JEE Mains · Maths · STD 12 - 1. relation and function
The domain of the definition of the function \(f\left( x \right) = \frac{1}{{4 - {x^2}}} + \log \,\left( {{x^3} - x} \right)\) is
- A \(\left( {1,2} \right) \cup \left( {2,\infty } \right)\)
- B \(\left( { - 1,0} \right) \cup \left( {1,2} \right) \cup \left( {3,\infty } \right)\)
- C \(\left( { - 1,0} \right) \cup \left( {1,2} \right) \cup \left( {2,\infty } \right)\)
- D \(\left( { - 2, - 1} \right) \cup \left( { - 1,0} \right) \cup \left( {2,\infty } \right)\)
Answer & Solution
Correct Answer
(C) \(\left( { - 1,0} \right) \cup \left( {1,2} \right) \cup \left( {2,\infty } \right)\)
Step-by-step Solution
Detailed explanation
\(f\left( x \right) = \frac{1}{{4 - {x^2}}} + {\log _{10}}\left( {{x^3} - x} \right)\) Let \({f_1} = \frac{1}{{4 - {x^2}}}\) and \({f_2} = {\log _{10}}\left( {{x^3} - x} \right)\) \( \Rightarrow 4 - {x^2} \ne 0\,\,\,\,\,\,\,\,\,\,\,\,\,{x^3} - x > 0\)…
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