Question

फलन $f(x) = \dfrac{x^{100}}{100} + \dfrac{x^{99}}{99} + ... + \dfrac{x^2}{2} + x + 1$ के लिए सिद्ध कीजिए कि $f'(1) = 100 f'(0)$

Answer 1

Answer:

Step-by-step explanation:

दिया गया है -

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

$f'(x) = \dfrac{d}{dx} (\dfrac{x^{100}}{100} )+ \dfrac{d}{dx}(\dfrac{x^{99}}{99}) + ... + \dfrac{d}{dx}(\dfrac{x^2}{2}) + \dfrac{d}{dx}(x) + \dfrac{d}{dx}(1)\\\\\f'(x)=\dfrac{1}{100}\dfrac{d}{dx}(x^{100})+\dfrac{1}{99}\dfrac{d}{dx}(x^{99})+......+\dfrac{1}{2}\dfrac{d}{dx}(x^2)+\dfrac{d}{dx}(x)+\dfrac{d}{dx}(1)\\\\f'(x)=\dfrac{1}{100}*100x^{99}+\dfrac{1}{99}*99x^{98}+......+\dfrac{1}{2}*2x+1+0\\\\f'(x)=x^{99}+x^{98}+....+x+1$

$f'(1)=1+1+.......+1+1=100$(∵100 पद है।)

$f'(1)=100*1=100f'(0)$  (∵ f'(0)  =  1 )

अतः $f'(1)=100f'(0)$