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Question

Find the range of
f(x)=sin1(x2+x+1)

A
[π3,π2]
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B
[π3,π]
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C
[π2,π]
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D
[π6,π2]
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Solution

The correct option is A [π3,π2]
Consider the equation
x2+x+1=0
It does not have any real roots as b24ac<0
Since b24ac<0, x2+x+1>0 for all xϵR..(i)
sin1(x2+x+1)ϵ(0,π2)...(i)
sin1θϵ[π2,π2]
Hence θϵ[1,1]
Since x2+x+1>0
Hence, 0<x2+x+1<1
x2+x+1<1 implies
x2+x<0
x(x+1)<0
x>1 and x<0
xϵ(1,0)...(ii)
Now f(x)=x2+x+1
f(x)=2x+1=0
f(x)=x=12
Hence f(x) attains a local minimum at x=12.
Now sin1(x2+x+1)x=12
=sin1(1412+1)
=sin1(34)
=sin1(32)
=π3...(iii)
Hence from i ii and iii
yϵ[π3,π2]

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