Question 14
Find a and b , at if the function given by f(x) = { ( 2+ , <1@2 +1, 1) is differentiable x = 1f(x) is differentiable at x = 1 if it is continuous.
So, first we check if it is continuous.
Check continuous
f is continuous at x = 1
if L.H.L = R.H.L = (1)
i.e. if lim (x 1^ ) ( ) = lim (x 1^+ ) ( ) = (1)
lim (x 1^ ) ( )
= lim (x 1^ ) ^2+
Putting x = 1
= a. (1)2 + b
= a + b
lim (x 1^+ ) ( )
= lim (x 1^+ ) 2 +1
Putting x = 1
= 2(1) +1
= 2 + 1
= 3
lim (x 1^+ ) ( )
= lim (x 1^+ ) 2 +1
Putting x = 1
= 2(1) +1
= 2 + 1
= 3
And (1)=2x+1=2(1)+1=3
Now,
lim (x 1^ ) ( ) = lim (x 1^+ ) ( ) = (1)
a + b = 3 = 3
Thus,
a + b = 3
Now, we check differentiability
Check differentiability
f(x) = { ( 2+ , <1@2 +1, 1)
f(x) is a differentiable at x = 1 if
LHD = RHD
( ) (h 0) ( ( ) ( ))/
= ( ) (h 0) ( (1) (1 ))/
= ( ) (h 0) ((2(1) + 1) ( (1 )^2+ ))/
= ( ) (h 0) (3 (1 )^2 )/
From (1), a + b = 3
Putting b = 3 a
= ( ) (h 0) (3 (1 )^2 (3 ))/
= ( ) (h 0) (3 (1 )^2 3 + )/
= ( ) (h 0) ( (1 )^2 )/
= ( ) (h 0) ( (1^2 + ^2 2 ))/
( ) (h 0) ( ( + ) ( ))/
= ( ) (h 0) ( (1 + ) (1))/
= ( ) (h 0) (2(1 + ) + 1 (2(1) + 1))/
= ( ) (h 0) (2 + 2 + 1 (2 + 1))/
= ( ) (h 0) (3 + 2 3)/
= ( ) (h 0) 2 /
= ( ) (h 0) (2)
= 2= ( ) (h 0) ( + ^2 + 2 )/
= ( ) (h 0) ( ^2 + 2 )/
= ( ) (h 0) +2
= a(0) + 2a
= 2a
= ( ) (h 0) ( + ^2 + 2 )/
= ( ) (h 0) ( ^2 + 2 )/
= ( ) (h 0) +2
= a(0) + 2a
= 2a
= ( ) (h 0) ( + ^2 + 2 )/
= ( ) (h 0) ( ^2 + 2 )/
= ( ) (h 0) +2
= a(0) + 2a
= 2a
= ( ) (h 0) ( + ^2 + 2 )/
= ( ) (h 0) ( ^2 + 2 )/
= ( ) (h 0) +2
= a(0) + 2a
= 2a
Since f(x) is differentiable at x = 1
LHD = RHD
2a = 2
a = 1
Putting a = 1 in (1)
a + b = 3
1 + b = 3
b = 3 1 = 2
So, a = 1, b = 2
Question 14
Determine the values of a and b such that the following function is continuous at x = 0:
f(x) = { (( + )/( ( +1) ), < <0 @ @2, =0@ @2 ( ^sin 1)/ , >0@ )
f is continuous at x = 0
if L.H.L = R.H.L = (0)
i.e. if lim (x 0^ ) ( ) = lim (x 0^+ ) ( ) = (0)
lim (x 0^ ) ( )
= lim (x 0^ ) ( + )/( ( + 1) )
Theory
If we put 0
It will become
( + )/( ( + 1) ) = (0 + sin 0)/sin ( +1) 0
= (0 + 0)/sin 0 = 0/0
Since it is of 0/0 form,
We use an identity
We use lim (x 0) ( )/ = 1
lim (x 0^+ ) ( )
= lim (x 0^+ ) 2 ( ^sin 1)/
Theory
If we put 0
It will become
2 ( ^sin 1)/ = 2( ^sin 0 1)/( 0)
= 2(1 1)/0 = 2 0/0
Since it is of 0/0 form,
We use an identity
We use lim (x 0) ( )/ = 1
and lim (x 0) ( ^ 1)/ = 1
Taking x common from numerator
= lim (x 0^ ) (1 + ( )/ )/( ( + 1) )
Multiplying and dividing (a + 1)x
in denominator
= lim (x 0^ ) (1 + ( )/ )/( ( + 1) ( + 1) /( + 1) )
= lim (x 0^ ) (1 + ( )/ )/(( + 1) ( ( + 1) )/( + 1) )
= lim (x 0^ ) ((1 + ( )/ ))/(( + 1) ( ( + 1) )/( + 1) )Multiplying and dividing sin bx
= lim (x 0^+ ) 2 ( ^sin 1)/ sin /sin
= lim (x 0^+ ) 2 ( ^sin 1)/sin sin /
Using lim (x 0) ( )/ = 1
and lim (x 0) ( ^ 1)/ = 1
= 2 1 1
= 2
= 1/((a + 1)) lim (x 0^ ) ((1 + ( )/ ))/( ( ( + 1) )/( + 1) )
Using lim (x 0) ( )/ = 1
= 1/((a + 1)) ((1 + 1))/( 1)
= 2/((a + 1))
And (0)=2
Now,
lim (x 0^ ) ( ) = lim (x 0^+ ) ( ) = (0)
2/((a + 1)) = 2 = 2
Thus,
2/( + 1) = 2
This is only possible when a = 0
And since there is no b in the equation,
b can be of any value
a = 0 and b can be of any value

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo

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