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JEE MATHEMATICS MCQ: FUNCTIONS

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Analyzing +4 | -1
The domain of definition of the function

f(x)=x⋅1+2(x+4)−0.52−(x+4)0.5+(x+4)0.5+4(x+4)0.5isf\left( x \right) = x \cdot \frac{{1 + 2{{\left( {x + 4} \right)}^{ - 0.5}}}}{{2 - {{\left( {x + 4} \right)}^{0.5}}}} + {\left( {x + 4} \right)^{0.5}} + 4{\left( {x + 4} \right)^{0.5}}\mathrm{is}
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Correct option: D
(−4,  0)∪(0,  ∞)\left( { - 4,\;0} \right) \cup \left( {0,\;\infty } \right)
The correct answer is D ((-4, 0) ∪ (0, ∞)) because it represents the values of x for which the square root is defined (x + 4 > 0) and the denominator is non-zero (x ≠ 0).

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