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

Find 'c' of the mean value theorem, if f(x)=x(x1)(x2);a=0,b=1/2

A
c=1+216
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
c=1216
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
c=1+63
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
c=163
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B c=1216
We have f(a)=f(0)=0 and f(b)=f(12)=38
f(b)f(a)ba=380120=34
Now
f(x)=x33x2+2xf(x)=3x26x+2f(c)=3c26c+2
Putting all these value in lagrange's mean value theorem
f(b)f(a)ba=f(c),(a<c,b)
We get 34=3c26c+2c=1±216
Hence c=1216 lies in the open interval (0,12)
Therefore it is the required value

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