Let the relation R on the set M = \1, 2, 3, ldots, 16\$M = \{1, 2, 3, \ldots, 16\}$ be given by R = \(x, y) : 4y = 5x - 3, x, y in M\$R = \{(x, y) : 4y = 5x - 3, x, y \in M\}$. Then the minimum number of elements required to be added in R, in order to make the relation symmetric, is equal to

### Related Formula For inverse trigonometric functions sin^-1(g(x))$\sin^{-1}(g(x))$ and cos^-1(h(x))$\cos^{-1}(h(x))$, the arguments must satisfy: -1 leq g(x) leq 1$$-1 \leq g(x) \leq 1$$ -1 leq h(x) leq 1$$-1 \leq h(x) \leq 1$$ ### Core Logic Given f(x) = cos^-1left(frac2x - 511 - 3xright) + sin^-1(2x^2 - 3x + 1)$f(x) = \cos^{-1}\left(\frac{2x - 5}{11 - 3x}\right) + \sin^{-1}(2x^2 - 3x + 1)$ We establish two simultaneous inequalities for the domain: 1) -1 leq frac2x - 511 - 3x leq 1$-1 \leq \frac{2x - 5}{11 - 3x} \leq 1$ 2) -1 leq 2x^2 - 3x + 1 leq 1$-1 \leq 2x^2 - 3x + 1 \leq 1$ ### Step 1: Solve the Quadratic Inequality From -1 leq 2x^2 - 3x + 1 leq 1$-1 \leq 2x^2 - 3x + 1 \leq 1$: Split into two parts: 2x^2 - 3x + 2 geq 0$2x^2 - 3x + 2 \geq 0$ (This is always true as discriminant D < 0, a > 0$D < 0, a > 0$) 2x^2 - 3x leq 0 Rightarrow x(2x - 3) leq 0$2x^2 - 3x \leq 0 \Rightarrow x(2x - 3) \leq 0$ x in left[0, frac32right] dots(i)$$x \in \left[0, \frac{3}{2}\right] \dots(i)$$ ### Step 2: Solve the Rational Inequality From -1 leq frac2x - 511 - 3x leq 1$-1 \leq \frac{2x - 5}{11 - 3x} \leq 1$: Part A: frac2x - 511 - 3x + 1 geq 0 Rightarrow frac2x - 5 + 11 - 3x11 - 3x geq 0 Rightarrow frac6 - x11 - 3x geq 0$\frac{2x - 5}{11 - 3x} + 1 \geq 0 \Rightarrow \frac{2x - 5 + 11 - 3x}{11 - 3x} \geq 0 \Rightarrow \frac{6 - x}{11 - 3x} \geq 0$

x in left(-infty, frac113right) cup [6, infty)$$x \in \left(-\infty, \frac{11}{3}\right) \cup [6, \infty)$$ Part B: frac2x - 511 - 3x - 1 leq 0 Rightarrow frac5x - 1611 - 3x leq 0 Rightarrow x in left(-infty, frac165right] cup left(frac113, inftyright)$\frac{2x - 5}{11 - 3x} - 1 \leq 0 \Rightarrow \frac{5x - 16}{11 - 3x} \leq 0 \Rightarrow x \in \left(-\infty, \frac{16}{5}\right] \cup \left(\frac{11}{3}, \infty\right)$ Intersection of Part A and Part B: $ Intersection of Part A and Part B: $x in left(-infty, frac165right] cup [6, infty) dots(ii)$x \in \left(-\infty, \frac{16}{5}\right] \cup [6, \infty) \dots(ii)$ ### Step 3: Final Intersection Taking the intersection of $ ### Step 3: Final Intersection Taking the intersection of $(i) and $ and $(ii): $: $x in left[0, frac32right]$x \in \left[0, \frac{3}{2}\right]$ Comparing this with $ Comparing this with $[\alpha, \beta], we have $, we have $\alpha = 0, \beta = \frac{3}{2}. Therefore, $. Therefore, $\alpha + 2\beta = 0 + 2\left(\frac{3}{2}\right) = 3 ### Pattern Recognition Whenever dealing with dual inverse trig terms, strictly isolate the bounding intervals $ ### Pattern Recognition Whenever dealing with dual inverse trig terms, strictly isolate the bounding intervals $[-1, 1]$ for each argument separately and use a number line intersection to find the strictest common region. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Functions Class 11 Maths: Linear Inequalities

### Related Formula textDomain of sin^-1(y) text is -1 leq y leq 1$$\text{Domain of } \sin^{-1}(y) \text{ is } -1 \leq y \leq 1$$ textDomain of frac1log(z) text is z > 0, z neq 1$$\text{Domain of } \frac{1}{\log(z)} \text{ is } z > 0, z \neq 1$$ ### Core Logic For f(x)$f(x)$ to be defined, two main conditions must hold: 1) -1 leq frac5 - x2x + 3 leq 1$-1 \leq \frac{5 - x}{2x + 3} \leq 1$ 2) 10 - x > 0$10 - x > 0$ and 10 - x neq 1$10 - x \neq 1$ ### Step 1: Solving the Logarithmic Domain 10 - x > 0 implies x < 10$10 - x > 0 \implies x < 10$ 10 - x neq 1 implies x neq 9$10 - x \neq 1 \implies x \neq 9$ ### Step 2: Solving the Inverse Sine Domain -1 leq frac5 - x2x + 3 leq 1$-1 \leq \frac{5 - x}{2x + 3} \leq 1$ This is equivalent to left|frac5 - x2x + 3right| leq 1$\left|\frac{5 - x}{2x + 3}\right| \leq 1$, which means: (5 - x)^2 leq (2x + 3)^2 quad text(assuming x neq -1.5)$$(5 - x)^2 \leq (2x + 3)^2 \quad \text{(assuming } x \neq -1.5)$$ (5 - x)^2 - (2x + 3)^2 leq 0$$(5 - x)^2 - (2x + 3)^2 \leq 0$$ Factor as difference of squares: (5 - x - (2x + 3))(5 - x + 2x + 3) leq 0$$(5 - x - (2x + 3))(5 - x + 2x + 3) \leq 0$$ (2 - 3x)(x + 8) leq 0$$(2 - 3x)(x + 8) \leq 0$$ Multiply by -1$-1$ and flip the inequality: (3x - 2)(x + 8) geq 0$$(3x - 2)(x + 8) \geq 0$$ So, x in (-infty, -8] cup left[frac23, inftyright)$x \in (-\infty, -8] \cup \left[\frac{2}{3}, \infty\right)$. ### Step 3: Intersection of Domains Intersecting with x < 10$x < 10$ and x neq 9$x \neq 9$: Domain = (-infty, -8] cup left[frac23, 10right) - \9\$= (-\infty, -8] \cup \left[\frac{2}{3}, 10\right) - \{9\}$ Comparing with the given form (-infty, alpha] cup [beta, gamma) - \delta\$(-\infty, \alpha] \cup [\beta, \gamma) - \{\delta\}$, we have: alpha = -8$\alpha = -8$, beta = frac23$\beta = \frac{2}{3}$, gamma = 10$\gamma = 10$, delta = 9$\delta = 9$. ### Step 4: Final Value Calculation Evaluate 6(alpha + beta + gamma + delta)$6(\alpha + \beta + \gamma + \delta)$: 6left(-8 + frac23 + 10 + 9right) = 6left(11 + frac23right) = 6left(frac353right) = 70$$6\left(-8 + \frac{2}{3} + 10 + 9\right) = 6\left(11 + \frac{2}{3}\right) = 6\left(\frac{35}{3}\right) = 70$$ ### Pattern Recognition Instead of solving two separate rational inequalities for leq 1$\leq 1$ and geq -1$\geq -1$, converting the absolute value fraction into a difference of squares quickly bypasses sign analysis pitfalls, delivering the required critical points effortlessly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Sets, Relations and Functions Class 12 Maths: Inverse Trigonometric Functions

### Related Formula textChain definition: a_i = 2 a_i+1 + 1 quad textfor i = 1, 2, dots, k$$\text{Chain definition: } a_{i} = 2 a_{i+1} + 1 \quad \text{for } i = 1, 2, \dots, k$$ ### Core Logic To find the longest sequence of connected pairs, we trace the relation backward starting from the smallest elements in A$A$. ### Step 1: Trace the relations backward To maximize k$k$, we want the chain of elements to go down as low as possible. Let the final element in the chain be a_k+1 in mathrmA$a_{k+1} \in \mathrm{A}$. Since a_k = 2 a_k+1 + 1$a_k = 2 a_{k+1} + 1$: - If a_k+1 = 1 implies a_k = 3$a_{k+1} = 1 \implies a_k = 3$ - If a_k+1 = 2 implies a_k = 5$a_{k+1} = 2 \implies a_k = 5$ Let's test the chain starting with a_k+1 = 1$a_{k+1} = 1$: - a_k = 2(1) + 1 = 3$a_k = 2(1) + 1 = 3$ - a_k-1 = 2(3) + 1 = 7$a_{k-1} = 2(3) + 1 = 7$ - a_k-2 = 2(7) + 1 = 15$a_{k-2} = 2(7) + 1 = 15$ - a_k-3 = 2(15) + 1 = 31$a_{k-3} = 2(15) + 1 = 31$ - a_k-4 = 2(31) + 1 = 63$a_{k-4} = 2(31) + 1 = 63$ - a_k-5 = 2(63) + 1 = 127$a_{k-5} = 2(63) + 1 = 127$ (but 127 notin mathrmA$127 \notin \mathrm{A}$!) Thus, the longest chain within the set A$A$ has 6 elements: \63, \, 31, \, 15, \, 7, \, 3, \, 1\$$\{63, \, 31, \, 15, \, 7, \, 3, \, 1\}$$ This chain corresponds to exactly 5 ordered pairs: (63, 31), \, (31, 15), \, (15, 7), \, (7, 3), \, (3, 1)$$(63, 31), \, (31, 15), \, (15, 7), \, (7, 3), \, (3, 1)$$ So the maximum number of pairs in the sequence is k = 5$k = 5$. ### Step 2: Check alternative chains If we start with a_k+1 = 2$a_{k+1} = 2$: - a_k+1 = 2$a_{k+1} = 2$ - a_k = 5$a_k = 5$ - a_k-1 = 11$a_{k-1} = 11$ - a_k-2 = 23$a_{k-2} = 23$ - a_k-3 = 47$a_{k-3} = 47$ - a_k-4 = 95$a_{k-4} = 95$ - a_k-5 = 191 > 100$a_{k-5} = 191 > 100$ Again, the maximum number of pairs is k = 5$k = 5$. Thus, the largest integer k$k$ is 5. ### Pattern Recognition Recursive scaling: Tracing exponential chains of the form x_n+1 = c x_n + d$x_{n+1} = c x_n + d$ shows that the elements grow very quickly. Calculating the limits of growth determines the maximum possible depth of the sequence. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Relations and Functions

### Related Formula textFor frac1sqrtg(x) text to be defined, we require: g(x) > 0$$\text{For } \frac{1}{\sqrt{g(x)}} \text{ to be defined, we require: } g(x) > 0$$ ### Core Logic We find the domains of the two constituent terms separately and then find their intersection. ### Step 1: Find the domain of the first term For the first term to be defined: 10 + 3x - x^2 > 0 implies x^2 - 3x - 10 < 0$$10 + 3x - x^2 > 0 \implies x^2 - 3x - 10 < 0$$ (x - 5)(x + 2) < 0 implies x in (-2, 5) quad text--- (1)$$(x - 5)(x + 2) < 0 \implies x \in (-2, 5) \quad \text{--- (1)}$$ ### Step 2: Find the domain of the second term For the second term to be defined: x + |x| > 0$x + |x| > 0$ - If x ge 0$x \ge 0$: x + x = 2x > 0 implies x > 0$x + x = 2x > 0 \implies x > 0$. - If x < 0$x < 0$: x - x = 0 ngtr 0$x - x = 0 \ngtr 0$. Thus, the domain of the second term is: x in (0, infty) quad text--- (2)$$x \in (0, \infty) \quad \text{--- (2)}$$ ### Step 3: Find intersection and calculate the final expression Intersecting domains (1) and (2): x in (-2, 5) cap (0, infty) implies x in (0, 5)$$x \in (-2, 5) \cap (0, \infty) \implies x \in (0, 5)$$ Comparing this with (a, b)$(a, b)$ gives a = 0$a = 0$ and b = 5$b = 5$. Now calculate the value: (1 + a)^2 + b^2 = (1 + 0)^2 + 5^2 = 1 + 25 = 26$$(1 + a)^2 + b^2 = (1 + 0)^2 + 5^2 = 1 + 25 = 26$$ ### Pattern Recognition Modulus domain constraint: The function x + |x|$x + |x|$ is non-zero only for positive values of x$x$. This is a standard math trick that collapses complex domains down to x > 0$x > 0$ instantly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Relations and Functions