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NDA 2017
Exam
NDA
Year
2017
Subject
Mathematics
Chapter
Continuity and Differentiability

NDA 2017 · Mathematics · Continuity and Differentiability

A function is defined as follows: f(x)=\ array cc - x x^2 , & x ≠ 0...

A function is defined as follows: f(\mathrm{x})=\left\{−xx2,x≠00,x=0\begin{array}{cc} -\frac{\mathrm{x}}{\sqrt{\mathrm{x}^2}}, & \mathrm{x} \neq 0 \\ 0, & \mathrm{x}=0 \end{array}\right. Which one of the following is correct in respect of the above function?

Options

  1. Af(x)f(\mathrm{x}) is continuous at x=0\mathrm{x}=0 but not differentiable at x=0\mathrm{x}=0
  2. Bf(x)f(x) is continuous as well as differentiable at x=0\mathrm{x}=0
  3. Cf(x)f(\mathrm{x}) is discontinuous at x=0\mathrm{x}=0
  4. DNone of the above

Correct answer

C. f(x)f(\mathrm{x}) is discontinuous at x=0\mathrm{x}=0

Step-by-step solution

f(x)=\left\{−xx2,x≠00,x=0\begin{array}{l}\frac{-x}{\sqrt{x^2}}, x \neq 0 \\ 0, x=0\end{array}\right.=\left\{−x∣x∣,x≠00,x=0.\begin{array}{l}\frac{-x}{|x|}, x \neq 0 \\ 0, x=0.\end{array}\right.\mathrm{f}(0+\mathrm{h})=\frac{-\mathrm{h}}{\mathrm{ h}}=-1 ; \mathrm{f}(0-\mathrm{h})=\frac{\mathrm{h}}{\mathrm{ h}}=1 So, f(x)\mathrm{f}(\mathrm{x}) is discontinuous at x=0\mathrm{x}=0.
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