हिंदी

The Domain of the Function F ( X ) = √ ( X + 1 ) ( X − 3 ) X − 2 is (A) [−1, 2) ∪ [3, ∞) (B) (−1, 2) ∪ [3, ∞) (C) [−1, 2] ∪ [3, ∞) (D) None of These - Mathematics

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प्रश्न

The domain of the function \[f\left( x \right) = \sqrt{\frac{\left( x + 1 \right) \left( x - 3 \right)}{x - 2}}\] is

  

विकल्प

  • (a) [−1, 2) ∪ [3, ∞)

  • (b) (−1, 2) ∪ [3, ∞)

  • (c) [−1, 2] ∪ [3, ∞)

  • (d) None of these

     
MCQ
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उत्तर

(a) [−1, 2) ∪ [3, ∞) 

\[f\left( x \right) = \sqrt{\frac{\left( x + 1 \right) \left( x - 3 \right)}{x - 2}}\]
\[\text{ For f(x) to be defined, }  \]
\[(x - 2) \neq 0\]
\[ \Rightarrow x \neq 2 . . . (1)\]
\[\text{ Also, } \]
\[\frac{(x + 1)(x - 3)}{(x - 2)} \geq 0\]
\[ \Rightarrow \frac{(x + 1)(x - 3)(x - 2)}{(x - 2 )^2} \geq 0\]
\[ \Rightarrow (x + 1)(x - 3)(x - 2) \geq 0\]
\[ \Rightarrow x \in [ - 1, 2) \cup [3, \infty ) . . . . . (2)\]
\[\text{ From (1) and (2),} \]
\[x \in [ - 1, 2) \cup [3, \infty ) \]

 

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अध्याय 3: Functions - Exercise 3.6 [पृष्ठ ४५]

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आरडी शर्मा Mathematics [English] Class 11
अध्याय 3 Functions
Exercise 3.6 | Q 33 | पृष्ठ ४५
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