Let $$(x, y)$$ be such that $${\sin ^{ - 1}}\left( {ax} \right) + {\cos ^{ - 1}}\left( y \right) + {\cos ^{ - 1}}\left( {bxy} \right) = {\pi \over 2}$$.

Column $$I$$
(A) If $$a=1$$ and $$b=0,$$ then $$(x, y)$$
(B) If $$a=1$$ and $$b=1,$$ then $$(x, y)$$
(C) If $$a=1$$ and $$b=2,$$ then $$(x, y)$$
(D) If $$a=2$$ and $$b=2,$$ then $$(x, y)$$

Column $$II$$
(p) lies on the circle $${x^2} + {y^2} = 1$$
(q) lies on $$\left( {{x^2} - 1} \right)\left( {{y^2} - 1} \right) = 0$$
(r) lies on $$y=x$$
(s) lies on $$\left( {4{x^2} - 1} \right)\left( {{y^2} - 1} \right) = 0$$

The tangent to the curve $$y = {e^x}$$ drawn at the point $$\left( {c,{e^c}} \right)$$ intersects the line joining the points $$\left( {c - 1,{e^{c - 1}}} \right)$$ and $$\left( {c + 1,{e^{c + 1}}} \right)$$

If a continuous function $$f$$ defined on the real line $$R$$, assumes positive and negative values in $$R$$ then the equation $$f(x)=0$$ has a root in $$R$$. For example, if it is known that a continuous function $$f$$ on $$R$$ is positive at some point and its minimum value is negative then the equation $$f(x)=0$$ has a root in $$R$$.
Consider $$f\left( x \right) = k{e^x} - x$$ for all real $$x$$ where $$k$$ is real constant.

The line $$y=x$$ meets $$y = k{e^x}$$ for $$k \le 0$$ at

If a continuous function $$f$$ defined on the real line $$R$$, assumes positive and negative values in $$R$$ then the equation $$f(x)=0$$ has a root in $$R$$. For example, if it is known that a continuous function $$f$$ on $$R$$ is positive at some point and its minimum value is negative then the equation $$f(x)=0$$ has a root in $$R$$.
Consider $$f\left( x \right) = k{e^x} - x$$ for all real $$x$$ where $$k$$ is real constant.

For $$k>0$$, the set of all values of $$k$$ for which $$k{e^x} - x = 0$$ has two distinct roots is