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

### Related Formula textFor ax^2+bx+c=0 text with a > 0 text, if a point k text lies strictly between the roots, f(k) < 0.$$\text{For } ax^2+bx+c=0 \text{ with } a > 0 \text{, if a point } k \text{ lies strictly between the roots, } f(k) < 0.$$ ### Core Logic Given f(x) = x^2 + 2ax + (3a + 10)$f(x) = x^2 + 2ax + (3a + 10)$. Since the coefficient of x^2$x^2$ is 1 > 0$1 > 0$, the parabola opens upward. For 1$1$ to lie between the roots alpha$\alpha$ and beta$\beta$, 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) = 1^2 + 2a(1) + 3a + 10 < 0$$f(1) = 1^2 + 2a(1) + 3a + 10 < 0$$ 1 + 2a + 3a + 10 < 0$$1 + 2a + 3a + 10 < 0$$ 5a + 11 < 0$5a + 11 < 0$ a < -frac115$$a < -\frac{11}{5}$$ So, a in left(-infty, -frac115right)$a \in \left(-\infty, -\frac{11}{5}\right)$. ### 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 Delta > 0$\Delta > 0$ manually because a f(k) < 0$a f(k) < 0$ guarantees real distinct roots. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Quadratic Equations