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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
0
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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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