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 , 1 4 ] B . &nbsp ( − π 2 , π 2 ) C . &nbsp [ − √ 3 , − 1 ] D . &nbsp [ 1 4 , ∞ )

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

Discussion Forum : Relations And Functions Ii

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 .

[- \sqrt{3}, \frac{1}{4}]

B .

(\frac{- π}{2}, \frac{π}{2})

C .

[- \sqrt{3}, - 1]

D .

[\frac{1}{4}, \infty)

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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Discussion Forum : Relations And Functions Ii

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