The correct options are
B (2,∞)
C (−∞,−4/3)
2cotx2sin2x2=2cosx2sinx2=sinx
∴f(x)=(4λ−3)(x+log5)+(λ−7)sinx
f′x=(4λ−3)+(λ−7)cosx=0 ...(1)∴cosx=4λ−3λ−7 ...(2)Now −1≤cosx≤1∴−1≤4λ−3λ−7≤1 Above gives us two
inequalities −1−4λ−3λ−7≤0
or 4λ−3λ−7≤0
−5λ+10λ−7≤0 or 3λ+4λ−7≤0 or 5(λ−2)λ−7≥0 or 3(λ+4/3)λ−7≤0 Above are the
conditions for f(x) to have critical points. But the function does not possess critical points. Therefore we must have 5(λ−2)λ−7<0 or
3(λ+4/3)λ−7>0 or 5(λ−2)(λ−7)(λ−7)2<0 or
3(λλ+4/3)(λ−7)(λ−7)2>0
∴λϵ(2,7) or λ<−4/3 or >7
∴λϵ(2,7) and λϵ(−∞,−4/3)or(7,∞)
∴λϵ(2,∞) or λϵ(−∞,−4/3)