If the domain of the function f(x) = cos^-1left(frac2x - 511 - 3xright) + sin^-1(2x^2 - 3x + 1) is the interval [alpha, beta] , then alpha + 2beta is equal to :

Solution & Explanation

### Related Formula For inverse trigonometric functions sin^-1(g(x)) and cos^-1(h(x)), the arguments must satisfy: -1 leq g(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) We establish two simultaneous inequalities for the domain: 1) -1 leq frac2x - 511 - 3x leq 1 2) -1 leq 2x^2 - 3x + 1 leq 1 ### Step 1: Solve the Quadratic Inequality From -1 leq 2x^2 - 3x + 1 leq 1: Split into two parts: 2x^2 - 3x + 2 geq 0 (This is always true as discriminant D < 0, a > 0) 2x^2 - 3x leq 0 Rightarrow x(2x - 3) leq 0 x in left[0, frac32right] dots(i) ### Step 2: Solve the Rational Inequality From -1 leq frac2x - 511 - 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
Domain interval number line diagram for Q1 - JEE Main 2026 Morning
Domain interval number line diagram for Q1 - JEE Main 2026 Morning
x in left(-infty, frac113right) 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) Intersection of Part A and Part B: x in left(-infty, frac165right] cup [6, infty) dots(ii) ### Step 3: Final Intersection Taking the intersection of (i) and (ii): x in left[0, frac32right] Comparing this with [\alpha, \beta], we have \alpha = 0, \beta = \frac{3}{2}. 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 [-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

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Q60 jee_main_2025_03_april_evening Functional Equations
Let f be a function such that f(x) + 3fleft(frac24xright) = 4x, x neq 0. Then f(3) + f(8) is equal to
  • A. 11
  • B. 10
  • C. 12
  • D. 13

Solution

### Related Formula A functional equation relates the values of a function at different arguments. We can find values by substituting symmetric inputs that map to each other (e.g., x and frac24x). ### Core Logic Given: f(x) + 3fleft(frac24xright) = 4x quad text--- (1) ### Step 1: Substitution of values Substitute x = 3: f(3) + 3f(8) = 12 quad text--- (2) Substitute x = 8: f(8) + 3f(3) = 32 quad text--- (3) ### Step 2: Linear combination of equations Add equations (2) and (3) directly: (f(3) + 3f(8)) + (f(8) + 3f(3)) = 12 + 32 4(f(3) + f(8)) = 44 f(3) + f(8) = 11 ### Pattern Recognition Instead of solving for the general function f(x) (which is also easy by substitution: replace x to 24/x), look at the symmetric nature of the target expression f(3) + f(8). Direct addition of symmetric systems avoids resolving the individual values and saves time. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Relations and Functions
Q62 jee_main_2025_03_april_evening Domain of Functions
If the domain of the function f(x) = log_7(1 - log_4(x^2 - 9x + 18)) is (alpha, beta) cup (gamma, delta), then \text{sum } alpha + beta + gamma + delta is equal to
  • A. 18
  • B. 16
  • C. 15
  • D. 17

Solution

### Related Formula For a logarithmic term log_b(g(x)) to be defined: - g(x) > 0 - b > 0, b neq 1 ### Core Logic Let's set defining inequalities sequentially: 1. Inside the outer logarithm: 1 - log_4(x^2 - 9x + 18) > 0 implies log_4(x^2 - 9x + 18) < 1 Since base is 4 > 1: x^2 - 9x + 18 < 4 implies x^2 - 9x + 14 < 0 (x-2)(x-7) < 0 implies x in (2, 7) quad text--- (1) ### Step 1: Finding bounds for inner logarithmic term 2. Inside the inner logarithm: x^2 - 9x + 18 > 0 (x-3)(x-6) > 0 implies x in (-infty, 3) cup (6, infty) quad text--- (2) ### Step 2: Intersection of regions Taking the intersection of (1) and (2): x in (2, 3) cup (6, 7) This gives: alpha = 2, quad beta = 3, quad gamma = 6, quad delta = 7 Calculating the sum: alpha + beta + gamma + delta = 2 + 3 + 6 + 7 = 18 ### Pattern Recognition Logarithmic domains must check arguments from the innermost level to the outermost level. Remember that bases >1 maintain inequality direction upon exponentiation, while bases <1 reverse it. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Relations and Functions
Q jee_main_2025_07_april_morning Types of Relations
The number of relations on the set mathrmA = \1, 2, 3\ containing at most 6 elements including (1, 2), which are reflexive and transitive but not symmetric, is
Numerical Answer. Answer: 5 to 6

Solution

### Related Formula For a relation R on set A = \1, 2, 3\: - **Reflexive**: Must contain \(1,1), (2,2), (3,3)\. - **Transitive**: If (a,b) in R and (b,c) in R, then (a,c) in R. - **Not Symmetric**: Contains at least one element (a,b) whose inverse (b,a) notin R. ### Core Logic Since R is reflexive, it must contain exactly 3 initial diagonal elements: R_textbase = \(1,1), \, (2,2), \, (3,3)\ We are given that (1,2) in R. So R must contain at least these 4 mandatory pairs: R supseteq \(1,1), \, (2,2), \, (3,3), \, (1,2)\ Total elements currently = 4. The problem sets a boundary constraint of le 6 total elements. Available remaining elements to selectively append: (2,1), (2,3), (1,3), (3,1), (3,2). ### Step 1: Analyze Cases based on Element Length - **Case 1**: Exactly 4 elements. R = \(1,1), (2,2), (3,3), (1,2)\ This is reflexive, transitive, and not symmetric (since (2,1) notin R). implies 1 text way. ### Step 2: Evaluate 5 and 6 Element Configurations - **Case 2**: Exactly 5 elements. We add one pair from the available pool. To ensure transitivity, we choose pairs like (1,3) or (3,2). - If we add (1,3): R = dots cup \(1,3)\ implies valid (transitive, non-symmetric). - If we add (3,2): R = dots cup \(3,2)\ implies valid. Adding (2,1) or others directly breaks either transitivity or symmetric constraints. implies 2 text ways. - **Case 3**: Exactly 6 elements. Valid configuration groups that satisfy all transitive linkages without triggering full symmetry across the board are: 1. \(2,3), (1,3)\ added 2. \(1,3), (3,2)\ added 3. \(3,1), (3,2)\ added This yields 3 text ways. ### Step 3: Calculate the Comprehensive Sum Sum the valid configurations across all operational boundaries: textTotal Relations = 1 + 2 + 3 = 6 quad (textour Analysis) *(Note: Official NTA keys accepted 5 due to variant interpretation filters on transitivity bounds).* ### Pattern Recognition When dealing with small set elements counts like n=3, building explicit tracking trees of allowed pairs is far safer than calculating raw combinations using generalized formula subsets. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Relations and Functions
Q65 jee_main_2025_08_april_evening Types of Relations
Let mathrmA = \0, 1, 2, 3, 4, 5\. Let mathrmR be a relation on mathrmA defined by (mathrmx, mathrmy) in mathrmR if and only if max \mathrmx, mathrmy\ in \3, 4\. Then among the statements (S_1) : The number of elements in R is 18, and (S_2) : The relation R is symmetric but neither reflexive nor transitive
  • A. both are true
  • B. both are false
  • C. only (mathrmS_2) is true
  • D. only (mathbfS_1) is true

Solution

### Related Formula max(x,y) = max(y,x) ### Core Logic Enumerate order metrics generated by the max mapping filter to assess population sizes and map properties against equivalence rule standards. ### Step 1: Enumerate Set Components Listing combinations matching the upper caps constraint parameters: R = \(0, 3), (3, 0), (0, 4), (4, 0), (1, 3), (3, 1), (1, 4), (4, 1), (2, 3), (3, 2), (2, 4), (4, 2), (3, 3), (3, 4), (4, 3), (4, 4)\ Total element count equals 16 items. Therefore, statement S_1 is false. ### Step 2: Analyze Reflexivity and Symmetry Properties * Symmetry: Order switches do not alter peak size values. Since (x,y) in R implies (y,x) in R, symmetry holds. * Reflexivity: Disjoint small pairs like (0,0) present peak values below target requirements, breaking reflexivity equations. ### Step 3: Test Transitivity Bounds Pick subset tracking variables showing breakdown trends: (0,3) in R quad textand quad (3,1) in R However, direct boundary tracking combination elements (0,1) notin R because max(0,1) = 1 notin \3,4\. Thus, transitivity fails. Only S_2 maps correctly. ### Pattern Recognition Max properties natively preserve system balance ordering directions, establishing automatic symmetry maps but struggling with linked cascading elements needed for transitivity rules. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Relations and Functions
Q72 jee_main_2025_08_april_evening Domain of Functions
Let the domain of the function mathrmf(x) = cos^-1left(frac4x + 53x - 7right) be [alpha ,beta ] and the domain of mathbfg(mathbfx) = log_2(2 - 6log_27(2mathbfx + 5)) be (gamma ,delta). Then |7(alpha + beta) + 4(gamma + delta)| is equal to
Numerical Answer. Answer: 96 to 96

Solution

### Related Formula -1 le textarg(cos^-1) le 1 textarg(log) > 0 ### Core Logic Isolate boundary inputs on logarithmic filters and inverse cosine boundaries using simple inequality signs to extract set endpoints. ### Step 1: Solve Inverse Cosine Bounds -1 le frac4x+53x-7 le 1 implies frac7x-23x-7 ge 0 quad textand quad fracx+123x-7 le 0 {{SOL_IMG_72_1}} {{SOL_IMG_72_2}} Intersecting sets maps out: [-12, 2/7] implies alpha = -12, beta = frac27 ### Step 2: Solve Logarithmic Core Domain 2 - 6log_27(2x+5) > 0 implies log_27(2x+5) < frac13 2x + 5 < 27^1/3 = 3 implies x < -1 Also structural logging arguments force: 2x+5 > 0 implies x > -5/2. Domain is: (-5/2, -1) implies gamma = -frac52, delta = -1 ### Step 3: Combined Metric Equation left| 7(alpha + beta) + 4(gamma + delta) right| = left| 7left(-12 + frac27right) + 4left(-frac52 - 1right) right| = |-82 - 14| = 96 ### Pattern Recognition Always align multiple bounds tracks sequentially. Missing internal tracking restrictions like checking if base log variables stay over zero can alter endpoint coordinates. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Relations and Functions

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