Question 1. Show that f(x) = |x – 3| is continuous but not differentiable at x = 3.
Solution:
f(3) = 3 – 3 = 0
=
= 0
= 0
Since LHL = RHL, f(x) is continuous at x = 3.
Now,
= –1
= 1
Since (LHD at x = 3) ≠ (RHD at x = 3)
f(x) is continuous but not differentiable at x =3.
Question 2. Show that f (x) = x1/3 is not differentiable at x = 0.
Solution:
(LHD at x = 0) =
= Undefined
(RHD at x = 0) =
= Undefined
Clearly LHD and RHD do not exist at 0.
f(x) is not differentiable at x = 0.
Question 3. Show that
Solution:
(LHD at x = 3) =
= 12
RHD at x = 3 =
= 12
Since LHL = RHL
f(x) is differentiable at x = 3.
Question 4. Show that the function f is defined as follows is continuous at x = 2, but not differentiable thereat:
Solution:
f(2) = 2(2)2 – 2 = 6
= 8 – 2
= 6
= 6
Clearly LHL = RHL at x = 2
Hence f(x) is differentiable at x = 2.
Question 5. Discuss the continuity and differentiability of the function f(x) = |x| + |x -1| in the interval of (-1, 2).
Solution:
(LHD at x = 0) =
= 2
(RHD at x = 0) =
= 0
Thus, f(x) is not differentiable at x = 0.
Question 6. Find whether the following function is differentiable at x = 1 and x = 2 or not.
Solution:
(LHD at x = 1) =
= 1
(RHD at x = 1) =
= –1
Clearly LHD ≠ RHD at x = 1
So f(x) is not differentiable at x = 1.
(LHD at x = 2) =
= –1
(RHD at x = 2) =
= –1
Clearly LHL = RHL at x = 2
Hence f(x) is differentiable at x = 2.
Question 7(i). Show that
Solution:
(LHD at x = 0) =
= 0 × k
= 0
(RHD at x = 0)
= 0 × k
= 0
Clearly LHL = RHL at x = 0
Hence f(x) is differentiable at x = 0.
Question 7(ii) Show that
Solution:
(LHD at x = 0)
= Not defined
(RHD at x = 0)
= Not defined
Clearly f(x) is not differentiable at x = 0.
Question 7(iii) Show that
Solution:
(LHD at x = 0)
= Not defined
(RHD at x = 0)
= Not defined
Clearly f(x) is not differentiable at x = 0.
Question 8. Find the value of a and b so that the function
Solution:
(LHD at x = 1) =
= 5
(RHD at x = 2) =
= b
Since f(x) is differentiable at x = 1,so
b = 5
Hence, 4 + a = b + 2
or, a = 7 – 4 = 3
Hence, a = 3 and b = 5.
Question 9. Show that the function
Solution:
(LHD at x = 1) =
= 0
(RHD at x =1) =
= –2
Since (LHD at x = 1) ≠ (RHD at x = 1)
f(x) is continuous but not differentiable at x =1.
Question 10. If
Solution:
We know f(x) is continuous at x = 1.
So, a – b = 1 …..(1)
(LHD at x = 1) =
Using (1), we get
= 2a
(RHD at x =1)
= –1
Since f(x) is differentiable, LHL = RHL
or, 2a = –1
a = –1/2
Substituting a = –1/2 in (1), we get,
b = –1/2 – 1
b = –3/2