Class 11Mathematics4-2.Quadratic Equations and InequationsQuadratic expressions and Position of roots
DifficultLet $\alpha$ and $\beta$ be the roots of the equation $x^{2}+2ax+(3a+10) = 0$ such that $\alpha < 1 < \beta$. Then the set of all possible values of $a$ is:
- A$(-\infty, -11/5) \cup (5, \infty)$
- B$(-\infty, -2) \cup (5, \infty)$
- C$(-\infty, -3)$
- D$(-\infty, -11/5)$
Explore More
Similar Questions
The real number $k$ for which the equation $2x^2 + 3x + k = 0$ has two distinct real roots in the interval $[0, 1]$.
DifficultJEE MAIN 2013
View SolutionIf $ax^2 + bx + c < 0$ for all $x \in R$ and the expressions $cx^2 + ax + b$ and $ax^2 + bx + c$ have their extreme values at the same point $x$,then for the expression $cx^2 + ax + b$:
DifficultAP EAMCET 2025
View SolutionIf one root of $x^2+px-q^2=0$, where $p$ and $q$ are real, is less than $2$ and the other is greater than $2$, then:
If the maximum value of $2x - 7 - ax^2$ cannot exceed $20$,then the minimum value of $a$ is
If $f(x) = \frac{1}{4x^2 + 2x + 1}$,then its maximum value is
Medium
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo