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Question

The inequality ∣∣x2sinx+cos2xex+ln2x∣∣<x2|sinx|+cos2xex+ln2x is true for x∈

A
(π,0)
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B
(0,π2)
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C
(π2,π)
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D
[2nπ,(2n+1)π] nN
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Solution

The correct option is A (π,0)
|a+b+c|<|a|+|b|+|c|
The above inequality is valid only if either one of a,b or c has different signs.
here, excos2x and ln2x are both positive,
hence inequality will be valid only iff
x2sinx<0
sinx<0
x(π,0)

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