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

If the function f(x) defined by f(x)=x100100+x9999+...+x22+x+1, then f'(0) is equal to:


A

100f'(0)

No worries! We‘ve got your back. Try BYJU‘S free classes today!
B

100

No worries! We‘ve got your back. Try BYJU‘S free classes today!
C

1

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D

-1

No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C

1


Explanation for the correct option

Given function, f(x)=x100100+x9999+...+x22+x+1

Upon differentiating, we get,

f'(x)=100x99100+99x9899+...+2x2+1+0 ddxxn=nxn-1

f'(x)=x99+x98...+x+1

Now we need to evaluate for x=0, so upon substituting we get,

f'(0)=0+0+...+0+1f'(0)=1

Hence, option (C), i.e. 1 is the correct answer.


flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Laws of Indices
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon