If ${sin}^{- 1} (sin 5) > x^{2} - 4 x$ , then

x \in (2 + \sqrt{9 - 2 π}, \infty)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

x \in (2 - \sqrt{9 - 2 π}, 2 + \sqrt{9 - 2 π})

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

x \in (2 - \sqrt{9 - 2 π}, \infty)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

none of these

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $x \in (2 - \sqrt{9 - 2 π}, 2 + \sqrt{9 - 2 π})$
$s i n^{- 1} (s i n (5)) > x^{2} - 4 x$
$s i n^{- 1} (s i n (5 - 2 π)) > x^{2} - 4 x$
$5 - 2 π > x^{2} - 4 x$
$x^{2} - 4 x - (5 - 2 π) < 0$
$(x - 2)^{2} - 4 < 5 - 2 π$
$(x - 2)^{2} < 9 - 2 π$
$- \sqrt{9 - 2 π} < x - 2 < \sqrt{9 - 2 π}$
$2 - \sqrt{9 - 2 π} < x < 2 + \sqrt{9 - 2 π}$

Suggest Corrections

Similar questions

3 f (tan x) + 2 f (cot x) = x + 3

x \in (0, \frac{π}{2})

15 f (\sqrt{3}) =

f (x) = \frac{cos x}{[\frac{x}{2 π}] + \frac{1}{2}}

\forall

x

\in

[- 2 π, 2 π]

[\cdot]

{log}_{\sqrt{3}} (sin x + 2 \sqrt{2} cos x) > 2, - 2 π \leq x \leq 2 π,

x

{cot}^{2} \frac{π}{9} + {cot}^{2} \frac{2 π}{9} + {cot}^{2} \frac{4 π}{9}

Smaller area enclosed by the circle x² + y² = 4 and the line x + y = 2 is

A. 2 (π – 2)

B. π – 2

C. 2π – 1

D. 2 (π + 2)

Join BYJU'S Learning Program