Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Inverse Trigonometric Functions

Question:

The set of values of λ for which $x^2 - λx + sin^{-1}(sin4) > 0 $ for all x ∈ R , is 

Options:

$\phi $

(-2, 2)

R

none of these

Correct Answer:

$\phi $

Explanation:

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.