The set of values of λ for which $x^2 - λx + sin^{-1}(sin4) > 0 $ for all x ∈ R , is
Answer & explanation
Correct answer: option 1
We have,
$sin^{-1}(sin4) = sin^{-1} (sin (\pi - 4)) = \pi - 4 $
$∴ x^2 - λx + sin^{-1}(sin4) > 0 $ for all x ∈ R
$ ⇒x^2 - λx + (\pi - 4) > 0 $ for all x ∈ R
$ ⇒ λ^2 - 4(\pi - 4) < 0 ⇒λ^2 + 16 - 4 \pi < 0 $
But , $λ^2 + 16 - 4 \pi > 0 $ for all x ∈ R.
So, there is no value of λ for which the given inequation holds true.