Let $$\,\,\,$$$$f\left( x \right) = 2 + \cos x$$ for all real $$X$$.

STATEMENT - 1: for eachreal $$t$$, there exists a point $$c$$ in $$\left[ {t,t + \pi } \right]$$ such that $$f'\left( c \right) = 0$$ because
STATEMENT - 2: $$f\left( t \right) = f\left( {t + 2\pi } \right)$$ for each real $$t$$.

Let $$ABCD$$ be a quadrilateral with area $$18$$, with side $$AB$$ parallel to the side $$CD$$ and $$2AB=CD$$. Let $$AD$$ be perpendicular to $$AB$$ and $$CD$$. If a circle is drawn inside the quadrilateral $$ABCD$$ touching all the sides, then its radius is

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 $$\left( {c + 1,{e^{c + 1}}} \right)$$