The sum of the real roots of the equation $\left|\begin{array}{ccc}\mathrm{x}& - 6& - 1\\ 2& - 3\mathrm{x}& \mathrm{x} - 3\\ - 3& 2\mathrm{x}& \mathrm{x} + 2\end{array}\right| = 0$, is equal to

• 0

• 6

• - 4

• 1

The tangent and normal to the ellipse 3x² + 5y² = 32 at the point P(2, 2) meet the x-axis at Q and R, respectively .Then the area (in sq. units) of the triangle PQR is :

• $\frac{68}{15}$

• $\frac{16}{3}$

• $\frac{14}{3}$

• $\frac{34}{15}$

If ${\cos }^{-1}\left(\mathrm{x}\right) - {\cos }^{-1}\left(\frac{\mathrm{y}}{2}\right) = \mathrm{\alpha }$, where $- 1 \le \mathrm{x} \le 1$, $- 2 \le \mathrm{y} \le 2, \mathrm{x} \le \frac{\mathrm{y}}{2}$ then for all x, y, 4x² - 4xy$\cos \left(\mathrm{\alpha }\right)$ + y² is equal to

• $4{\sin }^2\left(\mathrm{\alpha }\right)$

• $2{\sin }^2\left(\mathrm{\alpha }\right)$

• $4{\cos }^2\left(\mathrm{\alpha }\right) +\hspace{0.17em}2{\mathrm{x}}^2{\mathrm{y}}^2$

• $4{\sin }^2\left(\mathrm{\alpha }\right) - 2{\mathrm{x}}^2{\mathrm{y}}^2$

A spherical iron ball of radius 10cm is coated with a layer of ice of uniform thickness that melts a rate of 50cm³/min When the thickness of the ice 5cm, then the rate at which the thickness (in cm/min) of the ice decreases, is :

• $\frac{5}{6\mathrm{\pi }}$

• $\frac{1}{18\mathrm{\pi }}$

• $\frac{1}{9\mathrm{\pi }}$

• $\frac{1}{36\mathrm{\pi }}$

   

If $\int {\mathrm{x}}^5{\mathrm{e}}^{- {\mathrm{x}}^2}\mathrm{dx} = \mathrm{g}\left(\mathrm{x}\right){\mathrm{e}}^{- 2} + \mathrm{c}$, where c is a constant of interation then g(- 1) is equal to :

• - 1

• 1

• $\frac{1}{2}$

• $- \frac{5}{2}$

The area (in sq.units ) of the region bounded by the curves y = 2^x and $\mathrm{y} = \left|\mathrm{x} + 1\right|$,in the first quadrant is:

• $\frac{3}{2} - \frac{1}{{\log }_{\mathrm{e}}\left(2\right)}$

• ${\log }_{\mathrm{e}}\left(2\right) +\hspace{0.17em}\frac{3}{2}$

• $\frac{1}{2}$

• $\frac{3}{2}$

Let y = y(x) be the solution of the differential equation $\frac{\mathrm{dy}}{\mathrm{dx}} + \mathrm{ytan}\left(\mathrm{x}\right) = 2\mathrm{x} +\hspace{0.17em}{\mathrm{x}}^2\tan \left(\mathrm{x}\right), \mathrm{x} \in \left(- \frac{\mathrm{\pi }}{2}, \frac{{\displaystyle \mathrm{\pi }}}{{\displaystyle 2}}\right)$ such that if y(0) = 1, then

• $\mathrm{y}\text{'}\left(\frac{\mathrm{\pi }}{4}\right) - \mathrm{y}\text{'}\left(- \frac{\mathrm{\pi }}{4}\right) = \mathrm{\pi } - \sqrt{2}$

• $\mathrm{y}\left(\frac{\mathrm{\pi }}{4}\right) - \mathrm{y}\left(- \frac{\mathrm{\pi }}{4}\right) = \sqrt{2}$

• $\mathrm{y}\left(\frac{\mathrm{\pi }}{4}\right) + \mathrm{y}\left(- \frac{\mathrm{\pi }}{4}\right) = \frac{{\mathrm{\pi }}^2}{2} + 2$

• $\mathrm{y}\text{'}\left(\frac{\mathrm{\pi }}{4}\right) + \mathrm{y}\text{'}\left(- \frac{\mathrm{\pi }}{4}\right) = - \sqrt{2}$

If the plane 2x - y + 2z = 3 = 0 has the distances $\frac{1}{3} \mathrm{and} \frac{2}{3}$ units from the planes 4x - 2y + 4z + $\mathrm{\lambda }$ = 0  and 2x - y + 2z + $\mathrm{\mu }$, respectively then the maximum value of $\mathrm{\lambda } + \mathrm{\mu }$

• 5

• 15

• 9

• 13

9.

The integral ${\int }_{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}{\sec }^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\left({\csc }^{\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\left(\mathrm{x}\right)\right)\mathrm{dx}$ =

• 3^5/3 - 3^1/3

• 3^7/6 - 3^5/6

• 3^5/6 - 3^1/3

• 3^4/3 - 3^1/3

The distance of the point having position vector $- \hat{\mathrm{i}} + 2\hat{\mathrm{j}} + 6\hat{\mathrm{k}}$ from the straight line passing  through the point (2, 3, - 4) and parallel to the vector $6\hat{\mathrm{i}} + 3\hat{\mathrm{j}} - 4\hat{\mathrm{k}}$ is :

• $2\sqrt{13}$

• 7

• 6

• $4\sqrt{3}$