If x2+2x+3<cos−1(cos4)+2cot−1(cot5)∀x∈Z, then number of integral value(s) of x is
A
0
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B
2
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C
3
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D
4
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Solution
The correct option is C3 Given: x2+2x+3<cos−1(cos4)+2cot−1(cot5)
We know that cos−1(cosx)=2π−x;π<x<2π cot−1(cotx)=x−π;π<x<2π
Then inequality becomes x2+2x+3<2π−4+2×(5−π) ⇒x2+2x−3<0 ⇒(x+3)(x−1)<0 ⇒x∈(−3,1)
since x∈Z ∴x=−2,−1,0