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The Function F (X) = Tan X is Discontinuous on the Set (A) {N π : N ∈ Z} (B) {2n π : N ∈ Z} (C) { ( 2 N + 1 ) π 2 : N ∈ Z } (D) { N π 2 : N ∈ Z } - Mathematics

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प्रश्न

The function f (x) = tan x is discontinuous on the set

 

विकल्प

  • {n π : n ∈ Z}

  • {2n π : n ∈ Z}

  • \[\left\{ \left( 2n + 1 \right)\frac{\pi}{2}: n \in Z \right\}\]

  • \[\left\{ \frac{n\pi}{2}: n \in Z \right\}\]

MCQ
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उत्तर

\[\left\{ \left( 2n + 1 \right)\frac{\pi}{2}: n \in Z \right\}\]

When

\[\tan\left( 2n + 1 \right)\frac{\pi}{2} = \tan\left( n\pi + \frac{\pi}{2} \right) = - \cot\left( n\pi \right)\]
 , it is not defined at the integral points.
 
\[\left[ n \in Z \right]\]

Hence, 

\[f\left( x \right)\]  is discontinuous on the set 
\[\left\{ \left( 2n + 1 \right)\frac{\pi}{2}: n \in Z \right\}\].
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अध्याय 9: Continuity - Exercise 9.4 [पृष्ठ ४५]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 9 Continuity
Exercise 9.4 | Q 24 | पृष्ठ ४५

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