Start Free Mock LoginAlready registered? Login here

Mathematics - Quadratic Equation Question with Solution | TestHub

MathematicsQuadratic EquationGraph and Sign of QuadraticMedium2 minQB

MathematicsMediumsingle choice

(L-1)The set of values of 'a' for which $x^{2} + ax + sin^{- 1} (x^{2} - 4 x + 5) + cos^{- 1} (x^{2} - 4 x + 5) = 0$ has at least one real solution is given by

Options:

Answer:

Solution:

Charly $x^{2} - 4 x + 5 = (x - 2)^{2} + 1$ , lies $b ∣ w - 1, 1. \Rightarrow x = 2$ is the only point of the domain,

It must be the solution. $∴ 4 + 2 a + \frac{π}{2} = 0 \Rightarrow a \Rightarrow \frac{- π - 8}{4}$

Stream:JEESubject:MathematicsTopic:Quadratic EquationSubtopic:Graph and Sign of Quadratic

⏱ 2mℹ️ Source: QB

Doubts & Discussion

Loading discussions...