FINDING UNKNOWN COEFFICIENTS IN CONTINUOUS FUCNTION

Problem 1 :

Find the value of a if the function is continuous.

types-of-discontinutyq13.png

Solution :

Lim x->1-  f(x) = Lim x->1- 7x - 2

Applying the limit, we get

Lim x->1-  f(x) = 7(1) - 2

= 5 ----(1)

Lim x->1+  f(x) = Lim x->1+ ax2

Applying the limit, we get

Lim x->1+  f(x) = a(1)2

= a----(2)

(1) = (2)

a = 5

Problem 2 :

Find the value of a if the function is continuous.

types-of-discontinutyq14.png

Solution :

Lim x->2-  f(x) = Lim x->2- ax2

Applying the limit, we get

Lim x->2-  f(x) = a(2)2

= 4a ----(1)

Lim x->2+  f(x) = Lim x->2+ 2x + a

Applying the limit, we get

Lim x->2+  f(x) = 2(2) + a

= 4 + a----(2)

(1) = (2)

4a = 4 + a

4a - a = 4

3a = 4

a = 4/3

Problem 3 :

types-of-discontinutyq15.png

Solution :

lim x->1-  f(x) = Lim x->1- x + 1

Applying the limit, we get

Lim x->1-  f(x) = 1 + 1

= 2

Lim x->1+  f(x) = Lim x->1+ ax + b

Applying the limit, we get

Lim x->1+  f(x) = a(1) + b

= a + b

Equating left hand limit and right hand limit, we get

a + b = 2 ----(1)

lim x->2-  f(x) = Lim x->2- ax + b

Applying the limit, we get

Lim x->2-  f(x) = a(2) + b

= 2a + b

Lim x->2+  f(x) = Lim x->2+ 3x

Applying the limit, we get

Lim x->2+  f(x) = 3(2)

= 6

Equating the left hand limit and right hand limit, we get

2a + b = 6 ----(2)

(1) - (2)

a + b - (2a + b) = 2 - 6

-a = -4

a = 4

Applying the value of a in (1), we get

2(4) + b = 6

b = 6 - 8

b = -2

Problem 4 :

Find the values of a and b, if the function is continuous at x = 4.

finding-unknown-in-continuityq4

Solution :

lim x4- f(x) = lim x4-ax2-xApplying the limit, we get= a(4)2-4= 16a - 4 ----(1)lim x4+ f(x) = lim x4+x3+bxApplying the limit, we get= (4)3+b(4)= 64 + 4b ----(2)lim x4 f(x) = lim x4 6= 6 ----(3)lim x4- f(x) lim x4+ f(x) lim x4 f(x)

(1) = (3)

16a - 4 = 6

16a = 6 + 4

16a = 10

a = 10/16

a = 5/8

(2) = (3)

64 + 4b = 6

4b = 6 - 64

4b = -58

b = -58/4

b = -29/2

Problem 5 :

Find the values of a and b, if the function is continuous for all x.

finding-unknown-in-continuityq5.png

Solution :

lim x0- f(x) = lim x0-axApplying the limit, we get= a(0)+= 2b ----(1)lim x0+ f(x) = lim x0+2a+b=2a+b ----(2)lim x2- f(x) = lim x2-2a+b= 2a+b ----(3)lim x2+ f(x) = lim x2+3x-5=3(2)-5----(4)lim x0- f(x) lim x0+ f(x) lim x0 f(x) lim x2- f(x) lim x2+ f(x) lim x2 f(x)

(1) = (2)

2b = 2a + b

2b - b = 2a

b = 2a

(3) = (4)

2a + b = 1

Applying the value of b here, we get

2a + 2a = 1

4a = 1

a = 1/4

Applying the value of a in b = 2a

b = 2(1/4)

b = 1/2

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