Let the domain of the function f(x) = log_3 log_5 (7 - log_2 (x^2 - 10x + 85)) + sin^-1 left(left|frac3x-717-xright|right)$f(x) = \log_3 \log_5 (7 - \log_2 (x^2 - 10x + 85)) + \sin^{-1} \left(\left|\frac{3x-7}{17-x}\right|\right)$ be (alpha, beta]$(\alpha, \beta]$. Then alpha + beta$\alpha + \beta$ 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 For a set with n$n$ elements, the number of relations that are both reflexive and symmetric is given by: 2^fracn(n-1)2$$2^{\frac{n(n-1)}{2}}$$ ### Core Logic A reflexive relation must contain all diagonal pairs (x,x)$(x,x)$. There is only 1$1$ way to assign these elements (they MUST be present). A symmetric relation requires that if (x,y)$(x,y)$ is present, (y,x)$(y,x)$ must also be present. Thus, we only have the freedom to choose whether to include the unordered pairs \x, y\$\{x, y\}$ where x neq y$x \neq y$. ### Step 1: Calculate the available independent pairs Number of distinct elements n = 4$n = 4$. Total number of pairs in the cartesian product is n^2 = 16$n^2 = 16$. Number of diagonal pairs (reflexive necessity) = n = 4$= n = 4$. Remaining non-diagonal pairs = 16 - 4 = 12$= 16 - 4 = 12$. Since symmetry pairs them up (a,b) leftrightarrow (b,a)$(a,b) \leftrightarrow (b,a)$, there are exactly frac122 = 6$\frac{12}{2} = 6$ independent choices. ### Step 2: Calculate total relations Each of the 6 independent pairs can either be included or excluded (2$2$ choices). Total relations = 1^4 times 2^6 = 64$= 1^4 \times 2^6 = 64$. ### Pattern Recognition Memorize the combinatorics of binary relations for n$n$ elements: Total = 2^n^2$= 2^{n^2}$, Reflexive = 2^n(n-1)$= 2^{n(n-1)}$, Symmetric = 2^n(n+1)/2$= 2^{n(n+1)/2}$, Reflexive & Symmetric = 2^n(n-1)/2$= 2^{n(n-1)/2}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Sets and Relations

### Related Formula |X| leq a iff -a leq X leq a$$|X| \leq a \iff -a \leq X \leq a$$ |X| > a iff X < -a text or X > a$$|X| > a \iff X < -a \text{ or } X > a$$ ### Core Logic Expand both set definitions onto the real number line to find explicit intervals for A$A$ and B$B$, then apply set operations to verify the options. ### Step 1: Simplify Set A |x^2 - 10| leq 6$$|x^2 - 10| \leq 6$$ -6 leq x^2 - 10 leq 6$$-6 \leq x^2 - 10 \leq 6$$ 4 leq x^2 leq 16$$4 \leq x^2 \leq 16$$ This yields x in [-4, -2] cup [2, 4]$x \in [-4, -2] \cup [2, 4]$. So, A = [-4, -2] cup [2, 4]$A = [-4, -2] \cup [2, 4]$. ### Step 2: Simplify Set B |x - 2| > 1$|x - 2| > 1$ x - 2 < -1 text or x - 2 > 1$$x - 2 < -1 \text{ or } x - 2 > 1$$ x < 1 text or x > 3$$x < 1 \text{ or } x > 3$$ So, B = (-infty, 1) cup (3, infty)$B = (-\infty, 1) \cup (3, \infty)$. ### Step 3: Evaluate Options A cup B = (-infty, 1) cup [2, infty)$A \cup B = (-\infty, 1) \cup [2, \infty)$ (Option 1 is wrong, has 1]$1]$) A cap B = [-4, -2] cup (3, 4]$A \cap B = [-4, -2] \cup (3, 4]$ (Option 3 is wrong, has [3,4]$[3,4]$) A - B = A cap B^c$A - B = A \cap B^c$. B^c = [1, 3]$B^c = [1, 3]$. A cap [1, 3] = [2, 3]$A \cap [1, 3] = [2, 3]$. (Option 2 is wrong, has [2, 3)$[2, 3)$) B - A = B cap A^c$B - A = B \cap A^c$. A^c = (-infty, -4) cup (-2, 2) cup (4, infty)$A^c = (-\infty, -4) \cup (-2, 2) \cup (4, \infty)$. B cap A^c = (-infty, -4) cup (-2, 1) cup (4, infty)$B \cap A^c = (-\infty, -4) \cup (-2, 1) \cup (4, \infty)$. This matches Option 4 perfectly. ### Pattern Recognition Draw inequalities involving absolute values directly onto a single number line graph to perform union and intersection operations without logic gaps. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Sets and Relations

### Related Formula textA relation R text is symmetric if (x,y) in R implies (y,x) in R.$$\text{A relation } R \text{ is symmetric if } (x,y) \in R \implies (y,x) \in R.$$ ### Core Logic Check condition y geq frac2x3$y \geq \frac{2x}{3}$ for each x in A = \2, 3, 5, 7, 9\$x \in A = \{2, 3, 5, 7, 9\}$ to generate ordered pairs (x,y)$(x,y)$. Count the pairs to get ell$\ell$. Then find asymmetric pairs to get m$m$. ### Step 1: Enumerate elements of R x = 2 implies y geq 4/3 = 1.33 implies y in \2, 3, 5, 7, 9\$x = 2 \implies y \geq 4/3 = 1.33 \implies y \in \{2, 3, 5, 7, 9\}$ (5 elements) x = 3 implies y geq 6/3 = 2 implies y in \2, 3, 5, 7, 9\$x = 3 \implies y \geq 6/3 = 2 \implies y \in \{2, 3, 5, 7, 9\}$ (5 elements) x = 5 implies y geq 10/3 = 3.33 implies y in \5, 7, 9\$x = 5 \implies y \geq 10/3 = 3.33 \implies y \in \{5, 7, 9\}$ (3 elements) x = 7 implies y geq 14/3 = 4.66 implies y in \5, 7, 9\$x = 7 \implies y \geq 14/3 = 4.66 \implies y \in \{5, 7, 9\}$ (3 elements) x = 9 implies y geq 18/3 = 6 implies y in \7, 9\$x = 9 \implies y \geq 18/3 = 6 \implies y \in \{7, 9\}$ (2 elements) Total elements in R$R$ is ell = 5 + 5 + 3 + 3 + 2 = 18$\ell = 5 + 5 + 3 + 3 + 2 = 18$. ### Step 2: Determine Missing Symmetric Pairs We need to check which reverse pairs (y,x)$(y,x)$ are missing. (2, 5) in R$(2, 5) \in R$, but (5, 2) notin R$(5, 2) \notin R$. (2, 7) in R$(2, 7) \in R$, but (7, 2) notin R$(7, 2) \notin R$. (2, 9) in R$(2, 9) \in R$, but (9, 2) notin R$(9, 2) \notin R$. (3, 5) in R$(3, 5) \in R$, but (5, 3) notin R$(5, 3) \notin R$. (3, 7) in R$(3, 7) \in R$, but (7, 3) notin R$(7, 3) \notin R$. (3, 9) in R$(3, 9) \in R$, but (9, 3) notin R$(9, 3) \notin R$. (5, 9) in R$(5, 9) \in R$, but (9, 5) notin R$(9, 5) \notin R$. These are m = 7$m = 7$ pairs that need to be added. Thus, ell + m = 18 + 7 = 25$\ell + m = 18 + 7 = 25$. ### Pattern Recognition When counting pairs for small sets, list them manually by rows. To make symmetric, any non-diagonal pair (x,y)$(x,y)$ that is present while (y,x)$(y,x)$ is absent counts as 1$1$ missing element. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Sets and Relations

### Related Formula textA relation R text is symmetric if (a,b) in R implies (b,a) in R$$\text{A relation } R \text{ is symmetric if } (a,b) \in R \implies (b,a) \in R$$ ### Core Logic Given 4y = 5x - 3 implies y = frac5x - 34$4y = 5x - 3 \implies y = \frac{5x - 3}{4}$. We evaluate this for x in \1, 2, ldots, 16\$x \in \{1, 2, \ldots, 16\}$ such that y$y$ is also an integer in the same set. - If x = 3 implies y = frac15 - 34 = 3 implies (3,3) in R$x = 3 \implies y = \frac{15 - 3}{4} = 3 \implies (3,3) \in R$ - If x = 7 implies y = frac35 - 34 = 8 implies (7,8) in R$x = 7 \implies y = \frac{35 - 3}{4} = 8 \implies (7,8) \in R$ - If x = 11 implies y = frac55 - 34 = 13 implies (11,13) in R$x = 11 \implies y = \frac{55 - 3}{4} = 13 \implies (11,13) \in R$ - If x = 15 implies y = frac75 - 34 = 18 notin M$x = 15 \implies y = \frac{75 - 3}{4} = 18 \notin M$ Therefore, R = \(3,3), (7,8), (11,13)\$R = \{(3,3), (7,8), (11,13)\}$ ### Step 1: Identifying Missing Symmetric Elements To make the relation symmetric, for every (x,y) in R$(x,y) \in R$, the pair (y,x)$(y,x)$ must also belong to R$R$. - (3,3)$(3,3)$ is symmetric to itself. - For (7,8)$(7,8)$, we must add (8,7)$(8,7)$. - For (11,13)$(11,13)$, we must add (13,11)$(13,11)$. Thus, the required elements to be added are (8,7)$(8,7)$ and (13,11)$(13,11)$, which totals 2 elements. ### Pattern Recognition When evaluating linear Diophantine equations over a small finite set, simply substitute modular values (here, 5x - 3 equiv 0 pmod 4$5x - 3 \equiv 0 \pmod 4$) to find the explicit ordered pairs, then mechanically apply the equivalence property criteria. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Sets, Relations and Functions