Question

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

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

Answer 1

Answer: (D)

We have, $f\left(x\right)=1-x+x^{2}-x^{3}...-x^{99}+x^{100}\quad \ldots(i)$
Differentiating $(i)$ w.r.t. $x$, we get
$f'\left(x\right)=0-1+2x-3x^{2}+...-99x^{98}+100x^{99}$
$\Rightarrow f'\left(x\right)=-1+2x-3x^{2}+...-99x^{98}+100x^{99}$
$\therefore f'\left(1\right)=-1+2-3+...-99+100$
$= \left(- 1 - 3 - 5... - 99\right) + \left(2 + 4 +... + 100\right)$
$= -\left(1+ 3 + 5 +... + 99\right) + \left(2 + 4 +... + 100\right)$
$=-\frac{50}{2}\left\{2+\left(49\times2\right)\right\}+\frac{50}{2}\left\{\left(2\times2\right)+\left(49\times2\right)\right\}$
$=-2500+2550=50$