(D) Given the quadratic equation $f(x) = x^{2}+2ax+(3a+10) = 0$. Since the roots $\alpha$ and $\beta$ satisfy $\alpha < 1 < \beta$, the value of the f

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

(D) Given the quadratic equation $f(x) = x^{2}+2ax+(3a+10) = 0$. Since the roots $\alpha$ and $\beta$ satisfy $\alpha Substituting $x=1$ into the equation: $f(1) = (1)^{2} + 2a(1) + (3a+10) $1 + 2a + 3a + 10 $5a + 11 $5a $a Therefore, the set of all possible values of $a$ is $(-\infty, -11/5)$.

Class 11 Mathematics 2 Complex Numbers and Quadratic Equations: Quadratic Equations Quadratic expressions and Position of roots

Difficult

Let $\alpha$ 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:

JEE MAIN 2026 All JEE MAIN

A
$(-\infty, -11/5) \cup (5, \infty)$
B
$(-\infty, -2) \cup (5, \infty)$
C
$(-\infty, -3)$
D
$(-\infty, -11/5)$

Explore More

Browse All

Question Bank

All Subjects

Class 11 Questions

Class 11

Mathematics Questions

Chapter

2 Complex Numbers and Quadratic Equations: Quadratic Equations

Subtopic

Quadratic expressions and Position of roots

Previous Year

JEE MAIN 2026 Paper All JEE MAIN years →

For what interval of $m$ do all roots of the quadratic equation $x^2 - 2mx + m^2 - 1 = 0$ lie between $-2$ and $4$?

Medium

View Solution

If $\alpha, \beta$ are the roots of $x^2 - 3x + a = 0, a \in R$ and $\alpha < 1 < \beta$,then :-

Medium

View Solution

The values of $a$ for which the equation $2x^2 - 2(2a + 1)x + a(a + 1) = 0$ has one root less than $a$ and the other root greater than $a$ are given by:

Medium

View Solution

If non-zero real numbers $b$ and $c$ are such that $\min \,f(x) > \max \,g(x)$,where $f(x) = x^2 + 2bx + 2c^2$ and $g(x) = -x^2 - 2cx + b^2$ for $x \in R$; then $\left| \frac{c}{b} \right|$ lies in the interval

DifficultJEE MAIN 2014

View Solution

If the roots of the equation $x^2 - 2ax + a^2 + a - 3 = 0$ are real and less than $3$,then

DifficultIIT 1999

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

Let $\alpha$ 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:

Similar Questions

For what interval of $m$ do all roots of the quadratic equation $x^2 - 2mx + m^2 - 1 = 0$ lie between $-2$ and $4$?

If $\alpha, \beta$ are the roots of $x^2 - 3x + a = 0, a \in R$ and $\alpha < 1 < \beta$,then :-

The values of $a$ for which the equation $2x^2 - 2(2a + 1)x + a(a + 1) = 0$ has one root less than $a$ and the other root greater than $a$ are given by:

If non-zero real numbers $b$ and $c$ are such that $\min \,f(x) > \max \,g(x)$,where $f(x) = x^2 + 2bx + 2c^2$ and $g(x) = -x^2 - 2cx + b^2$ for $x \in R$; then $\left| \frac{c}{b} \right|$ lies in the interval

If the roots of the equation $x^2 - 2ax + a^2 + a - 3 = 0$ are real and less than $3$,then

Vedclass Test Series

Exam Paper Generator

Online Exam Module