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:

Question

Let $f$ be a twice differentiable function such that $f(x) = \int_0^x \tan(t-x) dt - \int_0^x f(t) \tan t dt, x \in (-\frac{\pi}{2}, \frac{\pi}{2})$. Then $f''(\frac{\pi}{6}) + 12 f'(-\frac{\pi}{6}) + f(\frac{\pi}{6})$ is equal to ________.

Answer 1

Identify $\phi(x)$ by analyzing the function composition.
The function $g$ maps the set $\{1, e, e^2, e^3\}$ to $\{1, 1/2, 1/3, 1/4\}$ increasingly. Thus, $g$ maps the second largest element of its domain to the second largest element of its codomain: $g(e^2) = 1/2$. This gives $g^{-1}(1/2) = e^2$.
The function $f$ maps $\{1, 2, 3, 4\}$ to $\{1, e, e^2, e^3\}$ decreasingly. Thus, $f$ maps the second smallest element of its domain to the second largest element of its codomain: $f(2) = e^2$. This gives $f^{-1}(e^2) = 2$.
Therefore, the base of the exponent in $\phi(x)$ is $2$, making $\phi(x) = 2^x$.

The area is bounded between $y = 2^x$ and $y = x^2$ from $x=0$ to $x=1$.
Area = $\int_0^1 2^x dx - \int_0^1 x^2 dx$
Using the basic integration formulas $\int a^x dx = \frac{a^x}{\ln a}$ and $\int x^n dx = \frac{x^{n+1}}{n+1}$:
Area = $[\frac{2^x}{\ln 2}]_0^1 - [\frac{x^3}{3}]_0^1$
Area = $(\frac{2}{\ln 2} - \frac{1}{\ln 2}) - (\frac{1}{3} - 0)$
Area = $\frac{1}{\ln 2} - \frac{1}{3} = \frac{3 - \ln 2}{3 \ln 2}$.

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The Correct Option is A

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