JEE Main 202621 Jan, Evening shiftMathematicsEasyComplex Numbers and Quadratic Equations264× since 2002Let α\alphaα and β\betaβ be the roots of the equation x2+2ax+(3a+10)=0x^2+2ax+(3a+10)=0x2+2ax+(3a+10)=0 such that α<1<β\alpha<1<\betaα<1<β. Then the set of all possible values of a is :AOption A. (−∞,−2)∪(5,∞)(-\infty, -2)\cup(5, \infty)(−∞,−2)∪(5,∞)BOption B. (−∞,−115)∪(5,∞)\left(-\infty, \frac{-11}{5}\right)\cup(5, \infty)(−∞,5−11)∪(5,∞)COption C. (−∞,−115)\left(-\infty, \frac{-11}{5}\right)(−∞,5−11)DOption D. (−∞,−3)(-\infty, -3)(−∞,−3)CheckSkip · show solution
