Commerce (English Medium) Class 12 - CBSE Question Bank Solutions for Mathematics

A medical company has factories at two places, A and B. From these places, supply is made to each of its three agencies situated at P, Q and R. The monthly requirements of the agencies are respectively 40, 40 and 50 packets of the medicines, while the production capacity of the factories, A and B, are 60 and 70 packets respectively. The transportation cost per packet from the factories to the agencies are given below:

By graphical method, the solution of linear programming problem

The region represented by the inequation system x, y ≥ 0, y ≤ 6, x + y ≤ 3 is 

The point at which the maximum value of x + y subject to the constraints x + 2y ≤ 70, 2x + y ≤ 95, x ≥ 0, y ≥ 0 is obtained, is ______.

The value of objective function is maximum under linear constraints ______.

If l_i, m_i, n_i, i = 1, 2, 3 denote the direction cosines of three mutually perpendicular vectors in space, prove that AA^T = I, where \[A = \begin{bmatrix}l_1 & m_1 & n_1 \\ l_2 & m_2 & n_2 \\ l_3 & m_3 & n_3\end{bmatrix}\]

If\[A = \begin{bmatrix}2 & 3 \\ 4 & 5\end{bmatrix}\]prove that A − A^T is a skew-symmetric matrix.

The solution of the differential equation \[\frac{dy}{dx} = \frac{y}{x} + \frac{\phi\left( \frac{y}{x} \right)}{\phi'\left( \frac{y}{x} \right)}\] is

\[\frac{dy}{dx} = \frac{\sin x + x \cos x}{y\left( 2 \log y + 1 \right)}\]

x (e^2y − 1) dy + (x² − 1) e^y dx = 0

\[\frac{dy}{dx} + 1 = e^{x + y}\]

\[\frac{dy}{dx} = \left( x + y \right)^2\]

cos (x + y) dy = dx

\[\frac{dy}{dx} + \frac{y}{x} = \frac{y^2}{x^2}\]

\[\frac{dy}{dx} = \frac{y\left( x - y \right)}{x\left( x + y \right)}\]

(x + y − 1) dy = (x + y) dx

\[\frac{dy}{dx} - y \cot x = cosec\ x\]

\[\frac{dy}{dx} - y \tan x = - 2 \sin x\]

\[\frac{dy}{dx} - y \tan x = e^x \sec x\]

\[\frac{dy}{dx} - y \tan x = e^x\]