Advertisement Remove all ads

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

Advertisement Remove all ads
Advertisement Remove all ads
Advertisement Remove all ads
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

Advertisement Remove all ads

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
  Is there an error in this question or solution?

APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 30 Derivatives
Q 7 | Page 48

Video TutorialsVIEW ALL [1]

Advertisement Remove all ads
Share
Notifications

View all notifications


      Forgot password?
View in app×