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Question

The function f(x)=1+x(sinx)cosx, 0<xπ2


A

is continuous on 0,π2

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B

is strictly increasing on 0,π2

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C

is strictly decreasing on 0,π2

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D

None of these

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Solution

The correct option is A

is continuous on 0,π2


Explanation for the correct option

Given function, f(x)=1+x(sinx)cosx

Since, 0<xπ2

0cosx<1

So, [cosx]=0

f(x)=1

So, f(x) is a constant function and is neither an increasing nor decreasing function.

Thus, f(x) is a continuous function.

Hence, correct option is A


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