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]∀n∈N
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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)