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Limits and Derivatives — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsLimits and Derivatives1 Mark
MCQ
Choose the correct answer. If $\text{f}(\text{x})=\text{x}-[\text{x}],\in\text{R}$ then $\text{f}'\big(\frac{1}{2}\big)$ is equal to:
  • A
    $\frac{3}{2}$
  • ✓
    $1$
  • C
    $0$
  • D
    $-1$
✓

Answer

Correct option: B.
$1$
Given $f(x) = x - [x]$
we have ti first check for differentiability of $f(x)$ at $\text{x}=\frac{1}{2}$
$\therefore \text{Lf}'\Big(\frac{1}{2}\Big)=\text{LHD}=\lim\limits_{\text{h} \rightarrow 0}\frac{\text{f}\big[\frac{1}{2}-\text{h}\big]-\text{f}\big[\frac{1}{2}\big]}{-\text{h}}$
$=\lim\limits_{\text{h} \rightarrow 0}\frac{\big(\frac{1}{2}-\text{h}\big)-\big[\frac{1}{2}-\text{h}\big]-\frac{1}{2}+\big[\frac{1}{2}\big]}{-\text{h}}$
$=\lim\limits_{\text{h} \rightarrow 0}\frac{\frac{1}{2}-\text{h}-0-\frac{1}{2}+0}{-\text{h}}$
$=\frac{-\text{h}}{-\text{h}}$
$=1$
$\therefore \text{Rf}'\Big(\frac{1}{2}\Big)=\text{RHD}=\lim\limits_{\text{h} \rightarrow 0}\frac{\text{f}\big[\frac{1}{2}-\text{h}\big]-\text{f}\big[\frac{1}{2}\big]}{\text{h}}$
$=\lim\limits_{\text{h} \rightarrow 0}\frac{\big(\frac{1}{2}+\text{h}\big)-\big[\frac{1}{2}+\text{h}\big]-\frac{1}{2}+\big[\frac{1}{2}\big]}{-\text{h}}$
$=\lim\limits_{\text{h} \rightarrow 0}\frac{\frac{1}{2}+\text{h}-1-\frac{1}{2}+1}{\text{h}}$
$=\frac{\text{h}}{\text{h}}$
$=1$
Since, $\text{LHD = RHD}$
$\text{f}'\big(\frac{1}{2}\big)=1$

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