Maths For F.Sc part 2 Exercises 1.4

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					   MathCity.org                                                   Exercise 1.4 (Solutions)
                                                         Calculus and Analytic Geometry, MATHEMATICS 12
   Merging man and maths                                  Available online @ http://www.mathcity.org, Version: 1.0.0


Question # 1:
       (i )   f ( x ) = 2x2 + x − 5                                c =1
        lim f ( x ) = lim ( 2 x 2 + x − 5 ) = 2 (1) + 1 − 5 = 2 + 1 − 5 = −2
                                                              2
           −             −
        x →1               x →1

        lim f ( x ) = lim ( 2 x 2 + x − 5 ) = 2 (1) + 1 − 5 = 2 + 1 − 5 = −2
                                                              2
           +             +
        x →1               x →1

        ⇒         lim f ( x ) = lim f ( x ) = −2                          ∴          lim f ( x ) = −2
                  x →1−1           +
                                         x →1                                         x →1

                                  x −9
                                    2
        ( ii )    f ( x) =                                C = −3
                                   x −3
                             x 2 − 9 xlim− ( x − 9 ) ( −3) − 9 9 − 9 0
                                               2          2

         lim f ( x ) = lim−         = →−3
                                                     =        =     =    =0
        x →−3−         x →−3 x − 3     lim− ( x − 3)   −3 − 3   −6    −6
                                                     x →−3


                                       x 2 − 9 xlim+ ( x − 9 ) ( −3) − 9 9 − 9 0
                                                         2          2

        Now        lim f ( x ) = lim+         = →−3
                                                               =        =     =    =0
                  x →−3+         x →−3 x − 3     lim+ ( x − 3)   −3 − 3   −6    −6
                                                              x →−3

        ⇒         lim f ( x ) = lim+ f ( x ) = 0                          ∴           lim f ( x ) = 0
                  x →−3−                  x →−3                                       x →−3

        ( iii )   f ( x) = x − 5                          C =5
        lim f ( x ) = lim− x − 5                                                                     x − 5 = ± ( x − 5)
        x →5−              x →5

                                                                                     − ( x − 5)                    + ( x − 5)
                                                                          −∞                             5                      +∞

        = lim  − ( x − 5 )  = − lim ( x − 5 ) = − ( 5 − 5 ) = 0
                           
          x →5 −                  x →5

        lim f ( x ) = lim x − 5 = lim ( x − 5 ) = 5 − 5 = 0
        x →5+            +
                           x →5      +
                                                  x →5



        ⇒         lim f ( x ) = lim+ f ( x ) = 0
                  x →5−                  x →5

                                  lim f ( x ) = 0
                                  x →5


Question # 2:
       Discuss the continuity of                          f ( x ) at x = c
                        2 x + 5        if                    x≤2
                        
       (i )   f ( x ) = 4 x + 1        if                    x>2
                                                             c=2
                        
       We haveto discuss the continuity of                    f ( x)      at x = 2
        (a)       f ( 2 ) = 2 ( 2) + 5 = 4 + 5 = 9                        .................... (1)
        (b )      lim f ( x ) = ?
                  x→2

                                                   f ( x) = 2x + 5                           f ( x) = 4x +1
                                                −∞                           2                               +∞
        lim f ( x ) = lim ( 2 x + 5 ) = 2 ( 2 ) + 5 = 4 + 5 = 9
        x →2 −             x→2

        and       lim f ( x ) = lim ( 4 x + 1) = 4 ( 2 ) + 1 = 8 + 1 = 9
                  x →2 +                 x→2

       ∴          lim f ( x ) = 9 ............... ( 2 )
                  x→2
                                                                                                                         FSc-II/Ex 1.4-2

       (c)       from     (1) and ( 2 )                           we get
       lim f ( x ) = f ( 2 )
       x→2

       ∴         f ( x ) is continous at x = 2
                           3 x − 1                      if x < 1
                           
       ( ii )    f ( x ) = 4                            x =1 c = 2
                           2 x                          x >1
                           
       if        c=2                        f (c) = f ( 2)
       is not defined so given function is discontinous
       ( ii )                  Correction
                           3 x − 1                      if x < 1
                           
                 f ( x ) = 4                            if x = 1
                           2 x                          if x > 1
                           
       c =1      ( correction )
                                                                                      f (1) = 4
                                                       −∞                                 1                        +∞
                                                          f ( x ) = 3x −1                                  f ( x) = 2x
       (a)       f (1) = 4                            ( given )
       (b )      lim f ( x ) = ?
                 x →1

       lim f ( x ) = lim ( 3 x − 1) = 3 (1) − 1 = 2
          −
       x →1             x →1

       and       lim f ( x ) = lim ( 2 x ) = 2 (1) = 2
                    +
                 x →1               x →1

       ⇒         lim f ( x ) = lim f ( x ) = 2
                    −             +
                 x →1               x →1

       ∴         lim f ( x ) = 2 ............................. ( 2 )
                 x →1

       (c)       From     (1) and ( 2 )                           we get
                 lim f ( x ) ≠ f (1)
                  x →1

       ∴          f ( x ) is discontinous at                       x =1
                                        3 x − 1                      if x < 1
                                        
       ( iii )                 f ( x) =                                                    c =1
                                        2 x                          if x > 1
                                        
       (a)       f (1)         is not defined
       ∴         f ( x ) is discontinous at x = 1

Question # 3:
       Given that
                 3 x    if                  x ≤ −2
                  2
       f ( x ) =  x − 1 if                  −2 < x < 2
                 3      if                  x≥2
                 
           −∞                                                                                         +∞
        f ( x ) = 3x           −2                   f ( x) = x −1 2
                                                                                        2          f ( x) = 3
       (i )      We check continuity at x = 2
       (a)       f ( 2) = 3                ................................ (1)   ( given )




                                                                                              Available online at http://www.MathCity.org
FSc-II/Ex 1.4-3

       (b )          lim f ( x ) = ?
                      x→2

        lim f ( x ) = lim ( x 2 − 1) = ( 2 ) − 1 = 4 − 1 = 3
                                                         2
           −                   x→2
        x →2

        lim f ( x ) = lim ( 3) = 3
        x →2 +                 x→2

       ⇒              lim f ( x ) = lim+ f ( x ) = 3
                      x →2 −             x→2

       ∴             lim f ( x ) = 3 ....................... ( 2 )
                      x→2

       (c)            From (1) and ( 2 ) ,                   we get
       lim f ( x ) = f ( 2 )
        x→2

       ∴              f ( x ) is continuous at                           x=2
       ( ii )         At         x = −2
       (a)            f ( −2 ) = 3 ( −2 ) = −6 ............................ (1)
       (b )           lim f ( x ) = ?
                      x →−2

        lim f ( x ) = lim ( 3 x ) = 3 ( −2 ) = −6
        x →−2−                 x →−2

                      lim f ( x ) = lim ( x 2 − 1) = ( −2 ) − 1 = 4 − 1 = 3
                                                                          2
        and
                      x →−2+              x →−2

       ⇒              lim f ( x ) ≠ lim+ f ( x )                                  ⇒ lim f ( x )   does not exist
                      x →−2−              x →−2                                      x →−2

      ∴        f ( x ) is discontinuous at                               x = −2
Question # 4:
       Given that
                x + 2                            x ≤ −1
       f ( x) = 
                c + 2                            x > −1
       c=?
                      −∞                                                                            +∞
                              f ( x) = x + 2                     −1                   f ( x) = c + 2
       Q              lim f ( x )               exists
                      x →−1

       ∴              lim f ( x ) = lim+ f ( x )
                      x →−1−              x →−1

       ⇒              lim ( x + 2 ) = lim ( c + 2 )
                      x →−1                x →−1

       ⇒            −1 + 2 = c + 2
       ⇒            1 = c+2
       ⇒            c = 1− 2      ⇒                          c = −1

Question # 5:
       (i )
                           mx       if              x<3
                           
                 f ( x ) = n        if              x=3
                           −2 x + 9 if              x>3
                           
               here                  f ( 3) = n              ( given )
       Q              f ( x ) is continuous at                           x =3
       ∴              lim f ( x ) = lim f ( x ) = f ( 3)
                      x →3−            +
                                         x →3

       ⇒             lim ( mx ) = lim ( −2 x + 9 ) = n
                      x →3              x →3




                                                                                             Available online at http://www.MathCity.org
                                                                                                                          FSc-II/Ex 1.4-4
       ⇒         ( m )( 3) = −2 ( 3) + 9 = n
       ⇒         3m = −6 + 9 = n
       ⇒         3m = 3 = n
       ⇒         3m = 3         ,                    n=3
       ⇒         m =1           ,                    n=3
                          mx     if                  x<4
       ( ii )    f ( x) =  2
                          x      if                     x≥4
                  f ( 4 ) = ( 4 ) = 16
                                      2
       here
      Q           f ( x ) is continuous at                               x=4
      Q           lim f ( x ) = lim+ f ( x ) = f ( 4 )
                 x →4 −                   x→4

       ⇒         lim ( mx ) = lim ( x 2 ) = 16
                  x→4                 x→4

       ⇒         4m = ( 4 ) = 16
                                 2


       ⇒         4m = 16 = 16                        ⇒                   4m = 16
       ⇒         m=4

Question # 6:

       Given that
                 2x + 5 − x + 7
                                                                         x≠2
       f ( x) =      x−2
                                                                         x=2
                       K
                      K =?
       here f ( 2 ) = K      given
      Q           f ( x ) is continuous at                               x=2
      Q          lim f ( x ) = f ( 2 )
                  x→2

                2x + 5 − x + 7                                                 2x + 5 − x + 7   2x + 5 + x + 7
       lim                     =K                    ⇒                   lim                                   =K
       x→2          x−2                                                  x→2       x−2          2x + 5 + x + 7

                           (                ) −(             )
                                                2                2
                                2x + 5              x+7                                       ( 2 x + 5) − ( x + 7 )
       ⇒                                                                 =K     ⇒                                    =K
                          ( x − 2) (                                 )                  ( x − 2) ( 2x + 5 + x + 7 )
                 lim
                  x→2
                                           2x + 5 + x + 7

                               2x + 5 − x − 7                                                        ( x − 2)
       ⇒                                                     =K                 ⇒                                    =K
                  ( x − 2) (         2x + 5 + x + 7      )                              ( x − 2) ( 2x + 5 + x + 7 )
                                  1                                                        1
       ⇒         lim                       =K                            ⇒                                 =K
                  x→2       2x + 5 + x + 7                                                            
                                                                                 lim  2 x + 5 + x + 7 
                                                                                 x →2

                                 1                                                  1
       ⇒                                            =K                   ⇒              =K
                    2 ( 2) + 5 + 2 + 7                                             9+ 9
                  1
       ⇒              =K
                 3+3
                     1
       ⇒         K=
                     6

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Description: Maths Solved Exercises for F.Sc part 2