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Question

If f(x)=2sin2x+2sinx+3sin2x+sinx+1, then the number of integer(s) in the range of f is

A
7
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B
3
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C
1
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D
4
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Solution

The correct option is C 1
f(x)=2sin2x+2sinx+3sin2x+sinx+1
=2+1sin2x+sinx+1
=2+1(sinx+12)2+34
Clearly, domain of f is R.

Now, fmin=2+194+34=73
and fmax=2+13/4=103

Hence, range of f is [73,103]
Only one integral value i.e., 3.

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