Let |M| denote the determinant of a square ma | vedclass

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(A) First,evaluate the determinant $f(	heta)$. The first determinant is $\frac{1}{2} 	imes [1(1+\sin^2 	heta) - \sin 	heta(-\sin 	heta + \sin 	h

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$A, C$

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(A) First,evaluate the determinant $f( heta)$. The first determinant is $\frac{1}{2} imes [1(1+\sin^2 heta) - \sin heta(-\sin heta + \sin heta) + 1(\sin^2 heta + 1)] = \frac{1}{2} imes [1+\sin^2 heta + 1+\sin^2 heta] = 1+\sin^2 heta$. The second determinant is a skew-symmetric matrix of odd order $(3 imes 3)$,so its value is $0$. Thus,$f( heta) = 1+\sin^2 heta$. Then $g( heta) = \sqrt{1+\sin^2 heta - 1} + \sqrt{1+\sin^2(\frac{\pi}{2}- heta) - 1} = \sqrt{\sin^2 heta} + \sqrt{\cos^2 heta} = |\sin heta| + |\cos heta|$. For $ heta \in [0, \frac{\pi}{2}]$,$g( heta) = \sin heta + \cos heta = \sqrt{2} \sin( heta + \frac{\pi}{4})$. The range of $g( heta)$ is $[1, \sqrt{2}]$. The roots of $p(x)$ are $1$ and $\sqrt{2}$. So $p(x) = k(x-1)(x-\sqrt{2})$. Given $p(2) = 2-\sqrt{2}$,we have $k(2-1)(2-\sqrt{2}) = 2-\sqrt{2} \implies k=1$. Thus $p(x) = (x-1)(x-\sqrt{2})$. $(A) \ p(\frac{3+\sqrt{2}}{4}) = (\frac{3+\sqrt{2}-4}{4})(\frac{3+\sqrt{2}-4\sqrt{2}}{4}) = (\frac{\sqrt{2}-1}{4})(\frac{3-3\sqrt{2}}{4}) < 0$ (True). $(B) \ p(\frac{1+3\sqrt{2}}{4}) = (\frac{1+3\sqrt{2}-4}{4})(\frac{1+3\sqrt{2}-4\sqrt{2}}{4}) = (\frac{3\sqrt{2}-3}{4})(\frac{1-\sqrt{2}}{4}) < 0$ (False). $(C) \ p(\frac{5\sqrt{2}-1}{4}) = (\frac{5\sqrt{2}-1-4}{4})(\frac{5\sqrt{2}-1-4\sqrt{2}}{4}) = (\frac{5\sqrt{2}-5}{4})(\frac{\sqrt{2}-1}{4}) > 0$ (True). $(D) \ p(\frac{5-\sqrt{2}}{4}) = (\frac{5-\sqrt{2}-4}{4})(\frac{5-\sqrt{2}-4\sqrt{2}}{4}) = (\frac{1-\sqrt{2}}{4})(\frac{5-5\sqrt{2}}{4}) > 0$ (False).

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