Mathematics - Functions Question with Solution | TestHub

MathematicsFunctionsTypes of Function (Mapping)Medium2 minQB

MathematicsMediumsingle choice

Let $f : R \to [0. \frac{π}{2})$ be defined by $f (x) = Tan^{- 1} (x^{2} + x + a)$ . Then the set of values of a for which f is onto is

Answer:

Solution:

$x^{2} + x + a = 0$ has a real solution

$\Rightarrow 1 - 4 a \geq 0$

Stream:JEESubject:MathematicsTopic:FunctionsSubtopic:Types of Function (Mapping)

⏱ 2mℹ️ Source: QB

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