Question 14 - CBSE Class 12 Sample Paper for 2018 Boards - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards
Last updated at April 16, 2024 by Teachoo

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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

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