Let $A = \{x:0 < x <π/2\}$ and f: R→ A be an onto function given by $f(x) = \tan^{-1}(x^2 + x + λ)$, where $λ$ is a constant. Then,
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → $λ≥1/4$
Since, $f: R→ A$ is an onto function. Therefore,
Range of f = A
$⇒0≤f(x)≤\frac{π}{2}$ for all $x∈R$
$⇒0 ≤ \tan^{-1}(x^2 + x + λ) ≤π/2$ for all $x∈R$
$⇒0 ≤x^2 + x + λ≤∞$ for all $x∈R$
$⇒x^2 + x + λ≥0$
$⇒1-4λ≤0⇒λ≥\frac{1}{4}$