Question

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

• A. 150
  
• B. -50
  
• C. -150
  
• D. 50

Answer 1

The correct option is D 50

$f\left(x\right)=1-x+x^2-x^3+...-x^{99}+x^{100},$
Differentiating both sides with respect to x, we get
$f^{\prime }\left(x\right)==\frac{d}{dx}(1-x+x^2-x^3+...-x^{99}+x^{100}$
$=\frac{d}{dx}\left(1\right)-\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(x^2\right)-\frac{d}{dx}\left(x^3\right)+...-\frac{d}{dx}\left(x^{99}\right)+\frac{d}{dx}\left(x^{100}\right)$
$=0-1+2x-3x^2+...-99x^{98}+100x^{99}$
$=-1+2x-3x^2+...-99x^{98}+100x^{99}$
Putting x=1, we get
$f^{\prime }\left(1\right)=-1+2-3+...-99+100$
$=(-1+2)+(-3+4)+(-5+6)+...+(-99+100)$
$=1+1+1+...+1$ (50 terms)
=50
Hence, the correct answer is option (d).