### Related Formula The number of one-one (injective) functions from a set D$D$ with n(D)$n(D)$ elements to a set C$C$ with n(C)$n(C)$ elements is given by: ^n(C)mathrmP_n(D) = fracn(C)!(n(C) - n(D))!${}^{n(C)}\mathrm{P}_{n(D)} = \frac{n(C)!}{(n(C) - n(D))!}$ ### Core Logic Step 1: Find the number of elements in set D = A cap B$D = A \cap B$. Solving the equations of circle A$A$ and ellipse B$B$ simultaneously: From A$A$, y^2 = 25 - x^2$y^2 = 25 - x^2$. Substitute this into B$B$: x^2 + 9(25 - x^2) = 144 implies -8x^2 = 144 - 225 = -81 implies x = pmfrac92sqrt2$x^2 + 9(25 - x^2) = 144 \implies -8x^2 = 144 - 225 = -81 \implies x = \pm\frac{9}{2\sqrt{2}}$ Correspondingly, y = pmfracsqrt1192sqrt2$y = \pm\frac{\sqrt{119}}{2\sqrt{2}}$. Thus, there are exactly 4 distinct intersection points, so n(D) = 4$n(D) = 4$.

### Step 1: Count elements in Set C Set C$C$ consists of integral lattice points (x,y)$(x,y)$ inside or on the circle x^2 + y^2 le 4$x^2 + y^2 \le 4$: Possible integer pairs are: (0,0), (pm1,0), (0,pm1), (pm2,0), (0,pm2), (pm1,pm1)$(0,0), (\pm1,0), (0,\pm1), (\pm2,0), (0,\pm2), (\pm1,\pm1)$. Counting them yields n(C) = 13$n(C) = 13$ elements. ### Step 2: Calculate Injective Functions The total number of one-one functions from D$D$ to C$C$ is: ^13mathrmP_4 = 13 times 12 times 11 times 10 = 17160${}^{13}\mathrm{P}_{4} = 13 \times 12 \times 11 \times 10 = 17160$ ### Pattern Recognition The intersection of a concentric circle and ellipse always yields 4 points if they cross completely. Break down the problem by counting the cardinality of domain and codomain independently before applying permutation formulas. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Conic Sections Class 12 Mathematics: Relations and Functions

### Related Formula Sum of first n$n$ squares: sum_r=1^n r^2 = fracn(n+1)(2n+1)6$\sum_{r=1}^{n} r^2 = \frac{n(n+1)(2n+1)}{6}$ ### Core Logic From functional relation f(x) - fleft(fracx2right) = fracx2$f(x) - f\left(\frac{x}{2}\right) = \frac{x}{2}$. Write a telescoping sequence by scaling variable down: fleft(fracx2right) - fleft(fracx4right) = fracx4$f\left(\frac{x}{2}\right) - f\left(\frac{x}{4}\right) = \frac{x}{4}$ fleft(fracx4right) - fleft(fracx8right) = fracx8$f\left(\frac{x}{4}\right) - f\left(\frac{x}{8}\right) = \frac{x}{8}$ dots$\dots$ fleft(fracx2^n-1right) - fleft(fracx2^nright) = fracx2^n$f\left(\frac{x}{2^{n-1}}\right) - f\left(\frac{x}{2^n}\right) = \frac{x}{2^n}$ ### Step 1: Evaluate the Limit Definition Summing all equations creates a telescoping sum on the left side: f(x) - fleft(fracx2^nright) = xleft(frac12 + frac14 + dots + frac12^nright) = xleft(1 - frac12^nright)$f(x) - f\left(\frac{x}{2^n}\right) = x\left(\frac{1}{2} + \frac{1}{4} + \dots + \frac{1}{2^n}\right) = x\left(1 - \frac{1}{2^n}\right)$ Taking the limit as n to infty$n \to \infty$: G(x) = lim_n to infty xleft(1 - frac12^nright) = x$G(x) = \lim_{n \to \infty} x\left(1 - \frac{1}{2^n}\right) = x$ ### Step 2: Final Sum Evaluation We need to compute sum_r=1^10 G(r^2) = sum_r=1^10 r^2$\sum_{r=1}^{10} G(r^2) = \sum_{r=1}^{10} r^2$: sum_r=1^10 r^2 = frac10 times 11 times 216 = 385$\sum_{r=1}^{10} r^2 = \frac{10 \times 11 \times 21}{6} = 385$ ### Pattern Recognition Linear iterative arguments of type f(2x)-f(x)=x$f(2x)-f(x)=x$ naturally condense into geometric progression properties via geometric series limits. Always look for telescoping patterns in infinite limits of difference terms. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Sequence and Series Class 12 Mathematics: Relations and Functions

### Related Formula An ellipse equation is structured as: frac(x-h)^2a^2 + frac(y-k)^2b^2 leq 1$\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} \leq 1$ ### Core Logic Analyzing set A$A$: |alpha - 1| le 4 implies -4 le alpha - 1 le 4 implies -3 le alpha le 5$|\alpha - 1| \le 4 \implies -4 \le \alpha - 1 \le 4 \implies -3 \le \alpha \le 5$ |beta - 5| le 6 implies -6 le beta - 5 le 6 implies -1 le beta le 11$|\beta - 5| \le 6 \implies -6 \le \beta - 5 \le 6 \implies -1 \le \beta \le 11$ Thus, region A$A$ forms a rectangle bounded between x in [-3, 5]$x \in [-3, 5]$ and y in [-1, 11]$y \in [-1, 11]$. Analyzing set B$B$: 16(alpha - 2)^2 + 9(beta - 6)^2 le 144$16(\alpha - 2)^2 + 9(\beta - 6)^2 \le 144$ Dividing by 144: frac(alpha - 2)^29 + frac(beta - 6)^216 le 1$\frac{(\alpha - 2)^2}{9} + \frac{(\beta - 6)^2}{16} \le 1$ This represents the interior and boundary of an ellipse centered at (2, 6)$(2, 6)$ with semi-minor axis a = 3$a = 3$ and semi-major axis b = 4$b = 4$. ### Step 1: Spatial Inclusion Check Let's check the extreme horizontal and vertical extents of the ellipse B$B$: Horizontal extent: x in [2 - 3, 2 + 3] = [-1, 5]$x \in [2 - 3, 2 + 3] = [-1, 5]$ Vertical extent: y in [6 - 4, 6 + 4] = [2, 10]$y \in [6 - 4, 6 + 4] = [2, 10]$ Comparing with the boundaries of rectangle A$A$ (x in [-3, 5]$x \in [-3, 5]$ and y in [-1, 11]$y \in [-1, 11]$): [-1, 5] subseteq [-3, 5]$[-1, 5] \subseteq [-3, 5]$ [2, 10] subseteq [-1, 11]$[2, 10] \subseteq [-1, 11]$

Since all points of the ellipse lie perfectly inside the rectangular region, we conclusively find that B subset A$B \subset A$. ### Pattern Recognition A bounding box check (finding h pm a$h \pm a$ and k pm b$k \pm b$) for conics is the fastest analytical shortcut to verify set inclusion without plotting extensive coordinates. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Sets, Relations and Functions Class 11 Mathematics: Conic Sections

### Related Formula For a quadratic equation Ax^2 + Bx + C = 0$Ax^2 + Bx + C = 0$ to yield real roots, its discriminant must satisfy: D = B^2 - 4AC ge 0$D = B^2 - 4AC \ge 0$ ### Core Logic Set y = frac5 - xx^2 - 3x + 2$y = \frac{5 - x}{x^2 - 3x + 2}$: y(x^2 - 3x + 2) = 5 - x$y(x^2 - 3x + 2) = 5 - x$ yx^2 - 3xy + 2y + x - 5 = 0$yx^2 - 3xy + 2y + x - 5 = 0$ Rearranging into a standard quadratic equation in terms of x$x$: yx^2 + (1 - 3y)x + (2y - 5) = 0$yx^2 + (1 - 3y)x + (2y - 5) = 0$ ### Step 1: Discriminant Method Case I: If y = 0$y = 0$, the equation simplifies to x - 5 = 0 implies x = 5$x - 5 = 0 \implies x = 5$, which is a valid part of the domain. Thus, 0$0$ belongs to the range. Case II: If y neq 0$y \neq 0$, for x$x$ to be real, D ge 0$D \ge 0$: (1 - 3y)^2 - 4(y)(2y - 5) ge 0$(1 - 3y)^2 - 4(y)(2y - 5) \ge 0$ 9y^2 + 1 - 6y - 8y^2 + 20y ge 0$9y^2 + 1 - 6y - 8y^2 + 20y \ge 0$ y^2 + 14y + 1 ge 0$y^2 + 14y + 1 \ge 0$ ### Step 2: Solving the Inequality Completing the square for y^2 + 14y + 1 ge 0$y^2 + 14y + 1 \ge 0$: (y + 7)^2 - 48 ge 0 implies (y + 7)^2 ge (4sqrt3)^2$(y + 7)^2 - 48 \ge 0 \implies (y + 7)^2 \ge (4\sqrt{3})^2$ This gives: y le -7 - 4sqrt3 quad textor quad y ge -7 + 4sqrt3$y \le -7 - 4\sqrt{3} \quad \text{or} \quad y \ge -7 + 4\sqrt{3}$ Comparing with the interval (-infty , alpha ] cup [ beta , infty)$(-\infty , \alpha ] \cup [ \beta , \infty)$: alpha = -7 - 4sqrt3$\alpha = -7 - 4\sqrt{3}$ beta = -7 + 4sqrt3$\beta = -7 + 4\sqrt{3}$ ### Step 3: Finding alpha^2 + beta^2 Using algebraic identities: alpha^2 + beta^2 = (-7 - 4sqrt3)^2 + (-7 + 4sqrt3)^2$\alpha^2 + \beta^2 = (-7 - 4\sqrt{3})^2 + (-7 + 4\sqrt{3})^2$ = 2(7^2 + (4sqrt3)^2) = 2(49 + 48) = 2(97) = 194$= 2(7^2 + (4\sqrt{3})^2) = 2(49 + 48) = 2(97) = 194$ ### Pattern Recognition For rational expressions of the form fractextLineartextQuadratic$\frac{\text{Linear}}{\text{Quadratic}}$, converting to a quadratic in x$x$ and forcing D ge 0$D \ge 0$ establishes the range boundaries elegantly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Sets, Relations and Functions