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The domain of the function f ( x ) = √ 2 − 2 x − x 2 is - Mathematics

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प्रश्न

The domain of the function

\[f\left( x \right) = \sqrt{2 - 2x - x^2}\] is
 

पर्याय

  • (a)  \[\left[ - \sqrt{3}, \sqrt{3} \right]\]

     

  • (b)  \[\left[ - 1 - \sqrt{3}, - 1 + \sqrt{3} \right]\]

     

  • (c) [−2, 2]

  • (d)  \[\left[ - 2 - \sqrt{3}, - 2 + \sqrt{3} \right]\]

     

MCQ

उत्तर

(b)  \[\left[ - 1 - \sqrt{3}, - 1 + \sqrt{3} \right]\]

\[f\left( x \right) = \sqrt{2 - 2x - x^2}\]
\[\text{ Since } , 2 - 2x - x^2 \geq 0, \]
\[ x^2 + 2x - 2 \leq 0\]
\[ \Rightarrow x^2 - 2x - 2 + 1 - 1 \leq 0\]
\[ \Rightarrow \left( x - 1 \right)^2 - \left( \sqrt{3} \right)^2 \leq 0\]
\[ \Rightarrow \left[ x - \left( - 1 - \sqrt{3} \right) \right]\left[ x - \left( - 1 + \sqrt{3} \right) \right] \leq 0\]
\[ \Rightarrow \left( - 1 - \sqrt{3} \right) \leq x \leq \left( - 1 + \sqrt{3} \right)\]
\[\text{ Thus, dom} (f) = \left[ - 1 - \sqrt{3}, - 1 + \sqrt{3} \right] . \]

 

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पाठ 3: Functions - Exercise 3.6 [पृष्ठ ४५]

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आरडी शर्मा Mathematics [English] Class 11
पाठ 3 Functions
Exercise 3.6 | Q 31 | पृष्ठ ४५

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