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

Reference Study Guides

Functions

More Functions Previous-Year Questions — Page 5

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
Q53 jee_main_2025_07_april_evening Set Inclusion and Regions
Let A = \(alpha ,beta)in mathbfRtimes mathbfR:|alpha -1|leq 4 text and |beta -5|leq 6\ and B = \(alpha , beta) in mathbfR times mathbfR: 16 (alpha - 2)^2 + 9 (beta - 6)^2 leq 144\. Then
  • A. B subset A
  • B. A cup B = \(x, y) : -4 leq x leq 4, -1 leq y leq 11\
  • C. neither A subset B nor B subset A
  • D. A subset B

Solution

### Related Formula An ellipse equation is structured as: frac(x-h)^2a^2 + frac(y-k)^2b^2 leq 1 ### Core Logic Analyzing set A: |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 Thus, region A forms a rectangle bounded between x in [-3, 5] and y in [-1, 11]. Analyzing set B: 16(alpha - 2)^2 + 9(beta - 6)^2 le 144 Dividing by 144: frac(alpha - 2)^29 + frac(beta - 6)^216 le 1 This represents the interior and boundary of an ellipse centered at (2, 6) with semi-minor axis a = 3 and semi-major axis b = 4. ### Step 1: Spatial Inclusion Check Let's check the extreme horizontal and vertical extents of the ellipse B: Horizontal extent: x in [2 - 3, 2 + 3] = [-1, 5] Vertical extent: y in [6 - 4, 6 + 4] = [2, 10] Comparing with the boundaries of rectangle A (x in [-3, 5] and y in [-1, 11]): [-1, 5] subseteq [-3, 5] [2, 10] subseteq [-1, 11]
Set Inclusion and Regions diagram for Q53 - JEE Main 2025 Evening
Set Inclusion and Regions diagram for Q53 - JEE Main 2025 Evening
Since all points of the ellipse lie perfectly inside the rectangular region, we conclusively find that B subset A. ### Pattern Recognition A bounding box check (finding h pm a and 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
Q54 jee_main_2025_07_april_evening Range of Rational Functions
If the range of the function f(x) = frac5 - xx^2 - 3x + 2, x neq 1, 2, is (-infty , alpha ] cup [ beta , infty), then alpha^2 +beta^2 is equal to :
  • A. 190
  • B. 192
  • C. 188
  • D. 194

Solution

### Related Formula For a quadratic equation Ax^2 + Bx + C = 0 to yield real roots, its discriminant must satisfy: D = B^2 - 4AC ge 0 ### Core Logic Set y = frac5 - xx^2 - 3x + 2: y(x^2 - 3x + 2) = 5 - x yx^2 - 3xy + 2y + x - 5 = 0 Rearranging into a standard quadratic equation in terms of x: yx^2 + (1 - 3y)x + (2y - 5) = 0 ### Step 1: Discriminant Method Case I: If y = 0, the equation simplifies to x - 5 = 0 implies x = 5, which is a valid part of the domain. Thus, 0 belongs to the range. Case II: If y neq 0, for x to be real, D ge 0: (1 - 3y)^2 - 4(y)(2y - 5) ge 0 9y^2 + 1 - 6y - 8y^2 + 20y ge 0 y^2 + 14y + 1 ge 0 ### Step 2: Solving the Inequality Completing the square for y^2 + 14y + 1 ge 0: (y + 7)^2 - 48 ge 0 implies (y + 7)^2 ge (4sqrt3)^2 This gives: y le -7 - 4sqrt3 quad textor quad y ge -7 + 4sqrt3 Comparing with the interval (-infty , alpha ] cup [ beta , infty): alpha = -7 - 4sqrt3 beta = -7 + 4sqrt3 ### Step 3: Finding alpha^2 + beta^2 Using algebraic identities: alpha^2 + beta^2 = (-7 - 4sqrt3)^2 + (-7 + 4sqrt3)^2 = 2(7^2 + (4sqrt3)^2) = 2(49 + 48) = 2(97) = 194 ### Pattern Recognition For rational expressions of the form fractextLineartextQuadratic, converting to a quadratic in x and forcing D ge 0 establishes the range boundaries elegantly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Sets, Relations and Functions
Q52 jee_main_2025_24_jan_evening One-One and Onto Functions
The function f:(-infty,infty)rightarrow(-infty,1), defined by f(x)=frac2^x-2^-x2^x+2^-x is: [cite: 3251, 3252]
  • A. textOne-one but not onto
  • B. textOnto but not one-one
  • C. textBoth one-one and onto
  • D. textNeither one-one nor onto

Solution

### 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: f(x) = frac2^2x - 12^2x + 1 = 1 - frac22^2x + 1 ### Step 1: Check One-One property Differentiating f(x) with respect to 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 Since 2^2x > 0 and ln 2 > 0, f'(x) > 0 always. Thus, 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 - 0 = 1 Thus, the range of the function is (-1, 1). Since the given co-domain is (-infty, 1) and textRange neq textCo-domain , the function is not onto. ### Pattern Recognition The expression given is a shifted form of the hyperbolic tangent function tanh(x ln 2). Hyperbolic tangent always maps to (-1, 1), making its restriction against (-infty, 1) non-surjective (not onto). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Relations and Functions
Q58 jee_main_2025_24_jan_evening Linear Programming and Inequalities in Two Variables
Let the points (frac112,alpha) lie on or inside the \triangle with sides x+y=11, x+2y=16 and 2x+3y=29 Then the product of the smallest and the largest values of a is equal to: [cite: 3284, 3285, 3286, 3287, 3288, 3289]
  • A. 22
  • B. 44
  • C. 33
  • D. 55

Solution

### Related Formula For a vertical line x = x_0 crossing a bounded region, the valid coordinates of y sit between the boundary lines intersecting that specific line vertical plane. ### Core Logic The point given is fixed at x = frac112 = 5.5[cite: 3285, 3925]. We evaluate the values of y along this vertical line segment across each boundary edge.
Linear Programming region graph for Q58 - JEE Main 2025 Evening
Linear Programming region graph for Q58 - JEE Main 2025 Evening
### Step 1: Evaluate Intersections Substitute x = frac112 into the three linear constraints: 1. From x + y = 11: frac112 + y = 11 Rightarrow y = 11 - 5.5 = 5.5 2. From x + 2y = 16: frac112 + 2y = 16 Rightarrow 2y = 16 - 5.5 = 10.5 Rightarrow y = 5.25 3. From 2x + 3y = 29: 2left(frac112right) + 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 along the section line is delimited by y = 5.5 and y = 6: alpha_min = frac112 = 5.5 alpha_max = 6 The product of the limits is : alpha_min cdot alpha_max = frac112 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 saves time during multi-line area problems. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Linear Inequalities Class 11 Mathematics: Straight Lines

More Functions Questions — jee_main_2026_21_jan_morning

Practice all Functions previous-year questions →

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%)
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.

Now Live Open
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

Exam Analysis Tools

Chapter weightage trends and topic-wise practice — all mapped from past shift papers.

Year-wise PYQ

Chapter Weightage & Trends

See high-yield chapters, year-wise counts and the frequency-based chance each topic repeats next exam.

Weightage Board Open
Targeted PYQs

Topic-wise Practice Hub

Every past-year question for a chapter pulled from all papers, with its key formulas alongside.

One chapter at a time Open
Latest from the Blog
View all →

Loading articles...