Tilak Maharashtra Vidyapeeth

(Deemed University)

 

QUESTION BANK

 

APPLIED MATHEMATICS - I

 

 

First Year Diploma (All)

 

UNIT 1 : ALGEBRA

 

I)                   ALGORITHM

 

1.        If 2x = 23y = 512 Find value of log zx3 where x = y3

2.      Find value of   a) log 3437     b) log 5Φ250      c) log 927    d) log 361     e) log 5125           f) log 10 1000   g) log 2Φ312

3.      Simplify   a) log214 – log 27      b) (log 34) * (log 481)      c) log 5243

 


Log 259

                                    d) log ab * log bc * log da

 

                                    e) log a2/bc + log b2/ca  +  log c2/ab

4.      Solve for x

 

a)      log 2(x+5) + log 2(x-2) = 3

b)      log x/ log 5 = log 25/ log 125

c)      (4 log 3) (log x) = log 27

 


log 9

 

                               d)   log X   =   log 64       

                                     log 4         log 16   

 

d)      logx2  + logx4  + logx8  = 6

 

5.      Prove that

 

1.      logxa3  *  logcb3 * logac3  = 27

2.      if log ( x+y )   =     1     ( log x + log y)

                    7               2 

 

       Prove that      x    +    y      = 47

                              Y         x

 

 

 

 

 

 

                                    3                                                   4

                   3.   log y           x          *   log Z y   * log x         z3    = 1

 

                   4. log   x2       +  log    y2         +   log    z2          =  0       

                               yz                         zx                      xy                  

 

 

    UNIT 2  DETERMINANTS

 

I)                   Solve determinant  equations

 

                2   3   1

       a)      6   x   2       = 0 

                            4   x  -2

 

                           4   9   2                 2   2x

                   b)     3   x   7         =      2   2x

                            8  1   6

 

                                                                                  

                         4x + 2       2x + 1                  1   2x   4x2

                  c)     x + 1        2x  - 1   = 0, in     1   4    16

                                                                      1    1     1

 

 

                           1   x   x2                   1    1

                  d)     1   2   4      =      2    2

                          1   3    9

 

 

 

 

II)                 With out expanding prove that

 


               4   5    7                5   3   11

      a)       2   3    1        +     4    2   9       =  0

                           9   11  13              7    1   13

 

 

                          4    5     7                4   4   3

                 B)     2    3     1     =        -2   1   5

                          9   11   13              -1  -3   7

 

 

 

 

                 5    1   3                10    1   3

      c)        6    2   5      +        19    2   5    = 0

                 9    1   7                26    1   7

 

 

 

III)              Without expanding find the value of

 


                 6    1    9

       a)       9    4    7   

                18   3   27

 

 

                 x-y    y-z    z-x

      b)        y-z    z-x    x-y

                 z-x     x-y    y-z

 

 


                  6    9   12

     c)          2    3    4

                  5    6   13

 

 


                 3    4    5

                 7    8    9

    d)          1    1    1

 

 

                  61    54    57  

    e)          52     55    58

                 53     56    59

 

 

 

V) Solution of Simultaneous equation using CRAMER’S Rule

i)        4x - 3y = 2              ii)   2x - 3y + 1 = 0

3x + 5y = -13                5y – 4x = 3

   

  ii)    1  +   3   = 5             iv)    3

          x       y                            x - 2

          3  -   4   = 2

          x       y

 

   V) 2x + y + z  = 1                        vi) x + y + z = 6

        X + Y = 2z = 0                            3x + 3y + z = 12

      5x + 3y = 3z = 2                           2x + 3y + z = 14

 

 

 

vii) x + y + z = 1                           viii) x + y + z = 4

      2x + 3y + z =4                               2x + y + z = 1

      4x + 1y + z = 16                            x – y  + z = -3

 

ix) x + y + z = 6                            x) x – 2y + 3z = 4

    2x + y + 2z = -2                            2x + y – 3z = 5

     x + y –3z = -6                               x – y – 2z = -3

 

 

III) PARTIAL FRACTION         

       Find partial fraction

 

   1) x+1                                           2) (x+3) (x+1)

      (x-z)(x-3)                                       x(x+4) (x+2)

 

 3)           x                                      4)        x

     (x-1) (x-2) (x-3)                                x2 + x-z

 

5)       x-5                                         6)           1

     x3 + x2 –6x                                          x2 + 3x + z

 

7)    5x + 1                                      8)          3x-z

        x2+x-z                                               (x-z)(x=3)  

 

9)   x2-2x+7                                  10)     3x+2

       (x+1) (x2 –1)2                                 (x + 1) (x2-1)

 

11) 3x2-4x                                    12)   x2+1

 


        (x-1)3                                            x3 + 1

 

 

Binomial Theorem

 

i)   Expand using Binomial Theorem

     a)  (2x+3y)4      b)     2x      -   3              c)  (x-2y)2

                                      3          2x

 

 

ii) Using Binominal thermo prove that

                                  5                                  5

    a)                                                                                                     

               2     + 1             -            2     -1          = 82

 

 

                                   5                                 5

  b)            3      +  1         -           3     -  1         =  152

 

 

 

iii) a) Find 5th term of (x = 2y)8

     b) Find 4th term of    x3     -  2         9

                                                2           x22

                                                                                                                      11

    c) Find the term independent of x in the expansion of     x3  +   m                is 1320

 


                                                                                                           x8

                 find m.

                                                                           2x3 -  b

iv) a) Find the middle term in expansion of       a       x

 

 

b)      Find the middle terms in expansion of  ( 3x – x3 /  9 )9

 


v) a) using binomial thermo find approximate value of       9.  18 .

 

    b) Find approximate value of    1       using binominal theorem. 

                                          

                                                        99

                                                                                               

  c) Find approximate value of     3    997        &               408

 

 

 

  UNIT 2       TRIGNOMETRY

 

5)    prove pollinating

 

      i) Cosec2θ  - Cos2θ . Cosec2θ  = 1

 

     ii) Sec2 θ  +  Cosec2 θ  = Sec2 θ Cosec2 θ 

  

                   1 -  sin θ         =  Sec θ – Tan θ

    iii)    

                   1 + sin θ

 

   iv)       Sin θ                  1 – Sin θ

                                 +                          =  2   Sec θ  - Tan θ

              1+ Cos θ                  Cos θ

 

 

v) Cos2 θ     1  + Tan2 θ        = 1

 

vii) Sin θ + Cot θ  Cos θ   = Cosec θ

 

viii) Sec2 θ – Sin2 θ Sec2 θ = 1

 


ix)                Sec2 θ + Cosec2 θ   = tan θ + Cot θ

 

 

x)          Sec θ + 1      =  Cot θ + Cosec θ  

             Sec θ - 1

 

 

6) Trigonometric Ratios of Allied, Compound, Multiple angle

 

2) Find the value of or show that

  i) Tan2 45 + Cos 60 – Cosec2 30 + Sin2 0 = -5/2

 


  ii) Cos 450 Cos 600  - Sin 45  Sin 60 = -          3  - 1

                                                                             

                                                                         2       2

 

  iii)     tan 420  + tan 300   = 0

           1- tan 420 tan 300

 

iv) sin230  cos 60    + cos2 45  tan 45

 

v) tan245 – cosec230  + 2 sin 30 = -2

 

v) if tan θ = 1/            find value of     cosec2 θ – sec2 θ

                           7