Commerce (English Medium) कक्षा १२ - CBSE Question Bank Solutions

Evaluate:  \[\int 2^x  \text{ dx }\]

Evaluate: \[\int\frac{x^3 - x^2 + x - 1}{x - 1} \text{ dx }\]

Evaluate:\[\int\frac{e\tan^{- 1} x}{1 + x^2} \text{ dx }\]

Evaluate: \[\int\frac{1}{\sqrt{1 - x^2}} \text{ dx }\]

Write the value of\[\int\sec x \left( \sec x + \tan x \right)\text{  dx }\]

Evaluate: \[\int\frac{1}{x^2 + 16}\text{ dx }\]

Evaluate: \[\int\left( 1 - x \right)\sqrt{x}\text{  dx }\]

Evaluate: \[\int\frac{x + \cos6x}{3 x^2 + \sin6x}\text{ dx }\]

Evaluate:  \[\int\frac{2}{1 - \cos2x}\text{ dx }\]

Evaluate:

\[\int \cos^{-1} \left(\sin x \right) \text{dx}\]

Evaluate:

`∫ (1)/(sin^2 x cos^2 x) dx`

Evaluate : \[\int\frac{1}{x(1 + \log x)} \text{ dx}\]

Find the coordinates of the point where the line  \[\frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 2}{2}\]   intersects the plane x − y + z − 5 = 0. Also, find the angle between the line and the plane. 

Find the distance of the point (−1, −5, −10) from the point of intersection of the line \[\vec{r} = \left( 2 \hat{i}  - \hat{j} + 2 \hat{k}  \right) + \lambda\left( 3 \hat{i}+ 4 \hat{j} + 2 \hat{k}  \right)\] and the plane  \[\vec{r} . \left( \hat{i}  - \hat{j}  + \hat{k} \right) = 5 .\]

Find the distance of the point (2, 12, 5) from the point of intersection of the line \[\vec{r} = 2 \hat{i}  - 4 \hat{j}+ 2 \hat{k}  + \lambda\left( 3 \hat{i}  + 4 \hat{j}  + 2 \hat{k} \right)\] and \[\vec{r} . \left( \hat{i}  - 2 \hat{j}  + \hat{k}  \right) = 0\]

Find the distance of the point P(−1, −5, −10) from the point of intersection of the line joining the points A(2, −1, 2) and B(5, 3, 4) with the plane  \[x - y + z = 5\] . 

Find the distance of the point P(3, 4, 4) from the point, where the line joining the points A(3, –4, –5) and B(2, –3, 1) intersects the plane 2x + y + z = 7.

Find the distance of the point (1, -5, 9) from the plane

Find the equation of the plane containing the line \[\frac{x + 1}{- 3} = \frac{y - 3}{2} = \frac{z + 2}{1}\]  and the point (0, 7, −7) and show that the line  \[\frac{x}{1} = \frac{y - 7}{- 3} = \frac{z + 7}{2}\] also lies in the same plane.