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Let F ( X ) = ⎧ ⎨ ⎩ 1 , X ≤ − 1 | X | , − 1 < X < 1 0 , X ≥ 1 Then, F is (A) Continuous at X = − 1 (B) Differentiable at X = − 1 (C) Everywhere Continuous - Mathematics

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Question

Let \[f\left( x \right) = \begin{cases}1 , & x \leq - 1 \\ \left| x \right|, & - 1 < x < 1 \\ 0 , & x \geq 1\end{cases}\] Then, f is 

Options

  • continuous at x = − 1

  • differentiable at x = − 1

  • everywhere continuous

  • everywhere differentiable

MCQ
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Solution

(b) differentiable at x = − 1 

\[f\left( x \right) = \begin{cases}1 , & x \leq - 1 \\ \left| x \right|, & - 1 < x < 1 \\ 0 , & x \geq 1\end{cases}\]

Differentiabilty at x = − 1

(LHD x = − 1)

\[{lim}_{x \to - 1^-} \frac{f(x) - f( - 1)}{x + 1}\]
\[ = {lim}_{x \to - 1} \frac{f(x) - f( - 1)}{x + 1}\]
\[ = {lim}_{x \to - 1} \frac{1 - 1}{- 1 + 1}\]
\[ = 0\]

(RHD x = − 1)

\[= {lim}_{x \to - 1^+} \frac{f (x) - f( - 1)}{x + 1} \]
\[ = {lim}_{x \to - 1} \frac{f (x) - f( - 1)}{x + 1} \]
\[ = {lim}_{x \to - 1} \frac{f (x) - f( - 1)}{x + 1}\]
\[ = {lim}_{x \to - 1} \frac{|x| - | - 1|}{x + 1}\]
\[ = \frac{1 - 1|}{- 1 + 1}\]
\[ = 0 \]

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Chapter 10: Differentiability - Exercise 10.4 [Page 20]

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RD Sharma Mathematics [English] Class 12
Chapter 10 Differentiability
Exercise 10.4 | Q 26 | Page 20

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