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Question

If the real-valued function f(x)=px+sinx is a bijective function then the set of possible values of pR is

A
R{0}
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B
R
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C
(0,+)
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D
|p|>1
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Solution

The correct option is A |p|>1
f(x)=px+sinx
f(x)=p+cosx
For given function to be bijective,
f(x)>0xR|p|>1

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