HSC Commerce: Marketing and Salesmanship १२ वीं कक्षा - Maharashtra State Board Question Bank Solutions for Mathematics and Statistics

Evaluate `int(3x-2)/((x+1)^2(x+3))  dx`

Complete the following activity to divide 84 into two parts such that the product of one part and square of the other is maximum.

Solution: Let one part be x. Then the other part is 84 - x

Letf (x) = x² (84 - x) = 84x² - x³

∴ f'(x) = `square`

and f''(x) = `square`

For extreme values, f'(x) = 0

∴ x = `square  "or"    square`

f(x) attains maximum at x = `square`

Hence, the two parts of 84 are 56 and 28.

Solve the differential equation (x² + y²) dx - 2xy dy = 0 by completing the following activity.

Solution: (x² + y²) dx - 2xy dy = 0

∴ `dy/dx=(x^2+y^2)/(2xy)`                      ...(1)

Puty = vx

∴ `dy/dx=square`

∴ equation (1) becomes

`x(dv)/dx = square`

∴ `square  dv = dx/x`

On integrating, we get

`int(2v)/(1-v^2) dv =intdx/x`

∴ `-log|1-v^2|=log|x|+c_1`

∴ `log|x| + log|1-v^2|=logc       ...["where" - c_1 = log c]`

∴ x(1 - v²) = c

By putting the value of v, the general solution of the D.E. is `square`= cx

`inte^(xloga).e^x dx` is ______

The area enclosed by the parabola x² = 4y and its latus rectum is `8/(6m)` sq units. Then the value of m is ______.

The integrating factor of `ylogy.dx/dy+x-logy=0` is ______.

`int logx  dx = x(1+logx)+c`

`int(xe^x)/((1+x)^2)  dx` = ______

Find `dy/dx,"if"  y=x^x+(logx)^x`

If Mr. Rane order x chairs at the price p = (2x² - 12x - 192) per chair. How many chairs should he order so that the cost of deal is minimum?

Solution: Let Mr. Rane order x chairs.

Then the total price of x chairs = p·x = (2x² - 12x- 192)x

= 2x³ - 12x² - 192x

Let f(x) = 2x³ - 12x² - 192x

∴ f'(x) = `square` and f''(x) = `square`

f'(x ) = 0 gives x = `square` and f''(8) = `square` > 0

∴ f is minimum when x = 8

Hence, Mr. Rane should order 8 chairs for minimum cost of deal.

`int(f'(x))/sqrt(f(x)) dx = 2sqrt(f(x))+c`

Find the area of the region lying in the first quadrant and bounded by y = 4x², x = 0,y = 2 and y = 4.

The rectangle has area of 50 cm². Complete the following activity to find its dimensions for least perimeter.

Solution: Let x cm and y cm be the length and breadth of a rectangle.

Then its area is xy = 50

∴ `y =50/x`

Perimeter of rectangle `=2(x+y)=2(x+50/x)`

Let f(x) `=2(x+50/x)`

Then f'(x) = `square` and f''(x) = `square`

Now,f'(x) = 0, if x = `square`

But x is not negative.

∴ `x = root(5)(2)   "and" f^('')(root(5)(2))=square>0`

∴ by the second derivative test f is minimum at x = `root(5)(2)`

When x = `root(5)(2),y=50/root(5)(2)=root(5)(2)`

∴ `x=root(5)(2)  "cm" , y = root(5)(2)  "cm"`

Hence, rectangle is a square of side `root(5)(2)  "cm"`

Solve the following

`int_0^1 e^(x^2) x^3 dx`

Find `dy/dx` if, y = `x^(e^x)`

Find `dy/dx if, y =  x^(e^x)`

Evaluate the following.

`intx^3  e^(x^2) dx`

Find the area of the regions bounded by the line y = −2x, the X-axis and the lines x = −1 and x = 2.

Complete the following activity:

`int_0^2 dx/(4 + x - x^2) `

= `int_0^2 dx/(-x^2 + square + square)`

= `int_0^2 dx/(-x^2 + x + 1/4 - square + 4)`

= `int_0^2 dx/ ((x- 1/2)^2 - (square)^2)`

= `1/sqrt17 log((20 + 4sqrt17)/(20 - 4sqrt17))`

Three new machines M1, M2, M3 are to be installed in a machine shop. There are four vacant places A, B, C, D. Due to limited space, machine M2 can not be placed at B. The cost matrix (in hundred rupees) is as follows:

Determine the optimum assignment schedule and find the minimum cost.