The minimum value of the function f(x) = x^{1 | vedclass

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(B) The function is defined as $f(x) = x^{10} + x^2 + \frac{1}{x^{12}} + \frac{1}{1 + \sec^{-1} x}$.For the term $x^{10} + x^2 + \frac{1}{x^{12}}$,we

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$\frac{3\pi + 4}{\pi + 1}$

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(B) The function is defined as $f(x) = x^{10} + x^2 + \frac{1}{x^{12}} + \frac{1}{1 + \sec^{-1} x}$.For the term $x^{10} + x^2 + \frac{1}{x^{12}}$,we apply the Arithmetic Mean-Geometric Mean $(AM \geq GM)$ inequality:$\frac{x^{10} + x^2 + \frac{1}{x^{12}}}{3} \geq \sqrt[3]{x^{10} \cdot x^2 \cdot \frac{1}{x^{12}}} = \sqrt[3]{1} = 1$.Thus,$x^{10} + x^2 + \frac{1}{x^{12}} \geq 3$. Equality holds when $x^{10} = x^2 = \frac{1}{x^{12}}$,which implies $x^2 = 1$,so $x = 1$ or $x = -1$.Now consider the domain of $\sec^{-1} x$,which is $(-\infty, -1] \cup [1, \infty)$.To minimize $f(x)$,we need to minimize the term $\frac{1}{1 + \sec^{-1} x}$.If $x = 1$,$\sec^{-1}(1) = 0$,so the term is $\frac{1}{1+0} = 1$.If $x = -1$,$\sec^{-1}(-1) = \pi$,so the term is $\frac{1}{1+\pi}$.Since $\frac{1}{1+\pi} Therefore,$f(-1) = 3 + \frac{1}{1+\pi} = \frac{3(1+\pi) + 1}{1+\pi} = \frac{3\pi + 4}{\pi + 1}$.

The minimum value of the function $f(x) = x^{10} + x^2 + \frac{1}{x^{12}} + \frac{1}{1 + \sec^{-1} x}$ is

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