If f(x)=1−sinxsin2x,x≠π2 is continuous at x=π2, then the value of f(π2) is
A
2
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B
12
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C
0
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D
1
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Solution
The correct option is C0 For continuity of f(x) at x=π2, we must have f(π2)=limx→π21−sinxsin2x(00form)
Using L Hospital's Rule =limx→π2−cosx2cos2x=−cosπ22cosπ=0