The set of values of λ for which $x^2 - λx + sin^{-1}(sin4) > 0 $ for all x ∈ R , is |
$\phi $ (-2, 2) R none of these |
$\phi $ |
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. |