Limits and Derivatives — MATHS STD 11 Science — Question

4. 50

Solution:

$\text{f}(\text{x})=1-\text{x}+\text{x}^2-\text{x}^3+\dots-\text{x}^{99}+\text{x}^{100}$

Differentiate both the sides with respect to x, we get 

$\text{f}'(\text{x})=\frac{\text{d}}{\text{dx}}(1-\text{x}+\text{x}^2-\text{x}^3+\dots-\text{x}^{99}+\text{x}^{100})$

$=\frac{\text{d}}{\text{dx}}(1)-\frac{\text{d}}{\text{dx}}(\text{x})+\frac{\text{d}}{\text{dx}}(\text{x}^2)+\frac{\text{d}}{\text{dx}}(\text{x}^3)+\dots-\frac{\text{d}}{\text{dx}}(\text{x}^{99})+\frac{\text{d}}{\text{dx}}(\text{x}^{100})$

$=0-1+\text{2x}-\text{3x}^2+\dots-\text{99x}^{98}+100\text{x}^{99}$

Putting x = 1, we get 

$\text{f}'(1)=-1+2-3+\dots-99+100$

$=(-1+2)+(-3+4)+(-5+6)+\dots+(-99+100)$

$=1+1+1+\dots+1(50$terms$)$

$=50$