II Internal Test - Solved Problems

II Internal Test

Mathematical Methods – II

PART – A

1.     If  y = log x, then dy/dx = ?

       a)      1/x                        b) x                 c) 1                  d) 0

2.      For maxima of a function Y = f (x), dy/dx = ?

       a)      Equal to zero        b) Greater than zero     c)Less than zero       d) Equal to one

PART – B

 3.      Explain the rule of differentiation with example

4.      Determine the marginal utility of x, if  x = 3 for the total utility function  U= 5x² + 2x + 9

PART – C

 5.      Solve the following equations using Cramer’s rule

2x – 4y + 3z = 3

4x – 6y + 5z = 2

-2x + y – z   = 1

6.      Find dy/dx,  if

              1)      Y = (2x³ + 9) (x² + 3x)

                   
                          x - 1
              2) Y = ---------
                          x² + 1                                       

 

ANSWERS

1.      a) 1/x

2.      a) Equal to zero

3.      If u = f(x), v=g(x) and w=h(x) are the functions of x, then the product of two functions is y=uv, and the product of three functions is y=uvw.

Derivative of the product of two functions

dy         dv           du ---- = u ------ + v-------       =          uv’ + vu’       dx         dx           dx

 

     Derivative of the product of three functions

dy           dw             dv                du ---- = uv ------ + uw-------  + vw -------      uvw’ + uwv’ + vwu’ dx           dx              dx                dx

For example

y=(3x²+1) (x3+2x)

 dy           dv           du

---- =   u ------ + v-------          

       dx           dx           dx

= (3x²+1) (3x²+2) + (x³+2x) (6x)

            = (9x⁴+6x²+3x²+2) + (6x⁴+12x²)

            = 9x⁴+6x²+3x²+2+6x⁴+12x²

            = 15x⁴+21x²+2

4.      U=5x²+2x+9

du          

MU =  ----   

dx

MU= 10x+2

If x=3

MU=10(3)+2

      =30+2

MU=32

        [ 2    -4   3 ]    [ x ]         [ 3 ]

5.                  | 4   -6    5   |   |  y |   =     | 2  |

        [-2   1   -1  ]   [  z ]         [ 1 ]

      | ∆x |                           | ∆y |                        | ∆z |

x= -----                 y= -----                z= -----

      | ∆ |                             | ∆ |                          | ∆ |

                  

            [  2    -4   3  ]   

∆   =    |   4    -6    5  | 

          [ -2    1   -1  ]  

                 |   -6    5  |       |   4     5  |      |   4    -6  |

| ∆ |  =  2   |    1   -1  | +4  |  -2   -1  | +3 |  -2     1  |

        = 2(6-5) + 4(-4+10) +3(4-12)

        = 2(1) +4(6) +3(-8)

        = 2+24+-24

| ∆ | = 2

            

              [  3    -4   3  ]   

∆x   =     |  2    -6    5  | 

            [  1    1   -1  ]  

                |   -6    5  |       |   2     5 |      |   2    -6  |

| ∆x | = 3  |    1   -1  | +4  |   1   -1  | +3 |  1      1  |

         = 3(6-5) + 4(-2-5) +3(2+6)

         =3(1)+4(-7)+3(8)

         =3-28+24

| ∆x | = -1

 

             [   2   3     3  ]   

∆y   =   |   4   2     5 | 

             [  -2  1    -1  ]

                |   2    5  |       |   4     5  |       |  4     2  |

| ∆y | = 2  |   1   -1  | -3   |  -2   -1  | +3 | -2     1  |

             = 2(-2-5) -3(-4-10) +3(4+4)

             =2(-7)-3(6)+3(8)

             =-14-18+24

            =-32-24

  | ∆y |  = -8

            [  2     -4    3]   

∆z   =  | 4     -6    2| 

            [ -2     1     1]

                |  -6    2  |       |   4     2  |      |  4    -6  |

| ∆z | = 2  |   1     1  | +4  |  -2    1  | +3 | -2     1  |

             = 2(-6-2) +4(4+4) +3(4-12)

             =2(-8)+4(8)+3(-8)

             =-16+32-24

            =-40+32

| ∆z | = -8

      | ∆x |     -1                             | ∆y |   -8                             | ∆z |     -8

x= ----- =---- = -0.5              y= ----- = ---- = -4               z= ----- = ---- = -4

      | ∆ |       2                              | ∆ |     2                             | ∆ |        2