Let P be a point in the first octant, whose image Q in the plane x + y = 3 (that is, the line segment PQ is perpendicular to the plane x + y = 3 and the mid-point of PQ lies in the plane x + y = 3) lies on the Z-axis. Let the distance of P from the X-axis be 5. If R is the image of P in the XY-plane, then the length of PR is ...............

Consider the cube in the first octant with sides OP, OQ and OR of length 1, along the X-axis, Y-axis and Z-axis, respectively, where O(0, 0, 0) is the origin. Let $$S\left( {{1 \over 2},{1 \over 2},{1 \over 2}} \right)$$ be the centre of the cube and T be the vertex of the cube opposite to the origin O such that S lies on the diagonal OT. If p = SP, q = SQ, r = SR and t = ST, then the value of |(p $$ \times $$ q) $$ \times $$ (r $$ \times $$ t)| is ............

Let $$X = {({}^{10}{C_1})^2} + 2{({}^{10}{C_2})^2} + 3{({}^{10}{C_3})^2} + ... + 10{({}^{10}{C_{10}})^2}$$,

where $${}^{10}{C_r}$$, r $$ \in $${1, 2, ..., 10} denote binomial coefficients. Then, the value of $${1 \over {1430}}X$$ is ..........

Let $${E_1} = \left\{ {x \in R:x \ne 1\,and\,{x \over {x - 1}} > 0} \right\}$$ and


$${E_2} = \left\{ \matrix{ x \in {E_1}:{\sin ^{ - 1}}\left( {{{\log }_e}\left( {{x \over {x - 1}}} \right)} \right) \hfill \cr is\,a\,real\,number \hfill \cr} \right\}$$

(Here, the inverse trigonometric function $${\sin ^{ - 1}}$$ x assumes values in $$\left[ { - {\pi \over 2},{\pi \over 2}} \right]$$.).

Let f : E₁ $$ \to $$ R be the function defined by f(x) = $${{{\log }_e}\left( {{x \over {x - 1}}} \right)}$$ and g : E₂ $$ \to $$ R be the function defined by g(x) = $${\sin ^{ - 1}}\left( {{{\log }_e}\left( {{x \over {x - 1}}} \right)} \right)$$.

LIST-I	LIST-II
P. The range of $f$ is	1. $\left( -\infty, \frac{1}{1-e} \right] \cup \left[ \frac{e}{e-1}, \infty \right)$
Q. The range of $g$ contains	2. $(0, 1)$
R. The domain of $f$ contains	3. $\left[ -\frac{1}{2}, \frac{1}{2} \right]$
S. The domain of $g$ is	4. $(-\infty, 0) \cup (0, \infty)$
	5. $\left( -\infty, \frac{e}{e-1} \right)$
	6. $(-\infty, 0) \cup \left( \frac{1}{2}, \frac{e}{e-1} \right]$

The correct option is :