PUC Science 2nd PUC Class 12 - Karnataka Board PUC Question Bank Solutions for Mathematics

The vector cos α cos β \[\hat{i}\] + cos α sin β \[\hat{j}\] + sin α \[\hat{k}\] is a

A company sells two different products A and B. The two products are produced in a common production process and are sold in two different markets. The production process has a total capacity of 45000 man-hours. It takes 5 hours to produce a unit of A and 3 hours to produce a unit of B. The market has been surveyed and company officials feel that the maximum number of units of A that can be sold is 7000 and that of B is 10,000. If the profit is Rs 60 per unit for the product A and Rs 40 per unit for the product B, how many units of each product should be sold to maximize profit? Formulate the problem as LPP.

A toy company manufactures two types of dolls, A and B. Market tests and available resources have indicated that the combined production level should not exceed 1200 dolls per week and the demand for dolls of type B is at most half of that for dolls of type A. Further, the production level of dolls of type A can exceed three times the production of dolls of other type by at most 600 units. If the company makes profit of ₹ 12 and ₹ 16 per doll respectively on dolls A and B, how many of each should be produced weekly in order to maximise the profit?

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

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

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

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

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

(1 + y + x² y) dx + (x + x³) dy = 0

(x² + 1) dy + (2y − 1) dx = 0

`y sec^2 x + (y + 7) tan x(dy)/(dx)=0`

`(2ax+x^2)(dy)/(dx)=a^2+2ax`