If −2x2+2x+α<cosec−1(cosec 6)+tan−1(tan(−5))∀x∈R, then possible value(s) of α can be (where α∈Z)
A
−3
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B
0
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C
−2
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D
1
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Solution
The correct option is C−2 Given: −2x2+2x+α<cosec−1(cosec 6)+tan−1(tan(−5))
we know that cosec−1(cosecx)=x−2π;3π2<x<2π tan−1(tanx)=x+2π;−2π<x<−3π2
Then inequality becomes −2x2+2x+α<6−2π−5+2π ⇒−2x2+2x+α−1<0 ∵ax2+bx+c<0and a<0⇒D<0 ⇒4+4×2(α−1)<0 ⇒α<12