Statement II: If f(x)=|sinx|+|cosx|, then 1≤f(x)≤√2.
A
Statement I is true, Statement II is true. Statement II is a correct explanation for Statement I
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B
Statement I is true, Statement II is true. Statement II is not a correct explanation for Statement I
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C
Statement I is true, Statement II is false
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D
Statement I is false, Statement II is true.
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Solution
The correct option is DStatement I is false, Statement II is true. We know that 1≤|sinx|+|cosx|≤√2 ⇒[|sinx|+|cosx|]=1 I=4π∫0[|sinx|+|cosx|]dx=4π∫01⋅dx=4π
Thus, Statement I is false, Statement II is true.