Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • For the function f x = x ^ 100/100 + x^99/99

5.   For the function f (x) = \frac{x^{100}}{100}+ \frac{x ^ {99}}{99} + ....+ \frac{x ^2 }{2}+ x+ 1 Prove that f '(1) =100 f '(0).

Answers (1)

best_answer

f (x) = \frac{x^{100}}{100}+ \frac{x ^ {99}}{99} + ....+ \frac{x ^2 }{2}+ x+ 1

As we know, the property,

f'(x^n)=nx^{n-1}

applying that property we get 

f '(x) = 100\frac{x^{99}}{100}+ 99\frac{x ^ {98}}{99} + ....+ 2\frac{x }{2}+ 1+ 0

f '(x) = x^{99}+x^{98}+......x+1

Now.

f '(0) = 0^{99}+0^{98}+......0+1=1

f '(1) = 1^{99}+1^{98}+......1+1=100

So,

Here

 1\times 100=100

f'(0)\times 100=f'(1)

Hence Proved.

Posted by

Pankaj Sanodiya

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Similar Questions

Trending Articles/News

anam-khan

CBSE board exams 2025 from February 15; Class 10, 12 subject-wise marks distribution
anam-khan

CBSE Date Sheet 2025: Class 10, 12 board exam dates soon on cbse.gov.in; previous trends, sample papers
anam-khan

CBSE Board Exams 2025: 75% attendance must for students to be eligible, reiterates board

Latest Question