Let mathrmA = \1, 2, 3, dots, 100\$\mathrm{A} = \{1, 2, 3, \dots, 100\}$ and mathrmR$\mathrm{R}$ be a relation on mathrmA$\mathrm{A}$ such that mathrmR = \(a, b) : a = 2b + 1\$\mathrm{R} = \{(a, b) : a = 2b + 1\}$. Let (a_1, a_2), (a_2, a_3), (a_3, a_4), dots, (a_k, a_k+1)$(a_1, a_2), (a_2, a_3), (a_3, a_4), \dots, (a_k, a_{k+1})$ be a sequence of k$k$ elements of mathrmR$\mathrm{R}$ such that the second entry of an ordered pair is equal to the first entry of the next ordered pair. Then the largest integer k$k$, for which such a sequence exists, is equal to:

### 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

### Related Formula A function is one-one if its derivative is strictly monotonic (always positive or always negative) across its domain. It is onto if its range equals its co-domain. ### Core Logic Rewrite the function by multiplying the numerator and denominator by 2^x$2^x$: f(x) = frac2^2x - 12^2x + 1 = 1 - frac22^2x + 1$$f(x) = \frac{2^{2x} - 1}{2^{2x} + 1} = 1 - \frac{2}{2^{2x} + 1}$$ ### Step 1: Check One-One property Differentiating f(x)$f(x)$ with respect to x$x$ : f'(x) = frac2(2^2x + 1)^2 cdot 2 cdot 2^2x cdot ln 2 = frac4 cdot 2^2x cdot ln 2(2^2x + 1)^2$$f'(x) = \frac{2}{(2^{2x} + 1)^2} \cdot 2 \cdot 2^{2x} \cdot \ln 2 = \frac{4 \cdot 2^{2x} \cdot \ln 2}{(2^{2x} + 1)^2}$$ Since 2^2x > 0$2^{2x} > 0$ and ln 2 > 0$\ln 2 > 0$, f'(x) > 0$f'(x) > 0$ always. Thus, f(x)$f(x)$ is strictly increasing, confirming it is a one-one function. ### Step 2: Check Onto property Analyze the limits at boundaries [cite: 3880, 3881]: lim_x to -infty f(x) = 1 - frac20 + 1 = -1$$\lim_{x \to -\infty} f(x) = 1 - \frac{2}{0 + 1} = -1$$ lim_x to infty f(x) = 1 - 0 = 1$$\lim_{x \to \infty} f(x) = 1 - 0 = 1$$ Thus, the range of the function is (-1, 1)$(-1, 1)$. Since the given co-domain is (-infty, 1)$(-\infty, 1)$ and textRange neq textCo-domain$\text{Range} \neq \text{Co-domain}$ , the function is not onto. ### Pattern Recognition The expression given is a shifted form of the hyperbolic tangent function tanh(x ln 2)$\tanh(x \ln 2)$. Hyperbolic tangent always maps to (-1, 1)$(-1, 1)$, making its restriction against (-infty, 1)$(-\infty, 1)$ non-surjective (not onto). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Relations and Functions

### Related Formula For a vertical line x = x_0$x = x_0$ crossing a bounded region, the valid coordinates of y$y$ sit between the boundary lines intersecting that specific line vertical plane. ### Core Logic The point given is fixed at x = frac112 = 5.5$x = \frac{11}{2} = 5.5$[cite: 3285, 3925]. We evaluate the values of y$y$ along this vertical line segment across each boundary edge.

### Step 1: Evaluate Intersections Substitute x = frac112$x = \frac{11}{2}$ into the three linear constraints: 1. From x + y = 11$x + y = 11$: frac112 + y = 11 Rightarrow y = 11 - 5.5 = 5.5$$\frac{11}{2} + y = 11 \Rightarrow y = 11 - 5.5 = 5.5$$ 2. From x + 2y = 16$x + 2y = 16$: frac112 + 2y = 16 Rightarrow 2y = 16 - 5.5 = 10.5 Rightarrow y = 5.25$$\frac{11}{2} + 2y = 16 \Rightarrow 2y = 16 - 5.5 = 10.5 \Rightarrow y = 5.25$$ 3. From 2x + 3y = 29$2x + 3y = 29$: 2left(frac112right) + 3y = 29 Rightarrow 11 + 3y = 29 Rightarrow 3y = 18 Rightarrow y = 6$$2\left(\frac{11}{2}\right) + 3y = 29 \Rightarrow 11 + 3y = 29 \Rightarrow 3y = 18 \Rightarrow y = 6$$ ### Step 2: Define Extrema and Multiply Checking the internal region of the \triangle bounded by these lines [cite: 3286, 3287], the valid range for alpha$\alpha$ along the section line is delimited by y = 5.5$y = 5.5$ and y = 6$y = 6$: alpha_min = frac112 = 5.5$$\alpha_{\min} = \frac{11}{2} = 5.5$$ alpha_max = 6$$\alpha_{\max} = 6$$ The product of the limits is : alpha_min cdot alpha_max = frac112 times 6 = 33$$\alpha_{\min} \cdot \alpha_{\max} = \frac{11}{2} \times 6 = 33$$ ### Pattern Recognition Instead of drawing full coordinate diagrams or computing all three vertex points, evaluating values directly at the fixed coordinate constraint x = 5.5$x = 5.5$ saves time during multi-line area problems. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Linear Inequalities Class 11 Mathematics: Straight Lines