Let $$y(x) = (1 + x)(1 + {x^2})(1 + {x^4})(1 + {x^8})(1 + {x^{16}})$$. Then $$y' - y''$$ at $$x = - 1$$ is equal to

Consider the lines $$L_1$$ and $$L_2$$ given by

$${L_1}:{{x - 1} \over 2} = {{y - 3} \over 1} = {{z - 2} \over 2}$$

$${L_2}:{{x - 2} \over 1} = {{y - 2} \over 2} = {{z - 3} \over 3}$$.

A line $$L_3$$ having direction ratios 1, $$-$$1, $$-$$2, intersects $$L_1$$ and $$L_2$$ at the points $$P$$ and $$Q$$ respectively. Then the length of line segment $$PQ$$ is

The points of intersection of the line $$ax + by = 0,(a \ne b)$$ and the circle $${x^2} + {y^2} - 2x = 0$$ are $$A(\alpha ,0)$$ and $$B(1,\beta )$$. The image of the circle with AB as a diameter in the line $$x + y + 2 = 0$$ is :

Let $$\mathrm{z_1=2+3i}$$ and $$\mathrm{z_2=3+4i}$$. The set $$\mathrm{S = \left\{ {z \in \mathbb{C}:{{\left| {z - {z_1}} \right|}^2} - {{\left| {z - {z_2}} \right|}^2} = {{\left| {{z_1} - {z_2}} \right|}^2}} \right\}}$$ represents a