The set of values of λ for which x2−λx+sin−1(sin4)>0 for all x∈R, is
ϕ
(-2, 2)
R
none of these
We have,
sin−1(sin4)=sin−1(sin(π−4))=π−4.∴x2−λx+sin−1(sin4)>0 for all x∈R⇒x2−λx+(π−4)>0 for all x∈R⇒λ2−4(π−4)<0⇒λ2+16−4π<0
But, λ2+16−4π>0 for all λ∈R.
So, there is no value of λ for which the given in equation holds true.