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Limits and Derivatives — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsLimits and Derivatives1 Mark
MCQ
If $\text{f}(\text{x})=1-\text{x}+\text{x}^2-\text{x}^3+\dots-\text{x}^{99}+\text{x}^{100},$then $f'(1)$ equals
  • A
    $150$
  • B
    $-50$
  • C
    $-150$
  • ✓
    $50$
✓

Answer

Correct option: D.
$50$
$\text{f}(\text{x})=1-\text{x}+\text{x}^2-\text{x}^3+\dots-\text{x}^{99}+\text{x}^{100}$
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$)$
$=50$

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