Least value of n for which ( n − 2 ) x 2 + 8 x + n + 4 > sin − 1 sin ( 12 ) + cos − 1 cos ( 12 ) , ∀ x ∈ R where n ∈ N

sin − 1 sin ( 12 ) = − ( 4 π − 12 ) cos − 1 cos ( 12 ) = 4 π − 12 ( n − 2 ) x 2 + 8 x + n + 4 > 0∀ x ∈ R

Mathematics - Inverse Trigonometric Function Question with Solution | TestHub

MathematicsInverse Trigonometric FunctionProperties of ITFMedium2 minQB

MathematicsMediuminteger

Least value of n for which $(n - 2) x^{2} + 8 x + n + 4 > sin^{- 1} sin (12) + cos^{- 1} cos (12), \forall x \in R$ where $n \in N$

Answer:

Solution:

$sin^{- 1} sin (12) = - (4 π - 12) cos^{- 1} cos (12) = 4 π - 12$

$(n - 2) x^{2} + 8 x + n + 4 > 0\forall x \in R$

Stream:JEESubject:MathematicsTopic:Inverse Trigonometric FunctionSubtopic:Properties of ITF

⏱ 2mℹ️ Source: QB

Loading discussions...