Let alpha and beta be the roots of equation x2 + 2ax + (3a + 10) = 0 such that alpha < 1 < beta. Then the set of all possible values of a is:

$For ax2+bx+c=0 with a > 0 , if a point k lies strictly between the roots, f(k) 0, the parabola opens upward. For 1 to lie between the roots alpha and beta, the value of the function at x = 1 must be strictly less than 0. f(1) 0, this simplifies to f(k) 0 manually because a f(k) < 0 guarantees real distinct roots. Class 11 Maths: Quadratic Equations

Let alpha and beta be the roots of equation x2 + 2ax + (3a + 10) = 0 such that alpha < 1 < beta. Then the set of all possible values of a is:

$For ax2+bx+c=0 with a > 0 , if a point k lies strictly between the roots, f(k) 0, the parabola opens upward. For 1 to lie between the roots alpha and beta, the value of the function at x = 1 must be strictly less than 0. f(1) 0, this simplifies to f(k) 0 manually because a f(k) < 0 guarantees real distinct roots. Class 11 Maths: Quadratic Equations

Let alpha and beta be the roots of equation x2 + 2ax + (3a + 10) = 0 such that alpha < 1 < beta. Then the set of all possible values of a is:

$For ax2+bx+c=0 with a > 0 , if a point k lies strictly between the roots, f(k) 0, the parabola opens upward. For 1 to lie between the roots alpha and beta, the value of the function at x = 1 must be strictly less than 0. f(1) 0, this simplifies to f(k) 0 manually because a f(k) < 0 guarantees real distinct roots. Class 11 Maths: Quadratic Equations

Let alpha and beta be the roots of equation x2 + 2ax + (3a + 10) = 0 such that alpha < 1 < beta. Then the set of all possible values of a is:

$For ax2+bx+c=0 with a > 0 , if a point k lies strictly between the roots, f(k) 0, the parabola opens upward. For 1 to lie between the roots alpha and beta, the value of the function at x = 1 must be strictly less than 0. f(1) 0, this simplifies to f(k) 0 manually because a f(k) < 0 guarantees real distinct roots. Class 11 Maths: Quadratic Equations

JEE Main · Mathematics ↑ Rising

Quadratic Equations appeared 26 times across 3 years — 3% of Mathematics. This question is from Location of Roots.

Year	2026	2025	2024	Total
Questions	11	10	5	26

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Let $α$ $\alpha$ and $β$ $\beta$ be the roots of equation $x² + 2ax + (3a + 10) = 0$ $x^{2} + 2ax + (3a + 10) = 0$ such that $α < 1 < β$ $\alpha < 1 < \beta$. Then the set of all possible values of $a$ $a$ is:

Q: Let alpha and beta be the roots of equation x2 + 2ax + (3a + 10) = 0 such that alpha < 1 < beta. Then the set of all possible values of a is:

$For ax2+bx+c=0 with a > 0 , if a point k lies strictly between the roots, f(k) 0, the parabola opens upward. For 1 to lie between the roots alpha and beta, the value of the function at x = 1 must be strictly less than 0. f(1) 0, this simplifies to f(k) 0 manually because a f(k) < 0 guarantees real distinct roots. Class 11 Maths: Quadratic Equations

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

Solution & Explanation

Related Formula

For ax²+bx+c=0 with a > 0 , if a point k lies strictly between the roots, f(k) < 0.

Core Logic

Given $f(x) = x² + 2ax + (3a + 10)$ $f(x) = x^2 + 2ax + (3a + 10)$. Since the coefficient of $x²$ $x^2$ is $1 > 0$ $1 > 0$, the parabola opens upward. For $1$ $1$ to lie between the roots $α$ $\alpha$ and $β$ $\beta$, the value of the function at $x = 1$ $x = 1$ must be strictly less than $0$ $0$. $f(1) < 0$ $f(1) < 0$

Step 1: Evaluate Inequality

f(1) = 1² + 2a(1) + 3a + 10 < 0

1 + 2a + 3a + 10 < 0

$5a + 11 < 0$ $5a + 11 < 0$

a < -(11)/(5)

So, $a in (-\infty, -(11)/(5))$ .

Pattern Recognition

When a specified value $k$ $k$ lies between roots, $a \cdot f(k) < 0$ $a \cdot f(k) < 0$. If $a>0$ $a>0$, this simplifies to $f(k) < 0$ $f(k) < 0$. No need to check discriminant $Δ > 0$ $\Delta > 0$ manually because $a f(k) < 0$ $a f(k) < 0$ guarantees real distinct roots.

Chapter Mix

Class 11 Maths: Quadratic Equations

Reference Study Guides

Quadratic Equations

Let $α$ $\alpha$ and $β$ $\beta$ be the roots of equation $x² + 2ax + (3a + 10) = 0$ $x^{2} + 2ax + (3a + 10) = 0$ such that $α < 1 < β$ $\alpha < 1 < \beta$. Then the set of all possible values of $a$ $a$ is:

Solution & Explanation

Related Formula

Core Logic

Step 1: Evaluate Inequality

Pattern Recognition

Chapter Mix

Reference Study Guides

More Quadratic Equations Previous-Year Questions — Page 6

Solution

Core Logic

Step 1: Constraint on Numerator

Step 2: Conclusion

Chapter Mix

More Quadratic Equations Questions — jee_main_2026_21_jan_evening

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Let α$\alpha$ and β$\beta$ be the roots of equation x² + 2ax + (3a + 10) = 0$x^{2} + 2ax + (3a + 10) = 0$ such that α < 1 < β$\alpha < 1 < \beta$. Then the set of all possible values of a$a$ is:

Solution & Explanation

Related Formula

Core Logic

Step 1: Evaluate Inequality

Pattern Recognition

Chapter Mix

Reference Study Guides

More Quadratic Equations Previous-Year Questions — Page 6

Solution

Core Logic

Step 1: Constraint on Numerator

Step 2: Conclusion

Chapter Mix

More Quadratic Equations Questions — jee_main_2026_21_jan_evening

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Let $α$ $\alpha$ and $β$ $\beta$ be the roots of equation $x² + 2ax + (3a + 10) = 0$ $x^{2} + 2ax + (3a + 10) = 0$ such that $α < 1 < β$ $\alpha < 1 < \beta$. Then the set of all possible values of $a$ $a$ is: