Let alpha, beta in mathbbN$\alpha, \beta \in \mathbb{N}$ be roots of equation x^2 - 70x + lambda = 0$x^2 - 70x + \lambda = 0$ where fraclambda2, fraclambda3 notin mathbbN$\frac{\lambda}{2}, \frac{\lambda}{3} \notin \mathbb{N}$. If lambda$\lambda$ assumes the minimum possible value, then frac(sqrtalpha - 1 + sqrtbeta - 1)(lambda + 35)|alpha - beta|$\frac{(\sqrt{\alpha - 1} + \sqrt{\beta - 1})(\lambda + 35)}{|\alpha - \beta|}$ is equal to:

### Related Formula x^2 + x + 1 = 0 Rightarrow x = omega, omega^2$$x^2 + x + 1 = 0 \Rightarrow x = \omega, \omega^2$$ Properties of cube roots of unity: omega^3 = 1$\omega^3 = 1$ and 1 + omega + omega^2 = 0$1 + \omega + \omega^2 = 0$. ### Core Logic Let alpha = omega$\alpha = \omega$. Then frac1x = frac1omega = omega^2$\frac{1}{x} = \frac{1}{\omega} = \omega^2$. The series is sum_k=1^25 (omega^k + omega^2k)^4$\sum_{k=1}^{25} (\omega^k + \omega^{2k})^4$. Evaluate the term T_k = (omega^k + omega^2k)^4$T_k = (\omega^k + \omega^{2k})^4$ based on the modulo of k$k$ with 3. ### Step 1: Cyclic Evaluation Case 1: k = 3m$k = 3m$ (multiples of 3) T_3m = (omega^3m + omega^6m)^4 = (1 + 1)^4 = 2^4 = 16$$T_{3m} = (\omega^{3m} + \omega^{6m})^4 = (1 + 1)^4 = 2^4 = 16$$ There are 8$8$ such multiples up to 25 (3, 6, 9, dots, 24$3, 6, 9, \dots, 24$). Case 2: k neq 3m$k \neq 3m$ (non-multiples of 3) For k = 1, 2, 4, 5, dots$k = 1, 2, 4, 5, \dots$ omega^k + omega^2k$\omega^k + \omega^{2k}$ will always be omega + omega^2$\omega + \omega^2$ or omega^2 + omega$\omega^2 + \omega$. Since 1 + omega + omega^2 = 0$1 + \omega + \omega^2 = 0$, we have omega + omega^2 = -1$\omega + \omega^2 = -1$. T_k neq 3m = (-1)^4 = 1$$T_{k \neq 3m} = (-1)^4 = 1$$ There are 25 - 8 = 17$25 - 8 = 17$ such non-multiples up to 25. ### Step 2: Total Sum textSum = 17 times 1 + 8 times 16$$\text{Sum} = 17 \times 1 + 8 \times 16$$ = 17 + 128 = 145$$= 17 + 128 = 145$$ ### Pattern Recognition Powers of x+1/x$x+1/x$ when x$x$ solves x^2 pm x + 1 = 0$x^2 \pm x + 1 = 0$ perfectly orbit around periods of 3 or 6. Isolate the 3m$3m$ resonant beats (which hit pure scalars like 1+1=2$1+1=2$) versus the out-of-phase beats (which collapse to -1$-1$ or 1$1$ via basic omega$\omega$ identities). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Complex Numbers and Quadratic Equations

### Related Formula textFor a quadratic equation A x^2 + B x + C = 0 text to have two positive real roots:$$\text{For a quadratic equation } A x^2 + B x + C = 0 \text{ to have two positive real roots:}$$ text1. Real roots: D = B^2 - 4AC ge 0$$\text{1. Real roots: } D = B^2 - 4AC \ge 0$$ text2. Sum of roots: -fracBA > 0$$\text{2. Sum of roots: } -\frac{B}{A} > 0$$ text3. Product of roots: fracCA > 0$$\text{3. Product of roots: } \frac{C}{A} > 0$$ ### Core Logic We write down three systems of inequalities based on real and positive root conditions, find their intersection, and map the boundaries to solve for the parameters. ### Step 1: Apply the discriminant condition (Real roots) For real roots, the discriminant D ge 0$D \ge 0$: D = left[ 2(a - 3) right]^2 - 4(1 - a)(9) ge 0$$D = \left[ 2(a - 3) \right]^2 - 4(1 - a)(9) \ge 0$$ 4(a^2 - 6a + 9) - 36(1 - a) ge 0$$4(a^2 - 6a + 9) - 36(1 - a) \ge 0$$ (a^2 - 6a + 9) - 9(1 - a) ge 0$$(a^2 - 6a + 9) - 9(1 - a) \ge 0$$ a^2 - 6a + 9 - 9 + 9 a ge 0$$a^2 - 6a + 9 - 9 + 9 a \ge 0$$ a^2 + 3a ge 0 implies a(a + 3) ge 0$$a^2 + 3a \ge 0 \implies a(a + 3) \ge 0$$ Thus, the interval is: a in (-infty, -3] cup [0, infty) quad text--- (1)$$a \in (-\infty, -3] \cup [0, \infty) \quad \text{--- (1)}$$ ### Step 2: Apply the sum of roots condition (Positive sum) For positive roots, the sum of roots must be positive: -fracBA = frac-2(a - 3)1 - a = frac2(a - 3)a - 1 > 0$$-\frac{B}{A} = \frac{-2(a - 3)}{1 - a} = \frac{2(a - 3)}{a - 1} > 0$$ Using the wavy curve method for fraca-3a-1 > 0$\frac{a-3}{a-1} > 0$: a in (-infty, 1) cup (3, infty) quad text--- (2)$$a \in (-\infty, 1) \cup (3, \infty) \quad \text{--- (2)}$$ ### Step 3: Apply the product of roots condition (Positive product) For positive roots, the product of roots must be positive: fracCA = frac91 - a > 0 implies 1 - a > 0 implies a < 1$$\frac{C}{A} = \frac{9}{1 - a} > 0 \implies 1 - a > 0 \implies a < 1$$ Thus, the interval is: a in (-infty, 1) quad text--- (3)$$a \in (-\infty, 1) \quad \text{--- (3)}$$ ### Step 4: Find the intersection of all conditions Intersecting equations (1), (2), and (3): - First, intersect (2) and (3): ( (-infty, 1) cup (3, infty) ) cap (-infty, 1) = (-infty, 1)$$( (-\infty, 1) \cup (3, \infty) ) \cap (-\infty, 1) = (-\infty, 1)$$ - Next, intersect with (1): ( (-infty, -3] cup [0, infty) ) cap (-infty, 1) = (-infty, -3] cup [0, 1)$$( (-\infty, -3] \cup [0, \infty) ) \cap (-\infty, 1) = (-\infty, -3] \cup [0, 1)$$ Comparing this with (-infty, -alpha] cup [beta, gamma)$(-\infty, -\alpha] \cup [\beta, \gamma)$: - alpha = 3$\alpha = 3$ - beta = 0$\beta = 0$ - gamma = 1$\gamma = 1$ Now calculate the target sum: 2alpha + beta + gamma = 2(3) + 0 + 1 = 7$$2\alpha + \beta + \gamma = 2(3) + 0 + 1 = 7$$ ### Pattern Recognition Location of roots: When both roots are positive, checking sum and product signs along with D ge 0$D \ge 0$ is the standard and fastest set of inequalities, avoiding complex vertex projections. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Practical Chemistry

### Related Formula For a complex number lying on the unit circle: |z| = 1 implies zoverlinez = 1 implies overlinez = frac1z$$|z| = 1 \implies z\overline{z} = 1 \implies \overline{z} = \frac{1}{z}$$ Maximum distance from a point P$P$ to a circle with center C$C$ and radius r$r$ is: d_max = PC + r$$d_{\max} = PC + r$$ ### Core Logic Simplify the algebraic condition using overlinez = 1/z$\overline{z} = 1/z$ to uniquely determine the value of the parameter k$k$, then compute geometric distances. ### Step 1: Solve for k Cross-multiply the given expression: 2 + k^2z = kz(k + overlinez) = k^2z + kzoverlinez$$2 + k^2z = kz(k + \overline{z}) = k^2z + kz\overline{z}$$ Since zoverlinez = |z|^2 = 1$z\overline{z} = |z|^2 = 1$: 2 + k^2z = k^2z + k(1) implies k = 2$$2 + k^2z = k^2z + k(1) \implies k = 2$$ ### Step 2: Locate Point and Circle Parameters Substitute k=2$k=2$ into the target point expression P = k + ik^2$P = k + ik^2$: P = 2 + 4i equiv (2,4)$$P = 2 + 4i \equiv (2,4)$$ The circle equation is |z - (1 + 2i)| = 1$|z - (1 + 2i)| = 1$, which represents a circle centered at C = (1, 2)$C = (1, 2)$ with radius r = 1$r = 1$. ### Step 3: Compute Geometric Distances Find the Euclidean distance between P(2,4)$P(2,4)$ and center C(1,2)$C(1,2)$: PC = sqrt(2-1)^2 + (4-2)^2 = sqrt1 + 4 = sqrt5$$PC = \sqrt{(2-1)^2 + (4-2)^2} = \sqrt{1 + 4} = \sqrt{5}$$ The maximum distance from the point to the circle boundary is: d_max = PC + r = sqrt5 + 1$$d_{\max} = PC + r = \sqrt{5} + 1$$ ### Pattern Recognition Notice how k^2z$k^2z$ cancels perfectly on both sides during expansion due to the unique property of uni-modular complex numbers (zoverlinez=1$z\overline{z}=1$), rendering the calculation of k$k$ trivial. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Complex Numbers and Quadratic Equations

### Related Formula Newton's Sums for the roots of a quadratic equation ax^2 + bx + c = 0$ax^2 + bx + c = 0$: a P_n + b P_n-1 + c P_n-2 = 0$$a P_n + b P_{n-1} + c P_{n-2} = 0$$ ### Core Logic Observe the recurrence relation from the given numerical values of P_n$P_n$ to construct the base quadratic equation satisfied by alpha$\alpha$ and beta$\beta$, then invert the roots. ### Step 1: Identify the Linear Recurrence Relation Compare the provided sequence values: P_8 + P_9 = 47 + 76 = 123 = P_10$$P_8 + P_9 = 47 + 76 = 123 = P_{10}$$ This fits the general sequence relation: P_n = P_n-1 + P_n-2 implies P_n - P_n-1 - P_n-2 = 0$$P_n = P_{n-1} + P_{n-2} \implies P_n - P_{n-1} - P_{n-2} = 0$$ ### Step 2: Construct the Base Quadratic Equation The characteristic equation corresponding to this recurrence relation is: x^2 - x - 1 = 0$x^2 - x - 1 = 0$ Thus, alpha$\alpha$ and \(\beta\) are roots of x^2 - x - 1 = 0$x^2 - x - 1 = 0$, giving sum alpha+beta = 1$\alpha+\beta = 1$ and product alphabeta = -1$\alpha\beta = -1$ (which matches P_1 = alpha+beta = 1$P_1 = \alpha+\beta = 1$). ### Step 3: Construct Equation with Reciprocal Roots To find the equation with roots frac1alpha$\frac{1}{\alpha}$ and frac1beta$\frac{1}{\beta}$, apply the transformations: textSum of new roots = frac1alpha + frac1beta = fracalpha + betaalphabeta = frac1-1 = -1$$\text{Sum of new roots} = \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{1}{-1} = -1$$ textProduct of new roots = frac1alphabeta = frac1-1 = -1$$\text{Product of new roots} = \frac{1}{\alpha\beta} = \frac{1}{-1} = -1$$ The new quadratic equation is: x^2 - (textSum)x + (textProduct) = 0 implies x^2 - (-1)x + (-1) = 0 implies x^2 + x - 1 = 0$$x^2 - (\text{Sum})x + (\text{Product}) = 0 \implies x^2 - (-1)x + (-1) = 0 \implies x^2 + x - 1 = 0$$ ### Pattern Recognition The recurrence pattern P_n = P_n-1 + P_n-2$P_n = P_{n-1} + P_{n-2}$ is the Fibonacci sequence recurrence line. Its roots generate the Golden Ratio layout from x^2-x-1=0$x^2-x-1=0$. Inverting roots swaps the coefficients of x^2$x^2$ and the constant term, yielding x^2+x-1=0$x^2+x-1=0$ instantly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Complex Numbers and Quadratic Equations Class 11 Mathematics: Sequences and Series

### Related Formula For any equilateral triangle with vertices z_1, z_2, z_3$z_1, z_2, z_3$: z_1^2 + z_2^2 + z_3^2 = z_1 z_2 + z_2 z_3 + z_3 z_1 quad text--- (1)$$z_1^2 + z_2^2 + z_3^2 = z_1 z_2 + z_2 z_3 + z_3 z_1 \quad \text{--- (1)}$$ Centroid equation: z_0 = fracz_1 + z_2 + z_33 implies z_1 + z_2 + z_3 = 3z_0$$z_0 = \frac{z_1 + z_2 + z_3}{3} \implies z_1 + z_2 + z_3 = 3z_0$$ ### Core Logic Let's expand the target sum: sum_k=1^3 (z_k - z_0)^2 = (z_1 - z_0)^2 + (z_2 - z_0)^2 + (z_3 - z_0)^2$$\sum_{k=1}^{3} (z_k - z_0)^2 = (z_1 - z_0)^2 + (z_2 - z_0)^2 + (z_3 - z_0)^2$$ ### Step 1: Expansion and Algebraic Grouping Expanding each quadratic term: = (z_1^2 + z_2^2 + z_3^2) - 2z_0(z_1 + z_2 + z_3) + 3z_0^2$$= (z_1^2 + z_2^2 + z_3^2) - 2z_0(z_1 + z_2 + z_3) + 3z_0^2$$ Substitute z_1 + z_2 + z_3 = 3z_0$z_1 + z_2 + z_3 = 3z_0$: = (z_1^2 + z_2^2 + z_3^2) - 2z_0(3z_0) + 3z_0^2$$= (z_1^2 + z_2^2 + z_3^2) - 2z_0(3z_0) + 3z_0^2$$ = z_1^2 + z_2^2 + z_3^2 - 3z_0^2$$= z_1^2 + z_2^2 + z_3^2 - 3z_0^2$$ ### Step 2: Resolving using Equilateral Condition Substitute 3z_0^2 = 3 left(fracz_1+z_2+z_33right)^2 = frac(z_1+z_2+z_3)^23$3z_0^2 = 3 \left(\frac{z_1+z_2+z_3}{3}\right)^2 = \frac{(z_1+z_2+z_3)^2}{3}$: = (z_1^2 + z_2^2 + z_3^2) - fracz_1^2 + z_2^2 + z_3^2 + 2(z_1z_2 + z_2z_3 + z_3z_1)3$$= (z_1^2 + z_2^2 + z_3^2) - \frac{z_1^2 + z_2^2 + z_3^2 + 2(z_1z_2 + z_2z_3 + z_3z_1)}{3}$$ = frac2(z_1^2 + z_2^2 + z_3^2) - 2(z_1z_2 + z_2z_3 + z_3z_1)3$$= \frac{2(z_1^2 + z_2^2 + z_3^2) - 2(z_1z_2 + z_2z_3 + z_3z_1)}{3}$$ = frac23(z_1^2 + z_2^2 + z_3^2 - (z_1 z_2 + z_2 z_3 + z_3 z_1))$$= \frac{2}{3}(z_1^2 + z_2^2 + z_3^2 - (z_1 z_2 + z_2 z_3 + z_3 z_1))$$ Using condition (1) for equilateral triangles, the terms inside the parentheses equal 0$0$. Thus: = 0$= 0$ ### Pattern Recognition This is a standard invariant of equilateral triangles. Any translation to center of mass coordinates leaves the shape invariant, making the sum of squares of coordinate vectors relative to the centroid equal to zero. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Complex Numbers