The function f x = π 4 + tan - 1 x , x ≤ 1 1 2 x - 1 , x > 1 is :

Mathematics - Continuity - Differentiability Question with Solution | TestHub

MathematicsContinuity - DifferentiabilityDifferentiabilityHard2 minPYQ_2020

MathematicsHardsingle choice

The function $f (x) = \{\begin{cases} \frac{π}{4} + \tan^{- 1} x, & |x| \leq 1 \\ \frac{1}{2} (|x| - 1), & |x| > 1 \end{cases}$ is :

Answer:

Solution:

$f (x) = \{\begin{cases} \frac{π}{4} + \tan^{- 1} x, & |x| \leq 1 \\ \frac{1}{2} (|x| - 1), & |x| > 1 \end{cases}$

Making graph

Clearly, function is continuous on $R - \{1\}$ and differentiable on $R - \{- 1, 1\}$ .

Stream:JEESubject:MathematicsTopic:Continuity - DifferentiabilitySubtopic:Differentiability

⏱ 2mℹ️ Source: PYQ_2020

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