\frac14,\quad g(e)=\frac13,\quad g(e^2)=\frac12,\quad g(e^3)=1 $$

Therefore,

$$ g^{-1}\left(\frac12\right)=e^2 $$

Now,

$$ f^{-1}\left(g^{-1}\left(\frac12\right)\right)=f^{-1}(e^2)=2 $$

So,

$$ \phi(x)=\left[f^{-1}\left\{g^{-1}\left(\frac12\right)\right\}\right]^x=2^x $$

Thus the region is

$$ R=\{(x,y):x^2\le y\le 2^x,\ 0\le x\le 1\} $$

Its area is

$$ \int_0^1 (2^x-x^2)\,dx $$

Now,

$$ \int 2^x\,dx=\frac{2^x}{\ln 2} $$

So,

$$ \int_0^1 2^x\,dx=\left[\frac{2^x}{\ln 2}\right]_0^1=\frac{2-1}{\ln 2}=\frac{1}{\ln 2} $$

Also,

$$ \int_0^1 x^2\,dx=\left[\frac{x^3}{3}\right]_0^1=\frac13 $$

Hence area

$$ =\frac{1}{\ln 2}-\frac13 $$

Taking LCM,

$$ =\frac{3-\ln 2}{3\ln 2} $$

Therefore, the required area is

$$ \boxed{\frac{3-\log_e 2}{3\log_e 2}} $$

So the correct option is $\boxed{\text{A}}$.

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