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Question

If f′′(x)>0 and f(1)=0 such that g(x)=f(cot2x+2cotx+2) where 0<x<π then the interval in which g(x) is decreasing is

A
(0,π)
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B
(π2,π)
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C
(3π4,π)
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D
(0,3π4)
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Solution

The correct option is D (0,3π4)

g(x)
=f((1+cot(x))2+1).(2(1+cot(x))).cosec2x
Hence
g(x)<0 for g(x) to be a decreasing function.
f((1+cot(x))2+1).(2(1+cot(x))).cosec2x<0
Or
f((1+cot(x))2+1).(2(1+cot(x))).cosec2x>0
Now
cosec2x is always greater than 0.
Hence
f((1+cot(x))2+1).(2(1+cot(x)))>0
Or
(2(1+cot(x)))>0 ...(i) and f((1+cot(x))2+1)>0 ...(ii)
Or
2(1+cot(x))>0
cot(x)<1
x<ππ4
Or
x<3π4 ...(a)
Now consider ii,
f((1+cotx)2+1)
Let x=3π4
cot(x)=1
Hence
f(1)x=3π4
=0
Hence
x=3π4 is a critical point for f(x).
Thus
g(x) is decreasing on the interval (0,3π4)... from a.


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