Given below is a graph of a function f(x) then 0 I. f(x) exists everywhere for c in(1,3) II. lim x 2 f(x) exist.; lim x 0^ + f(x)=lim x 0^ - f(x) Which of the following options is correct? [SolveLancer Test] (a) Only I and II are true. (b) Only II and III are true. (c) I, II, III all are false. (d) I, II, III all are true.
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Answer:Correct option is
B
3 only
Left hand limit=lim
x→1
−
f(x)
=
h→0
lim
f(1−h)
=
h→0
lim
2+(1−h)=2+1=3
Right hand limit=
x→1
+
lim
f(x)
=
h→0
lim
f(1+h)
=
h→0
lim
2+(1+h)=2+1=3
So, the limit exists at x=1
Left hand limit=
x→0
−
lim
f(x)
=
h→0
lim
f(0−h)
=
h→0
lim
2−(0−h)=2−0=2
Right hand limit=
x→0
+
lim
f(x)
=
h→0
lim
f(0+h)
=
h→0
lim
2+(0+h)=2+0=2
So, LHL=RHL=f(0)
So, f(x) is continuous at x=0
Right hand limit=
h→0
lim
h
f(0+h)−f(0)
=
h→0
lim
h
(2+(0+h))−2
=
h→0
lim
h
h
=1
Left hand limit=
h→0
lim
−h
f(0−h)−f(0)
=
h→0
lim
−h
(2−(0−h))−(2)
=
h→0
lim
−h
h
=−1
LHL
=RHL
So, f(x) is not differentiable at x=0
The answer is option (B)
Step-by-step explanation:
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