Let alpha, beta$\alpha, \beta$ be the roots of the quadratic equation 12x^2 - 20x + 3lambda = 0$12x^2 - 20x + 3\lambda = 0$, lambda in mathbbZ$\lambda \in \mathbb{Z}$. If frac12 le |beta - alpha| le frac32$\frac{1}{2} \le |\beta - \alpha| \le \frac{3}{2}$, then the sum of all possible values of lambda$\lambda$ is:

### Related Formula X^2 = |X|^2$X^2 = |X|^2$ Quadratic factorization: t^2 - 5t + 6 = (t-2)(t-3)$t^2 - 5t + 6 = (t-2)(t-3)$ ### Core Logic Rewrite the equation taking |x - 1| = t$|x - 1| = t$, where t geq 0$t \geq 0$. Since (x-1)^2 = |x-1|^2$(x-1)^2 = |x-1|^2$, the equation becomes: t^2 - 5t + 6 = 0$$t^2 - 5t + 6 = 0$$ ### Step 1: Solve for modulus (t - 2)(t - 3) = 0 Rightarrow t = 2, 3$$(t - 2)(t - 3) = 0 \Rightarrow t = 2, 3$$ Since both roots are positive, both provide valid solutions for the modulus. |x - 1| = 2 quad textand quad |x - 1| = 3$|x - 1| = 2 \quad \text{and} \quad |x - 1| = 3$ ### Step 2: Unpack x values From |x - 1| = 2$|x - 1| = 2$: x - 1 = 2 Rightarrow x = 3$$x - 1 = 2 \Rightarrow x = 3$$ x - 1 = -2 Rightarrow x = -1$$x - 1 = -2 \Rightarrow x = -1$$ From |x - 1| = 3$|x - 1| = 3$: x - 1 = 3 Rightarrow x = 4$$x - 1 = 3 \Rightarrow x = 4$$ x - 1 = -3 Rightarrow x = -2$$x - 1 = -3 \Rightarrow x = -2$$ The roots are 3, -1, 4, -2$3, -1, 4, -2$. ### Step 3: Sum of Roots textSum = 3 + (-1) + 4 + (-2) = 4$$\text{Sum} = 3 + (-1) + 4 + (-2) = 4$$ ### Pattern Recognition In symmetric modulus equations f(|x-a|) = 0$f(|x-a|) = 0$, every valid root t$t$ generates twin solutions (a+t)$(a+t)$ and (a-t)$(a-t)$. The sum of each pair is perfectly 2a$2a$. If there are n$n$ distinct valid positive roots for t$t$, the sum of all x$x$-roots is exactly n times 2a$n \times 2a$. Here, n=2, a=1 Rightarrow 2 times 2(1) = 4$n=2, a=1 \Rightarrow 2 \times 2(1) = 4$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Quadratic Equations

### Related Formula sum_r=1^n (2r-1) = n^2$$\sum_{r=1}^{n} (2r-1) = n^2$$ sum_r=1^n r(r-1) = fracn(n^2-1)3$$\sum_{r=1}^{n} r(r-1) = \frac{n(n^2-1)}{3}$$ ### Core Logic Rewrite the series in summation form: sum_r=1^n(x+2r-2)(x+2r)=frac8n3$$\sum_{r=1}^{n}(x+2r-2)(x+2r)=\frac{8n}{3}$$ nx^2 + 2xsum_r=1^n(2r-1) + 4sum_r=1^nr(r-1) = frac8n3$$nx^2 + 2x\sum_{r=1}^{n}(2r-1) + 4\sum_{r=1}^{n}r(r-1) = \frac{8n}{3}$$ ### Step 1: Simplify the Equation Substitute the standard summation formulas: nx^2 + 2x(n^2) + frac4n(n^2-1)3 - frac8n3 = 0$$nx^2 + 2x(n^2) + \frac{4n(n^2-1)}{3} - \frac{8n}{3} = 0$$ Divide the entire equation by n$n$: x^2 + 2nx + frac4(n^2-1)3 - frac83 = 0$$x^2 + 2nx + \frac{4(n^2-1)}{3} - \frac{8}{3} = 0$$ Let the roots be alpha$\alpha$ and beta$\beta$. ### Step 2: Apply the Condition for Roots Since the roots are two consecutive even integers, their difference is 2$2$. |alpha - beta| = 2 implies fracsqrtD|a| = 2 implies D = 4$$|\alpha - \beta| = 2 \implies \frac{\sqrt{D}}{|a|} = 2 \implies D = 4$$ (2n)^2 - 4(1)left( frac4(n^2-1)3 - frac83 right) = 4$$(2n)^2 - 4(1)\left( \frac{4(n^2-1)}{3} - \frac{8}{3} \right) = 4$$ 4n^2 - frac16(n^2-1)3 + frac323 = 4$$4n^2 - \frac{16(n^2-1)}{3} + \frac{32}{3} = 4$$ n^2 - frac4n^23 = -3$$n^2 - \frac{4n^2}{3} = -3$$ frac-n^23 = -3 implies n^2 = 9$$\frac{-n^2}{3} = -3 \implies n^2 = 9$$ Since n$n$ is a positive integer, n = 3$n = 3$. ### Pattern Recognition When dealing with equations where roots have a specific difference k$k$, instantly use D = a^2k^2$D = a^2k^2$. Here k=2$k=2$, so D=4a^2$D=4a^2$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Quadratic Equations Class 11 Maths: Sequence and Series

### 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

### Related Formula |y| = begincases y, & y geq 0 \\ -y, & y < 0 endcases$$|y| = \begin{cases} y, & y \geq 0 \\ -y, & y < 0 \end{cases}$$ ### Core Logic Break the domain into intervals based on the critical points of the absolute values, which are x = -4$x = -4$ and x = -2$x = -2$. Intervals to test: Case I: x < -4$x < -4$ Case II: -4 leq x < -2$-4 \leq x < -2$ Case III: x geq -2$x \geq -2$ ### Step 1: Analyzing Case I For x < -4$x < -4$: Both (x+4)$(x+4)$ and (x+2)$(x+2)$ are negative. The equation becomes: x(-(x + 4)) + 3(-(x + 2)) + 10 = 0$$x(-(x + 4)) + 3(-(x + 2)) + 10 = 0$$ -x^2 - 4x - 3x - 6 + 10 = 0$$-x^2 - 4x - 3x - 6 + 10 = 0$$ x^2 + 7x - 4 = 0$$x^2 + 7x - 4 = 0$$ Roots are x = frac-7 pm sqrt49 + 162 = frac-7 pm sqrt652$x = \frac{-7 \pm \sqrt{49 + 16}}{2} = \frac{-7 \pm \sqrt{65}}{2}$. Check if they fall in the interval x < -4$x < -4$: sqrt65 approx 8.06$\sqrt{65} \approx 8.06$ x_1 = frac-7 + 8.062 approx 0.53$x_1 = \frac{-7 + 8.06}{2} \approx 0.53$ (Rejected, not <-4$<-4$) x_2 = frac-7 - 8.062 approx -7.53$x_2 = \frac{-7 - 8.06}{2} \approx -7.53$ (Accepted) So, 1 valid solution here. ### Step 2: Analyzing Case II For -4 leq x < -2$-4 \leq x < -2$: (x+4) geq 0$(x+4) \geq 0$ and (x+2) < 0$(x+2) < 0$. The equation becomes: x(x + 4) + 3(-(x + 2)) + 10 = 0$$x(x + 4) + 3(-(x + 2)) + 10 = 0$$ x^2 + 4x - 3x - 6 + 10 = 0$$x^2 + 4x - 3x - 6 + 10 = 0$$ x^2 + x + 4 = 0$x^2 + x + 4 = 0$ Discriminant D = 1^2 - 4(1)(4) = 1 - 16 = -15 < 0$D = 1^2 - 4(1)(4) = 1 - 16 = -15 < 0$. No real roots in this interval. ### Step 3: Analyzing Case III For x geq -2$x \geq -2$: Both (x+4)$(x+4)$ and (x+2)$(x+2)$ are positive. The equation becomes: x(x + 4) + 3(x + 2) + 10 = 0$$x(x + 4) + 3(x + 2) + 10 = 0$$ x^2 + 4x + 3x + 6 + 10 = 0$$x^2 + 4x + 3x + 6 + 10 = 0$$ x^2 + 7x + 16 = 0$$x^2 + 7x + 16 = 0$$ Discriminant D = 49 - 64 = -15 < 0$D = 49 - 64 = -15 < 0$. No real roots in this interval. ### Step 4: Final Count Combining the results from all cases, there is exactly 1 distinct real solution. ### Pattern Recognition For sum-of-absolute-value equations, systematically partition the number line using the roots of the arguments. Discard roots generated by the quadratics that fall outside their respective assumed interval. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Quadratic Equations