Question 14
Last updated at Dec. 21, 2017 by Teachoo

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Question 14
Find ‘a’ and ‘b’ , at if the function given by f(x) = {█(𝑎𝑥2+𝑏, 𝑖𝑓 𝑥<1@2𝑥+1, 𝑖𝑓 𝑥 ≥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 = 𝑓(1)
i.e. if lim┬(x→1^− ) 𝑓(𝑥) = lim┬(x→1^+ ) 𝑓(𝑥) = 𝑓(1)
lim┬(x→1^− ) 𝑓(𝑥)
= lim┬(x→1^− ) 𝑎𝑥^2+𝑏
Putting x = 1
= a. (1)2 + b
= a + b
lim┬(x→1^+ ) 𝑓(𝑥)
= lim┬(x→1^+ ) 2𝑥+1
Putting x = 1
= 2(1) +1
= 2 + 1
= 3
lim┬(x→1^+ ) 𝑓(𝑥)
= lim┬(x→1^+ ) 2𝑥+1
Putting x = 1
= 2(1) +1
= 2 + 1
= 3
And 𝑓(1)=2x+1=2(1)+1=3
Now,
lim┬(x→1^− ) 𝑓(𝑥) = lim┬(x→1^+ ) 𝑓(𝑥) = 𝑓(1)
a + b = 3 = 3
Thus,
a + b = 3
Now, we check differentiability
Check differentiability
f(x) = {█(𝑎𝑥2+𝑏, 𝑖𝑓 𝑥<1@2𝑥+1, 𝑖𝑓 𝑥 ≥1)┤
f(x) is a differentiable at x = 1 if
LHD = RHD
(𝑙𝑖𝑚)┬(h→0) (𝑓(𝑥) − 𝑓(𝑥 − ℎ))/ℎ
= (𝑙𝑖𝑚)┬(h→0) (𝑓(1) − 𝑓(1 − ℎ))/ℎ
= (𝑙𝑖𝑚)┬(h→0) ((2(1) + 1) − (𝑎(1−ℎ)^2+𝑏))/ℎ
= (𝑙𝑖𝑚)┬(h→0) (3 − 𝑎(1 − ℎ)^2 − 𝑏)/ℎ
From (1), a + b = 3
Putting b = 3 – a
= (𝑙𝑖𝑚)┬(h→0) (3 − 𝑎(1 − ℎ)^2 − (3 − 𝑎))/ℎ
= (𝑙𝑖𝑚)┬(h→0) (3 − 𝑎(1 − ℎ)^2 − 3 + 𝑎)/ℎ
= (𝑙𝑖𝑚)┬(h→0) (𝑎 − 𝑎(1 − ℎ)^2 )/ℎ
= (𝑙𝑖𝑚)┬(h→0) (𝑎 −𝑎(1^2 + ℎ^2 − 2ℎ))/ℎ
(𝑙𝑖𝑚)┬(h→0) (𝑓(𝑥 + ℎ) − 𝑓(𝑥))/ℎ
= (𝑙𝑖𝑚)┬(h→0) (𝑓(1 + ℎ) − 𝑓(1))/ℎ
= (𝑙𝑖𝑚)┬(h→0) (2(1 + ℎ) + 1 − (2(1) + 1))/ℎ
= (𝑙𝑖𝑚)┬(h→0) (2 + 2ℎ + 1 − (2 + 1))/ℎ
= (𝑙𝑖𝑚)┬(h→0) (3 + 2ℎ − 3)/ℎ
= (𝑙𝑖𝑚)┬(h→0) 2ℎ/ℎ
= (𝑙𝑖𝑚)┬(h→0) (2)
= 2= (𝑙𝑖𝑚)┬(h→0) (𝑎 − 𝑎 + 𝑎ℎ^2 + 2𝑎ℎ)/ℎ
= (𝑙𝑖𝑚)┬(h→0) (𝑎ℎ^2 + 2𝑎ℎ)/ℎ
= (𝑙𝑖𝑚)┬(h→0) 𝑎ℎ+2𝑎
= a(0) + 2a
= 2a
= (𝑙𝑖𝑚)┬(h→0) (𝑎 − 𝑎 + 𝑎ℎ^2 + 2𝑎ℎ)/ℎ
= (𝑙𝑖𝑚)┬(h→0) (𝑎ℎ^2 + 2𝑎ℎ)/ℎ
= (𝑙𝑖𝑚)┬(h→0) 𝑎ℎ+2𝑎
= a(0) + 2a
= 2a
= (𝑙𝑖𝑚)┬(h→0) (𝑎 − 𝑎 + 𝑎ℎ^2 + 2𝑎ℎ)/ℎ
= (𝑙𝑖𝑚)┬(h→0) (𝑎ℎ^2 + 2𝑎ℎ)/ℎ
= (𝑙𝑖𝑚)┬(h→0) 𝑎ℎ+2𝑎
= a(0) + 2a
= 2a
= (𝑙𝑖𝑚)┬(h→0) (𝑎 − 𝑎 + 𝑎ℎ^2 + 2𝑎ℎ)/ℎ
= (𝑙𝑖𝑚)┬(h→0) (𝑎ℎ^2 + 2𝑎ℎ)/ℎ
= (𝑙𝑖𝑚)┬(h→0) 𝑎ℎ+2𝑎
= 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) = {█((𝑥 + 𝑠𝑖𝑛 𝑥)/(𝑠𝑖𝑛 (𝑎+1)𝑥), 𝑖𝑓−𝜋<𝑥<0 @ @2, 𝑖𝑓 𝑥=0@ @2 (𝑒^sin𝑏𝑥 − 1)/𝑏𝑥,𝑖𝑓 𝑥>0@ )┤
f is continuous at x = 0
if L.H.L = R.H.L = 𝑓(0)
i.e. if lim┬(x→0^− ) 𝑓(𝑥) = lim┬(x→0^+ ) 𝑓(𝑥) = 𝑓(0)
lim┬(x→0^− ) 𝑓(𝑥)
= lim┬(x→0^− ) (𝑥 + 𝑠𝑖𝑛 𝑥)/(𝑠𝑖𝑛 (𝑎 + 1)𝑥)
Theory
If we put 0
It will become
(𝑥 + 𝑠𝑖𝑛 𝑥)/(𝑠𝑖𝑛 (𝑎 + 1)𝑥) = (0 + sin0)/sin〖 (𝑎+1) ×0〗
= (0 + 0)/sin0 = 0/0
Since it is of 0/0 form,
We use an identity
We use lim┬(x→0) (𝑠𝑖𝑛 𝑥)/𝑥 = 1
lim┬(x→0^+ ) 𝑓(𝑥)
= lim┬(x→0^+ ) 2 (𝑒^sin𝑏𝑥 − 1)/𝑏𝑥
Theory
If we put 0
It will become
2 (𝑒^sin𝑏𝑥 − 1)/𝑏𝑥 = 2(𝑒^sin0 − 1)/(𝑏 × 0)
= 2(1 − 1)/0 = 2 × 0/0
Since it is of 0/0 form,
We use an identity
We use lim┬(x→0) (𝑠𝑖𝑛 𝑥)/𝑥 = 1
and lim┬(x→0) (𝑒^𝑥 − 1)/𝑥 = 1
Taking x common from numerator
= lim┬(x→0^− ) 𝑥(1 + (𝑠𝑖𝑛 𝑥)/𝑥)/(𝑠𝑖𝑛 (𝑎 + 1)𝑥)
Multiplying and dividing (a + 1)x
in denominator
= lim┬(x→0^− ) 𝑥(1 + (𝑠𝑖𝑛 𝑥)/𝑥)/(𝑠𝑖𝑛 (𝑎 + 1)𝑥 × (𝑎 + 1)𝑥/(𝑎 + 1)𝑥)
= lim┬(x→0^− ) 𝑥(1 + (𝑠𝑖𝑛 𝑥)/𝑥)/((𝑎 + 1) 𝑥 × (𝑠𝑖𝑛 (𝑎 + 1)𝑥)/(𝑎 + 1)𝑥)
= lim┬(x→0^− ) ((1 + (𝑠𝑖𝑛 𝑥)/𝑥))/((𝑎 + 1) × (𝑠𝑖𝑛 (𝑎 + 1)𝑥)/(𝑎 + 1)𝑥)Multiplying and dividing sin bx
= lim┬(x→0^+ ) 2 (𝑒^sin𝑏𝑥 − 1)/𝑏𝑥 × sin𝑏𝑥/sin𝑏𝑥
= lim┬(x→0^+ ) 2 (𝑒^sin𝑏𝑥 − 1)/sin𝑏𝑥 × sin𝑏𝑥/𝑏𝑥
Using lim┬(x→0) (𝑠𝑖𝑛 𝑥)/𝑥 = 1
and lim┬(x→0) (𝑒^𝑥 − 1)/𝑥 = 1
= 2 ×1 ×1
= 2
= 1/((a + 1)) lim┬(x→0^− ) ((1 + (𝑠𝑖𝑛 𝑥)/𝑥))/( (𝑠𝑖𝑛 (𝑎 + 1)𝑥)/(𝑎 + 1)𝑥)
Using lim┬(x→0) (𝑠𝑖𝑛 𝑥)/𝑥 = 1
= 1/((a + 1)) ×((1 + 1))/( 1)
= 2/((a + 1))
And 𝑓(0)=2
Now,
lim┬(x→0^− ) 𝑓(𝑥) = lim┬(x→0^+ ) 𝑓(𝑥) = 𝑓(0)
2/((a + 1)) = 2 = 2
Thus,
2/(𝑎 + 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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