Question

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

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

Answer 1

Answer: (D)

We have, $f\left(x\right)=1-x+x^2-x^3...-x^{99}+x^{100}\dots (i)$
Differentiating $(i)$ w.r.t. $x$, we get
$f^{\prime }\left(x\right)=0-1+2x-3x^2+...-99x^{98}+100x^{99}$
$\Rightarrow f^{\prime }\left(x\right)=-1+2x-3x^2+...-99x^{98}+100x^{99}$
$\therefore f^{\prime }\left(1\right)=-1+2-3+...-99+100$
$=\left(-1-3-\mathrm{5...}-99\right)+\left(2+4+...+100\right)$
$=-\left(1+3+5+...+99\right)+\left(2+4+...+100\right)$
$=-\frac{50}{2}\left\{2+\left(49\times 2\right)\right\}+\frac{50}{2}\left\{\left(2\times 2\right)+\left(49\times 2\right)\right\}$
$=-2500+2550=50$