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For the Function F ( X ) = X 100 100 + X 99 99 + . . . + X 2 2 + X + 1 .

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प्रश्न

For the function \[f(x) = \frac{x^{100}}{100} + \frac{x^{99}}{99} + . . . + \frac{x^2}{2} + x + 1 .\]

 
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उत्तर

\[f'\left( x \right) = \frac{d}{dx}\left( \frac{x^{100}}{100} + \frac{x^{99}}{99} + . . . + \frac{x^2}{2} + x + 1 \right)\]
\[ = \frac{1}{100}\left( 100 x^{99} \right) + \frac{1}{99}\left( 99 x^{98} \right) + . . . + \frac{1}{2}\left( 2x \right) + 1 + 0\]
\[ = x^{99} + x^{98} + . . . + x + 1\]
\[f'\left( 1 \right) = 1^{99} + 1^{98} + . . . + 1 + 1\]
\[ = 99 + 1\]
\[ = 100\]
\[f'\left( 0 \right) = 0 + 0 + . . . + 0 + 1\]
\[ = 1\]
\[RHS = 100 f'\left( 0 \right)\]
\[ = 100\left( 1 \right)\]
\[ = 100\]
\[ = f'\left( 1 \right)\]
\[ = LHS\]
\[ \therefore f'\left( 1 \right) = 100 f'\left( 0 \right)\]

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अध्याय 30: Derivatives - Exercise 30.3 [पृष्ठ ३४]

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आर.डी. शर्मा Mathematics [English] Class 11
अध्याय 30 Derivatives
Exercise 30.3 | Q 26 | पृष्ठ ३४

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