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Question
Mark the correct alternative in of the following:
If\[f\left( x \right) = 1 - x + x^2 - x^3 + . . . - x^{99} + x^{100}\]then \[f'\left( 1 \right)\]
Options
150
−50
−150
50
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Solution
\[f\left( x \right) = 1 - x + x^2 - x^3 + . . . - x^{99} + x^{100}\]
Differentiating both sides with respect to x, we get
\[f'\left( x \right) = \frac{d}{dx}\left( 1 - x + x^2 - x^3 + . . . - x^{99} + x^{100} \right)\]
\[ = \frac{d}{dx}\left( 1 \right) - \frac{d}{dx}\left( x \right) + \frac{d}{dx}\left( x^2 \right) - \frac{d}{dx}\left( x^3 \right) + . . . - \frac{d}{dx}\left( x^{99} \right) + \frac{d}{dx}\left( x^{100} \right)\]
\[ = 0 - 1 + 2x - 3 x^2 + . . . - 99 x^{98} + 100 x^{99} \]
\[ = - 1 + 2x - 3 x^2 + . . . - 99 x^{98} + 100 x^{99}\]
Putting x = 1, we get
\[f'\left( 1 \right) = - 1 + 2 - 3 + . . . - 99 + 100\]
\[ = \left( - 1 + 2 \right) + \left( - 3 + 4 \right) + \left( - 5 + 6 \right) + . . . + \left( - 99 + 100 \right)\]
\[ = 1 + 1 + 1 + . . . + 1 \left( 50 \text{ terms } \right)\]
\[ = 50\]
Hence, the correct answer is option (d).
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