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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Mathematics
Chapter
Continuity and Differentiability

NTA Abhyas 2020 · Mathematics · Continuity and Differentiability

The function f(x) = max\ (1 - x), (1 + x), 2\ x ∈ ℝ is

The function f(x)=max{(1−x),(1+x),2}∀x∈Rf\left(x\right) = max\left\{\left(1 - x\right), \left(1 + x\right), 2\right\} \forall x \in \mathbb{R} is

Options

  1. Adiscontinuous at exactly two points
  2. Bdifferentiable ∀x∈R\forall x \in \mathbb{R}
  3. Cdifferentiable ∀x∈R−{−1,1}\forall x \in \mathbb{R} - \left\{- 1, 1\right\}
  4. Dcontinuous ∀x∈R−{0,1,−1}\forall x \in \mathbb{R} - \left\{0, 1, - 1\right\}

Correct answer

C. differentiable ∀x∈R−{−1,1}\forall x \in \mathbb{R} - \left\{- 1, 1\right\}

Step-by-step solution

f\left(x\right) = \left\{\begin{cases}1 - x & : & x \le - 1 \\ 2 & : & - 1 1\end{cases}\right)<br />Continuity atx = - 1<br/><br />f\left(- 1\right) = 1 - \left(- 1\right) = 2<br/><br />f\left(- {1}^{-}\right) = 1 - \left(- 1\right) = 2<br/><br />f \left(- {1}^{+}\right) =22
∵ f\left(-1\right) = f\left(- {1}^{-}\right) =f(−1+)<br/>f\left(- {1}^{+}\right)<br />∴continuousatcontinuous atx = - 1<br/> <br />~f\left(1\right) = 2, f\left({1}^{-}\right) =2<br/><br />f \left({1}^{+}\right) = 1 + 1 = 2<br/><br />∵ f\left({1}^{-}\right) = f\left(1\right) = f\left({1}^{+}\right)</p><p></p><p>∴continuousatcontinuous atx = 1<br/>Fordifferentiability,<br/><br />For differentiability,<br />f^{′} \left(x\right) = \left\{{−1x1\begin{cases}- 1 & x 1\end{cases}\right)
at x=−1x = - 1,

f′(−1−)=−1,f(−1+)=0f^{'}\left(- {1}^{-}\right) = - 1, f\left(- {1}^{+}\right) = 0
∵f′(−1−)≠f′(−1+)⇒∵ f^{'}\left(- {1}^{−}\right) \ne f^{'}\left(- {1}^{+}\right) ⇒non-differentiable
at x=1x = 1,

f′(1−)=0,f′(1+)=1f^{'}\left({1}^{-}\right) = 0, f^{'}\left({1}^{+}\right) = 1
∵f′(1−)≠f′(1+)⇒∵ f^{'}\left({1}^{−}\right) \ne f^{'}\left({1}^{+}\right) ⇒non-differentiable

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