Question

फलन f(x) = `x^100/100 + x^99/99 + ....... + x^2/2 + x + 1` के लिए सिद्ध कीजिए कि f'(1) = 100 f'(0).

Answer 1

दिया गया फलन है

`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)`

प्रमेय `d/(dx)(x^n)` = `nx^(n - 1)` का उपयोग करने पर, हम प्राप्त करते हैं

`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`

x = 0 पर,

f'(0) = 1

x = 1 पर,

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

इस प्रकार, f'(1) = 100 × f'(0)