# Mark the Correct Alternative in of the Following: If F ( X ) = X 100 + X 99 + . . . + X + 1 Then F ′ ( 1 ) is Equal to - Mathematics

MCQ

Mark the correct alternative in of the following:
If $f\left( x \right) = x^{100} + x^{99} + . . . + x + 1$  then $f'\left( 1 \right)$ is equal to

#### Options

• 5050

•  5049

• 5051

• 50051

#### Solution

$f\left( x \right) = x^{100} + x^{99} + . . . + x + 1$

Differentiating both sides with respect to x, we get $f'\left( x \right) = \frac{d}{dx}\left( x^{100} + x^{99} + . . . + x + 1 \right)$
$= \frac{d}{dx}\left( x^{100} \right) + \frac{d}{dx}\left( x^{99} \right) + . . . + \frac{d}{dx}\left( x^2 \right) + \frac{d}{dx}\left( x \right) + \frac{d}{dx}\left( 1 \right)$
$= 100 x^{99} + 99 x^{98} + . . . + 2x + 1 + 0 \left( y = x^n \Rightarrow \frac{dy}{dx} = n x^{n - 1} \right)$
$= 100 x^{99} + 99 x^{98} + . . . + 2x + 1$

Putting x = 1, we get

$f'\left( 1 \right) = 100 + 99 + 98 + . . . + 2 + 1$
$= \frac{100\left( 100 + 1 \right)}{2} \left( S_n = \frac{n\left( n + 1 \right)}{2} \right)$
$= 50 \times 101$
$= 5050$

Hence, the correct answer is option (a).

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 7 | Page 48