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If $ sin^{-1}(sin 5) > x^2 - 4x , $ then |
$x ∈ ( 2 + \sqrt{9-2\pi}, ∞ )$ $x ∈ ( 2 - \sqrt{9-2\pi}, 2+\sqrt{9-2\pi} )$ $x ∈ ( 2 - \sqrt{9-2\pi}, ∞ )$ none of these |
$x ∈ ( 2 - \sqrt{9-2\pi}, 2+\sqrt{9-2\pi} )$ |
We have, $ sin^{-1}(sin 5) > x^2 - 4x $ $⇒ sin^{-1}(sin(5-2\pi)) > x^2 - 4 x$ $⇒ 5 - 2 \pi > x^2 - 4x $ $⇒ x^2 - 4x - 5 + 2 \pi < 0 ⇒2 - \sqrt{9-2 \pi} < x < 2 + \sqrt{9-2 \pi}$ |
