Advertisement Remove all ads

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

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

Mark the correct alternative in  of the following:

If\[f\left( x \right) = 1 - x + x^2 - x^3 + . . . - x^{99} + x^{100}\]then \[f'\left( 1 \right)\] 

Options

  •  150       

  • −50                   

  • −150            

  • 50 

Advertisement Remove all ads

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
  Is there an error in this question or solution?

APPEARS IN

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

Video TutorialsVIEW ALL [1]

Advertisement Remove all ads
Share
Notifications

View all notifications


      Forgot password?
View in app×