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Question

Let R be the set of real numbers and f:R→R be given by f(x)=√|x|−log(1+|x|). We now make the following assertions:I. There exists a real number A such that f(x)≤A for all x.II. There exists a real number B such that f(x)≥B for all x.

A
I is true and II is false
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B
I is false and II is true
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C
I and II both are true
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D
I and II both are false
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Solution

The correct option is B I is false and II is truef(x) will be minimum when √x is 0.⇒f(x)min=√|x|−log(1+|x|)=0−log1=0Minimum value of f(x) is 0, when x=0Thus, f(x)≥0And maximum value of f(x) is not defined.Therefore, B exist but A does not exist.

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