What is the domain of (f(x)=\sqrt{x^2+4x-12})?फलन (f(x)=\sqrt{x^2+4x-12}) का डोमेन क्या है?
Answer and explanationउत्तर और व्याख्या
Correct answer:सही उत्तर: ((-\infty,-6]\cup[2,\infty))((-\infty,-6]\cup[2,\infty))
The inside expression is (x^2+4x-12=(x+6)(x-2)) and it must be (\ge0). In exams choose the outer intervals for an upward quadratic.अंदर की राशि (x^2+4x-12=(x+6)(x-2)) है और उसे (\ge0) होना चाहिए। परीक्षा में upward quadratic के लिए बाहरी अंतराल चुनें।
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