If the two lines x + (a - 1)y = 1 and 2x + a²y = 1  are perpendicular, then the distance of their point of intersection from the origin is

• $\frac{2}{5}$

• $\frac{2}{\sqrt{5}}$

• $\sqrt{\frac{2}{5}}$

• $\frac{\sqrt{2}}{5}$

12.

Let P be the plane, which contains the line of intersection of the planes, x + y + z = 6 and 2x + 3y + z + 5 = 0 and it is perpendicular to the xy - plane. Then the distance of the point (0, 0, 256) from P is equal to

• $63\sqrt{5}$

• $\raisebox{1ex}{$17$}\!\left/ \!\raisebox{-1ex}{$\sqrt{5}$}\right.$

• $205\sqrt{5}$

• $\raisebox{1ex}{$11$}\!\left/ \!\raisebox{-1ex}{$\sqrt{5}$}\right.$

If $\int {\mathrm{e}}^{\sec \left(\mathrm{x}\right)}\left(\sec \left(\mathrm{x}\right)\tan \left(\mathrm{x}\right)\mathrm{f}\left(\mathrm{x}\right) + \left(\sec \left(\mathrm{x}\right)\tan \left(\mathrm{x}\right) + {\sec }^2\left(\mathrm{x}\right)\right)\right)\mathrm{dx} = {\mathrm{e}}^{\sec \left(\mathrm{x}\right)}\mathrm{f}\left(\mathrm{x}\right) +\hspace{0.17em}\mathrm{C}$, then a possible choice of f(x) is :

• $\mathrm{xsec}\left(\mathrm{x}\right) + \tan \left(\mathrm{x}\right) + \frac{1}{2}$

• $\sec \left(\mathrm{x}\right) + \tan \left(\mathrm{x}\right) - \frac{1}{2}$

• $\sec \left(\mathrm{x}\right) + \tan \left(\mathrm{x}\right) + \frac{1}{2}$

• $\sec \left(\mathrm{x}\right) - \tan \left(\mathrm{x}\right) - \frac{1}{2}$