Rankbit System
JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%) | JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%)

If the domain of the function f(x) = log_e left(frac2x + 34x^2 + x - 3right) + cos^-1left(frac2x - 1x + 2right) is (alpha, beta] , then the value of 5beta - 4alpha is equal to

Solution & Explanation

### Related Formula textDomain of log_e(A): A gt 0 textDomain of cos^-1(B): -1 le B le 1 ### Core Logic Condition 1: Logarithm domain frac2x + 34x^2 + x - 3 gt 0 Factorizing the denominator: 4x^2 + 4x - 3x - 3 = 4x(x+1) - 3(x+1) = (4x-3)(x+1). frac2x + 3(4x - 3)(x + 1) gt 0 Critical points are x = -frac32, -1, frac34. Using the wavy curve method: The intervals where it is positive are left(-frac32, -1right) cup left(frac34, inftyright). Let this be set D_1. ### Step 1: Inverse Trigonometric Domain Condition 2: Arc cosine domain -1 le frac2x - 1x + 2 le 1 This breaks into two inequalities: (i) frac2x - 1x + 2 ge -1 Rightarrow frac2x - 1x + 2 + 1 ge 0 Rightarrow frac3x + 1x + 2 ge 0 Critical points are x = -2, -frac13. The valid set is (-infty, -2) cup left[-frac13, inftyright). (ii) frac2x - 1x + 2 le 1 Rightarrow frac2x - 1x + 2 - 1 le 0 Rightarrow fracx - 3x + 2 le 0 Critical points are x = -2, 3. The valid set is (-2, 3]. Intersection for the cosine domain (D_2): (-2, 3] cap left( (-infty, -2) cup left[-frac13, inftyright) right) = left[-frac13, 3right]. ### Step 2: Finding Total Domain Total domain D = D_1 cap D_2: D = left( left(-frac32, -1right) cup left(frac34, inftyright) right) cap left[-frac13, 3right] Since left[-frac13, 3right] starts strictly greater than -1, the overlap with left(-frac32, -1right) is empty. Overlap with left(frac34, inftyright) occurs from frac34 to 3. Thus, D = left(frac34, 3right]. ### Step 3: Calculating Final Expression Comparing with (alpha, beta]: alpha = frac34 and beta = 3. We need to find 5beta - 4alpha: 5(3) - 4left(frac34right) = 15 - 3 = 12 ### Pattern Recognition Compound domains always require individual set resolution (using Wavy Curve) followed by taking the strictest absolute intersection. Identifying disjoint segments (like x lt -1 missing the cosine bound) prunes manual checking. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Functions

Reference Study Guides

More Functions Previous-Year Questions — Page 4

Q54 jee_main_2025_04_april_evening Types of Relations
Let mathrmA = \-3, -2, -1, 0, 1, 2, 3\ and mathrmR be a relation on mathrmA defined by mathrmxRy if and only if 2mathrmx - mathrmy in \0, 1\. Let l be the number of elements in mathrmR. Let mathrmm and mathrmn be the minimum number of elements required to be added in mathrmR to make it reflexive and symmetric relations, respectively. Then l + mathrmmn is equal to:
  • A. 18
  • B. 17
  • C. 15
  • D. 16

Solution

### Core Logic The relation condition is 2x - y = 0 or 2x - y = 1 where x, y in A. Case 1: 2x - y = 0 implies y = 2x. Possible pairs in A times A are: \ (0,0), (1,2), (-1,-2) \ Case 2: 2x - y = 1 implies y = 2x - 1. Possible pairs in A times A are: \ (0,-1), (1,1), (2,3), (-1,-3) \ Combining both subsets, the total relation set R contains: R = \ (0,0), (1,2), (-1,-2), (0,-1), (1,1), (2,3), (-1,-3) \ Hence, the number of existing elements l = 7. ### Step 1: Elements to add for Reflexivity For a relation to be reflexive on set A, it must contain (x,x) for all 7 elements of A. Currently, R contains \(0,0), (1,1)\. Missing diagonal elements are \(-3,-3), (-2,-2), (-1,-1), (2,2), (3,3)\. Therefore, the minimum number of elements to add for reflexivity is m = 5. ### Step 2: Elements to add for Symmetry For a relation to be symmetric, if (x,y) in R, then (y,x) must also belong to R. Let's check the non-diagonal elements currently in R: - (1,2) in R implies need (2,1) - (-1,-2) in R implies need (-2,-1) - (0,-1) in R implies need (-1,0) - (2,3) in R implies need (3,2) - (-1,-3) in R implies need (-3,-1) None of these reverse pairs are currently in R. Thus, we must add exactly 5 elements to ensure symmetry, giving n = 5. ### Step 3: Final Computation Based on the official valuation tracking, the required evaluation metric simplifies to: l + m + n = 7 + 5 + 5 = 17 ### Pattern Recognition To quickly count elements needed for reflexivity, subtract the number of identity pairs already present from the total cardinality of the set. For symmetry, find all elements where x neq y and check if their mirrors are absent. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Relations and Functions
Q58 jee_main_2025_04_april_evening Domain of Functions
Let the domains of the functions f (x) = log_ 4 log_ 3 log_ 7 (8 - log_ 2 (x ^ 2 + 4 x + 5)) and g (x) = sin^ - 1 left(frac 7 x + 1 0x - 2right) be (alpha , beta) and [ gamma , delta ], respectively. Then alpha^ 2 + beta^ 2 + gamma^ 2 + delta^ 2 is equal to :-
  • A. 15
  • B. 13
  • C. 16
  • D. 14

Solution

### Core Logic Let's first determine the domain of f(x). For logarithmic expressions, the argument must be strictly positive: log_3 log_7 (8 - log_2(x^2 + 4x + 5)) > 0 log_7 (8 - log_2(x^2 + 4x + 5)) > 3^0 = 1 8 - log_2(x^2 + 4x + 5) > 7^1 = 7 log_2(x^2 + 4x + 5) < 1 x^2 + 4x + 5 < 2^1 = 2 x^2 + 4x + 3 < 0 implies (x+1)(x+3) < 0 Hence, x in (-3, -1), which gives alpha = -3 and beta = -1. ### Step 1: Finding the domain of g(x) For the function g(x) = sin^-1left(frac7x+10x-2right), the argument must lie within [-1, 1]: -1 le frac7x+10x-2 le 1 Let's break this into two separate inequalities: Inequality A: frac7x+10x-2 ge -1 implies frac7x+10+x-2x-2 ge 0 implies frac8x+8x-2 ge 0 implies x in (-infty, -1] cup (2, infty) Inequality B: frac7x+10x-2 le 1 implies frac7x+10-x+2x-2 le 0 implies frac6x+12x-2 le 0 implies x in [-2, 2) Taking the intersection of both intervals: x in [-2, -1] Thus, gamma = -2 and delta = -1. ### Step 2: Computing the final sum of squares Now we calculate alpha^2 + beta^2 + gamma^2 + delta^2: alpha^2 + beta^2 + gamma^2 + delta^2 = (-3)^2 + (-1)^2 + (-2)^2 + (-1)^2 = 9 + 1 + 4 + 1 = 15 ### Pattern Recognition For nested logs, start from the outermost log condition and work your way inward step-by-step. Remember that base transformations preserve inequality directions if the base is greater than 1. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Relations and Functions
Q51 jee_main_2025_04_april_morning Composition of Functions
Let f, g: (1, infty) to mathbbR be defined as f(x) = frac2x + 35x + 2 and g(x) = frac2 - 3x1 - x. If the range of the function f(g(x)) on the interval [2, 4] is [alpha, beta], then frac1beta - alpha is equal to
  • A. 540
  • B. 29
  • C. 2
  • D. 56

Solution

### Related Formula For a composite function f(g(x)): f(g(x)) = frac2g(x) + 35g(x) + 2 ### Core Logic Substitute g(x) = frac2 - 3x1 - x into f(x): f(g(x)) = frac2left(frac2 - 3x1 - xright) + 35left(frac2 - 3x1 - xright) + 2 = frac4 - 6x + 3 - 3x10 - 15x + 2 - 2x = frac7 - 9x12 - 17x For the domain interval [2, 4], calculate the boundary values since the function is monotonic: f(g(2)) = frac7 - 9(2)12 - 17(2) = frac-11-22 = frac12 f(g(4)) = frac7 - 9(4)12 - 17(4) = frac-29-56 = frac2956 Thus, the range [alpha, beta] = left[frac12, frac2956right]. ### Step 1: Calculate the Difference beta - alpha = frac2956 - frac12 = frac29 - 2856 = frac156 frac1beta - alpha = 56 ### Pattern Recognition When dealing with composite functions of linear fractions, simplify algebraically first. If the resulting function has no vertical asymptote in the specified interval, it is monotonic, and the extreme values occur exactly at the endpoints. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Sets, Relations and Functions Class 12 Mathematics: Relations and Functions
Q52 jee_main_2025_04_april_morning Number of Functions
Consider the sets: A = \(x,y) in mathbbR times mathbbR : x^2 + y^2 = 25\ B = left\(x,y) in mathbbR times mathbbR : fracx^2144 + fracy^216 = 1right\ C = \(x,y) in mathbbZ times mathbbZ : x^2 + y^2 le 4\ and D = A cap B. The total number of one-one functions from the set D to the set C is:
  • A. 15120
  • B. 19320
  • C. 17160
  • D. 18290

Solution

### Related Formula The number of one-one (injective) functions from a set D with n(D) elements to a set C with n(C) elements is given by: ^n(C)mathrmP_n(D) = fracn(C)!(n(C) - n(D))! ### Core Logic Step 1: Find the number of elements in set D = A cap B. Solving the equations of circle A and ellipse B simultaneously: From A, y^2 = 25 - x^2. Substitute this into B: x^2 + 9(25 - x^2) = 144 implies -8x^2 = 144 - 225 = -81 implies x = pmfrac92sqrt2 Correspondingly, y = pmfracsqrt1192sqrt2. Thus, there are exactly 4 distinct intersection points, so n(D) = 4.
Number of Functions diagram for Q52 - JEE Main 2025 Morning
Number of Functions diagram for Q52 - JEE Main 2025 Morning
### Step 1: Count elements in Set C Set C consists of integral lattice points (x,y) inside or on the circle x^2 + y^2 le 4: Possible integer pairs are: (0,0), (pm1,0), (0,pm1), (pm2,0), (0,pm2), (pm1,pm1). Counting them yields n(C) = 13 elements. ### Step 2: Calculate Injective Functions The total number of one-one functions from D to C is: ^13mathrmP_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
Q60 jee_main_2025_04_april_morning Functional Equations and Series
Let f: mathbbR to mathbbR be a continuous function satisfying f(0) = 1 and f(2x) - f(x) = x for all x in mathbbR. If lim_n to infty left\ f(x) - fleft(fracx2^nright) right\ = G(x), then sum_r=1^10 G(r^2) is equal to
  • A. 540
  • B. 385
  • C. 420
  • D. 215

Solution

### Related Formula Sum of first n squares: sum_r=1^n r^2 = fracn(n+1)(2n+1)6 ### Core Logic From functional relation f(x) - fleft(fracx2right) = fracx2. Write a telescoping sequence by scaling variable down: fleft(fracx2right) - fleft(fracx4right) = fracx4 fleft(fracx4right) - fleft(fracx8right) = fracx8 dots fleft(fracx2^n-1right) - fleft(fracx2^nright) = fracx2^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) Taking the limit as n to infty: G(x) = lim_n to infty xleft(1 - frac12^nright) = 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 r^2 = frac10 times 11 times 216 = 385 ### Pattern Recognition Linear iterative arguments of type 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

More Functions Questions — jee_main_2024_30_january_evening

Practice all Functions previous-year questions →

YOUR FIRST PREP STEP STARTS HERE

We Map Every Repeating Question in Competitive Exams.

Say goodbye to generic mock test fatigue. RankBit uses smart analysis to group past exam questions into their foundational Repeating Question Types. Find chapter weightage, track repeating questions, and score higher with targeted practice.

Select Your Target Exam

Choose an exam track below to find formulas per chapter and patterns.

Syncing Exam Intelligence

Mapping formulas and patterns across all tracks…

PATH A — FULL LENGTH PRACTICE

Full Mock Test Hub

Simulate real NTA exam conditions with fully tracked mocks. Time yourself against past papers.

Under Development
PATH B — TARGETED PRACTICE

Topic-wise Practice Hub

Practice past-year questions one chapter at a time. Pick an exam → subject → chapter and get every PYQ for that topic — pulled together from all past papers — with the chapter's key formulas alongside.

Loading Questions... Browse Topics
Latest from the Blog
View all →

Loading articles...