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(D) Let $f(x) = \cos (	an ^{-1}(\sin (\cos ^{-1} x))) + \sin (\cot ^{-1}(\cos (\sin ^{-1} x)))$.Since $\sin (\cos ^{-1} x) = \sqrt{1-x^2}$ and $\cos

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$m^2+M$

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(D) Let $f(x) = \cos (	an ^{-1}(\sin (\cos ^{-1} x))) + \sin (\cot ^{-1}(\cos (\sin ^{-1} x)))$.Since $\sin (\cos ^{-1} x) = \sqrt{1-x^2}$ and $\cos (\sin ^{-1} x) = \sqrt{1-x^2}$,we have:$f(x) = \cos (	an ^{-1}(\sqrt{1-x^2})) + \sin (\cot ^{-1}(\sqrt{1-x^2}))$.Let $	heta = 	an ^{-1}(\sqrt{1-x^2})$,then $	an 	heta = \sqrt{1-x^2}$.Then $\cos 	heta = \frac{1}{\sqrt{1+	an^2 	heta}} = \frac{1}{\sqrt{1+(1-x^2)}} = \frac{1}{\sqrt{2-x^2}}$.Similarly,let $\phi = \cot ^{-1}(\sqrt{1-x^2})$,then $\cot \phi = \sqrt{1-x^2}$.Then $\sin \phi = \frac{1}{\sqrt{1+\cot^2 \phi}} = \frac{1}{\sqrt{1+(1-x^2)}} = \frac{1}{\sqrt{2-x^2}}$.Thus,$f(x) = \frac{1}{\sqrt{2-x^2}} + \frac{1}{\sqrt{2-x^2}} = \frac{2}{\sqrt{2-x^2}}$.Since $x \in [-1, 1]$,$x^2 \in [0, 1]$,so $2-x^2 \in [1, 2]$.Thus,$\sqrt{2-x^2} \in [1, \sqrt{2}]$,and $f(x) \in [\sqrt{2}, 2]$.So $m = \sqrt{2}$ and $M = 2$.The equation is $\operatorname{sgn} (|x - 1| - 2) = \ln |x - 2|$.By analyzing the graph of $y = \operatorname{sgn} (|x - 1| - 2)$ and $y = \ln |x - 2|$,we find that the curves intersect at $4$ points.

Column $I$	Column $II$
$(A)$ If $a=1$ and $b=0$,then $(x, y)$	$(p)$ lies on the circle $x^2+y^2=1$
$(B)$ If $a=1$ and $b=1$,then $(x, y)$	$(q)$ lies on $(x^2-1)(y^2-1)=0$
$(C)$ If $a=1$ and $b=2$,then $(x, y)$	$(r)$ lies on $y=x$
$(D)$ If $a=2$ and $b=2$,then $(x, y)$	$(s)$ lies on $(4x^2-1)(y^2-1)=0$

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