Question

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

   

Answer 1

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

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

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