The minimum value of the function f(x) = x^ 10 + x^2 + 1 x^ 12 + … - Vidyadip

MCQ

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

A
$\frac{{\pi\ +\ 4}}{{\pi\ +\ 1}}$
✓
$\frac{{3\pi\ +\ 4}}{{\pi\ +\ 1}}$
C
$\frac{{\pi\ +\ 4}}{{3\pi\ +\ 1}}$
D
$3$

✓

Answer

Correct option: B.

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

b
$x^{10}+x^{2}+\frac{1}{x^{12}} \geq 3$ $(AM.GM)$

Equality occurs of $x=\pm 1$

But ${\sec ^{ - 1}}{\rm{x}}$ takes maximum at $\mathrm{x}=-$

$f_{\min }(\mathrm{x})=f(-1)=3+\frac{1}{\pi+1}$

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