# Mark the correct alternative in of the following: If f ( x ) = 1 − x + x 2 − x 3 + . . . − x 99 + x 100 then f ′ ( 1 ) - Mathematics

MCQ

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

#### 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).

Concept: The Concept of Derivative - Algebra of Derivative of Functions
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#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 30 Derivatives
Q 4 | Page 48