# 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