Question

The domain of definition of  \[f\left( x \right) = \sqrt{\frac{x + 3}{\left( 2 - x \right) \left( x - 5 \right)}}\] is 

  

विकल्प

• (a) (−∞, −3] ∪ (2, 5)

• (b) (−∞, −3) ∪ (2, 5)

• (c) (−∞, −3) ∪ [2, 5]

• (d) None of these

   

Answer 1

(a) (−∞, −3] ∪ (2, 5) 

\[f\left( x \right) = \sqrt{\frac{x + 3}{\left( 2 - x \right) \left( x - 5 \right)}}\]

\[ \text{ For f(x) to be defined ,}  \]

\[\left( 2 - x \right)\left( x - 5 \right) \neq 0\]

\[ \Rightarrow x \neq 2, 5 . . . . (1)\]

\[\text{ Also } , \frac{\left( x + 3 \right)}{\left( 2 - x \right)\left( x - 5 \right)} \geq 0\]

\[ \Rightarrow \frac{\left( x + 3 \right)\left( 2 - x \right)\left( x - 5 \right)}{\left( 2 - x \right)^2 \left( x - 5 \right)^2} \geq 0\]

\[ \Rightarrow \left( x + 3 \right)\left( x - 2 \right)\left( x - 5 \right) \leq 0\]

\[ \Rightarrow x \in ( - \infty , - 3] \cup (2, 5) . . . . (2)\]

\[\text{ From (1) and (2),}  \]

\[x \in ( - \infty , - 3] \cup (2, 5)\]