Question

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 :

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

Hint:

Whenever a fixed value $k$ lies between the roots of $f(x) = Ax^2 + Bx + C$, simply use $A \cdot f(k)<0$.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
For a quadratic $f(x) = x^2 + 2ax + (3a + 10)$, if the roots $\alpha$ and $\beta$ are such that $\alpha<1<\beta$, then the value of the function at $x = 1$ must be negative, as the parabola opens upwards.

Step 2: Detailed Explanation:
The condition for roots to lie on either side of $k$ is $a \cdot f(k)<0$. Here $a = 1>0$ and $k = 1$.
\[ f(1)<0 \]
\[ (1)^2 + 2a(1) + 3a + 10<0 \]
\[ 1 + 5a + 10<0 \]
\[ 5a + 11<0 \implies a<-\frac{11}{5} \]
Thus, $a \in (-\infty, -\frac{11}{5})$.
Note: The condition $f(1)<0$ for an upward-opening parabola automatically ensures that the discriminant $D>0$ (real and distinct roots exist).

Step 3: Final Answer:
The set of values is $(-\infty, -\frac{11}{5})$.