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Question

The range of the function, f(x)=(1+sec1x)(1+cos1x) is

A
(,)
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B
(,0][4,)
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C
{0,(1+π)2}
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D
[1,(1+π)2]
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Solution

The correct option is D [1,(1+π)2]
f(x)=(1+sec1(x))(1+cos1(x))
Here the limiting component is cos1(x), since the domain of cos1(x) is [1,1].
Therefore,
f(1)=(1+0)(1+0)
=1
f(1)=(1+π)(1+π)
=(1+π)2
Hence range of f(x)=[1,(1+π)2]

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