फलन (f(x)=\sqrt{x-2+4x-12}) का डोमेन क्या है?
What is the domain of (f(x)=\sqrt{x-2+4x-12})?
Explanation opens after your attempt
A. (\(-\infty,-6]\cup[2,\infty\))
Concept
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.
Why this answer is correct
The correct answer is A. (\(-\infty,-6]\cup[2,\infty\)). 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.
Exam Tip
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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