wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

In which of the following functions Rolle's theorem is applicable?

A
f(x)={x,0x<10,x=1 on [0,1]
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
f(x)=sinxx,πx<00,x=0 on [π,0]
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
f(x)=x2x6x1 on [2,3]
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
f(x)=x32x25x+6x1,ifx16,ifx=1 on [2,3]
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

The correct option is D f(x)=x32x25x+6x1,ifx16,ifx=1 on [2,3]
Option A, f(x)={x,0x<10,x=1 on [0,1]
We will check continuity at x=1
LHL=limx1f(x)=limh0f(1h)=1
Also, f(1)=0
Since, LHLf(1)
f(x) is discontinuous at x=1
So, Rolle's theorem does not hold for option A.
Option B,f(x)=sinxx,πx<00,x=0 on [π,0]
We know that limx0sinxx=1
Also, f(0)=0
f(x) is discontinuous at x=0
So, Rolle's theorem does not hold for option B.
Option C, f(x)=x2x6x1 on [2,3]
f(x)=(x3)(x+2)x1 on [2,3]
f(x) is not defined at x=1
So, Rolle's theorem does not hold for option C.
f(x)=x32x25x+6x1,ifx16,ifx=1 on [2,3]
f(x)=(x3)(x+2) at x1
Polynomial functions are continuous and differentiable everywhere.
Also, f(2)=f(3)=0
So, Rolle's theorem holds for option D.

flag
Suggest Corrections
thumbs-up
0
similar_icon
Similar questions
View More
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Theorems
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon