For the function: f (x) = x¹⁰⁰/100 + x⁹⁹/99 + .... + x²/2 + x + 1.
Prove that f' (1) = 100 f' (0)

Solution:

The given function is f (x) = x¹⁰⁰/100 + x⁹⁹/99 + .... + x²/2 + x + 1

f'(x) = d/dx [x¹⁰⁰/100 + x⁹⁹/99 + .... + x²/2 + x + 1]

f'(x) = d/dx (x¹⁰⁰/100) + d/dx (x⁹⁹/99) + .... + d/dx (x²/2) + d/dx (x) + d/dx (1) ....(1)

On using derivative formula d/dx (xⁿ) = nx^{n - 1}, we obtain

f'(x) = 100x⁹⁹/100 + 99x⁹⁸/99 + .... + 2x/2 + 1 + 0

= x⁹⁹ + x⁹⁸ + .... + x + 1

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

At x = 0,

f' (0) = 1

At x = 1,

f' (1) = (1)⁹⁹ + (1)⁹⁸ + .... + 1 + 1

= [1 + 1 + .... (100 terms) ]

= 1 x 100

= 100 f' (0)

Thus, f' (1) = 100 f' (0)

NCERT Solutions Class 11 Maths Chapter 13 Exercise 13.2 Question 5

For the function f (x) = x¹⁰⁰/100 + x⁹⁹/99 + .... + x²/2 + x + 1. Prove that f' (1) = 100 f' (0).

Summary:

For the function f (x) = x¹⁰⁰/100 + x⁹⁹/99 + .... + x²/2 + x + 1, we proved that f' (1) = 100 f' (0)