Question

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

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

Answer 1

The correct option is A $50$

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

To find : $f^{{}^{\prime }}\left(1\right)$

$\therefore f^{{}^{\prime }}\left(x\right)=-1+2x-3x^2+4x^3-.....-99x^{98}+100x^{99}$

$(\frac{d{x}^n}{dx}=n{x}^{n-1})$

Put, $x=1$, we get

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

$\Rightarrow f^{{}^{\prime }}\left(1\right)=\frac{50}{2}[2\times (-1)+(50-1\left)\right(-2\left)\right]+\frac{50}{2}[2\times 2+(50-1)\times 2]$
$(\begin{array}{c}\text{Sum of n terms of an A.P}\\ S_n=\frac{n}{2}(2a+(n-1\left)d\right)\end{array})$

$\Rightarrow f^{{}^{\prime }}\left(1\right)=-2500+2550$
$\Rightarrow f^{{}^{\prime }}\left(1\right)=50$

Hence, the correct option is (d)