If the extreme values of the function f(x)=(2 | vedclass

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(B) The function is of the form $f(x) = A \cos x + B \sin x + C$,where $A = 2 \sqrt{6} + 1$,$B = 2 \sqrt{2} - \sqrt{3}$,and $C = -6$. The extreme valu

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$12$

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(B) The function is of the form $f(x) = A \cos x + B \sin x + C$,where $A = 2 \sqrt{6} + 1$,$B = 2 \sqrt{2} - \sqrt{3}$,and $C = -6$. The extreme values of $A \cos x + B \sin x$ are $\pm \sqrt{A^2 + B^2}$. First,calculate $A^2 + B^2$: $A^2 = (2 \sqrt{6} + 1)^2 = 4(6) + 1 + 4 \sqrt{6} = 25 + 4 \sqrt{6}$. $B^2 = (2 \sqrt{2} - \sqrt{3})^2 = 4(2) + 3 - 4 \sqrt{6} = 11 - 4 \sqrt{6}$. $A^2 + B^2 = (25 + 4 \sqrt{6}) + (11 - 4 \sqrt{6}) = 36$. Thus,the range of $A \cos x + B \sin x$ is $[-6, 6]$. The range of $f(x)$ is $[-6-6, 6-6]$,which is $[-12, 0]$. Therefore,$m = -12$ and $M = 0$. We need to find $\sqrt{|M^2 - m^2|} = \sqrt{|0^2 - (-12)^2|} = \sqrt{|-144|} = \sqrt{144} = 12$.

If the extreme values of the function $f(x)=(2 \sqrt{6}+1) \cos x+(2 \sqrt{2}-\sqrt{3}) \sin x-6$ are $m$ and $M$,then $\sqrt{|M^2-m^2|}=$

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