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For the function f(x) = x100100+x9999+...+x22+x+1 Prove that f'(1) = 100 f'(0)

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प्रश्न

For the function

f(x) = `x^100/100 + x^99/99 + ...+ x^2/2 + x + 1`

Prove that f'(1) = 100 f'(0)

योग
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उत्तर

The given function is

`f(x) = x^100/100 + x^99/99 + ....... + x^2/2 + x + 1`

∴ `d/(dx) f(x) = [(x^100)/100 + (x^99)/99 + .... + (x^2)/2 + x + 1]`

`d/(dx) f(x) = d/(dx)(x^100/100) + d/(dx)(x^99/99) + ... + d/(dx) (x^2/2) + d/(dx)(x) + d/(dx)(1)`

On using theorem `d/(dx)(x^n)` = `nx^(n - 1)`, we obtain

`d/(dx) f(x)` = `(100x^99)/100 + (99^98)/99 + ... + (2x)/2 + 1 + 0`

= x99 + x98 + ..... + x + 1

∴ f'(x) = `x^99 + x^98 + ..... + x + 1`

At x = 0,

f'(0) = 1

At x = 1,

f'(1) = `1^99 + 1^98 + ... + 1 + 1 = [1 + 1 + ... + 1 + 1]_(100 "terms")` = 1 × 100 = 100

Thus, f'(1) = 100 × f'(0)

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अध्याय 12: Limits and Derivatives - EXERCISE 12.2 [पृष्ठ २४८]

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एनसीईआरटी Mathematics [English] Class 11
अध्याय 12 Limits and Derivatives
EXERCISE 12.2 | Q 5. | पृष्ठ २४८

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