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Question

If f(x)=x+sinx;g(x)=ex;u=c+1c;v=cc1;(c>1), then

A
fg(u)<fg(v)
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B
gf(u)<gf(v)
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C
gf(u)>gf(v)
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D
fg(u)>fg(v)
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Solution

The correct option is C gf(u)>gf(v)
u=c+1c
u=1c+1+c
and v=1c1+c
Clearly, u<v and they lie in (0,1)
f(x)=1+cosx
For x(0,1)
f(x)>0, so f(x) is increasing .
Therefore,
f(u)<f(v)
We know that g(x) is decreasing function, so
gf(u)>gf(v)

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