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Mark the Correct Alternative in of the Following: If F ( X ) = 1 + X + X 2 2 + . . . + X 100 100 Then F ′ ( 1 ) is Equal to

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Question

Mark the correct alternative in  of the following:
If\[f\left( x \right) = 1 + x + \frac{x^2}{2} + . . . + \frac{x^{100}}{100}\] then \[f'\left( 1 \right)\] is equal to 

Options

  • \[\frac{1}{100}\] 

  • 100         

  • 50        

MCQ
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Solution

\[f\left( x \right) = 1 + x + \frac{x^2}{2} + . . . + \frac{x^{100}}{100}\] 

Differentiating both sides with respect to x, we get 

\[f'\left( x \right) = \frac{d}{dx}\left( 1 + x + \frac{x^2}{2} + . . . + \frac{x^{100}}{100} \right)\]
\[ = \frac{d}{dx}\left( 1 \right) + \frac{d}{dx}\left( x \right) + \frac{d}{dx}\left( \frac{x^2}{2} \right) + . . . + \frac{d}{dx}\left( \frac{x^{100}}{100} \right)\]
\[ = \frac{d}{dx}\left( 1 \right) + \frac{d}{dx}\left( x \right) + \frac{1}{2}\frac{d}{dx}\left( x^2 \right) + . . . + \frac{1}{100}\frac{d}{dx}\left( x^{100} \right)\]
\[ = 0 + 1 + \frac{1}{2} \times 2x + . . . + \frac{1}{100} \times 100 x^{99} \left( y = x^n \Rightarrow \frac{dy}{dx} = n x^{n - 1} \right) \]
\[ = 1 + x + x^2 + . . . + x^{99}\]

Putting x = 1, we get

\[f'\left( 1 \right) = 1 + 1 + 1 + . . . + 1 \left( 100 \text{ terms } \right)\]
\[ = 100\]

Hence, the correct answer is option (b).

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Chapter 30: Derivatives - Exercise 30.7 [Page 48]

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R.D. Sharma Mathematics [English] Class 11
Chapter 30 Derivatives
Exercise 30.7 | Q 8 | Page 48

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