The range of fx-tan -1x2-x-a - x - is a subset of -0- -2 then range of a isA- -1-4- -B- -3-1- -2- -2

The correct option is A tan^ 1 (x^2+x+a)≥ 0 ⇒ x^2+x+a ≥ 0 ⇒ disc ≤ 0 ⇒ 1 4a ≤ 0 ⇒ a ≥1/4 ⇒ a ϵ [1/4,∞ ) ...

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Definition of Functions

The range of ...

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$T h e r a n g e o f f (x) = t a n^{- 1} (x^{2} + x + a) \forall x ϵ i s a s u b s e t o f [0, \frac{π}{2}) t h e n r a n g e o f a i s$

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The range of $f (x) = t a n^{- 1} (x^{2} + x + a)$ $\forall x ϵ$ R is a subset of $[0, \frac{π}{2})$ then the range of a is -

f (x) = t a n^{- 1} (x^{2} + x + a)

\forall x ϵ

[0, \frac{π}{2})

x ϵ [- 1, 1], then range of t a n^{- 1} (- x)

f (x) = {sin}^{- 1} (\frac{x^{2} + 1}{x^{2} + 2})

x ϵ [- 1, 1],

{tan}^{- 1} (- x)

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