Function $f (x) = 2 + 4 x^{2} + 6 x^{4} + 8 x^{6}$ has-

many maxima and many minima

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no maxima no minima

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only one minima

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only one maxima

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Solution

The correct option is C only one minima
$f (x) = 2 + 4 x^{2} + 6 x^{4} + 8 x^{6}$
Differentiating with respect to x, we get
$f^{'} (x) = 8 x + 24 x^{3} + 48 x^{5}$
$= 8 x (1 + 3 x^{2} + 6 x^{4})$
$= 0$
Then
$x = 0$ and $(6 x^{4} + 3 x^{2} + 1) = 0$
Let $x^{2} = t$
Then $6 t^{2} + 3 t + 1 = 0$
Now $D < 0$ for the above quadratic. Hence there is no real root.
Thus there is only one extremum at $x = 0.$

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