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Question

The domain of definition of function f(x)=1+2(x+4)0.52(x+4)0.5+(x+4)0.5+4(x+4)0.5 is

A
R
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B
(4,4)
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C
R+
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D
(4,0)(0,)
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Solution

The correct option is C (4,0)(0,)
We have,
f(x)=1+2(x+4)0.52(x+4)0.5+(x+4)0.5+4(x+4)0.5
f(x)=1+2x+42x+4+x+4+4x+4
Clearly, f(x) is defined for x+4>0 and x0. So, domain of f(x) is (4,0)(0,).

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