Question:
Let $e$ be the base of natural logarithm and let $f : \{1, 2, 3, 4\} \to \{1, e, e^2, e^3\}$ and $g : \{1, e, e^2, e^3\} \to \{1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}\}$ be two bijective functions such that $f$ is strictly decreasing and $g$ is strictly increasing. If $\phi(x) = \left[ f^{-1} \left\{ g^{-1} \left( \frac{1}{2} \right) \right\} \right]^x$, then the area of the region $R = \{ (x, y) : x^2 \le y \le \phi(x), 0 \le x \le 1 \}$ is:
Let $e$ be the base of natural logarithm and let $f : \{1, 2, 3, 4\} \to \{1, e, e^2, e^3\}$ and $g : \{1, e, e^2, e^3\} \to \{1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}\}$ be two bijective functions such that $f$ is strictly decreasing and $g$ is strictly increasing. If $\phi(x) = \left[ f^{-1} \left\{ g^{-1} \left( \frac{1}{2} \right) \right\} \right]^x$, then the area of the region $R = \{ (x, y) : x^2 \le y \le \phi(x), 0 \le x \le 1 \}$ is:
Show Hint
Determine the value of $g^{-1}(1/2)$ and $f^{-1}$ of that result using the given monotonicity. $\phi(x)$ will turn out to be $2^x$. Then integrate $2^x - x^2$.
Updated On: Apr 9, 2026
- $\frac{3 - \log_e(2)}{3 \log_e(2)}$
- $\frac{1}{3 \log_e(2)}$
- $3 + \log_e(2)$
- $\frac{3 + \log_e(2)}{2 + \log_e(3)}$
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The Correct Option is A
Solution and Explanation
Based on the properties of strictly monotonic bijective functions on discrete sets:
For $f : \{1, 2, 3, 4\} \to \{1, e, e^2, e^3\}$, if it is strictly decreasing, the mappings are:
$f(1) = e^3, f(2) = e^2, f(3) = e, f(4) = 1$.
For $g : \{1, e, e^2, e^3\} \to \{1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}\}$, if it is strictly increasing, the mappings are:
$g(1) = 1/4, g(e) = 1/3, g(e^2) = 1/2, g(e^3) = 1$.
Now, evaluate the function $\phi(x) = [f^{-1}\{g^{-1}(1/2)\}]^x$:
1. From $g(e^2) = 1/2$, we have $g^{-1}(1/2) = e^2$.
2. From $f(2) = e^2$, we have $f^{-1}(e^2) = 2$.
Thus, $\phi(x) = 2^x$.
The region $R$ is $x^2 \le y \le 2^x$ for $0 \le x \le 1$.
Area = $\int_0^1 (2^x - x^2) dx$
Area = $[\frac{2^x}{\ln 2} - \frac{x^3}{3}]_0^1 = (\frac{2}{\ln 2} - \frac{1}{3}) - (\frac{1}{\ln 2} - 0)$
Area = $\frac{1}{\ln 2} - \frac{1}{3} = \frac{3 - \ln 2}{3 \ln 2}$
This matches Option 1.
For $f : \{1, 2, 3, 4\} \to \{1, e, e^2, e^3\}$, if it is strictly decreasing, the mappings are:
$f(1) = e^3, f(2) = e^2, f(3) = e, f(4) = 1$.
For $g : \{1, e, e^2, e^3\} \to \{1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}\}$, if it is strictly increasing, the mappings are:
$g(1) = 1/4, g(e) = 1/3, g(e^2) = 1/2, g(e^3) = 1$.
Now, evaluate the function $\phi(x) = [f^{-1}\{g^{-1}(1/2)\}]^x$:
1. From $g(e^2) = 1/2$, we have $g^{-1}(1/2) = e^2$.
2. From $f(2) = e^2$, we have $f^{-1}(e^2) = 2$.
Thus, $\phi(x) = 2^x$.
The region $R$ is $x^2 \le y \le 2^x$ for $0 \le x \le 1$.
Area = $\int_0^1 (2^x - x^2) dx$
Area = $[\frac{2^x}{\ln 2} - \frac{x^3}{3}]_0^1 = (\frac{2}{\ln 2} - \frac{1}{3}) - (\frac{1}{\ln 2} - 0)$
Area = $\frac{1}{\ln 2} - \frac{1}{3} = \frac{3 - \ln 2}{3 \ln 2}$
This matches Option 1.
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