The range of f ( x ) = t a n − 1 ( x 2 + x + a ) ∀ x ϵ R is a subset of [ 0 , π 2 ) then the range of a is - Options: A . &nbsp[−√3,14] B . &nbsp(−π2,π2) C . &nbsp[−√3,−1] D . &nbsp[14,∞)

Answer: Option D : D t a n − 1 ( x 2 + x + a ) ≥ 0 ⇒ x 2 + x + a ≥ 0 ⇒ D ≤ 0 ⇒ 1 − 4 a ≤ 0 ⇒ a ≥ 1 4 ; D is discriminant of quadratic equation. ⇒ a ϵ [ 1 4 , ∞ )

The range of f(x)=tan−1(x2+x+a) ∀ xϵ R is a subset of&#

Question

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 -

Options:

A . [−√3,14]

B . (−π2,π2)

C . [−√3,−1]

D . [14,∞)

Answer: Option D
:
D

t a n^{- 1} (x^{2} + x + a) \geq 0 \Rightarrow x^{2} + x + a \geq 0

\Rightarrow D \leq 0 \Rightarrow 1 - 4 a \leq 0 \Rightarrow a \geq \frac{1}{4}

; D is discriminant of quadratic equation.

\Rightarrow a ϵ [\frac{1}{4}, \infty)

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