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Question

The extreme values of $4 cos invisible function application open parentheses x to the power of 2 end exponent close parentheses cos invisible function application open parentheses fraction numerator pi over denominator 3 end fraction plus x to the power of 2 end exponent close parentheses cos invisible function application open parentheses fraction numerator pi over denominator 3 end fraction minus x to the power of 2 end exponent close parentheses$ over $R$ , are

-1, 1
-2, 2
-3, 3
-4, 4

The correct answer is: -1, 1

Let $f open parentheses x close parentheses equals 4 cos invisible function application left parenthesis x to the power of 2 end exponent right parenthesis cos invisible function application open parentheses fraction numerator pi over denominator 3 end fraction plus x to the power of 2 end exponent close parentheses cos invisible function application open parentheses fraction numerator pi over denominator 3 end fraction minus x to the power of 2 end exponent close parentheses$
$equals 2 cos invisible function application open parentheses x to the power of 2 end exponent close parentheses open square brackets cos invisible function application open parentheses fraction numerator 2 pi over denominator 3 end fraction close parentheses plus cos invisible function application open parentheses 2 x to the power of 2 end exponent close parentheses close square brackets$

$open square brackets because blank 2 cos invisible function application A cos invisible function application B equals cos invisible function application open parentheses A plus B close parentheses plus cos invisible function application open parentheses A minus B close parentheses close square brackets$

$equals 2 cos invisible function application left parenthesis x to the power of 2 end exponent right parenthesis open square brackets negative fraction numerator 1 over denominator 2 end fraction plus cos invisible function application left parenthesis 2 x to the power of 2 end exponent right parenthesis close square brackets$

$equals negative cos invisible function application open parentheses x to the power of 2 end exponent close parentheses plus 2 cos invisible function application open parentheses x to the power of 2 end exponent close parentheses cos invisible function application open parentheses 2 x to the power of 2 end exponent close parentheses$

$equals negative cos invisible function application open parentheses x to the power of 2 end exponent close parentheses plus cos invisible function application open parentheses 3 x to the power of 2 end exponent close parentheses plus cos invisible function application open parentheses x to the power of 2 end exponent close parentheses$

$rightwards double arrow blank f open parentheses x close parentheses equals cos invisible function application left parenthesis 3 x to the power of 2 end exponent right parenthesis$ ...(i)

$rightwards double arrow blank f to the power of ´ end exponent open parentheses x close parentheses equals negative open square brackets sin invisible function application open parentheses 3 x to the power of 2 end exponent close parentheses close square brackets left parenthesis 6 x right parenthesis$

For extremum, put $f to the power of ´ end exponent open parentheses x close parentheses equals 0$

$rightwards double arrow blank minus sin invisible function application open parentheses 3 x to the power of 2 end exponent close parentheses open parentheses 6 x close parentheses equals 0$

$rightwards double arrow blank x equals 0 comma blank square root of pi$

Put $x equals 0 comma blank square root of pi$ in Eq. (i), we get

$f open parentheses 0 close parentheses equals cos invisible function application open parentheses 0 close parentheses equals 1$

And $f open parentheses pi close parentheses equals cos invisible function application left parenthesis 3 pi right parenthesis equals negative 1$

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