Question

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

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

Answer 1

The correct option is B $50$

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

The differentiation of the given function is :

$f^{\prime }\left(x\right)=\frac{d}{dx}(1-x+x^2-x^3+...-x^{99}+x^{100})$

$=\frac{d}{dx}\left(1\right)-\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(x^2\right)-\frac{d}{dx}\left(x^3\right)+...-\frac{d}{dx}\left(x^{99}\right)+\frac{d}{dx}\left(x^{100}\right)$

$=0-1+2x-3x^2+...-99x^{98}+100x^{99}$

$\Rightarrow f^{\prime }\left(x\right)=-1+2x-3x^2+...-99x^{98}+100x^{99}$

For $x=1$, we get:

$\Rightarrow f^{\prime }\left(1\right)=-1+2-3+.......-99+100$

$\Rightarrow f^{\prime }\left(1\right)=(-1+2)+(-3+4)+\left)\right(-5+6)+....+(-99+100)$

$\Rightarrow f^{\prime }\left(1\right)=1+1+1+......+1\left(50terms\right)$

$\therefore f^{\prime }\left(1\right)=50$

So, the correct option is (d).