Pre Boards

Practice to excel and get familiar with the paper pattern and the type of questions. Check you answers with answer keys provided.

Sample Papers

Download the PDF Sample Papers Free for off line practice and view the Solutions online.

CBSE Class 12 Mathematics Solved Question Paper 2008

Long Answer Type

21.

Evaluate: $\int_{0}^{π} \frac{xsinx}{1 + \cos^{2} x} dx$

22.

Solve the following differential equation:
(x² − y²⁾ dx + 2xy dy = 0 given that y = 1 when x = 1

23.

Solve the following differential equation:

$\frac{dy}{dx} = \frac{x (2 y - x)}{x (2 y + x)}$ , if y = 1 when x = 1

24.

Solve the following differential equation:

$\cos^{2} x \frac{dy}{dx} + y = tanx$

25.

$If \vec{a} = \hat{i} + \hat{j} + \hat{k} and \vec{b} = \hat{j} - \hat{k}, find a vector \vec{c} such that \vec{a} \times \vec{c} = \vec{b} and \vec{a} . \vec{c} = 3$

26.

If $\vec{a} + \vec{b} + \vec{c} = 0 and |\vec{a}| = 3, |\vec{b}| = 5 and |\vec{c}| = 7,$ show that the angle between $\vec{a}$ and $\vec{b}$ is 60⁰.

27.
Find the shortest distance between the following lines:
$\frac{x - 3}{1} = \frac{y - 5}{- 2} = \frac{z - 7}{1} and \frac{x + 1}{7} = \frac{y + 1}{- 6} = \frac{z + 1}{1}$

$\frac{x - 3}{1} = \frac{y - 5}{- 2} = \frac{z - 7}{1}$

The vector form of this equation is:

$\vec{r} = 3 \hat{i} + 5 \hat{j} + 7 \hat{k} + λ \hat{i} - 2 \hat{j} + \hat{k} \vec{r} = {\vec{a}}_{1} + λ {\vec{b}}_{1} . . . . . . . . . . . . (1) \frac{x + 1}{7} = \frac{y + 1}{- 6} = \frac{z + 1}{1}$

The vector form of this equation is:

$\vec{r} = - \hat{i} - \hat{j} - \hat{k} + λ 7 \hat{i} - 6 \hat{j} + \hat{k} Therefore, \vec{a_{1}} = 3 \hat{i} + 5 \hat{j} - 7 \hat{k} \vec{b_{1}} = \hat{i} - 2 \hat{j} + \hat{k} \vec{a_{2}} = - \hat{i} - \hat{j} - \hat{k} and \vec{b_{2}} = 7 \hat{i} - 6 \hat{j} + \hat{k}$

Now, the shortest distance between these two lines is given by:

$d = |\frac{\vec{b_{1}} x \vec{b_{2}} . \vec{a_{2}} - \vec{a_{1}}}{|\vec{b_{1}} x \vec{b_{2}}|}| \vec{b_{1}} x \vec{b_{2}} = [\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & - 2 & 1 \\ 7 & - 6 & 1 \end{matrix}] = \hat{i} (- 2 + 6) - \hat{j} (1 - 7) + \hat{k} (- 6 + 14) = 4 \hat{i} + 6 \hat{j} + 8 \hat{k} |\vec{b_{1}} x \vec{b_{2}}| = \sqrt{4^{2} + 6^{2} + 8^{2}} = \sqrt{116} \vec{a_{2}} - \vec{a_{1}} = - \hat{i} - \hat{j} - \hat{k} - 3 \hat{i} + 5 \hat{j} + 7 \hat{k} = - 4 \hat{i} - 6 \hat{j} - 8 \hat{k} ∴ d = |\frac{4 \hat{i} + 6 \hat{j} + 8 \hat{k} . - 4 \hat{i} - 6 \hat{j} - 8 \hat{k}}{\sqrt{116}}| = |\frac{- 16 - 36 - 64}{\sqrt{116}}| = |\frac{- 116}{\sqrt{116}}| = \sqrt{116}$