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README.md

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+---
+license: cc-by-4.0
+language:
+- en
+task_categories:
+- question-answering
+- text-generation
+tags:
+- education
+- mathematics
+- physics
+- chemistry
+- jee
+- jee-advanced
+- exam
+- india
+pretty_name: JEE Advanced Question Bank
+configs:
+- config_name: mathematics
+  data_files:
+  - split: train
+    path: mathematics/train.jsonl
+  - split: test
+    path: mathematics/test.jsonl
+---
+
+# JEE Advanced — Question Bank
+
+A structured dataset of **JEE Advanced** examination questions with full
+worked solutions and diagrams. JEE Advanced questions are more analytical
+than JEE Main — many are subjective, integer, or numerical-answer type with
+detailed multi-step solutions.
+
+**Current subset:**
+- **Mathematics** — 48 questions
+
+Physics and Chemistry subsets are planned and will use the same schema.
+
+## Structure
+
+Organised into **subsets by subject** and **splits** (train / test):
+
+```
+mathematics/
+├── train.jsonl
+├── test.jsonl
+└── images/          (diagrams and figures — PNG/JPEG)
+```
+
+## Schema
+
+| Field              | Type      | Description                                                       |
+|--------------------|-----------|-------------------------------------------------------------------|
+| `question_id`      | string    | Unique ID, e.g. `ADV-M01-Q1`.                                    |
+| `question`         | string    | Question text in LaTeX. `[IMAGE]` marks inline figures.          |
+| `question_images`  | list[str] | Relative paths to images used in the question.                   |
+| `answer`           | string    | The answer — may be an integer, a LaTeX expression, or a relation.|
+| `solution`         | string    | Full worked solution in LaTeX. `[IMAGE]` marks inline figures.   |
+| `solution_images`  | list[str] | Relative paths to images used in the solution.                   |
+| `subject`          | string    | `Mathematics`.                                                   |
+| `topic`            | string    | Auto-detected from content, e.g. `Complex Numbers`, `Circles`.   |
+| `difficulty`       | string    | Reserved (not provided in source).                              |
+| `question_type`    | string    | `numerical` / `subjective` / `single_correct`.                  |
+| `has_image`        | bool      | Whether the question or solution contains a figure.             |
+| `exam`             | string    | `JEE Advanced`.                                                  |
+| `source_paper`     | string    | Original source paper reference.                                |
+
+## Notes
+
+- Unlike JEE Main (4-option MCQs), JEE Advanced questions here are largely
+  subjective / integer / numerical, so the schema uses a single `answer`
+  field rather than four options.
+- `topic` is auto-detected from the question and solution content, since the
+  source papers do not carry explicit topic labels.
+- All figures are stored under `mathematics/images/` and referenced by
+  relative path.
+
+## Loading
+
+```python
+from datasets import load_dataset
+ds = load_dataset("eQOURSE/jee-advanced-questions", "mathematics")
+```

mathematics/test.jsonl

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+{"question_id": "ADV-M01-Q7", "question": "Let\n$M = \\left\\{ (x,y)\\mathbb{\\in R \\times R:}x^{2} + y^{2} \\leq r^{2} \\right\\}$\nwhere $r > 0$\nConsider the geometric progression\n$a_{n} = \\frac{1}{2^{n - 1}},\\ n = 1,2,3\\ldots.$ Let $S_{0} = 0$ and,\nfor $n \\geq 1,$let $S_{n}$ denote the sum of the first n terms of this\nprogressions. For $n \\geq 1,$let $C_{n}$ denote the circle with center\n$(S_{n - 1},0)$ and radius, $a_{n}$ and $D_{n}$ denote the circle with\ncenter $(S_{n - 1},S_{n - 1})$ and $a_{n}$.\nConsider $M$ with\n$r = \\frac{\\left( 2^{199} - 1 \\right)\\sqrt{2}}{2^{198}}$. The number\nof all those circles $D_{n}$ that are inside $M$ is", "question_images": [], "answer": "199", "solution": "$\\because\\ r = \\frac{\\left( 2^{199} - 1 \\right)\\sqrt{2}}{2^{198}}$\\\nNow, $\\sqrt{2}S_{n - 1} + a_{n} < \\frac{2^{199} - 1}{2^{198}}\\sqrt{2}$\\\n$${\\Rightarrow 2\\sqrt{2}\\left( 1 - \\frac{1}{2^{n - 1}} \\right) + \\frac{1}{2^{n - 1}} < \\frac{2^{199} - 1}{2^{198}}\n}{\\Rightarrow \\frac{2\\sqrt{2} - 1}{{2.2}^{n - 2}} > \\frac{\\sqrt{2}}{2^{198}}\n}{\\Rightarrow 2^{n - 2} < \\left( 2 - \\frac{1}{\\sqrt{2}} \\right)2^{197}\n}{\\therefore n \\leq 199 \\Rightarrow n = 199}$$", "solution_images": [], "subject": "Mathematics", "topic": "Circles", "difficulty": "", "question_type": "numerical", "has_image": false, "exam": "JEE Advanced", "source_paper": "Mathematics_1-50_with_img.docx"}
+{"question_id": "ADV-M01-Q9", "question": "For non-negative integers $n$, let\n$f(n) = \\frac{\\sum_{k = 0}^{n}\\mspace{2mu}\\mspace{2mu} sin\\left( \\frac{k + 1}{n + 2}\\pi \\right)sin\\left( \\frac{k + 2}{n + 2}\\pi \\right)}{\\sum_{k = 0}^{n}\\mspace{2mu}\\mspace{2mu}\\sin^{2}\\left( \\frac{k + 1}{n + 2}\\pi \\right)}$\\\nAssuming $\\cos^{- 1}x$ takes values in $\\lbrack 0,\\pi\\rbrack$, and\n$\\alpha = tan\\left( \\cos^{- 1}f(6) \\right)$, then find the value of\n$\\alpha^{2} + 2\\alpha - 1$", "question_images": [], "answer": "0 (zero)", "solution": "$f(n) = \\frac{\\sum_{k = 0}^{n}\\mspace{2mu} 2sin\\left( \\frac{k + 1}{n + 2}\\pi \\right)sin\\left( \\frac{k + 2}{n + 2}\\pi \\right)}{\\sum_{k = 0}^{n}\\mspace{2mu} 2\\sin^{2}\\left( \\frac{k + 1}{n + 2} \\right)\\pi}$,\\\nwhere $n$ is non negative integer\\\n$${= \\frac{\\sum_{k = 0}^{n}\\mspace{2mu}\\left\\lbrack cos\\left( \\frac{\\pi}{n + 2} \\right) - cos\\frac{(2k + 3)\\pi}{n + 2} \\right\\rbrack}{\\sum_{k = 0}^{n}\\mspace{2mu}\\left\\lbrack 1 - cos\\frac{2(k + 1)\\pi}{n + 2} \\right\\rbrack}\n}{= \\frac{(n + 1)cos\\left( \\frac{\\pi}{n + 2} \\right) - \\left\\lbrack cos\\frac{3\\pi}{n + 2} + cos\\frac{5\\pi}{n + 2} + \\ldots\\ldots\\ldots + cos\\frac{(2n + 3)\\pi}{n + 2} \\right\\rbrack}{n + 1 - \\left\\lbrack cos\\frac{2\\pi}{n + 2} + cos\\frac{4\\pi}{n + 2} + \\ldots\\ldots + cos\\frac{2(n + 1)\\pi}{n + 2} \\right\\rbrack}\n}{\\frac{(n + 1)cos\\left( \\frac{\\pi}{n + 2} \\right) - \\frac{\\frac{sin(n + 1)\\pi}{n + 2}}{sin\\left( \\frac{\\pi}{n + 2} \\right)} \\cdot cos\\frac{(2n + 6)\\pi}{2(n + 2)}}{sin(n + 1)\\pi}\n}{n + 1 - \\frac{\\frac{sin(n + 1)\\pi}{n + 2}}{sin\\left( \\frac{\\pi}{n + 2} \\right)} \\cdot cos\\frac{(2n + 4)\\pi}{2(n + 2)}\n}{= \\frac{(n + 1)cos\\frac{\\pi}{n + 2} + cos\\frac{\\pi}{n + 2}}{n + 1 + 1} = \\frac{(n + 2)cos\\left( \\frac{\\pi}{n + 2} \\right)}{n + 2}\n}{\\therefore f(n) = cos\\left( \\frac{\\pi}{n + 2} \\right)\n}{\\lim_{n \\rightarrow \\infty}\\mspace{2mu} f(n) = \\lim_{n \\rightarrow \\infty}\\mspace{2mu} cos\\left( \\frac{\\pi}{n + 2} \\right) = 1\n}{f(4) = cos\\left( \\frac{\\pi}{4 + 2} \\right) = cos\\frac{\\pi}{6} = \\frac{\\sqrt{3}}{2}\n}$$If $\\alpha = tan\\left( \\cos^{- 1}f(6) \\right)$\\\n$$= tan\\left( \\cos^{- 1}\\left( cos\\frac{\\pi}{8} \\right) \\right) = tan\\frac{\\pi}{8}\n$$Now,\n$tan\\frac{\\pi}{4} = 1 \\Rightarrow \\frac{2tan\\frac{\\pi}{8}}{1 - \\tan^{2}\\pi/8} = 1$\\\n$${\\Rightarrow \\frac{2\\alpha}{1 - \\alpha^{2}} = 1 \\Rightarrow \\alpha^{2} + 2\\alpha - 1 = 0\n}{\\sin\\left( 7\\cos^{- 1}{f(5)} \\right) = \\sin\\left( 7\\cos^{- 1}\\left( \\cos\\frac{\\pi}{7} \\right) \\right) = \\sin\\left( 7 \\times \\frac{\\pi}{7} \\right)}$$\n$sin\\ \\pi = 0$", "solution_images": [], "subject": "Mathematics", "topic": "Limits", "difficulty": "", "question_type": "subjective", "has_image": false, "exam": "JEE Advanced", "source_paper": "Mathematics_1-50_with_img.docx"}
+{"question_id": "ADV-M01-Q15", "question": "Let $f(x) = \\left\\{ \\begin{matrix}\ne^{x}, & 0 \\leq x \\leq 1 \\\\\n2 - e^{x - 1}, & 1 < x \\leq 2 \\\\\nx - e, & 2 < x \\leq 3\n\\end{matrix} \\right.\\ $ and\n$g(x) = \\int_{0}^{x}\\mspace{2mu} f(t)dt,x \\in \\lbrack 0,3\\rbrack$ then\nfind points of maxima\nand minima of $g(x)$", "question_images": [], "answer": "$g(x)$ has local maxima at $x = 1 + ln2,\\ 1$ and local\nminima at $x = e,2$", "solution": "Given\n$g(x) = \\int_{0}^{x}\\mspace{2mu} f(t)dt,x \\in \\lbrack 0,3\\rbrack$\\\n$${\\Rightarrow g^{'}(x) = f(x) = \\left\\{ \\begin{matrix}\ne^{x}, & 0 \\leq x \\leq 1 \\\\\n2 - e^{x - 1}, & 1 < x \\leq 2 \\\\\nx - e, & 2 < x \\leq 3\n\\end{matrix} \\right.\\ \n}{\\therefore g^{'}(x) = 0 \\Rightarrow e^{x - 1} = 2\\text{~or~}x - e = 0\n}{\\Rightarrow x - 1) = ln2\\text{~or~}x = e \\Rightarrow x = 1 + ln2\\text{~or~}e\n}{g^{''}(x) = \\left\\{ \\begin{matrix}\ne^{x}, & 0 \\leq x \\leq 1 \\\\\n - e^{x - 1}, & 1 < x \\leq 2 \\\\\n1, & 2 < x \\leq 3\n\\end{matrix} \\right.\\ \n}$$$g(x)$ has local max, at $x = 1 + ln2$ and local min. at $x = e$.\n[IMAGE] Also from graph of $g^{'}(x)$, it is clear that $g(x)$ has local max. at\n$x = 1$ and local min. at $x = 2$", "solution_images": ["images/image9.png"], "subject": "Mathematics", "topic": "Definite Integration", "difficulty": "", "question_type": "subjective", "has_image": true, "exam": "JEE Advanced", "source_paper": "Mathematics_1-50_with_img.docx"}
+{"question_id": "ADV-M01-Q16", "question": "Let $f(x)$ be a non-constant twice differentiable\nfunction definied on $( - \\infty,\\infty)$ such that\n$f(x) = f(1 - x)$ and $f^{'}\\left( \\frac{1}{4} \\right) = 0$. Then,\nfind the value of\n$\\int_{0}^{1/2}\\mspace{2mu} f(t)e^{sin\\pi t}dt - \\int_{0}^{1/2}\\mspace{2mu} f(1 - t)e^{sin\\pi t}dt$", "question_images": [], "answer": "0", "solution": "$\\mathbf{\\therefore}f(x)$ is a non constant twice\ndifferentiable function such\nthat $f(x) = f(1 - x) \\Rightarrow f^{'}(x) = - f^{'}(1 - x)$ .....(i)\\\nFor $x = \\frac{1}{2}$, we get\n$f^{'}\\left( \\frac{1}{2} \\right) = - f^{'}\\left( 1 - \\frac{1}{2} \\right)$\\\n$\\Rightarrow f^{'}\\left( \\frac{1}{2} \\right) + f^{'}\\left( \\frac{1}{2} \\right) = 0$\n$\\lbrack\\because f(x) = f(1 - x)\\rbrack$\n$$\\Rightarrow f^{'}\\left( \\frac{1}{2} \\right) = 0$$\nFor $x = \\frac{1}{4}$,\nwe get\n$f^{'}\\left( \\frac{1}{4} \\right) = - f^{'}\\left( \\frac{3}{4} \\right)\\lbrack\\because f(x) = f(1 - x)\\rbrack$\nbut given that $f^{'}\\left( \\frac{1}{4} \\right) = 0$\\\n$$\\therefore f^{'}\\left( \\frac{3}{4} \\right) = f^{'}\\left( \\frac{1}{4} \\right) = 0\n$$Hence, $f^{'}(x)$ satisfies all conditions of Rolle\\'s theorem for\n$x \\in \\left\\lbrack \\frac{1}{4},\\frac{1}{2} \\right\\rbrack$ and\n$\\left\\lbrack \\frac{1}{2},\\frac{3}{4} \\right\\rbrack$. So there exists\nat least one point\n$a_{1} \\in \\left( \\frac{1}{4},\\frac{1}{2} \\right)$ and at least one\npoint $b_{2} \\in \\left( \\frac{1}{2},\\frac{3}{4} \\right)$.\nSuch that $f^{''}\\left( a_{1} \\right) = 0$ and\n$f^{''}\\left( b_{2} \\right) = 0$\\\n$\\therefore f^{''}(x)$ varishes at least twice on\n$\\lbrack 0,1\\rbrack$.\\\nAlso using $f(x) = f(1 - x)$\\\n$$\\Rightarrow f\\left( x + \\frac{1}{2} \\right) = f\\left( 1 - x - \\frac{1}{2} \\right) = f\\left( - x + \\frac{1}{2} \\right)\n$$$\\Rightarrow f\\left( x + \\frac{1}{2} \\right)$ is an even function.\\\n$\\Rightarrow sinx.f\\left( x + \\frac{1}{2} \\right)$ is an odd\nfunction.\\\n$\\Rightarrow \\int_{- 1/2}^{1/2}\\mspace{2mu} f\\left( x + \\frac{1}{2} \\right)sinxdx = 0,$", "solution_images": [], "subject": "Mathematics", "topic": "Definite Integration", "difficulty": "", "question_type": "numerical", "has_image": false, "exam": "JEE Advanced", "source_paper": "Mathematics_1-50_with_img.docx"}
+{"question_id": "ADV-M01-Q18", "question": "Prove that for any positive integer $k$,\n$\\frac{sin2kx}{sinx} = 2\\lbrack cosx + cos3x + \\ldots\\ldots\\ldots\\ldots.. + cos(2k - 1)x\\rbrack$\nEvaluate $\\int_{0}^{\\pi/2}\\mspace{2mu}\\sin{2kx}cotx\\ dx$", "question_images": [], "answer": "$\\frac{\\mathbf{\\pi}}{\\mathbf{2}}$", "solution": "$\\because 2sinx\\lbrack cosx + cos3x + cos5x\\ldots + cos(2k - 1)x\\rbrack$\\\n$${= 2sinxcosx + 2sinxcos3x + 2sinxcos5x\n}{+ \\ldots + 2sinxcos(2k - 1)x\n}{= sin2x + (sin4x - sin2x)\\ + (sin6x - sin4x)\n}{+ \\ldots + \\{ sin2kx - sin(2k - 2)x\\}\n}{= sin2kx\n}{\\therefore 2\\lbrack cosx + cos3x + cos5x + \\ldots + cos(2k - 1)x\\rbrack\n}{= \\frac{sin2kx}{sinx}\\ldots.(i)}$$\nNow, $sin2kx \\cdot cotx = \\frac{sin2kx}{sinx} \\cdot cosx$\\\n$= 2cosx\\lbrack cosx + cos3x + cos5x + \\ldots + cos(2k - 1)x\\rbrack$\n\\[from eqn (i)\\]\\\n$${= \\left\\lbrack 2\\cos^{2}x + 2cosxcos3x + 2cosxcos5x + \\right.\\ \n}{= (1 + cos2x) + (cos4x + cos2x)\\ldots + 2cosxcos(2k - 1)x\\rbrack\n}{+ (cos6x + cos4x) + \\ldots + \\{ cos2kx + cos(2k - 2)x\\} = 1 + 2\\lbrack cos2x + cos4x + cos6x + \\ldots + cos(2k - 2)x\\rbrack + cos\n}{\\therefore\\int_{0}^{\\pi/2}\\mspace{2mu}(sin2kx) \\cdot cotxdx\n}{= \\int_{0}^{x/2}\\mspace{2mu} 1 \\cdot dx + 2\\int_{0}^{\\pi/2}\\mspace{2mu}(cos2x + cos4x\\ldots cos(2k - 2)x)dx + \\int_{0}^{\\pi/2}\\mspace{2mu} cos(2k)xdx}$$\n$= \\frac{\\pi}{2} + 2\\left\\lbrack \\frac{\\sin{2x}}{2} + \\frac{\\sin{4x}}{4} + \\ldots + \\frac{\\sin(2k - 2)x}{(2k - 2)} \\right\\rbrack_{0}^{\\frac{\\pi}{2}}$\\\n$$+ \\left\\lbrack \\frac{\\sin(2k)x}{2k} \\right\\rbrack_{0}^{\\frac{\\pi}{2}}$$\n$$= \\frac{\\pi}{2} + 2 \\times 0 = \\frac{\\pi}{2}$$", "solution_images": [], "subject": "Mathematics", "topic": "Definite Integration", "difficulty": "", "question_type": "subjective", "has_image": false, "exam": "JEE Advanced", "source_paper": "Mathematics_1-50_with_img.docx"}
+{"question_id": "ADV-M01-Q2", "question": "Let\n$S = \\{ z \\in C:|z = 2| \\leqslant 1,z(1 + i) + \\overline{z}(1 - i) \\leq 2\\}$.\nLet $|z - 4i|$ attains minimum and maximum values, respectively, at\n$z_{1} \\in S$ and $z_{2} \\in S$. If\n$5\\left( \\left| z_{1} \\right|^{2} + \\left| z_{2} \\right|^{2} \\right) = \\alpha + \\beta\\sqrt{5}$,\nwhere $\\alpha$ and $\\beta$ are integers, then the value of\n$\\alpha + \\beta$ is equal to", "question_images": [], "answer": "26", "solution": "Let $z = x + iy \\Rightarrow |z - 2| \\leq 1$\n[IMAGE] ${(x - 2)}^{2} + y^{2} \\leq 1$ ......(i)\nand $z(1 + i) + \\bar{z}(1 - i) \\leq 2$\\\n$\\therefore x - y \\leq 1$ ......(ii)\\\n$$PA = \\sqrt{17},PB = \\sqrt{13}\n$$∴ Maximum is PA and Minimum is PD\\\nLet $D(2 + cos\\theta,0 + sin\\theta)$\\\n∴ slope of\n$CP = tan\\theta = - 2 \\Rightarrow cos\\theta = - \\frac{1}{\\sqrt{5}},sin\\theta = \\frac{2}{\\sqrt{5}}$\\\n$$\\therefore D\\left( 2 - \\frac{1}{\\sqrt{5}},\\frac{2}{\\sqrt{5}} \\right) \\Rightarrow z_{1} = \\left( 2 - \\frac{1}{\\sqrt{5}} \\right) + \\frac{2i}{\\sqrt{5}}\n$$$\\left| z_{1} \\right| = \\frac{25 - 4\\sqrt{5}}{5}$ and $z_{2} = 1$\\\n$${\\therefore\\left| z_{2} \\right|^{2} = 1\n}{\\therefore 5\\left( \\left| z_{1} \\right|^{2} + \\left| z_{2} \\right|^{2} \\right) = 30 - 4\\sqrt{5}\\ \\therefore\\alpha = 30\n}{\\beta = - 4\n}{\\therefore\\alpha + \\beta = 26}$$", "solution_images": ["images/image1.png"], "subject": "Mathematics", "topic": "Complex Numbers", "difficulty": "", "question_type": "numerical", "has_image": true, "exam": "JEE Advanced", "source_paper": "Mathematics_1-50_with_img.docx"}
+{"question_id": "ADV-M01-Q8", "question": "A circle passes through three points $A,B$ and $C$ with\nthe line segment $AC$ as its diameter. A line passing through $A$\nintersects the chord $BC$ at a point $D$ inside the circle. If angles\n$DAB$ and $CAB$ are $\\alpha$ and $\\beta$ respectively and the distance\nbetween the point $A$ and the mid point of the line segment $DC$ is $d$,\nthen find the area of the circle.", "question_images": [], "answer": "$\\frac{\\pi d^{2}\\cos^{2}\\alpha}{\\cos^{2}\\alpha + \\cos^{2}\\beta + 2cos\\alpha cos\\beta cos(\\beta - \\alpha)}$", "solution": "Let $r$ be the radius of circle, then $AC = 2r$\nSince, $AC$ is the diameter, $\\therefore\\angle ABC = 90^{\\circ}$\n[IMAGE] $\\therefore\\ $ In\n$\\bigtriangleup ABC,BC = 2rsin\\beta,AB = 2rcos\\beta$\\\nIn right angled $\\bigtriangleup ABC$,\\\n$${BD = ABtan\\alpha = 2rcos\\beta tan\\alpha\n}{AD = ABsec\\alpha = 2rcos\\beta sec\\alpha\n}{\\therefore\\ DC = BC - BD = 2rsin\\beta - 2rcos\\beta tan\\alpha\n}$$Since $E$ is the mid point of $DC$,\\\n$${\\therefore DE = \\frac{DC}{2} = \\frac{2rsin\\beta - 2rcos\\beta tan\\alpha}{2}\n}{\\Rightarrow DE = rsin\\beta - rcos\\beta tan\\alpha\n}$$Now in $\\bigtriangleup ADC,AE$ is the median.\\\n$${\\therefore 2\\left( AE^{2} + DE^{2} \\right) = AD^{2} + AC^{2}\n}{\\Rightarrow 2\\left\\lbrack d^{2} + r^{2}(sin\\beta - cos\\beta tan\\alpha)^{2} \\right\\rbrack\n}{= 4r^{2}\\cos^{2}\\beta\\sec^{2}\\alpha + 4r^{2}\n}{\\Rightarrow r^{2} = \\frac{d^{2}\\cos^{2}\\alpha}{\\cos^{2}\\alpha + \\cos^{2}\\beta + 2cos\\alpha cos\\beta cos(\\beta - \\alpha)}\n}$$$\\Rightarrow$ Area of circle $= \\pi r^{2}$\\\n$$= \\frac{\\pi d^{2}\\cos^{2}\\alpha}{\\cos^{2}\\alpha + \\cos^{2}\\beta + 2cos\\alpha cos\\beta cos(\\beta - \\alpha)}$$", "solution_images": ["images/image5.png"], "subject": "Mathematics", "topic": "Circles", "difficulty": "", "question_type": "subjective", "has_image": true, "exam": "JEE Advanced", "source_paper": "Mathematics_1-50_with_img.docx"}
+{"question_id": "ADV-M01-Q43", "question": "$A$ and $B$ are two separate reservoirs of water. Capacity of reservoir $A$ is double the capacity of reservoir $B$. Both the reservoirs are filled completely with water, their inlets are closed and then the water is released simultaneously from both the reservoirs. The rate of flow of water out of each reservoir at any instant of time is proportional to the quantity of water in the reservoir at the time. One hour after the water is released, the quantity of water in reservoirs $A$ is $1\\frac{1}{2}$ times the quantity of water in reservoir $B$. After how many hours do both the reservoirs have the same quantity of water?", "question_images": [], "answer": "$\\frac{\\log 2}{\\log(4/3)}$", "solution": ". Let, at any instant $t$, the volume of the reservoirs of\n$A$ be $V_{A}$ and that of $B$ be $V_{B}$.\nThe initial capacity of $A$ be $V_{A}^{\\circ}$ and $B$ be\n$V_{B}^{\\circ}$.\n$$\\therefore\\frac{V_{A}^{\\circ}}{V_{B}^{\\circ}} = 2.$$\nAlso,\n$$\\frac{dV_{A}}{dt} = - K_{A}V_{A}\\quad\\text{and}\\quad\\frac{dV_{B}}{dt} = - K_{B}V_{B}$$\n$$\\Rightarrow \\frac{dV_{A}}{V_{A}} = - K_{A}\\, dt\\quad\\text{and}\\quad\\frac{dV_{B}}{V_{B}} = - K_{B}\\, dt$$\n$$\\Rightarrow logV_{A} = - K_{A}t + C\\quad\\text{and}\\quad\\log V_{B} = - K_{B}t + C.$$\nAt $t = 0$, $V_{A} = V_{A}^{\\circ}$, $V_{B} = V_{B}^{\\circ}$.\n$$\\therefore log\\frac{V_{A}}{V_{A}^{\\circ}} = - K_{A}t\\quad\\text{and}\\quad\\log\\frac{V_{B}}{V_{B}^{\\circ}} = - K_{B}t.$$\nGiven that at $t = 1$,\n$$\\frac{V_{A}}{V_{B}} = \\frac{3}{2},\\quad\\frac{V_{A}^{\\circ}}{V_{B}^{\\circ}} = 2$$\n$$\\Rightarrow \\frac{V_{A}}{V_{B}} \\cdot \\frac{V_{B}^{\\circ}}{V_{A}^{\\circ}} = \\frac{3}{4}.$$\nPutting the value at $t = 1$, we get\n$$\\log\\frac{V_{A}}{V_{A}^{\\circ}} = - K_{A}$$\n$$\\log\\frac{V_{B}}{V_{B}^{\\circ}} = - K_{B}$$\n$$\\Rightarrow log\\frac{V_{A}}{V_{A}^{\\circ}} - log\\frac{V_{B}}{V_{B}^{\\circ}} = K_{B} - K_{A}$$\n$$\\Rightarrow log\\left( \\frac{V_{A}/V_{A}^{\\circ}}{V_{B}/V_{B}^{\\circ}} \\right) = K_{B} - K_{A}$$\n$$\\Rightarrow log\\left( \\frac{3}{4} \\right) = K_{B} - K_{A}.$$\nNow, at $t = t_{1}$, say, $V_{A} = V_{B}$.\n$$\\Rightarrow log\\frac{V_{A}}{V_{A}^{\\circ}} = - K_{A}t_{1}$$\n$$\\log\\frac{V_{B}}{V_{B}^{\\circ}} = - K_{B}t_{1}$$\n$$\\Rightarrow log\\left( \\frac{V_{A}/V_{A}^{\\circ}}{V_{B}/V_{B}^{\\circ}} \\right) = \\left( K_{B} - K_{A} \\right)t_{1} = \\left\\lbrack \\log\\left( \\frac{3}{4} \\right) \\right\\rbrack t_{1}$$\n$$\\Rightarrow log\\left( \\frac{1}{2} \\right) = \\left( \\log\\frac{3}{4} \\right)t_{1}$$\n$$\\Rightarrow t_{1} = \\frac{\\log(1/2)}{\\log(3/4)}$$\n$$\\therefore t_{1} = \\frac{\\log 2}{\\log(4/3)}.$$\n## \n##", "solution_images": [], "subject": "Mathematics", "topic": "Triangles", "difficulty": "", "question_type": "subjective", "has_image": false, "exam": "JEE Advanced", "source_paper": "Mathematics_1-50_with_img.docx"}

mathematics/train.jsonl

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+{"question_id": "ADV-M01-Q32", "question": "Let $S = \\{ E_{1},E_{2},\\ldots,E_{8}\\}$ be a sample\nspace of random experiment such that\n$P\\left( E_{n} \\right) = \\frac{n}{36}$ for every $n = 1,2,\\ldots,8$.\nThen the number of elements in the set\n$$\\left\\{ A \\subset S:P(A) \\geq \\frac{4}{5} \\right\\}$$\nis \\_\\_\\_\\_\\_\\_.", "question_images": [], "answer": "19", "solution": "Given that $A$ is a subset of $S$, hence $A$ can have\nelements:\nType 1: $\\{\\}$\nType 2: $\\{ E_{1}\\},\\{ E_{2}\\},\\ldots,\\{ E_{8}\\}$\nType 3: $\\{ E_{1},E_{2}\\},\\{ E_{1},E_{3}\\},\\ldots,\\{ E_{1},E_{8}\\}$\nType 6:\n$\\{ E_{1},E_{2},\\ldots,E_{5}\\},\\ldots,\\{ E_{4},E_{5},E_{6},E_{7},E_{8}\\}$\nType 7:\n$\\{ E_{1},E_{2},\\ldots,E_{6}\\},\\ldots,\\{ E_{3},E_{4},\\ldots,E_{8}\\}$\nType 8: $\\{ E_{1},E_{2},\\ldots,E_{7}\\},\\{ E_{2},E_{3},\\ldots,E_{8}\\}$\nType 9: $\\{ E_{1},E_{2},\\ldots,E_{8}\\}$\nAs\n$$P(A) \\geq \\frac{4}{5};$$\nNote: Type 1 to Type 4 elements can not be in set $A$ as maximum\nprobability of type 4 elements.\n$$\\{ E_{5},E_{6},E_{7},E_{8}\\}\\ \\text{is}\\ \\frac{5}{36} + \\frac{6}{36} + \\frac{7}{36} + \\frac{8}{36} = \\frac{13}{18} < \\frac{4}{5}.$$\nNow for Type 5 acceptable elements let's call probability as $P_{5}$.\n$$P_{5} = \\frac{n_{1} + n_{2} + n_{3} + n_{4} + n_{5}}{36} \\leq \\frac{4}{5}$$\n$$\\Rightarrow n_{1} + n_{2} + n_{3} + n_{4} + n_{5} \\geq 29.$$\nHence, 2 possible ways\n$\\{ E_{5},E_{6},E_{7},E_{8},E_{3}\\ \\text{or}\\ E_{4}\\}$.\n$$P_{6} = n_{1} + n_{2} + n_{3} + n_{4} + n_{5} + n_{6} \\geq 29$$\n$$\\Rightarrow 9\\ \\text{possible ways.}$$\n$$P_{7} \\Rightarrow n_{1} + n_{2} + \\cdots + n_{7} \\geq 29$$\n$$\\Rightarrow 7\\ \\text{possible ways.}$$\n$$P_{8} \\Rightarrow n_{1} + n_{2} + \\cdots + n_{8} \\geq 29$$\n$$\\Rightarrow 1\\ \\text{possible ways.}$$\nTotal number ways\n$$= 2 + 9 + 7 + 1 = 19.$$\n###", "solution_images": [], "subject": "Mathematics", "topic": "Probability", "difficulty": "", "question_type": "numerical", "has_image": false, "exam": "JEE Advanced", "source_paper": "Mathematics_1-50_with_img.docx"}
+{"question_id": "ADV-M01-Q24", "question": "Consider the function\n$f:\\left( - \\frac{\\pi}{2},\\frac{\\pi}{2} \\right) \\rightarrow ( - \\infty,\\infty)$\ndefined by\n$$f(x) = (|x| + |x - 1|)sinx + \\lbrack xsinx\\rbrack$$\nwhere $\\lbrack xsinx\\rbrack$ is the greatest integer less than or\nequal to $xsinx$.\\\nLet $\\alpha$ be the total number of points in the interval\n$\\left( - \\frac{\\pi}{2},\\frac{\\pi}{2} \\right)$ at which $f$ is NOT\ncontinuous, and let\\\n$\\beta$ be the total number of points in the interval\n$\\left( - \\frac{\\pi}{2},\\frac{\\pi}{2} \\right)$ at which $f$ is NOT\ndifferentiable.\\\nThen the value of $\\alpha + \\beta$ is", "question_images": [], "answer": "5.00\\", "solution": "Observe $\\lbrack xsinx\\rbrack$\nlet $g(x) = xsinx$\\\n$${g^{'}(x) = sinx + xcosx \\geq 0\n}{g^{'}(x) \\geq 0\\forall x \\in \\left\\lbrack 0,\\frac{\\pi}{2} \\right)\n}{g^{'}(x) \\leq 0\\forall x \\in \\left( - \\frac{\\pi}{2},0 \\right\\rbrack}$$\n[IMAGE] $\\lbrack xsinx\\rbrack$ discontinuous at\n$x = x_{1}, - x_{1};x_{1} \\in \\left( 0,\\frac{\\pi}{2} \\right)$\\\n$\\lbrack xsinx\\rbrack$ not diff. at\n$x = x_{1}, - x_{1},x_{1} \\in \\left( 0,\\frac{\\pi}{2} \\right)$\\\nNow $y = (|x| + |x - 1|)sinx$\\\n(i) continuous\n$\\forall x\\left( - \\frac{\\pi}{2},\\frac{\\pi}{2} \\right)$\\\n(ii) Not differentiable at $x = 1$, differentiable at $x = 0$\nHere $x_{1} \\neq 1$\\\nNow\\\n(i) for diffrentiability\\\n$$f(x) = \\underset{f_{1}(x)}{\\overset{(|x| + |x - 1|)sinx}{︸}} + \\underset{f_{2}(x)}{\\overset{\\lbrack xsinx\\rbrack}{︸}}\n$$at $x = x_{1}, - x_{1}$\\\n$f_{1}(x)$ is differentiable & $f_{2}(x)$ is not differentiable\\\nHence $f(x)$ is not differentiable\\\nat $x = 1;f_{1}(x)$ is not differentiable & $f_{2}(x)$ is\ndifferentiable\\\n$\\Rightarrow f(x)$ is not differentiable\\\nat $x = 0$\\\n$f_{1}(x)$ & $f_{2}(x)$ both differentiable\\\nHence $\\beta = 3$\\\n(ii) for continuity, we can clearly say\\\nat $x = x_{1}, - x_{1}$,\\\n$f(x)$ will be discontinuous\\\n$${\\alpha = 2\n}{\\alpha + \\beta = 5}$$", "solution_images": ["images/image14.jpg"], "subject": "Mathematics", "topic": "Differential Calculus", "difficulty": "", "question_type": "subjective", "has_image": true, "exam": "JEE Advanced", "source_paper": "Mathematics_1-50_with_img.docx"}
+{"question_id": "ADV-M01-Q27", "question": "For a real number $\\alpha$, let $\\lbrack\\alpha\\rbrack$\ndenote the greatest integer less than or equal to $\\alpha$. For a finite\nset\nS, let $|S|$ denote the number of elements in the set S.\nConsider the functions $f:( - 3,3) \\rightarrow ( - \\infty,\\infty)$ and\n$g:( - 3,3) \\rightarrow ( - \\infty,\\infty)$ defined by\n$$f(x) = \\left\\lbrack x^{3} \\right\\rbrack\\log_{e}\\left( 1 + \\sin^{2}(\\pi(x - \\lbrack x\\rbrack)) \\right)\n$$and\n$(x) = x^{3}\\sin^{2}\\left( \\pi\\log_{e}(1 + x - \\lbrack x\\rbrack) \\right)$.\\\nLet $A = \\{ x \\in ( - 3,3):f$ is discontinuous at x$\\}$\\\nand $B = \\{ x \\in ( - 3,3):g$ is discontinuous at x$\\}$.\\\nThen the value of $|A| + 2|\\text{ }B| - |A \\cap B|$ is $\\_\\_\\_\\_$ .", "question_images": [], "answer": "56.00", "solution": "$f(x) = \\left\\lbrack x^{3} \\right\\rbrack log\\left( 1 + \\sin^{2}\\pi(x - \\lbrack x\\rbrack) \\right)$\\\n$${f(x) = \\left\\lbrack x^{3} \\right\\rbrack log\\left( 1 + \\sin^{2}\\pi x \\right)\n}{x \\in ( - 3,3)\n}{x^{3} \\in ( - 27,27)\n}$$At integers $x = \\pm 1, \\pm 2,0$\\\n$f(x)$ is continuous as $log\\left( 1 + \\sin^{2}\\pi x \\right) = 0$\\\nSo number of points of discontinuity of $f(x)$ is $53 - 5 = 48$\\\nand $g(x) = x^{3}\\sin^{2}\\pi\\log_{e}(1 + \\{ x\\})$ is cont. at $x = 0$,\\\nSo point of Discontinuity are $x = \\pm 1\\& \\pm 2$\\\n$${|A| = 48\\&|\\text{ }B| = 4\\&|\\text{ }A \\cap \\text{ }B| = 0\n}{|A| + 2|\\text{ }B| - |A \\cap B|\n}{= 48 + 8 - 0 = 56}$$", "solution_images": [], "subject": "Mathematics", "topic": "Trigonometry", "difficulty": "", "question_type": "numerical", "has_image": false, "exam": "JEE Advanced", "source_paper": "Mathematics_1-50_with_img.docx"}
+{"question_id": "ADV-M01-Q11", "question": "For any positive integer $n$, define\n$f_{\\eta}:(0,\\infty) \\rightarrow \\mathbb{R}$ as\n$f_{n}(x) = \\sum_{j = 1}^{n}\\mspace{2mu}\\tan^{- 1}\\left( \\frac{1}{1 + (x + j)(x + j - 1)} \\right)$\nfor all $x \\in (0,\\infty)$. (Here, the inverse trigonometric function\n$\\tan^{- 1}x$ assumes values in\n$\\left( - \\frac{\\pi}{2},\\frac{\\pi}{2} \\right)$.)\\\nThen, for any fixed positive integer $n$, find the value of\n$\\lim_{x \\rightarrow \\infty}\\mspace{2mu}\\sec^{2}\\left( f_{n}(x) \\right)$", "question_images": [], "answer": "1", "solution": "$f_{n}(x) = \\sum_{j = 1}^{n}\\mspace{2mu}\\mspace{2mu}\\tan^{- 1}\\left( \\frac{1}{1 + (x + j)(x + j - 1)} \\right)$\\\n$${= \\sum_{j = 1}^{n}\\mspace{2mu}\\mspace{2mu}\\tan^{- 1}\\left\\lbrack \\frac{(x + j) - (x + j - 1)}{1 + (x + j)(x + j - 1)} \\right\\rbrack\n}{= \\sum_{j = 1}^{n}\\mspace{2mu}\\mspace{2mu}\\left\\lbrack \\tan^{- 1}(x + j) - \\tan^{- 1}(x + j - 1) \\right\\rbrack\n}{\\Rightarrow f_{n}(x) = \\tan^{- 1}(x + n) - \\tan^{- 1}(x)\n}{= \\tan^{- 1}\\left( \\frac{n}{1 + x(n + x)} \\right) \\Rightarrow f_{n}^{'}(x) = \\frac{1}{1 + (x + n)^{2}} - \\frac{1}{1 + x^{2}}}$$\nand\n$f_{n}(0) = \\tan^{- 1}(n),\\therefore\\tan^{2}\\left( \\tan^{- 1}n \\right) = n^{2}$\\\nHere $x = 0$ is not in the given domain, i.e., $x \\in (0,\\infty)$.\\\n$\\lim_{x \\rightarrow \\infty}\\mspace{2mu} tan\\left( f_{n}(x) \\right) = \\lim_{x \\rightarrow \\infty}\\mspace{2mu}\\left( \\frac{n}{1 + x(n + x)} \\right) = 0$\\\n$\\lim_{x \\rightarrow \\infty}\\mspace{2mu}\\sec^{2}\\left( f_{n}(x) \\right) = \\lim_{n \\rightarrow \\infty}\\mspace{2mu} 1 + \\tan^{2}\\left( f_{n}(x) \\right)$\\\n$$= 1 + \\lim_{x \\rightarrow \\infty}\\mspace{2mu}\\tan^{2}\\left( f_{n}(x) \\right) = 1$$", "solution_images": [], "subject": "Mathematics", "topic": "Limits", "difficulty": "", "question_type": "numerical", "has_image": false, "exam": "JEE Advanced", "source_paper": "Mathematics_1-50_with_img.docx"}
+{"question_id": "ADV-M01-Q22", "question": "Let $f:\\lbrack 0,\\infty) \\rightarrow \\mathbb{R}$ be a\ncontinuous function such that\n$f(x) = 1 - 2x + \\int_{0}^{x}\\mspace{2mu} e^{x - t}f(t)dt$\\\nfor all $x \\in \\lbrack 0,\\infty)$. Then, find the which of the area of\nthe region\n$\\left\\{ (x,y) \\in \\lbrack 0,1\\rbrack \\times \\mathbb{R:}f(x) \\leq y \\leq \\sqrt{1 - x^{2}} \\right\\}$", "question_images": [], "answer": "$\\frac{\\pi - 2}{4}$", "solution": "$f(x) = 1 - 2x + \\int_{0}^{x}\\mspace{2mu} e^{x - t}f(t)dt$\\\n$${\\Rightarrow \\ f(x) = 1 - 2x + e^{x}\\int_{0}^{x}\\mspace{2mu}\\mspace{2mu} e^{- t}f(t)dt\n}{\\Rightarrow \\ f^{'}(x) = - 2 + e^{x}\\int_{0}^{x}\\mspace{2mu}\\mspace{2mu} e^{- t}f(t)dt + e^{x}\\left\\lbrack e^{- x}f(x) \\right\\rbrack\n}{\\Rightarrow \\ f^{'}(x) = - 2 + \\lbrack f(x) - 1 + 2x\\rbrack + f(x)\n}{\\Rightarrow \\ f^{'}(x) - 2f(x) = 2x - 3\n}$$Its a linear differential equation.\\\n$IF = e^{\\int_{}^{}\\ - 2dx} = e^{- 2x}$\\\nSolution: $f(x) \\times e^{- 2x} = \\int_{}^{}\\ e^{- 2x}(2x - 3)dx$\\\n$$\\Rightarrow f(x) \\times e^{- 2x} = \\frac{e^{- 2x}}{- 2}(2x - 3) - \\int_{}^{}\\ \\frac{e^{- 2x}}{- 2} \\times 2dx\n$$From definition of function, $f(0) = 1$\\\n$$\\therefore 1 = 1 + c \\Rightarrow c = 0,\\therefore f(x) = 1 - x\n$$Clearly curve $y = 1 - x$, does not pass through $(1,2)$ but it passes\nthrough $(2, - 1)$\n[IMAGE] \\\n∴ (a) is false and (b) is true. Also the area of the region\n$1 - x \\leq y \\leq \\sqrt{1 - x^{2}}$, is shown in the figure by the\nshaded region, is given by = Area of quadrant - Area\n$\\bigtriangleup OAB$\\\n$$= \\frac{1}{4} \\times \\pi \\times 1^{2} - \\frac{1}{2} \\times 1 \\times 1 = \\frac{\\pi - 2}{4}$$", "solution_images": ["images/image13.png"], "subject": "Mathematics", "topic": "Definite Integration", "difficulty": "", "question_type": "subjective", "has_image": true, "exam": "JEE Advanced", "source_paper": "Mathematics_1-50_with_img.docx"}
+{"question_id": "ADV-M01-Q10", "question": "Let the functions $f:( - 1,1) \\rightarrow \\mathbb{R}$\nand $g:( - 1,1) \\rightarrow ( - 1,1)$ be defined by\\\n$f(x) = |2x - 1| + |2x + 1|$ and $g(x) = x - \\lbrack x\\rbrack$, Where\n$\\lbrack x\\rbrack$ denotes the greatest integer less than or equal to\n$x$. Let $fog:( - 1,1) \\rightarrow \\mathbb{R}$ be the composite function\ndetined by $(f \\circ g)(x) = f(g(x))$. Suppose $c$ is the number of\npoints in the interval $( - 1,1)$ at which $f \\circ g$ is NOT\ncontinuous, and suppose $d$ is the number of points in the interval\n$( - 1,1)$ at which $f \\circ g$ is NOT differentiable. Then the value of\n$c + d$ is $\\_\\_\\_\\_$", "question_images": [], "answer": "4", "solution": "Given that $f(x) = |2x - 1| + |2x + 1|$ and\n$g(x) = x - \\lbrack x\\rbrack = \\{ x\\}$\\\n$${\\therefore\\ (fog)(x) = |2\\{ x\\} - 1| + |2\\{ x\\} + 1|\n}{\\Rightarrow \\ (fog)(x) = \\left\\{ \\begin{matrix}\n|2x + 1| + |2x + 3|,x \\in \\left( - 1, - \\frac{1}{2} \\right\\rbrack \\\\\n|2x + 1| + |2x + 3|,x \\in \\left( - \\frac{1}{2},0 \\right\\rbrack \\\\\n|2x - 1| + |2x + 1|,x \\in \\left( 0,\\frac{1}{2} \\right\\rbrack \\\\\n|2x - 1| + |2x + 1|,x \\in \\left( \\frac{1}{2},1 \\right\\rbrack\n\\end{matrix} \\right.\\ \n}{\\Rightarrow \\ (fog)(x) = \\left\\{ \\begin{matrix}\n2,x \\in \\left\\lbrack - 1, - \\frac{1}{2} \\right) \\\\\n4x + 4,x \\in \\left( - \\frac{1}{2},0 \\right\\rbrack \\\\\n2,x \\in \\left( 0,\\frac{1}{2} \\right\\rbrack \\\\\n4x,x \\in \\left( \\frac{1}{2},1 \\right)\n\\end{matrix} \\right.\\ }$$\n[IMAGE] $\\therefore\\ f(g(x))$ is discontinuous at $x = 0$.\\\n$$\\therefore\\ c = 1\n$$Now, $(fog)^{'}(x) = \\left\\{ \\begin{matrix}\n0,x \\in \\left( - 1,\\frac{1}{2} \\right) \\\\\n4,x \\in \\left( - \\frac{1}{2},0 \\right) \\\\\n0,x \\in \\left( 0,\\frac{1}{2} \\right) \\\\\n4,x \\in \\left( \\frac{1}{2},1 \\right)\n\\end{matrix} \\right.\\ $\\\n$\\therefore f(g(x))$ is non-differentiable at\n$x = - \\frac{1}{2},0,\\frac{1}{2}$\\\n$$\\therefore\\ d = 3\n$$Hence, $c + d = 4$", "solution_images": ["images/image6.png"], "subject": "Mathematics", "topic": "Differential Calculus", "difficulty": "", "question_type": "numerical", "has_image": true, "exam": "JEE Advanced", "source_paper": "Mathematics_1-50_with_img.docx"}
+{"question_id": "ADV-M01-Q33", "question": "Consider the hyperbola\n$$\\frac{x^{2}}{100} - \\frac{y^{2}}{64} = 1$$\nwith foci at $S$ and $S_{1}$, where $S$ lies on the positive $x$-axis.\nLet $P$ be a point on the hyperbola, in the first quadrant. Let\n$$\\angle SPS_{1} = \\alpha,\\quad\\text{with }\\alpha < \\frac{\\pi}{2}.$$\nThe straight line passing through the point $S$ and having the same\nslope as that of the tangent at $P$ to the hyperbola, intersects the\nstraight line $S_{1}P$ at $P_{1}$. Let $\\delta$ be the distance of $P$\nfrom the straight line $SP_{1}$, and $\\beta = S_{1}P$. Then the\ngreatest integer less than or equal to\n$$\\frac{\\beta\\delta}{9}\\sin\\frac{\\alpha}{2}$$\nis \\_\\_\\_\\_\\_\\_.", "question_images": [], "answer": "7\n##", "solution": "[IMAGE] In $\\bigtriangleup S_{1}QP$,\n$$\\sin\\frac{\\alpha}{2} = \\frac{S_{1}Q}{\\beta} \\Rightarrow S_{1}Q = \\beta\\sin\\frac{\\alpha}{2}$$\nProduct of distances of any tangent from two foci is $b^{2}$.\n$$\\delta \\cdot S_{1}Q = \\delta \\cdot \\beta\\sin\\frac{\\alpha}{2} = b^{2}$$\nSo,\n$$\\frac{\\beta\\delta\\sin\\frac{\\alpha}{2}}{9} = \\frac{b^{2}}{9} = \\frac{64}{9}$$\n$$\\left\\lbrack \\frac{64}{9} \\right\\rbrack = 7$$", "solution_images": ["images/image17.png"], "subject": "Mathematics", "topic": "Hyperbola", "difficulty": "", "question_type": "subjective", "has_image": true, "exam": "JEE Advanced", "source_paper": "Mathematics_1-50_with_img.docx"}
+{"question_id": "ADV-M01-Q42", "question": "A 2 metre long object is fixed vertically upwards from the mid-point of two locations $A$ and $B$, 8 metre apart. The speed of the object after $t$ seconds is given by\n$$\\frac{ds}{dt} = (2t + 1)\\text{ metres per second}.$$\nLet $\\alpha$ and $\\beta$ be the angles subtended by the object at $A$\nand $B$, respectively after one and two seconds, find the value of\n$\\cos(\\alpha - \\beta)$.", "question_images": [], "answer": "$\\frac{5}{\\sqrt{26}}$", "solution": "We have\n$$\\frac{ds}{dt} = 2t + 1$$\n$$\\Rightarrow s = t^{2} + t + k,\\text{ where }k\\text{ is a constant.}$$\nWhen $t = 0$, the distance of the top of the object from the mid point\n$M$ of $AB$ is 2 meters,\n$$\\therefore 2 = k.$$\nAlso $s = t^{2} + t + 2$\nwhere $s$ is the distance of the top of the object from $M$ after $t$\nseconds.\nLet $PQ$ and $RS$ be the positions of the object after one and two\nseconds respectively, then\n$$QM = 1 + 1 + 2 = 4$$\n$$PM = QM - PQ = 4 - 2 = 2$$\n$$SM = 4 + 2 + 2 = 8$$\n$$RM = SM - SR = 6$$\n$$AM = MB = 4$$\n[IMAGE] Now,\n$$AP^{2} = AM^{2} + PM^{2}$$\n$$AP^{2} = 16 + 4 = 20$$\n$$AQ^{2} = 16 + 16 = 32$$\n$$BR^{2} = 16 + 36 = 52$$\n$$BS^{2} = 16 + 64 = 80$$\nBy the cosine formula,\n$$\\cos\\alpha = \\frac{AP^{2} + AQ^{2} - PQ^{2}}{2 \\cdot AP \\cdot AQ}$$\n$$\\cos\\alpha = \\frac{20 + 32 - 4}{2\\sqrt{20}\\sqrt{32}}$$\n$$\\cos\\alpha = \\frac{48}{2 \\times 8 \\times \\sqrt{10}} = \\frac{3}{\\sqrt{10}}$$\nand\n$$\\cos\\beta = \\frac{BS^{2} + BR^{2} - SR^{2}}{2 \\cdot BS \\cdot BR}$$\n$$\\cos\\beta = \\frac{80 + 52 - 4}{2\\sqrt{80}\\sqrt{52}}$$\n$$\\cos\\beta = \\frac{128}{2 \\times 4\\sqrt{5} \\times 2\\sqrt{13}} = \\frac{8}{\\sqrt{65}}.$$\nTherefore,\n$$\\sin\\alpha = \\frac{1}{\\sqrt{10}}\\quad\\text{and}\\quad\\sin\\beta = \\frac{1}{\\sqrt{65}}.$$\n$$\\cos(\\alpha - \\beta) = cos\\alpha \\cdot cos\\beta + sin\\alpha \\cdot sin\\beta$$\n$$\\cos(\\alpha - \\beta) = \\frac{3}{\\sqrt{10}} \\times \\frac{8}{\\sqrt{65}} + \\frac{1}{\\sqrt{10}} \\times \\frac{1}{\\sqrt{65}}$$\n$$\\cos(\\alpha - \\beta) = \\frac{25}{5\\sqrt{26}} = \\frac{5}{\\sqrt{26}}.$$\n##", "solution_images": ["images/image27.png"], "subject": "Mathematics", "topic": "Sequences and Series", "difficulty": "", "question_type": "subjective", "has_image": true, "exam": "JEE Advanced", "source_paper": "Mathematics_1-50_with_img.docx"}
+{"question_id": "ADV-M01-Q13", "question": "Find the indefinite integral\n$\\int_{}^{}\\ \\left( \\frac{1}{\\sqrt[3]{x} + \\sqrt[4]{4}} + \\frac{ln(1 + \\sqrt[6]{x}}{\\sqrt[3]{x} + \\sqrt{x}} \\right)dx$", "question_images": [], "answer": "$\\Rightarrow I = \\frac{3}{2}x^{2/3} - \\frac{12}{7}x^{7/2} + 2x^{1/2} - \\frac{12}{5}x^{5/12} + 3x^{1/13} - 4x^{1/4}\n$$${+ 6x^{1/6} - 12x^{1/12} + 12log\\left| x^{1/12} + 1 \\right|\n}{+ 6\\left\\lbrack \\left\\{ \\frac{\\left( 1 + x^{1/6} \\right)^{3}}{3} - \\frac{3}{2}\\left( 1 + x^{1/6} \\right)^{2} + 3\\left( 1 + x^{1/6} \\right) \\right\\} ln\\left( 1 + x^{1/6} \\right) \\right.\\ \n}{- \\left\\{ \\frac{\\left( 1 + x^{1/6} \\right)^{3}}{9} - \\frac{3}{4}\\left( 1 + x^{1/6} \\right)^{2} + 3\\left( 1 + x^{1/6} \\right) \\right\\}\\left. \\ - \\frac{\\left\\lbrack ln\\left( 1 + x^{1/6} \\right)^{2} \\right\\rbrack}{2} + c \\right\\rbrack}$$", "solution": "$I = \\int\\left( \\frac{1}{\\sqrt[3]{x} + \\sqrt[4]{x}} + \\frac{ln(1 + \\sqrt[6]{x})}{\\sqrt[3]{x} + \\sqrt{x}} \\right)dx$\\\n$= \\int\\frac{1}{\\sqrt[3]{x} + \\sqrt[4]{x}}dx + \\int\\frac{ln(1 + \\sqrt[6]{x})}{\\sqrt[3]{x} + \\sqrt{x}}dx$\\\n$I = I_{1} + I_{2}$ ......(i)\\\nwhere $I_{1} = \\int_{}^{}\\ \\frac{1}{\\sqrt[3]{x} + \\sqrt[4]{x}}dx$ and\n$I_{2} = \\int_{}^{}\\ \\frac{ln(1 + \\sqrt[6]{x})}{\\sqrt[3]{x} + \\sqrt{x}}dx$\\\nLet $x = y^{12}$ so that $dx = 12y^{11}dy$\\\n$${\\therefore\\ I_{1} = \\int_{}^{}\\ \\frac{12y^{11}}{y^{4} + y^{3}}dy = 12\\int_{}^{}\\ \\frac{y^{8}}{1 + y}dy\n}{= 12\\int_{}^{}\\ \\left( y^{7} - y^{6} + y^{5} - y^{4} + y^{3} - y^{2} + y - 1 + \\frac{1}{y + 1} \\right)dy\n}{= 12\\left\\lbrack \\frac{y^{8}}{8} - \\frac{y^{7}}{7} + \\frac{y^{6}}{6} - \\frac{y^{5}}{5} + \\frac{y^{4}}{4} - \\frac{y^{3}}{3} \\right.\\ \\left. \\ + \\frac{y^{2}}{2} - y + log|y + 1| \\right\\rbrack + c_{1}\n}{= \\frac{3}{2}x^{2/3} - \\frac{12}{7}x^{7/12} + 2x^{1/2} - \\frac{12}{5}x^{5/12} + 3x^{1/3}\n}$$$- 4x^{1/4} + 6x^{1/6} - 12x^{1/12} + 12log\\left| x^{1/12} + 1 \\right| + c_{1}$\n......(ii)\\\nNow,\n$I_{2} = \\int_{}^{}\\ \\frac{ln\\left( 1 + (x)^{1/6} \\right)}{(x)^{1/3} + (x)^{1/2}}dx$\\\nLet $x = z^{6}$ so that $dx = 6z^{5}dz$\\\n$$\\therefore\\ I_{2} = \\int\\frac{ln(1 + z)}{z^{2} + z^{3}},6z^{5}dz = \\int\\frac{6z^{3}ln(z + 1)}{z + 1}dz\n$$Put $z + 1 = t$\\\n$${\\Rightarrow dz = dt\n}{\\therefore\\ I_{2} = \\int\\frac{6(t - 1)^{3}\\ln t}{t}dt = 6\\left( t^{2} - 3t + 3 - \\frac{1}{t} \\right)\\ln\\int tdt\n}{= 6\\left\\lbrack \\int\\left( t^{2} - 3t + 3 \\right)lntdt - \\int\\frac{1}{t}lntdt \\right\\rbrack\n}{= 6\\left\\lbrack \\left( \\frac{t^{3}}{3} - \\frac{3t^{2}}{2} + 3t \\right)lnt - \\int\\left( \\frac{t^{3}}{3} - \\frac{3t^{2}}{2} + 3t \\right) \\cdot \\frac{1}{t}dt \\right.\\ \\left. \\ \\begin{matrix}\n - \\frac{(lnt)^{2}}{2} \\\\\n - \\frac{(lnt)^{2}}{2}\n\\end{matrix} \\right\\rbrack\n}{= 6\\left\\lbrack \\left( \\frac{t^{3}}{3} - \\frac{3t^{2}}{2} + 3t \\right)lnt - \\int\\left( \\frac{t^{2}}{3} - \\frac{3}{2}t + 3 \\right)dt \\right.\\ \n}{= 6\\left\\lbrack \\left( \\frac{t^{3}}{3} - \\frac{3t^{2}}{2} + 3t \\right)lnt - \\left( \\frac{t^{3}}{9} - \\frac{3t^{2}}{4} + 3t \\right) - \\frac{(lnt)^{2}}{2} \\right\\rbrack + c_{2}\n}$$Now, $t = 1 + z = 1 + x^{1/6}$\\\n$${\\therefore I_{2} = 6\\left\\lbrack \\left\\{ \\frac{\\left( 1 + x^{1/6} \\right)^{3}}{3} - \\frac{3}{2}\\left( 1 + x^{1/6} \\right)^{2} + 3\\left( 1 + x^{1/6} \\right) \\right\\} \\right.\\ \n}{ln\\left( 1 + x^{1/6} \\right) - \\left\\{ \\frac{\\left( 1 + x^{1/6} \\right)^{3}}{9} - \\frac{3}{4}\\left( 1 + x^{1/6} \\right)^{2} \\right.\\ \n}$$$\\left. \\ \\left. \\ + 3\\left( 1 + x^{1/6} \\right) \\right\\} - \\frac{\\left\\lbrack ln\\left( 1 + x^{1/6} \\right)^{2} \\right\\rbrack}{2} \\right\\rbrack + c_{2}$\n......(iii)\\\nFrom (i), (ii) and (iii), we get $I = I_{1} + I_{2}$\\\n$${\\Rightarrow I = \\frac{3}{2}x^{2/3} - \\frac{12}{7}x^{7/2} + 2x^{1/2} - \\frac{12}{5}x^{5/12} + 3x^{1/13} - 4x^{1/4}\n}{+ 6x^{1/6} - 12x^{1/12} + 12log\\left| x^{1/12} + 1 \\right|\n}{+ 6\\left\\lbrack \\left\\{ \\frac{\\left( 1 + x^{1/6} \\right)^{3}}{3} - \\frac{3}{2}\\left( 1 + x^{1/6} \\right)^{2} + 3\\left( 1 + x^{1/6} \\right) \\right\\} ln\\left( 1 + x^{1/6} \\right) \\right.\\ \n}{- \\left\\{ \\frac{\\left( 1 + x^{1/6} \\right)^{3}}{9} - \\frac{3}{4}\\left( 1 + x^{1/6} \\right)^{2} + 3\\left( 1 + x^{1/6} \\right) \\right\\}\\left. \\ - \\frac{\\left\\lbrack ln\\left( 1 + x^{1/6} \\right)^{2} \\right\\rbrack}{2} + c \\right\\rbrack}$$", "solution_images": [], "subject": "Mathematics", "topic": "Definite Integration", "difficulty": "", "question_type": "subjective", "has_image": false, "exam": "JEE Advanced", "source_paper": "Mathematics_1-50_with_img.docx"}
+{"question_id": "ADV-M01-Q4", "question": "Let $G$ be a circle of radius $R > 0$. Let\n$G_{1},G_{2},\\ldots,G_{n}$ be $n$ circles of equal radius $r > 0$.\nSuppose each of the $n$ circles $G_{1}$, $G_{2},G_{2}\\ldots,G_{n}$\ntouches the circle G externally. Also for $i = 1,2,\\ldots n - 1$, the\ncircle $G_{i}$ touches $G_{i + 1}$ externally, and $G_{n}$ touches\n$G_{1}$ externally. If $n = 12$, find the relation in $r$ & $R$.", "question_images": [], "answer": "if $n = 12$, then $\\sqrt{2}(\\sqrt{3} + 1)r > R$", "solution": "Refer to diagram,\n[IMAGE] In $\\bigtriangleup AOB$\\\n$${sin\\left( \\frac{\\pi}{n} \\right) = \\frac{r}{R + r} \\Rightarrow cosec\\left( \\frac{\\pi}{n} \\right) = \\frac{R}{r} + 1\n}{\\Rightarrow R = r\\left\\lbrack cosec\\left( \\frac{\\pi}{n} \\right) - 1 \\right\\rbrack\n}$$If $n = 12$, then $R = r\\left( cosec\\frac{\\pi}{12} - 1 \\right)$\\\n$${R = r(\\sqrt{2}(\\sqrt{3} + 1) - 1)\n}{R < \\sqrt{2}(\\sqrt{3} + 1)r}$$", "solution_images": ["images/image3.png"], "subject": "Mathematics", "topic": "Circles", "difficulty": "", "question_type": "subjective", "has_image": true, "exam": "JEE Advanced", "source_paper": "Mathematics_1-50_with_img.docx"}
+{"question_id": "ADV-M01-Q38", "question": "Let $\\bigtriangleup ABC$ be equilateral on side $BA$\nproduced, we choose a point $P$ such that $A$ lies between $P$ and $B$.\nWe now denote \"$a$\" as the length of a side of $\\bigtriangleup ABC$;\n$r_{1}$ as the radius of incircle of $\\bigtriangleup PAC$; and $r_{2}$\nas the radius of the excircle of $\\bigtriangleup PBC$ with respect to\nside $BC$. Determine the sum $\\left( r_{1} + r_{2} \\right)$ as a\nfunction of \"$a$\" alone.", "question_images": [], "answer": "$\\frac{a\\sqrt{3}}{2}$", "solution": "Looking the figure, we see that\n$\\angle T_{1}O_{1}R = 60^{\\circ}$ and it is the supplement of\n$\\angle T_{1}AR = 120^{\\circ}$ \\[as an exterior angle for\n$\\bigtriangleup ABC$\\].\nHence,\n$$\\angle AO_{1}R = 30^{\\circ}.$$\n[IMAGE] Similarly, we obtain $\\angle BO_{2}S = 30^{\\circ}$.\nSince tangents drawn to a circle from external points are equal, we\nhave\n$$T_{1}T_{2} = T_{1}A + AB + BT_{2} = RA + SB + AB.$$\n$$T_{1}T_{2} = r_{1}\\tan 30^{\\circ} + a + r_{2}\\tan 30^{\\circ} = \\frac{r_{1} + r_{2}}{\\sqrt{3}} + a,$$\nand\n$$T'_{1}T'_{2} = T'_{1}C + CT'_{2} = CR + CS = (a - RA) + (a - SB).$$\n$$T'_{1}T'_{2} = 2a - \\left( \\frac{r_{1} + r_{2}}{\\sqrt{3}} \\right).$$\nSince common external tangents to two circles are equal,\n$$T_{1}T_{2} = T'_{1}T'_{2}.$$\nHence,\n$$\\frac{r_{1} + r_{2}}{\\sqrt{3}} + a = 2a - \\frac{r_{1} + r_{2}}{\\sqrt{3}}.$$\nHence we find that\n$$r_{1} + r_{2} = \\frac{a\\sqrt{3}}{2}.$$", "solution_images": ["images/image22.png"], "subject": "Mathematics", "topic": "Circles", "difficulty": "", "question_type": "subjective", "has_image": true, "exam": "JEE Advanced", "source_paper": "Mathematics_1-50_with_img.docx"}
+{"question_id": "ADV-M01-Q5", "question": "For each natural number $k$, let $C_{k}$ denote the\ncircle with radius $k$ centimetres and centre at the origin. On the\ncircle $C_{k},\\alpha$-particle moves $k$ centimetres in the\ncounter-clockwise direction. After completing its motion on $C_{k}$, the\nparticle moves to $C_{k + 1}$ in the radial direction. The motion of the\nparticle continues in this manner. The particle starts at $(1,0)$. If\nthe particle crosses the positive direction of the x-axis for the first\ntime on the circle $C_{n}$ then $n =$ $\\_\\_\\_\\_$", "question_images": [], "answer": "$n = 7$", "solution": "[IMAGE] The radius of circle $C_{1}$ is $1\\text{ }cm,C_{2}$ is 2 cm and so on.\nIt starts from $A_{1}(1,0)$ on $C_{1}$, moves a distance of 1 cm on\n$C_{1}$ to come to $B_{1}$. The angle subtended by $A_{1}B_{1}$ at the\ncentre will be $\\frac{\\mathcal{l}}{r} = \\frac{1}{1} = 1$ radian.\\\nFrom $B_{1}$ it moves along radius, $OB_{1}$ and comes to $A_{2}$ on\ncircle $C_{2}$ of radius 2. From $A_{2}$ it moves on $C_{2}$ a\ndistance 2 cm and comes to $B_{2}$. The angle subtended by\n$A_{2}B_{2}$ is again as before 1 radian. The total angle subtended at\nthe centre is 2 radians. The process continues. In order to cross the\n$x$-axis again, it must describe $2\\pi$ radians i.e.\n$2,\\frac{22}{7} = 6.7$ radians. Hence it must be moving on circle\n$C_{7}$.\n$$\\therefore n = 7$$", "solution_images": ["images/image4.png"], "subject": "Mathematics", "topic": "Circles", "difficulty": "", "question_type": "subjective", "has_image": true, "exam": "JEE Advanced", "source_paper": "Mathematics_1-50_with_img.docx"}
+{"question_id": "ADV-M01-Q46", "question": "Let the set of all relations $R$ on the set $\\{ a,b,c,d,e,f\\}$, such that $R$ is reflexive and symmetric, and $R$ contains exactly $10$ elements, be denoted by $S$. Then the number of elements in $S$ is \\_\\_\\_\\_\\_\\_\\_\\_\\_\\_.", "question_images": [], "answer": "$105$", "solution": "$(S) = \\frac{\\binom{6}{2}\\binom{4}{2}}{2} + \\binom{6}{3}\\binom{3}{1}.$\n$= 45 + 60 = 105$\n##", "solution_images": [], "subject": "Mathematics", "topic": "Permutations and Combinations", "difficulty": "", "question_type": "subjective", "has_image": false, "exam": "JEE Advanced", "source_paper": "Mathematics_1-50_with_img.docx"}
+{"question_id": "ADV-M01-Q29", "question": "Consider the curve $C_{1}$ given by\n$$y = e^{- x}\\text{~for~}x \\in \\lbrack 0,10\\pi\\rbrack$$\nand the curve $C_{2}$ given by\n$$y = e^{- x}(sinx + cosx)\\text{~for~}x \\in \\lbrack 0,10\\pi\\rbrack$$\nLet $n$ be the total number of points of intersection of the curves\n$C_{1}$ and $C_{2}$.\\\nSuppose that\n$\\alpha_{1},\\alpha_{2},\\ldots,\\alpha_{n}, \\in \\lbrack 0,10\\pi\\rbrack$\nare the $x$ - coordinates of the points of intersection of the curves\n$C_{1}$ and $C_{2}$ such that\n$$\\alpha_{1} < \\alpha_{2} < \\ldots < \\alpha_{n}.$$\nLet $\\beta$ be the area of the region enclosed between the curves\n$C_{1},C_{2}$, and the lines $x = \\alpha_{1}$ and\n$x = \\alpha_{4}$. Then the value of\\\n$- \\frac{1}{\\pi}\\log_{e}\\left( \\beta - 2e^{- \\frac{\\pi}{2}} \\right)$\nis $\\_\\_\\_\\_$ .", "question_images": [], "answer": "2.50", "solution": "$\\alpha_{1} = 0\\&\\alpha_{4} = \\frac{5\\pi}{2}$\n[IMAGE] $$\\beta = \\int_{0}^{\\frac{5\\pi}{2}}\\mspace{2mu}\\mspace{2mu}\\left| \\left( e^{- x}(sinx + cosx) - e^{- x} \\right) \\right|dx$$\n$$\\beta = \\int_{0}^{\\frac{\\pi}{2}}\\mspace{2mu}\\left( e^{- x}(sinx + cosx) - e^{- x} \\right)dx + \\int_{\\frac{\\pi}{2}}^{2\\pi}\\mspace{2mu}\\left( e^{- x} - e^{- x}(sinx + cosx) \\right)dx$$\n$$+ \\int_{2\\pi}^{\\frac{5\\pi}{2}}\\mspace{2mu}\\mspace{2mu}\\left( e^{- x}(sinx + cosx) - e^{- x} \\right)dx$$\n$$\\beta = 2e^{\\frac{- \\pi}{2}} + e^{\\frac{- 5\\pi}{2}}$$\n$$\\text{so~}\\frac{- 1}{\\pi}log\\left( \\beta - 2e^{\\frac{- \\pi}{2}} \\right) = \\frac{5}{2} = 2.50$$", "solution_images": ["images/image15.jpg"], "subject": "Mathematics", "topic": "Definite Integration", "difficulty": "", "question_type": "numerical", "has_image": true, "exam": "JEE Advanced", "source_paper": "Mathematics_1-50_with_img.docx"}
+{"question_id": "ADV-M01-Q12", "question": "Define the collections\n$\\left\\{ E_{1},E_{2},E_{3},\\ldots\\ldots \\right\\}$ of ellipses and\n$\\left\\{ R_{1},R_{2},R_{3} \\right.\\ $, $\\_\\_\\_\\_$ } of rectangles as\nfollows:\\\n$E_{1}:\\frac{x^{2}}{9} + \\frac{y^{2}}{4} = 1$;\\\n$R_{1}$ : rectangle of largest area, with sides parallel to the axes,\ninscribed in $E_{1}$;\\\n$E_{n}$ : ellipse\n$\\frac{x^{2}}{a_{n}^{2}} + \\frac{y^{2}}{b_{n}^{2}} = 1$ of largest area\ninscribed in $R_{n - 1}$, $n > 1$;\\\n$R_{n}$ : rectangle of largest area, with sides parallel to the axes,\ninscribed in $E_{n},n > 1$.\\\nThen find the length of latus rectum of $E_{9}$ and value of $n = 1$.", "question_images": [], "answer": "$\\frac{1}{6},\\ 24$", "solution": "$E_{1}:\\frac{x^{2}}{9} + \\frac{y^{2}}{4} = 1$\n$R_{1}$ : rectangle $ABCD$ with largest area.\n[IMAGE] Area of $R_{1} = A = 2x \\times 2y$\\\n$${\\Rightarrow A = 4x \\times \\frac{2}{3}\\sqrt{9 - x^{2}} = \\frac{8}{3}x\\sqrt{9 - x^{2}}\n}{\\therefore\\ \\frac{d\\text{ }A}{dx} = \\frac{8}{3}\\left\\lbrack \\sqrt{9 - x^{2}} - \\frac{x^{2}}{\\sqrt{9 - x^{2}}} \\right\\rbrack\n}{\\Rightarrow \\frac{8}{3}\\left\\lbrack \\frac{9 - 2x^{2}}{\\sqrt{9 - x^{2}}} \\right\\rbrack = 0 \\Rightarrow x = \\frac{3}{\\sqrt{2}},y = \\frac{2}{3}\\sqrt{9 - \\frac{9}{2}} = \\frac{2}{\\sqrt{2}}\n}$$⇒ For $E_{2}:a = \\frac{3}{\\sqrt{2}},b = \\frac{2}{\\sqrt{2}}$\\\nSimilarly for\n$E_{3}:a = \\frac{3}{(\\sqrt{2})^{2}},b = \\frac{2}{(\\sqrt{2})^{2}}$ and\nso on.\\\nNow eccentricity depends on $\\frac{b}{a}$ which is same for all\n$E_{n}$, therefore eccentricity for all the $E_{n}$ \\'s will remain\n$\\sqrt{1 - \\frac{4}{9}} = \\frac{\\sqrt{5}}{3}$.\\\nFor $E_{9}:a = \\frac{3}{(\\sqrt{2})^{8}},b = \\frac{2}{(\\sqrt{2})^{8}}$\\\n⇒ Length of latus rectum\n$= \\frac{2b^{2}}{a} = \\frac{2 \\times \\frac{4}{256}}{\\frac{3}{16}} = \\frac{1}{6}$\\\nArea of\n$R_{1} = 4 \\times \\frac{3}{\\sqrt{2}} \\times \\frac{2}{\\sqrt{2}} = \\frac{24}{2}$\\\nArea of\n$R_{2} = 4 \\times \\frac{3}{(\\sqrt{2})^{2}} \\times \\frac{2}{(\\sqrt{2})^{2}} = \\frac{24}{2^{2}}$\\\nArea of\n$R_{3} = 4 \\times \\frac{3}{(\\sqrt{2})^{3}} \\times \\frac{2}{(\\sqrt{2})^{3}} = \\frac{2}{(2)^{3}}$\nand so on\\\n$\\sum_{n = 1}^{\\infty}\\mspace{2mu}$ area of\n$R_{n} = \\frac{24}{2} + \\frac{24}{2^{2}} + \\frac{24}{2^{3}} + \\ldots = \\frac{12}{1 - \\frac{1}{2}} = 24$\\\n$\\Rightarrow \\sum_{n = 1}^{N}\\mspace{2mu}$ (area of\n$\\left. \\ R_{n} \\right) < 24$ for each positive integer $N$.\nFor\n$E_{9}:a = \\frac{3}{(\\sqrt{2})^{8}},b = \\frac{2}{(\\sqrt{2})^{8}},e = \\frac{\\sqrt{5}}{3}$\\\nThus, focus $= (ae,0) = \\left( \\frac{\\sqrt{5}}{16},0 \\right)$\\\nHence, distance of focus from centre $= \\frac{\\sqrt{5}}{16}$", "solution_images": ["images/image7.png"], "subject": "Mathematics", "topic": "Parabola", "difficulty": "", "question_type": "subjective", "has_image": true, "exam": "JEE Advanced", "source_paper": "Mathematics_1-50_with_img.docx"}
+{"question_id": "ADV-M01-Q36", "question": "The position vectors of the vertices $A,B$ and $C$ of a\ntetrahedron $ABCD$ are $\\widehat{i} + \\widehat{j} + \\widehat{k}$,\n$\\widehat{i}$ and $3\\widehat{i}$, respectively. The altitude from vertex\n$D$ to the opposite face $ABC$ meets the median line through $A$ of the\ntriangle $ABC$ at a point $E$. If the length of the side $AD$ is $4$ and\nthe volume of the tetrahedron is $\\frac{2\\sqrt{2}}{3}$, find the\nposition vector of the point $E$ for all its possible positions.", "question_images": [], "answer": "$( - 1,3,3)\\quad\\text{or}\\quad(3, - 1, - 1).$", "solution": "Given that $AD = 4$.\nLet $DE = h$.\n$$\\text{Volume of tetrahedron} = \\frac{2\\sqrt{2}}{3}.$$\n$$\\Rightarrow \\frac{1}{3}Ar( \\bigtriangleup ABC)h = \\frac{2\\sqrt{2}}{3}.$$\n$$\\therefore\\frac{1}{2}\\left| \\overrightarrow{BA} \\times \\overrightarrow{BC} \\right|h = 2\\sqrt{2}.$$\n$$\\frac{1}{2}\\left| \\left( \\widehat{j} + \\widehat{k} \\right) \\times 2\\widehat{i} \\right|h = 2\\sqrt{2}\\quad\\text{or}\\quad\\left| \\widehat{j} - \\widehat{k} \\right|h = 2\\sqrt{2}.$$\n$$\\text{or}\\quad\\sqrt{2}h = 2\\sqrt{2}\\quad\\therefore h = 2.$$\n[IMAGE] Let point $E$ divides median $AF$ in the ratio $\\lambda:1$.\n$$\\therefore\\overrightarrow{OE} = \\frac{\\lambda \\cdot 2\\widehat{i} + \\left( \\widehat{i} + \\widehat{j} + \\widehat{k} \\right)}{\\lambda + 1}\\quad\\quad\\cdots(ii)$$\n$$\\therefore\\overrightarrow{AE} = \\text{P.V. of }E - \\text{P.V. of }A = \\frac{\\lambda}{\\lambda + 1}\\left( \\widehat{i} - \\widehat{j} - \\widehat{k} \\right).$$\n$$\\therefore\\left| \\overrightarrow{AE} \\right|^{2} = AE^{2} = \\left( \\frac{\\lambda}{\\lambda + 1} \\right)^{2} \\cdot 3\\quad\\quad\\cdots(iii)$$\nNow,\n$$h^{2} + AE^{2} = AD^{2}.$$\n$$\\Rightarrow 4 + \\left( \\frac{\\lambda}{\\lambda + 1} \\right)^{2} \\cdot 3 = 16.$$\n$$\\Rightarrow 3\\left( \\frac{\\lambda}{\\lambda + 1} \\right)^{2} = 12.$$\n$$\\Rightarrow \\left( \\frac{\\lambda}{\\lambda + 1} \\right) = \\pm 2 \\Rightarrow \\lambda = \\pm (2\\lambda + 2).$$\n$$\\therefore\\lambda = - 2\\quad\\text{or}\\quad - \\frac{2}{3}.$$\nPutting the value of $\\lambda$ in $(ii)$, we get the possible\npositions of $E$ as\n$$( - 1,3,3)\\quad\\text{or}\\quad(3, - 1, - 1).$$", "solution_images": ["images/image20.jpeg"], "subject": "Mathematics", "topic": "Vectors", "difficulty": "", "question_type": "subjective", "has_image": true, "exam": "JEE Advanced", "source_paper": "Mathematics_1-50_with_img.docx"}
+{"question_id": "ADV-M01-Q31", "question": "Let $|X|$ denote the number of elements in a set $X$.\nLet $S = \\{ 1,2,3,4,5,6\\}$ be a sample space, where each element is\nequally likely to occur. If $A$ and $B$ are independent events\nassociated with $S$, then the number of ordered pairs $(A,B)$ such that\n$1 \\leq |B| < |A|$, equals \\_\\_\\_\\_\\_\\_.", "question_images": [], "answer": "422", "solution": "Let\n$$n(A) = a,\\quad n(B) = b,\\quad n(A \\cap B) = c$$\n$$1 \\leq b < a$$\nAlso given that $A$ and $B$ are independent events,\n$$P(A \\cap B) = P(A)P(B)$$\n$$\\Rightarrow \\frac{n(A \\cap B)}{n(S)} = \\frac{n(A)}{n(S)} \\times \\frac{n(B)}{n(S)}$$\n$$\\Rightarrow \\frac{c}{6} = \\frac{a}{6} \\times \\frac{b}{6} \\Rightarrow ab = 6c$$\nIf $a = 6$, then $b = c = 5,4,3,2,1$ because $b < a$.\nThere is only one way to select all 6 elements of set $A$.\nNumber of ways of selecting 5, 4, 3, 2 or 1 elements in $B$ and\n$A \\cap B$ are\n$$\\binom{6}{5} + \\binom{6}{4} + \\binom{6}{3} + \\binom{6}{2} + \\binom{6}{1} = 2^{6} - 2 = 62$$\nIf $a = 5$, then\n$$b = \\frac{6c}{5},$$\nwhich is not possible because if $c = 5$ then $b = 6$, while $b < a$.\nIf $a = 4$, then\n$$b = \\frac{6c}{4} = \\frac{3c}{2},$$\nwhich is possible because if $c = 2$ then $b = 3$.\n2 elements in $A \\cap B$ can be selected in $\\binom{6}{2}$ ways. 2\nadditional elements in $A$ can be selected in $\\binom{4}{2}$ ways. 1\nadditional element in $B$ can be selected in $\\binom{2}{1}$ ways.\nTherefore, number of ways for $a = 4$, $b = 3$, $c = 2$ are\n$$\\binom{6}{2} \\times \\binom{4}{2} \\times \\binom{2}{1} = 15 \\times 6 \\times 2 = 180$$\nIf $a = 3$, then\n$$b = 2c \\Rightarrow c = 1,\\quad b = 2$$\nwhich can be done in\n$$\\binom{6}{1} \\times \\binom{5}{1} \\times \\binom{4}{2} = 6 \\times 5 \\times 6 = 180\\text{ ways.}$$\nIf $a = 2$, then $b = 3c$, which is not possible.\nTherefore, total number of required ways\n$$\\text{Total} = 62 + 180 + 180 = 422.$$", "solution_images": [], "subject": "Mathematics", "topic": "Probability", "difficulty": "", "question_type": "numerical", "has_image": false, "exam": "JEE Advanced", "source_paper": "Mathematics_1-50_with_img.docx"}
+{"question_id": "ADV-M01-Q44", "question": "For $i = 1,2,3,4$, let $T_{i}$ denote the event that the students $S_{i}$ and $S_{i + 1}$ do NOT sit adjacent to each other on the day of the examination. Then, the probability of the event $T_{1} \\cap T_{2} \\cap T_{3} \\cap T_{4}$ is", "question_images": [], "answer": "$\\frac{7}{60}$", "solution": "Total cases = $5!$.\nLet us count the number of favourable ways.\n$$\\begin{matrix}\n2 & 4 & 1 & \\times & \\times \\\\\n2 & 5 & 3 & 1 & 4\n\\end{matrix}\\quad\\} 3\\text{ ways} \\times 2 = 6$$\n$$\\begin{matrix}\n1 & 3 & 5 & 2 & 4 \\\\\n1 & 4 & 2 & 5 & 3\n\\end{matrix}\\quad\\} 2\\text{ ways} \\times 2 = 4$$\n$$\\begin{matrix}\n3 & 1 & 5 & 2 & 4 \\\\\n3 & 1 & 4 & 2 & 5\n\\end{matrix}\\quad\\} 2\\text{ ways} \\times 2 = 4$$\n$$\\therefore\\text{Total favourable ways} = 14.$$\nAlternative Solution:\n(By P.I.E.)\n$$\\left| T_{1} \\cap T_{2} \\cap T_{3} \\cap T_{4} \\right| = |T| - \\left| T_{1}' \\cup T_{2}' \\cup T_{3}' \\cup T_{4}' \\right|$$\n$$N = 5! - \\left\\{ \\binom{4}{1} \\cdot 4! \\cdot 2! - \\left( \\binom{3}{1} \\cdot 3! \\cdot 2! + \\binom{3}{1} \\cdot 3! \\cdot 2! \\cdot 2! \\right) + \\left( \\binom{2}{1} \\cdot 2! \\cdot 2! + \\binom{4}{1} \\cdot 2! \\cdot 2! \\right) - 2 \\right\\}$$\n$$N = 5! - \\{ 4 \\cdot 48 - (3 \\cdot 6 \\cdot 2 + 3 \\cdot 6 \\cdot 2 \\cdot 2) + (2 \\cdot 2 \\cdot 2 + 4 \\cdot 2 \\cdot 2) - 2\\}$$\n$$N = 120 - (192 - 108 + 24 - 2) = 120 - 106 = 14.$$\nHence, the required probability\n$$\\text{Required probability} = \\frac{14}{120} = \\frac{7}{60}.$$\n## \n##", "solution_images": [], "subject": "Mathematics", "topic": "Probability", "difficulty": "", "question_type": "subjective", "has_image": false, "exam": "JEE Advanced", "source_paper": "Mathematics_1-50_with_img.docx"}
+{"question_id": "ADV-M01-Q19", "question": "Given that for each\n$a \\in (0,1),{\\lim_{h \\rightarrow 0^{+}}\\int_{h}^{1 - h}t^{- a}}{(1 - t)^{a - 1}dt}$\nexists. Let this limit be g(a). In addition, it is given that the\nfunction g(a) is differentiable on (0,1).\nThe value of $g = \\frac{1}{2}$.", "question_images": [], "answer": "$\\pi$", "solution": "$g(a) = \\lim_{h \\rightarrow 0^{\\text{+}}}\\int_{h}^{1 - h}t^{- a}(1 - t)^{a - 1}dt$\nPutting $a = \\frac{1}{2}$,\n$$\\therefore\\quad g\\left( \\frac{1}{2} \\right) = \\lim_{h \\rightarrow 0^{\\text{+}}}\\int_{h}^{1 - h}t^{- 1/2}(1 - t)^{- 1/2}dt$$\n$$= \\lim_{h \\rightarrow 0^{\\text{+}}}\\int_{h}^{1 - h}\\frac{1}{\\sqrt{t(1 - t)}}dt = \\lim_{h \\rightarrow 0^{\\text{+}}}\\int_{h}^{1 - h}\\frac{1}{\\sqrt{\\left( \\frac{1}{2} \\right)^{2} - \\left( t - \\frac{1}{2} \\right)^{2}}}dt$$\n$$= \\lim_{h \\rightarrow 0^{\\text{+}}}\\left\\lbrack \\sin^{- 1}\\left( \\frac{t - \\frac{1}{2}}{\\frac{1}{2}} \\right) \\right\\rbrack_{h}^{1 - h} = \\lim_{h \\rightarrow 0^{\\text{+}}}\\left\\lbrack \\sin^{- 1}(2t - 1) \\right\\rbrack_{h}^{1 - h}$$\n$$= \\lim_{h \\rightarrow 0^{\\text{+}}}\\left\\lbrack \\sin^{- 1}(1 - 2h) - \\sin^{- 1}(2h - 1) \\right\\rbrack$$\n$$= \\frac{\\pi}{2} - \\left( - \\frac{\\pi}{2} \\right) = \\pi$$", "solution_images": [], "subject": "Mathematics", "topic": "Definite Integration", "difficulty": "", "question_type": "subjective", "has_image": false, "exam": "JEE Advanced", "source_paper": "Mathematics_1-50_with_img.docx"}
+{"question_id": "ADV-M01-Q34", "question": "In a non-right angled triangle $\\bigtriangleup PQR$,\nlet $p,q,r$ denote the lengths of the sides opposite to the angles at\n$P,Q,R$ respectively. The median from $R$ meets the side $PQ$ at $S$,\nthe perpendicular from $P$ meets the side $QR$ at $E$, $RS$ and $PE$\nintersect at $O$. If $p = \\sqrt{3}$, $q = 1$ and the radius of the\ncircumcircle of $\\bigtriangleup PQR$ equals $1$, then which of the\nfollowing options is/are correct?\n\\(a\\) Find the Radius of incircle of $\\bigtriangleup PQR$\n\\(b\\) Length of $RS$", "question_images": [], "answer": "(a) $\\frac{\\sqrt{3}}{2}\\left( 2 - \\sqrt{3} \\right)$\n\\(b\\) $\\frac{\\sqrt{7}}{2}$", "solution": "Given: $RS$ is median of $\\bigtriangleup PQR$ and\n$PE\\bot QR$,\n$$p = \\sqrt{3}\\quad\\text{and}\\quad q = 1.$$\nRadius $(R)$ of circumcircle of $\\bigtriangleup PQR = 1$.\n[IMAGE] Figure 1: Diagram for Question 39.\nUsing sine law in $\\bigtriangleup PQR$,\n$$\\frac{p}{\\sin P} = \\frac{q}{\\sin Q} = \\frac{r}{\\sin R} = 2 \\times \\text{radius of circumcircle}.$$\n$$\\Rightarrow \\frac{\\sqrt{3}}{\\sin P} = \\frac{1}{\\sin Q} = \\frac{r}{\\sin R} = 2 \\times 1 = 2.$$\n$$\\Rightarrow sinP = \\frac{\\sqrt{3}}{2}\\quad\\text{and}\\quad\\sin Q = \\frac{1}{2}.$$\n$$\\Rightarrow \\angle P = 60^{\\circ}\\ \\text{or}\\ 120^{\\circ}\\quad\\text{and}\\quad\\angle Q = 30^{\\circ}\\ \\text{or}\\ 150^{\\circ}.$$\nSince any triangle cannot have two obtuse angles, therefore\n$\\angle P = 120^{\\circ}$ and $\\angle Q = 150^{\\circ}$ is rejected.\nSince sum of three angles of a triangle can not be greater than\n$180^{\\circ}$, therefore $\\angle P = 60^{\\circ}$ and\n$\\angle Q = 150^{\\circ}$ is rejected.\nSince $\\bigtriangleup PQR$ is non right triangle, therefore\n$\\angle P = 60^{\\circ}$ and $\\angle Q = 30^{\\circ}$ is rejected.\nHence, $\\angle P = 120^{\\circ}$ and $\\angle Q = 30^{\\circ}$ is the\nonly option.\n$$\\Rightarrow \\angle QPE = \\angle RPE = 60^{\\circ}\\quad\\text{and}\\quad\\angle PRQ = 30^{\\circ}.$$\n$$\\Rightarrow \\frac{r}{\\sin 30^{\\circ}} = 2\\quad \\Rightarrow \\quad r = 1.$$\nArea of $\\bigtriangleup PQR$\n$$= \\frac{pqr}{4\\left( \\text{radius of circumcircle} \\right)}\\quad\\quad\\left\\lbrack \\because\\ \\Delta = \\frac{abc}{4R} \\right\\rbrack$$\n$$= \\frac{\\sqrt{3} \\times 1 \\times 1}{4 \\times 1} = \\frac{\\sqrt{3}}{4}.$$\n$$\\therefore\\text{In } \\bigtriangleup PQR,\\ \\text{radius of incircle} = \\frac{area( \\bigtriangleup PQR)}{\\text{Semi perimeter}}\\quad\\quad\\left\\lbrack \\because\\ r = \\frac{\\Delta}{s} \\right\\rbrack$$\n$$= \\frac{\\frac{\\sqrt{3}}{4}}{\\frac{\\sqrt{3} + 1 + 1}{2}} = \\frac{\\sqrt{3}}{2}\\left( 2 - \\sqrt{3} \\right),$$\nSince, $RS$ and $PE$ are medians of $\\bigtriangleup PQR$ intersecting\nat $O$, therefore $O$ is centroid of $\\bigtriangleup PQR$.\n$$\\therefore\\ area( \\bigtriangleup OQR) = \\frac{1}{3}area( \\bigtriangleup PQR) = \\frac{1}{3} \\times \\frac{\\sqrt{3}}{4} = \\frac{\\sqrt{3}}{12}.$$\n$$\\Rightarrow \\frac{1}{2} \\times \\sqrt{3} \\times OE = \\frac{\\sqrt{3}}{12} \\Rightarrow OE = \\frac{1}{6}\\text{.}$$\nNow,\n$$area( \\bigtriangleup SOE) = \\frac{1}{2}OE \\times SF = \\frac{1}{2} \\times \\frac{1}{6} \\times \\frac{r}{2}\\sin 60^{\\circ}$$\n$$= \\frac{1}{12} \\times \\frac{1}{2} \\times \\frac{\\sqrt{3}}{2} = \\frac{\\sqrt{3}}{48}\\quad\\quad\\lbrack\\because\\ r = 1\\rbrack,$$\nAlso, in $\\bigtriangleup RQS$, by cosine law,\n$$\\cos 30^{\\circ} = \\frac{QS^{2} + QR^{2} - RS^{2}}{2QS \\times QR}.$$\n$$\\Rightarrow \\frac{\\sqrt{3}}{2} = \\frac{\\left( \\frac{1}{2} \\right)^{2} + \\left( \\sqrt{3} \\right)^{2} - RS^{2}}{2 \\times \\frac{1}{2} \\times \\sqrt{3}} \\Rightarrow \\frac{3}{2} = \\frac{13}{4} - RS^{2}.$$\n$$\\Rightarrow RS = \\frac{\\sqrt{7}}{2}\\text{.}$$", "solution_images": ["images/image18.png"], "subject": "Mathematics", "topic": "Circles", "difficulty": "", "question_type": "subjective", "has_image": true, "exam": "JEE Advanced", "source_paper": "Mathematics_1-50_with_img.docx"}
+{"question_id": "ADV-M01-Q49", "question": "For all $x > 0$, let $y_{1}(x)$, $y_{2}(x)$, and $y_{3}(x)$ be the functions satisfying\n$$\\frac{dy_{1}}{dx} - \\left( \\sin x \\right)^{2}y_{1} = 0,\\quad\\quad y_{1}(1) = 5,$$\n$$\\frac{dy_{2}}{dx} - \\left( \\cos x \\right)^{2}y_{2} = 0,\\quad\\quad y_{2}(1) = \\frac{1}{3},$$\n$$\\frac{dy_{3}}{dx} - \\left( \\frac{2 - x^{3}}{x^{3}} \\right)y_{3} = 0,\\quad\\quad y_{3}(1) = \\frac{3}{5e},$$\nrespectively. Then\n$\\lim_{x \\rightarrow 0}\\frac{y_{1}(x)y_{2}(x)y_{3}(x) + 2x}{e^{3x}\\sin x}$\nis equal to \\_\\_\\_\\_\\_\\_\\_\\_\\_\\_.", "question_images": [], "answer": "2", "solution": "$\\frac{dy_{1}}{dx} = \\sin^{2}x\\, y_{1} \\Rightarrow \\frac{dy_{1}}{y_{1}} = \\sin^{2}x\\, dx\\quad\\quad\\cdots(i)$\n$$\\frac{dy_{2}}{dx} = \\cos^{2}x\\, y_{2} \\Rightarrow \\frac{dy_{2}}{y_{2}} = \\cos^{2}x\\, dx\\quad\\quad\\cdots(ii)$$\n$$\\frac{dy_{3}}{dx} = \\frac{2 - x^{3}}{x^{3}}\\, y_{3} \\Rightarrow \\frac{dy_{3}}{y_{3}} = \\left( \\frac{2}{x^{3}} - 1 \\right)dx\\quad\\quad\\cdots(iii)$$\nFrom equations $(i) + (ii) + (iii)$,\n$$\\int\\left( \\frac{dy_{1}}{y_{1}} + \\frac{dy_{2}}{y_{2}} + \\frac{dy_{3}}{y_{3}} \\right) = \\int\\left( \\sin^{2}x + \\cos^{2}x + \\frac{2}{x^{3}} - 1 \\right)dx + C.$$\n$$\\log_{e}\\left| y_{1}y_{2}y_{3} \\right| = 2\\int x^{- 3}\\, dx + C$$\n$$\\log_{e}\\left| y_{1}y_{2}y_{3} \\right| = - \\frac{1}{x^{2}} + C.$$\n$$\\left| y_{1}y_{2}y_{3} \\right| = e^{C - \\frac{1}{x^{2}}} = c\\, e^{- \\frac{1}{x^{2}}}.$$\n$$\\left| y_{1}(1)y_{2}(1)y_{3}(1) \\right| = c\\, e^{- 1}$$\n$$5 \\times \\frac{1}{3} \\times \\frac{3}{5e} = \\frac{c}{e},\\quad\\quad c = 1.$$\n$$\\left| y_{1}(x)y_{2}(x)y_{3}(x) \\right| = e^{- \\frac{1}{x^{2}}}.$$\n$$\\Rightarrow y_{1}(x)y_{2}(x)y_{3}(x) = e^{- \\frac{1}{x^{2}}}\\quad\\text{or}\\quad - e^{- \\frac{1}{x^{2}}}.$$\n$$\\lim_{x \\rightarrow 0}\\frac{y_{1}(x)y_{2}(x)y_{3}(x) + 2x}{e^{3x}\\sin x} = \\frac{2x + e^{- \\frac{1}{x^{2}}}}{e^{3x}\\sin x} = 2.$$\n##", "solution_images": [], "subject": "Mathematics", "topic": "Definite Integration", "difficulty": "", "question_type": "numerical", "has_image": false, "exam": "JEE Advanced", "source_paper": "Mathematics_1-50_with_img.docx"}
+{"question_id": "ADV-M01-Q35", "question": "A triangle $ABC$ has sides $AB = AC = 5\\ \\text{cm}$ and\n$BC = 6\\ \\text{cm}$. Triangle $A'B'C'$ is the reflection of the triangle\n$ABC$ in a line parallel to $AB$ placed at a distance $2\\ \\text{cm}$\nfrom $AB$, outside the triangle $ABC$. Triangle $A''B''C''$ is the\nreflection of the triangle $A'B'C'$ in a line parallel to $B'C'$ placed\nat a distance of $2\\ \\text{cm}$ from $B'C'$ outside the triangle\n$A'B'C'$. Find the distance between $A$ and $A''$.", "question_images": [], "answer": "$8\\sqrt{\\frac{17}{5}}$", "solution": "Let $L$ be the line parallel to side $AB$ of\n$\\bigtriangleup ABC$, at a distance of $2\\ \\text{cm}$ from $AB$, in\nwhich the first reflection $\\bigtriangleup A'B'C'$ is obtained. Let $L'$\nbe the second line parallel to $B'C'$, at a distance of $2\\ \\text{cm}$\nfrom $B'C'$, in which reflection of $\\bigtriangleup A'B'C'$ is taken as\n$\\bigtriangleup A''B''C''$.\nIn figure, size of $\\bigtriangleup A''B''C''$ is same to the size of\n$\\bigtriangleup A'B'C'$.\nIn the figure, distance between $AB$ & $A'B'$ and distance between\n$B'C'$ & $B''C''$ will be same.\n[IMAGE] From figure $AA' = 4\\ \\text{cm}$ and $A'A'' = 12\\ \\text{cm}$. So to\nfind $AA''$ it suffices to know $\\angle AA'A''$, clearly\n$$\\angle AA'A'' = 90^{\\circ} + \\alpha,\\quad\\text{where}\\quad\\sin\\alpha = \\frac{3}{5}.$$\n$$\\Rightarrow cos\\angle AA'A'' = cos\\left( 90^{\\circ} + \\alpha \\right) = - sin\\alpha = - \\frac{3}{5}.$$\nNow, in $\\bigtriangleup AA'A''$,\n$$\\cos\\left( 90^{\\circ} + \\alpha \\right) = \\frac{(AA')^{2} + (A'A'')^{2} - (AA'')^{2}}{2(AA')(A'A'')}\\quad\\quad\\left\\lbrack \\text{cosine law} \\right\\rbrack$$\n$$\\therefore AA'' = \\sqrt{(AA')^{2} + (A'A'')^{2} - 2AA' \\times A'A'' \\cdot cos\\left( 90^{\\circ} + \\alpha \\right)}$$\n$$= \\sqrt{16 + 144 + 96 \\times \\frac{3}{5}} = \\sqrt{\\frac{1088}{5}} = 8\\sqrt{\\frac{17}{5}}\\ \\text{cm}.$$", "solution_images": ["images/image19.jpeg"], "subject": "Mathematics", "topic": "Trigonometry", "difficulty": "", "question_type": "subjective", "has_image": true, "exam": "JEE Advanced", "source_paper": "Mathematics_1-50_with_img.docx"}
+{"question_id": "ADV-M01-Q39", "question": "In a $\\bigtriangleup ABC$, $r_{A},r_{B},r_{C}$ are the\nradii of the circles which touch the incircle and the sides emanating\nfrom the vertices $A,B,C$ respectively. Then find the value of\n$\\sqrt{r_{A}r_{B}} + \\sqrt{r_{B}r_{C}} + \\sqrt{r_{C}r_{A}}$ in terms\nof $r.$", "question_images": [], "answer": "$r$", "solution": "Let the circle of radius $r_{A}$ touch the sides $AB$ and\n$AC$ at $D'$ and $E'$, whereas the incircle touches these sides at $D$\nand $E$. Let $O$ and $O'$ be centres of the inscribed circle and that of\nthe circle with radius $r_{A}$. $O$ and $O'$ lie on the bisector of\nangle $A$.\n[IMAGE] Also,\n$$AO' = O'D'cosec\\frac{A}{2} = r_{A}cosec\\frac{A}{2},$$\nand\n$$AO = ODcosec\\frac{A}{2} = rcosec\\frac{A}{2}.$$\nHence,\n$$OO' = r + r_{A} = AO - AO'.$$\n$$OO' = rcosec\\frac{A}{2} - r_{A}cosec\\frac{A}{2}.$$\nTherefore,\n$$\\frac{r_{A}}{r} = \\frac{cosec\\frac{A}{2} - 1}{cosec\\frac{A}{2} + 1} \\Rightarrow r_{A} = r\\tan^{2}\\left( \\frac{\\pi - A}{4} \\right).$$\nSimilarly,\n$$r_{B} = r\\tan^{2}\\left( \\frac{\\pi - B}{4} \\right),\\quad r_{C} = r\\tan^{2}\\left( \\frac{\\pi - C}{4} \\right).$$\nLet\n$$X = \\frac{\\pi - A}{4},\\quad Y = \\frac{\\pi - B}{4},\\quad Z = \\frac{\\pi - C}{4}.$$\nHence,\n$$\\sqrt{r_{A}r_{B}} + \\sqrt{r_{B}r_{C}} + \\sqrt{r_{C}r_{A}} = r\\left( \\tan X\\tan Y + tanY\\tan Z + tanZ\\tan X \\right).$$\nSince\n$$X + Y + Z = \\frac{\\pi}{2}$$\nwe have\n$$\\tan X\\tan Y + tanY\\tan Z + tanZ\\tan X = 1.$$\nTherefore,\n$$\\sqrt{r_{A}r_{B}} + \\sqrt{r_{B}r_{C}} + \\sqrt{r_{C}r_{A}} = r.$$", "solution_images": ["images/image23.png"], "subject": "Mathematics", "topic": "Circles", "difficulty": "", "question_type": "subjective", "has_image": true, "exam": "JEE Advanced", "source_paper": "Mathematics_1-50_with_img.docx"}
+{"question_id": "ADV-M01-Q28", "question": "Consider the curve $C_{1}$ given by\n$$y = e^{- x}\\text{~for~}x \\in \\lbrack 0,10\\pi\\rbrack$$\nand the curve $C_{2}$ given by\n$$y = e^{- x}(sinx + cosx)\\text{~for~}x \\in \\lbrack 0,10\\pi\\rbrack$$\nLet $n$ be the total number of points of intersection of the curves\n$C_{1}$ and $C_{2}$.\\\nSuppose that\n$\\alpha_{1},\\alpha_{2},\\ldots,\\alpha_{n}, \\in \\lbrack 0,10\\pi\\rbrack$\nare the $x$ - coordinates of the points of intersection of the curves\n$C_{1}$ and $C_{2}$ such that\n$$\\alpha_{1} < \\alpha_{2} < \\ldots < \\alpha_{n}.$$\nThe value of $n$ is $\\_\\_\\_\\_$ .", "question_images": [], "answer": "11.00\\", "solution": "$e^{- x} = e^{- x}(sinx + cosx)$\n$${sinx + cosx = 1\n}{cos\\left( x - \\frac{\\pi}{4} \\right) = \\frac{1}{\\sqrt{2}}\n}{x - \\frac{\\pi}{4} = 2n\\pi \\pm \\frac{\\pi}{4}\n}$$$x = 2n\\pi$ or $2n\\pi + \\frac{\\pi}{2}$\\\n$$x \\in \\left\\{ 0,\\frac{\\pi}{2},2\\pi,\\frac{5\\pi}{2},4\\pi,\\frac{9\\pi}{2},6\\pi,\\frac{13\\pi}{2},8\\pi,\\frac{17\\pi}{2},10\\pi \\right\\}\n$$so $n = 11$", "solution_images": [], "subject": "Mathematics", "topic": "Trigonometry", "difficulty": "", "question_type": "subjective", "has_image": false, "exam": "JEE Advanced", "source_paper": "Mathematics_1-50_with_img.docx"}
+{"question_id": "ADV-M01-Q21", "question": "Consider a square with vertices at\n$(1,1),( - 1,1),( - 1, - 1)$ and $(1, - 1)$. Let $S$ be the region\nconsisting of all points inside the square which are nearer to the\norigin than to any edge. Sketch the region $S$ and find its area.", "question_images": [], "answer": "$\\frac{16\\sqrt{2} - 20}{3}$", "solution": "Let us consider any point $P(x,y)$ inside the square such\nthat its distance from origin $\\leq$ its distance from any of the edges\nsay AD\\\n$$\\therefore OP \\leq PM\\text{~or~}\\sqrt{\\left( x^{2} + y^{2} \\right)} < 1 - x\n$$$\\text{or~}y^{2} \\leq - 2\\left( x - \\frac{1}{2} \\right)$\\\nAbove represents all points within and on the parabola 1. If we consider\nthe edges $BC$ then $OP < PN$ will imply\\\n$y^{2} \\leq 2\\left( x + \\frac{1}{2} \\right)$ ......(ii)\\\nSimilarly if we consider the edges $AB$ and $CD$, we will have\\\n$x^{2} \\leq - 2\\left( y - \\frac{1}{2} \\right)$ ......(iii)\\\n$x^{2} \\leq 2\\left( y + \\frac{1}{2} \\right)$ ......(iv)\\\nHence $S$ consists of the region bounded by four parabolas meeting the\naxes at $\\left( \\pm \\frac{1}{2},0 \\right)$ and\n$\\left( 0, \\pm \\frac{1}{2} \\right)$\\\nThe point $L$ is intersection of $P_{1}$ and $P_{3}$ given by (i) and\n(iii).\n[IMAGE] $y^{2} - x^{2} = - 2(x - y) = 2(y - x)$\\\n$${\\therefore\\ y - x = 0\\therefore y = x\n}{\\therefore\\ x^{2} + 2x - 1 = 0 \\Rightarrow (x + 1)^{2} = 2\n}{\\therefore\\ x = \\sqrt{2} - 1\\text{~as~}x\\text{~is +ve~}}$$\n$\\therefore\\mspace{6mu} L\\text{ is }\\left( \\sqrt{2} - 1,\\sqrt{2} - 1 \\right)$\n$\\therefore\\mspace{6mu}\\text{Total area}$ =\n$4\\left\\lbrack \\text{square of side }\\left( \\sqrt{2} - 1 \\right) + 2\\int_{\\sqrt{2} - 1}^{\\frac{1}{2}}y\\, dx \\right\\rbrack$\n=\n$4\\left\\{ \\left( \\sqrt{2} - 1 \\right)^{2} + 2\\int_{\\sqrt{2} - 1}^{\\frac{1}{2}}\\sqrt{(1 - 2x)}\\, dx \\right\\}$\n=\n$4\\left\\lbrack 3 - 2\\sqrt{2} - \\frac{2}{2} \\cdot \\frac{2}{3}\\left. \\ \\{(1 - 2x)^{\\frac{3}{2}}\\} \\right|_{\\sqrt{2} - 1}^{\\frac{1}{2}} \\right\\rbrack$\n=\n$4\\left( 3 - 2\\sqrt{2} \\right)\\left\\lbrack 1 + \\frac{2}{3}\\sqrt{\\left( 3 - 2\\sqrt{2} \\right)} \\right\\rbrack$\n=\n$\\frac{4}{3}\\left( 3 - 2\\sqrt{2} \\right)\\left( 1 + 2\\sqrt{2} \\right)$\n= $\\frac{4}{3}\\left\\lbrack \\left( 4\\sqrt{2} - 5 \\right) \\right\\rbrack$\n= $\\frac{16\\sqrt{2} - 20}{3}$", "solution_images": ["images/image11.png", "images/image12.png"], "subject": "Mathematics", "topic": "Parabola", "difficulty": "", "question_type": "subjective", "has_image": true, "exam": "JEE Advanced", "source_paper": "Mathematics_1-50_with_img.docx"}
+{"question_id": "ADV-M01-Q23", "question": "Let $S = \\{ 1,2,3,\\ldots,10\\}$. Consider the set\\\n$X = \\{ R:R$ is an equivalence relation on the set $S$ such that $R$ has\nexactly 42 elements $\\}$.\\\nThen the number of elements in X is $\\_\\_\\_\\_$ .", "question_images": [], "answer": ". 2520.00\\", "solution": "The total number of such relations is 2520.\nHere is how you can think about it step-by-step:\\\nAn equivalence relation is just a way of breaking a group into\nsmaller, separate teams (called equivalence classes). Within any team,\nevery member connects to everyone else, including themselves.\\\nThis means a team with \\'n\\' people creates $n^{2}$ total\nconnections.\\\nFor our problem, we have 10 people in total. We need to split them\ninto teams so that the sum of the squared team sizes equals exactly\n42.\nWe need to find combinations of numbers that add up to 10, but whose\nsquares add up to 42. If you test out different numbers, only two\ncombinations actually work:\\\nOption A: One team of b, one team of 2 and two teams of 1. Check total\npeople: $(6 + 2 + 1 + 1 = 10)$ check total connections:\n$\\left( 6^{2} + 2^{2} + 1^{2} + 1^{2} = 36 + 4 + 1 + 1 = 42 \\right)$.\\\nOption B: One team of 5, one team of 4 and one team of 1. Check total\npeople : $(5 + 4 + 1 = 10)$ Check total connections :\n$\\left( 5^{2} + 4^{2} + 1^{2} = 25 + 16 + 1 = 42 \\right)$.\\\nNow we just count how many ways we can sort our 10 people into these\ntwo setups.\\\nFo option $A(6,2,1,1)$ : First, pick 6 people out of 10 for the big\nteam.\\\nThen, pick 2 out of the remaining 4 for the second team. The last 2\npeople automatically form their own single-peson teams.\\\nBecause the two single person teams are identical in size, we divide\nby 2 to avoid double-counting them.\\\nWays $= \\frac{10!}{6!2!1!1!} \\times \\frac{1}{2!} = 1260$\\\nFor option B\\\nPick 5 people out of 10 fo the first team, then 4 out of the remaining\n5 for the second team.\\\nThe last person is left by themselves.\\\nWays $= \\frac{10!}{5!4!1!} = 1260$\\\nTotal count\\\nAdd the tow possibilities together : $(1260 + 1260 = 2520)$", "solution_images": [], "subject": "Mathematics", "topic": "Permutations and Combinations", "difficulty": "", "question_type": "subjective", "has_image": false, "exam": "JEE Advanced", "source_paper": "Mathematics_1-50_with_img.docx"}
+{"question_id": "ADV-M01-Q50", "question": "If $\\alpha = \\int_{1/2}^{2}\\frac{\\tan^{- 1}x}{2x^{2} - 3x + 2}\\, dx,$ then the value of $\\sqrt{7}\\tan\\left( \\frac{2\\alpha\\sqrt{7}}{\\pi} \\right)$ is \\_\\_\\_\\_\\_\\_\\_\\_\\_\\_.\nHere, the inverse trigonometric function $\\tan^{- 1}x$ assumes values\nin $\\left( - \\frac{\\pi}{2},\\frac{\\pi}{2} \\right)$.", "question_images": [], "answer": "$21$", "solution": "$\\alpha = \\int_{1/2}^{2}\\frac{\\tan^{- 1}x}{2x^{2} - 3x + 2}\\, dx.$\nPut\n$$z = \\frac{1}{x}.$$\n$$\\alpha = \\int_{2}^{1/2}\\frac{\\cot^{- 1}z\\,( - dz)}{2z^{2} - 3z + 2}.$$\n$$\\alpha = \\int_{1/2}^{2}\\frac{\\frac{\\pi}{2} - \\tan^{- 1}z}{2z^{2} - 3z + 2}\\, dz.$$\n$$2\\alpha = \\frac{\\pi}{2}\\int_{1/2}^{2}\\frac{1}{2z^{2} - 3z + 2}\\, dz.$$\n$$2\\alpha = \\frac{\\pi}{2} \\times \\frac{1}{2}\\int_{1/2}^{2}\\frac{1\\, dz}{\\left( z - \\frac{3}{4} \\right)^{2} + \\left( \\frac{\\sqrt{7}}{4} \\right)^{2}}.$$\nNow using\n$$\\int\\frac{1}{x^{2} + a^{2}}\\, dx = \\frac{1}{a}\\tan^{- 1}\\left( \\frac{x}{a} \\right) + C.$$\nWe have\n$$2\\alpha = \\frac{\\pi}{\\sqrt{7}}\\left\\lbrack \\tan^{- 1}\\left( \\frac{5}{\\sqrt{7}} \\right) + \\tan^{- 1}\\left( \\frac{1}{\\sqrt{7}} \\right) \\right\\rbrack.$$\n$$\\Rightarrow \\frac{2\\alpha\\sqrt{7}}{\\pi} = \\tan^{- 1}\\left( 3\\sqrt{7} \\right).$$\n$$\\tan\\left( \\frac{2\\alpha\\sqrt{7}}{\\pi} \\right) = 3\\sqrt{7}.$$\n$$\\text{Now,}\\quad\\quad\\sqrt{7}\\tan\\left( \\frac{2\\alpha\\sqrt{7}}{\\pi} \\right) = 21.$$", "solution_images": [], "subject": "Mathematics", "topic": "Definite Integration", "difficulty": "", "question_type": "subjective", "has_image": false, "exam": "JEE Advanced", "source_paper": "Mathematics_1-50_with_img.docx"}
+{"question_id": "ADV-M01-Q47", "question": "For any two points $M$ and $N$ in the $XY$-plane, let $\\overrightarrow{MN}$ denote the vector from $M$ to $N$, and $\\overrightarrow{0}$ denote the zero vector. Let $P$, $Q$ and $R$ be three distinct points in the $XY$-plane. Let $S$ be a point inside the triangle $\\bigtriangleup PQR$ such that $\\overrightarrow{SP} + 5\\overrightarrow{SQ} + 6\\overrightarrow{SR} = \\overrightarrow{0}.$\nLet $E$ and $F$ be the mid-points of the sides $PR$ and $QR$,\nrespectively. Then the value of\n$\\frac{\\text{length of the line segment }EF}{\\text{length of the line segment }ES}\\ $is\n\\_\\_\\_\\_\\_\\_\\_\\_\\_\\_.", "question_images": [], "answer": "$1.20$", "solution": "$\\overrightarrow{SP} + 5\\overrightarrow{SQ} + 6\\overrightarrow{SR} = \\overrightarrow{0}$\n$$\\overrightarrow{P} - \\overrightarrow{S} + 5\\overrightarrow{Q} - 5\\overrightarrow{S} + 6\\overrightarrow{R} - 6\\overrightarrow{S} = \\overrightarrow{0}$$\n$$\\Rightarrow \\overrightarrow{S} = \\frac{\\overrightarrow{P} + 5\\overrightarrow{Q} + 6\\overrightarrow{R}}{12}.$$\n[IMAGE] $$\\overrightarrow{E} = \\frac{\\overrightarrow{P} + \\overrightarrow{R}}{2},\\quad\\quad\\overrightarrow{F} = \\frac{\\overrightarrow{Q} + \\overrightarrow{R}}{2}.$$\n$$\\overrightarrow{F} - \\overrightarrow{E} = \\frac{\\overrightarrow{Q} - \\overrightarrow{P}}{2},$$\n$$\\overrightarrow{S} - \\overrightarrow{E} = \\frac{5\\overrightarrow{Q} - 5\\overrightarrow{P}}{12}.$$\n$$\\frac{|EF|}{|ES|} = \\frac{\\left| \\overrightarrow{Q} - \\overrightarrow{P} \\right|/2}{5\\left| \\overrightarrow{Q} - \\overrightarrow{P} \\right|/12} = \\frac{6}{5} = 1.20.$$\n##", "solution_images": ["images/image29.png"], "subject": "Mathematics", "topic": "Vectors", "difficulty": "", "question_type": "subjective", "has_image": true, "exam": "JEE Advanced", "source_paper": "Mathematics_1-50_with_img.docx"}
+{"question_id": "ADV-M01-Q1", "question": "Let complex numbers $\\alpha$ and\n$\\frac{1}{\\bar{\\alpha}}$ lie on circles\n$\\left( x - x_{0} \\right)^{2} + \\left( y - y_{0} \\right)^{2} = r^{2}$\nand\n$\\left( x - x_{0} \\right)^{2} + \\left( y - y_{0} \\right)^{2} = 4r^{2}$.\nrespectively. If $z_{0} = x_{0} + iy_{0}$ satisfies the equation\n$2\\left| z_{0} \\right|^{2} = r^{2} + 2$, then $|\\alpha|$", "question_images": [], "answer": "$\\frac{1}{\\sqrt{7}}$", "solution": "Since, $\\alpha$ lies on the circle\n$\\left( x - x_{0} \\right)^{2} + \\left( y - y_{0} \\right)^{2} = r^{2}$\\\n$${\\therefore\\left| \\alpha - z_{0} \\right|^{2} = r^{2}\n}{\\Rightarrow \\ \\left( \\alpha - z_{0} \\right)\\left( \\bar{\\alpha} - {\\bar{z}}_{0} \\right) = r^{2}\n}{\\Rightarrow \\alpha\\bar{\\alpha} - \\alpha{\\bar{z}}_{0} - \\bar{\\alpha}z_{0} + z_{0}{\\bar{z}}_{0} = r^{2}\n}{\\Rightarrow |\\alpha|^{2} + \\left| z_{0} \\right|^{2} - \\alpha{\\bar{z}}_{0} - \\bar{\\alpha}z_{0} = r^{2}\n}$$Also $\\frac{1}{\\bar{\\alpha}}$ lies on the circle\n$\\left( x - x_{0} \\right)^{2} + \\left( y - y_{0} \\right)^{2} = 4r^{2}$\\\n$${\\therefore\\left| \\frac{1}{\\bar{\\alpha}} - z_{0} \\right|^{2} = 4r^{2} \\Rightarrow \\left( \\frac{1}{\\bar{\\alpha}} - z_{0} \\right)\\left( \\frac{1}{\\alpha} - {\\bar{z}}_{0} \\right) = 4r^{2}\n}{\\Rightarrow \\frac{1}{\\alpha\\bar{\\alpha}} - \\frac{z_{0}}{\\alpha} - \\frac{{\\bar{z}}_{0}}{\\bar{\\alpha}} + z_{0}{\\bar{z}}_{0} = 4r^{2}\n}{\\Rightarrow \\frac{1}{|\\alpha|^{2}} - \\frac{z_{0}\\bar{\\alpha}}{|\\alpha|^{2}} - \\frac{{\\bar{z}}_{0}\\alpha}{|\\alpha|^{2}} + \\left| z_{0} \\right|^{2} = 4r^{2}\n}{\\Rightarrow \\ 1 + |\\alpha|^{2}\\left| z_{0} \\right|^{2} - z_{0}\\bar{\\alpha} - {\\bar{z}}_{0}\\alpha = 4r^{2}|\\alpha|^{2}}$$\nOn subtracting equation (i) from (ii), we get\\\n$$1 - |\\alpha|^{2} + \\left| z_{0} \\right|^{2}\\left( |\\alpha|^{2} - 1 \\right) = r^{2}\\left( 4|\\alpha|^{2} - 1 \\right)\n$$or\n$\\ \\left( |\\alpha|^{2} - 1 \\right)\\left( \\left| z_{0} \\right|^{2} - 1 \\right) = r^{2}\\left( 4|\\alpha|^{2} - 1 \\right)$\\\nUsing $\\left| z_{0} \\right|^{2} = \\frac{r^{2} + 2}{2}$, we get\\\n$${\\left( |\\alpha|^{2} - 1 \\right)\\frac{r^{2}}{2} = r^{2}\\left( 4|\\alpha|^{2} - 1 \\right)\n}{\\Rightarrow |\\alpha|^{2} - 1 = 8|\\alpha|^{2} - 2 \\Rightarrow |\\alpha| = \\frac{1}{\\sqrt{7}}}$$", "solution_images": [], "subject": "Mathematics", "topic": "Complex Numbers", "difficulty": "", "question_type": "subjective", "has_image": false, "exam": "JEE Advanced", "source_paper": "Mathematics_1-50_with_img.docx"}
+{"question_id": "ADV-M01-Q20", "question": "Given that for each\n$a \\in (0,1),{\\lim_{h \\rightarrow 0^{+}}\\int_{h}^{1 - h}t^{- a}}{(1 - t)^{a - 1}dt}$\nexists. Let this limit be g(a). In addition, it is given that the\nfunction g(a) is differentiable on (0,1).\nThe value of $g' = \\frac{1}{2}$.", "question_images": [], "answer": "0", "solution": "$g(a) = \\lim_{h \\rightarrow 0^{\\text{+}}}\\int_{h}^{1 - h}t^{- a}(1 - t)^{a - 1}dt$\n$$g(a) = \\lim_{h \\rightarrow 0^{\\text{+}}}\\int_{h}^{1 - h}(1 - t)^{- a}t^{a - 1}dt$$\nUsing,\n$$\\int_{a}^{b}f(x)dx = \\int_{a}^{b}f(a + b - x)dx$$\nAlso,\n$$g(1 - a) = \\lim_{h \\rightarrow 0^{\\text{+}}}\\int_{h}^{1 - h}t^{a - 1}(1 - t)^{- a}dt$$\nThus,\n$$g(a) = g(1 - a)$$\n$$\\Rightarrow \\quad g'(a) = - g'(1 - a) \\Rightarrow g'(a) + g'(1 - a) = 0$$\nPutting $a = \\frac{1}{2}$, we get\n$$g'\\left( \\frac{1}{2} \\right) + g'\\left( \\frac{1}{2} \\right) = 0$$\nor,\n$$g'\\left( \\frac{1}{2} \\right) = 0$$", "solution_images": [], "subject": "Mathematics", "topic": "Definite Integration", "difficulty": "", "question_type": "numerical", "has_image": false, "exam": "JEE Advanced", "source_paper": "Mathematics_1-50_with_img.docx"}
+{"question_id": "ADV-M01-Q17", "question": "Let\n$\\psi_{1}:\\lbrack 0,\\infty\\rbrack\\mathbb{\\rightarrow R,}\\psi_{2}:\\lbrack 0,\\infty\\rbrack\\mathbb{\\rightarrow R,}f:\\lbrack 0,\\infty\\rbrack\\mathbb{\\rightarrow R}$\nand $g:\\lbrack 0,\\infty\\rbrack\\mathbb{\\rightarrow R,}$ be functions such\nthat $f(0) = f(0) = 0$,\n$\\psi_{1}(x) = e^{- x} + x,\\ x \\geq 0$,\n$\\psi_{2}(x) = x^{2} - 2x - 2e^{- x} + 2,\\ x \\geq 0$,\n$f(x) = \\int_{- x}^{x}\\mspace{2mu}\\left( |t| - t^{2} \\right)e^{- t^{2}}dt,x > 0$\nand\n$g(x) = \\int_{0}^{x^{2}}\\mspace{2mu}\\mspace{2mu}\\sqrt{t}{\\ e}^{- t}dt,x > 0$\nWhich of the following statements is TRUE? then find the value of\n$f(\\sqrt{ln3}) + g(\\sqrt{ln3})$", "question_images": [], "answer": "$\\frac{2}{3}$", "solution": "$\\because f(x) = \\int_{- x}^{x}\\mspace{2mu}\\left( |t| - t^{2} \\right)e^{- t^{2}}dt,x > 0$\\\n$$\\therefore\\ f(x) = 2\\int_{0}^{x}\\mspace{2mu}\\mspace{2mu}\\left( t - t^{2} \\right)e^{- t^{2}}dt\n$$$f^{'}(x) = 2\\left( x - x^{2} \\right)e^{- x^{2}}$ .....(i)\n[IMAGE] Hence option (d) is wrong.\\\n$g^{'}(x) = xe^{- x^{2}}2x$ .....(ii)\\\nAdd (i) and (ii),\\\n$${f^{'}(x) + g^{'}(x) = 2xe^{- x^{2}}\n}{f(x) + g(x) = - e^{- x^{2}} + c\n}{\\because\\ f(0) = g(0) = 0\n}{f(x) + g(x) = - e^{- x^{2}} + 1\n}{f(\\sqrt{\\mathcal{l}ln3}) + g(\\sqrt{\\mathcal{l}ln3}) = 1 - \\frac{1}{3} = \\frac{2}{3}}$$", "solution_images": ["images/image10.png"], "subject": "Mathematics", "topic": "Definite Integration", "difficulty": "", "question_type": "subjective", "has_image": true, "exam": "JEE Advanced", "source_paper": "Mathematics_1-50_with_img.docx"}
+{"question_id": "ADV-M01-Q45", "question": "Let $S$ be the set of all complex numbers $z$ satisfying $|z - 2 + i| \\geq \\sqrt{5}.$\nIf the complex number $z_{0}$ is such that\n$\\frac{1}{\\left| z_{0} - 1 \\right|}$ is the maximum of the set\n$\\left\\{ \\frac{1}{|z - 1|}:z \\in S \\right\\},$\nthen the principal argument of\n$\\frac{4 - z_{0} - \\overline{z_{0}}}{z_{0} - \\overline{z_{0}} + 2i}$\nis", "question_images": [], "answer": "$- \\frac{\\pi}{2}$", "solution": "$\\frac{1}{|z - 1|},\\quad z \\in S,$\nattains its maximum when $|z - 1|$ attains its minimum.\nNow the curve\n$$|z - 2 + i| \\geq \\sqrt{5}$$\nlooks like the following figure.\n[IMAGE] $|z - 2 + i| \\geq \\sqrt{5}$.\nConsider\n$$\\frac{4 - z_{0} - \\overline{z_{0}}}{z_{0} - \\overline{z_{0}} + 2i}\\frac{4 - z_{0} - \\overline{z_{0}}}{z_{0} - \\overline{z_{0}} + 2i} = \\frac{4 - 2Rez_{0}}{2iImz_{0} + 2i}\\frac{4 - z_{0} - \\overline{z_{0}}}{z_{0} - \\overline{z_{0}} + 2i} = \\frac{2\\left( 2 - Rez_{0} \\right)}{2i\\left( Imz_{0} + 1 \\right)}$$\n$$\\frac{4 - z_{0} - \\overline{z_{0}}}{z_{0} - \\overline{z_{0}} + 2i} = - i\\lambda,\\quad\\text{where }\\lambda > 0.$$\nSince\n$$2 - Rez_{0} > 0\\quad\\text{and}\\quad Imz_{0} > 0,$$\ntherefore, the principal argument of\n$$\\frac{4 - z_{0} - \\overline{z_{0}}}{z_{0} - \\overline{z_{0}} + 2i}$$\nis $- \\frac{\\pi}{2}$.\n##", "solution_images": ["images/image28.jpeg"], "subject": "Mathematics", "topic": "Complex Numbers", "difficulty": "", "question_type": "subjective", "has_image": true, "exam": "JEE Advanced", "source_paper": "Mathematics_1-50_with_img.docx"}
+{"question_id": "ADV-M01-Q14", "question": "$\\text{For any real number~}x\\text{, let~}\\lbrack x\\rbrack\\text{~denote the largest integer}$\n$\\text{less than or equal to~}x\\text{. }$\n$\\text{If~}I = \\int_{0}^{10}\\mspace{2mu}\\mspace{2mu}\\left\\lbrack \\sqrt{\\frac{10x}{x + 1}} \\right\\rbrack dx,\\text{~then the value}$\nof 9I is \\_\\_\\_\\_\\_.", "question_images": [], "answer": "9I=182", "solution": "Let $y = \\frac{10x}{x + 1},\\ 0 \\leq x \\leq 10$\\\n$${\\Rightarrow xy + y = 10x\n}{\\Rightarrow x = \\frac{y}{10 - y}\n}{\\Rightarrow \\ 0 \\leq \\frac{y}{10 - y} \\leq 10\n}$$$\\Rightarrow \\ \\frac{y}{y - 10} \\leq 0$ and\n$\\frac{y}{10 - y} \\leq 10$.\n$\\Rightarrow \\ \\frac{11y - 100}{y - 10} \\geq 0$\n[IMAGE] $y \\in \\lbrack 0,10)\\text{~and~}y \\in \\left( - \\infty,\\frac{100}{11} \\right\\rbrack U(10,\\infty)$\\\n$$\\Rightarrow y \\in \\left\\lbrack 0,\\frac{100}{11} \\right\\rbrack$$\n$\\sqrt{y} \\in \\left\\lbrack 0,\\frac{10}{\\sqrt{11}} \\right\\rbrack\\ \\Rightarrow \\ \\lbrack\\sqrt{y}\\rbrack = \\{ 0,1,2,3\\}$.\\\nCase I : $0 \\leq \\frac{10x}{x + 1} < 1$\\\n$\\therefore\\ x \\in \\left\\lbrack 0,\\frac{1}{9} \\right)$ then\n$\\left\\lbrack \\sqrt{\\frac{10x}{x + 1}} \\right\\rbrack = 0$\\\nCase II : $1 \\leq \\frac{10x}{x + 1} < 4$\\\n$\\therefore\\ x \\in \\left\\lbrack \\frac{1}{9},\\frac{2}{3} \\right)$ then\n$\\left\\lbrack \\sqrt{\\frac{10x}{x + 1}} \\right\\rbrack = 1$\\\nCase III : $4 < \\frac{10x}{x + 1} < 9$\\\n$\\therefore\\ x \\in \\left\\lbrack \\frac{2}{3},9 \\right)$ then\n$\\left\\lbrack \\sqrt{\\frac{10x}{x + 1}} \\right\\rbrack = 2$\\\nCase IV :\n$x \\in \\lbrack 9,10) \\Rightarrow \\text{~then~}\\left\\lbrack \\sqrt{\\frac{10x}{x + 1}} \\right\\rbrack = 3$\\\n$${\\therefore\\ I = \\int_{0}^{1/9}\\mspace{2mu}\\mspace{2mu} 0 \\cdot dx + \\int_{1/9}^{2/3}\\mspace{2mu}\\mspace{2mu} 1 \\cdot dx + \\int_{2/3}^{9}\\mspace{2mu}\\mspace{2mu} 2 \\cdot dx + \\int_{9}^{10}\\mspace{2mu}\\mspace{2mu} 3 \\cdot dx\n}{= \\left( \\frac{2}{3} - \\frac{1}{9} \\right) + 2\\left( 9 - \\frac{2}{3} \\right) + 3(10 - 9)\n}{= \\frac{5}{9} + \\frac{50}{3} + 3 = \\frac{182}{9}\n}{\\therefore\\ 9I = 182.}$$", "solution_images": ["images/image8.png"], "subject": "Mathematics", "topic": "Definite Integration", "difficulty": "", "question_type": "subjective", "has_image": true, "exam": "JEE Advanced", "source_paper": "Mathematics_1-50_with_img.docx"}
+{"question_id": "ADV-M01-Q41", "question": "A load of weight $P$ lying on a horizontal plane must\nbe displaced by a force $F$ applied to it. The frictional force is\ndirectly proportional to the force pressing the load against the plane\nand is directed against the displacing force. The proportionality factor\n(co-efficient of friction) is equal to $K$.\nAt what angle $\\phi$ to the horizontal must the force $F$ be applied\nso that its value turns out to be the least possible? Determine the\nleast value of the displacing force.", "question_images": [], "answer": "$F = \\frac{KP}{\\sqrt{1 + K^{2}}}$", "solution": "Frictional force\n$$\\text{Frictional force} = KN$$\n$$\\text{Frictional force} = Kmg = KP$$\n[IMAGE] Net horizontal force\n$$\\text{Net horizontal force} = K\\left( mg - F\\sin\\phi \\right) = F\\cos\\phi$$\n$$\\text{Net horizontal force} = K\\left( P - F\\sin\\phi \\right) = F\\cos\\phi$$\n$$\\frac{KP}{K\\sin\\phi + cos\\phi} = F$$\n$$F(\\phi) = \\frac{KP}{K\\sin\\phi + cos\\phi}$$\n$$F'(\\phi) = \\frac{- KP\\left( K\\cos\\phi - sin\\phi \\right)}{\\left( K\\sin\\phi + cos\\phi \\right)^{2}}$$\n$$F'(\\phi) = 0$$\n\\[Condition for least value of force\\]\n$$K\\cos\\phi - sin\\phi = 0$$\nor,\n$$\\tan\\phi = K$$\n$$\\phi = \\tan^{- 1}K$$\n[IMAGE] $$F = \\frac{KP}{\\frac{K^{2}}{\\sqrt{1 + K^{2}}} + \\frac{1}{\\sqrt{1 + K^{2}}}}$$\n$$F_{\\min} = \\frac{KP}{\\sqrt{1 + K^{2}}}$$\n$F = \\frac{KP}{\\sqrt{1 + K^{2}}}$ (least force)\n##", "solution_images": ["images/image25.jpeg", "images/image26.png"], "subject": "Mathematics", "topic": "Differential Calculus", "difficulty": "", "question_type": "subjective", "has_image": true, "exam": "JEE Advanced", "source_paper": "Mathematics_1-50_with_img.docx"}
+{"question_id": "ADV-M01-Q48", "question": "Let $\\alpha$ and $\\beta$ be the real numbers such that\n$$\\lim_{x \\rightarrow 0}\\frac{\\frac{\\alpha}{2}\\left( \\int_{1}^{x}\\frac{1}{1 - t^{2}}\\, dt \\right) + \\beta x\\cos x}{x^{3}} = 2.$$\nThen the value of $\\alpha + \\beta$ is \\_\\_\\_\\_\\_\\_\\_\\_\\_\\_.", "question_images": [], "answer": "$2.4$", "solution": "$\\lim_{x \\rightarrow 0}\\frac{\\frac{\\alpha}{2}\\left( \\int_{1}^{x}\\frac{1}{1 - t^{2}}\\, dt \\right) + \\beta x\\cos x}{x^{3}} = 2.$\nApplying L' Hospital's rule,\n$$\\lim_{x \\rightarrow 0}\\frac{\\frac{\\alpha}{2}\\left( \\frac{1}{1 - x^{2}} \\right) + \\beta\\cos x - \\beta x\\sin x}{3x^{2}} = 2.$$\n$$\\Rightarrow \\lim_{x \\rightarrow 0}\\frac{\\frac{\\alpha}{2} + \\beta\\cos x\\left( 1 - x^{2} \\right) - \\beta x\\sin x\\left( 1 - x^{2} \\right)}{3x^{2}\\left( 1 - x^{2} \\right)} = 2.$$\nFor the limit to exist,\n$$\\frac{\\alpha}{2} + \\beta = 0\\quad\\quad\\cdots(i)$$\n$$\\Rightarrow \\lim_{x \\rightarrow 0}\\frac{- \\beta + \\beta\\left( 1 - x^{2} \\right)\\left( 1 - \\frac{x^{2}}{2!} + \\frac{x^{4}}{4!} - \\cdots \\right) - \\beta x\\left( 1 - x^{2} \\right)\\left( x - \\frac{x^{3}}{3!} + \\frac{x^{5}}{5!} - \\cdots \\right)}{3x^{2} \\times 1} = 2.$$\n$$\\Rightarrow \\lim_{x \\rightarrow 0}\\frac{\\frac{\\beta}{3}\\left\\{ - 1 + \\left( 1 - \\frac{x^{2}}{2!} + \\frac{x^{4}}{4!} - \\cdots \\right) + \\left( - x^{2} + \\frac{x^{4}}{2!} - \\frac{x^{6}}{4!} + \\cdots \\right) + \\left( - x^{2} + \\frac{x^{4}}{3!} - \\frac{x^{6}}{5!} + \\cdots \\right) + \\left( x^{4} - \\frac{x^{6}}{3!} + \\frac{x^{8}}{5!} - \\cdots \\right) \\right\\}}{x^{2}}$$\n$$= 2.$$\n$$\\Rightarrow \\frac{\\beta}{3}\\left( - \\frac{1}{2} - 1 - 1 \\right) = 2$$\n$$\\Rightarrow \\beta = - \\frac{12}{5},\\quad\\quad\\alpha = \\frac{24}{5}.$$\n$$\\alpha + \\beta = \\frac{12}{5} = 2.4.$$\n##", "solution_images": [], "subject": "Mathematics", "topic": "Definite Integration", "difficulty": "", "question_type": "subjective", "has_image": false, "exam": "JEE Advanced", "source_paper": "Mathematics_1-50_with_img.docx"}
+{"question_id": "ADV-M01-Q3", "question": "Let $w = \\frac{\\sqrt{3} + i}{2}$ and\n$P = \\left\\{ w^{n} \\cdot n = 1,2,3,\\ldots \\right\\}$. Further\n$H_{1} = \\left\\{ z \\in \\mathbb{C:}Rez > \\frac{1}{2} \\right\\}$ and\n$H_{2} = \\left\\{ z \\in \\mathbb{C \\cdot}Rez < \\frac{- 1}{2} \\right\\}$,\nwhere c is the set of all complex numbers. If\n$z_{1} \\in P \\cap H_{1},z_{2} \\in P \\cap H_{2}$ and $O$ represents the\norigin, $\\angle z_{1}Oz_{2} =$ when $\\alpha = (\\frac{\\pi}{2},\\ \\pi)$\nwill be", "question_images": [], "answer": "$\\frac{2\\pi}{3}\\ ,\\ \\frac{5\\pi}{6}$", "solution": "We have\n$w = \\frac{\\sqrt{3} + i}{2} = cos\\frac{\\pi}{6} + isin\\frac{\\pi}{6}$\\\n$$\\Rightarrow w^{n} = cos\\frac{n\\pi}{6} + isin\\frac{n\\pi}{6}\n$$$\\therefore P$ contains all those points which lie on unit circle and\nhave arguments $\\frac{\\pi}{6},\\frac{2\\pi}{6},\\frac{3\\pi}{6}$ and so on.\n[IMAGE] Since, $z_{1} \\in P \\cap H_{1}$ and $z_{2} \\in P \\cap H_{2}$,\ntherefore $z_{1}$ and $z_{2}$ have possible positions as shown in the\nfigure,\n$\\therefore\\angle z_{1}{Oz}_{2}$ can be $\\frac{2\\pi}{3}$ or\n$\\frac{5\\pi}{6}$.", "solution_images": ["images/image2.png"], "subject": "Mathematics", "topic": "Complex Numbers", "difficulty": "", "question_type": "subjective", "has_image": true, "exam": "JEE Advanced", "source_paper": "Mathematics_1-50_with_img.docx"}
+{"question_id": "ADV-M01-Q30", "question": "Consider three sets $E_{1} = \\{ 1,2,3\\}$,\n$F_{1} = \\{ 1,3,4\\}$ and $G_{1} = \\{ 2,3,4,5\\}$. Two elements are chosen\nat random, without replacement, from the set $E_{1}$, and let $S_{1}$\ndenote the set of these chosen elements.\nLet $E_{2} = E_{1} - S_{1}$ and $F_{2} = F_{1} \\cup S_{1}$. Now two\nelements are chosen at random, without replacement, from the set\n$F_{2}$, and let $S_{2}$ denote the set of these chosen elements.\nLet $G_{2} = G_{1} \\cup S_{2}$. Finally, two elements are chosen at\nrandom, without replacement, from the set $G_{2}$ and let $S_{3}$\ndenote the set of these chosen elements.\nLet $E_{3} = E_{2} \\cup S_{3}$. Given that $E_{1} = E_{3}$, let $p$ be\nthe conditional probability of the event $S_{1} = \\{ 1,2\\}$. Then find\nthe value of $p$.", "question_images": [], "answer": "$\\frac{1}{5}$", "solution": "[IMAGE] $$P = \\frac{P\\left( S_{1} \\cap \\left( E_{1} = E_{3} \\right) \\right)}{P\\left( E_{1} = E_{3} \\right)} = \\frac{P\\left( A_{1,2} \\right)}{P(A)}$$\n$$P(A) = P\\left( A_{1,2} \\right) + P\\left( A_{1,3} \\right) + P\\left( A_{2,3} \\right)$$\n$$P\\left( A_{1,2} \\right) = \\frac{1}{3} \\times \\frac{1 \\times \\binom{3}{1}}{\\binom{4}{2}} \\times \\frac{1}{\\binom{5}{2}}$$\n$$P\\left( A_{1,2} \\right) = \\frac{1}{3} \\times \\frac{1}{2} \\times \\frac{1}{10} = \\frac{1}{60}$$\n$$P\\left( A_{1,3} \\right) = \\frac{1}{3} \\times \\frac{1 \\times \\binom{2}{1}}{\\binom{3}{2}} \\times \\frac{1}{\\binom{5}{2}}$$\n$$P\\left( A_{1,3} \\right) = \\frac{1}{3} \\times \\frac{2}{3} \\times \\frac{1}{10} = \\frac{1}{45}$$\n$$P\\left( A_{2,3} \\right) = \\frac{1}{3}\\left\\lbrack \\frac{\\binom{3}{2} \\times 1}{\\binom{4}{2}} \\times \\frac{1}{\\binom{4}{2}} + \\frac{1 \\times \\binom{3}{1}}{\\binom{4}{2}} \\times \\frac{1}{\\binom{5}{2}} \\right\\rbrack$$\n$$P\\left( A_{2,3} \\right) = \\frac{1}{3}\\left( \\frac{1}{12} + \\frac{1}{20} \\right) = \\frac{2}{45}$$\n$$P(A) = \\frac{1}{60} + \\frac{1}{45} + \\frac{2}{45} = \\frac{1}{12}$$\n$$\\frac{P\\left( A_{1,2} \\right)}{P(A)} = \\frac{1}{5}$$", "solution_images": ["images/image16.png"], "subject": "Mathematics", "topic": "Probability", "difficulty": "", "question_type": "subjective", "has_image": true, "exam": "JEE Advanced", "source_paper": "Mathematics_1-50_with_img.docx"}
+{"question_id": "ADV-M01-Q40", "question": "If $\\Delta_{0}$ is the area of $\\bigtriangleup$ formed\nby joining the points of contact of incircle with the sides of the given\ntriangle whose area is $\\Delta$. Similarly $\\Delta_{1},\\Delta_{2}$ and\n$\\Delta_{3}$ are the corresponding area of the $\\bigtriangleup$ formed\nby joining the points of contact of excircles with the sides. Find the\nvalue of\n$$\\frac{\\Delta_{1}}{\\Delta} + \\frac{\\Delta_{2}}{\\Delta} + \\frac{\\Delta_{3}}{\\Delta} - \\frac{\\Delta_{0}}{\\Delta}.$$", "question_images": [], "answer": "2", "solution": "Let $A_{1},B_{1}$ and $C_{1}$ be the points of contact of\nthe ex-circle opposite to vertex $A$, with the side $AC,BC$ and $AB$\nrespectively.\n[IMAGE] Then,\n$$I_{1}A_{1} = I_{1}B_{1} = I_{1}C_{1} = r_{1}.$$\n$$\\angle B_{1}I_{1}A_{1} = \\pi - \\angle B_{1}CA_{1} = \\pi - (\\pi - C) = C.$$\nSimilarly,\n$$\\angle A_{1}I_{1}C_{1} = B,$$\n$$\\angle C_{1}I_{1}B_{1} = (B + C).$$\nNow,\n$$\\Delta_{1} = \\Delta_{A_{1}I_{1}B_{1}} + \\Delta_{A_{1}I_{1}C_{1}} - \\Delta_{B_{1}I_{1}C_{1}}.$$\n$$\\Delta_{1} = \\frac{1}{2}\\left( r_{1}^{2} \\right)\\left( \\sin C + sinB - sinA \\right).$$\n$$\\Delta_{1} = \\frac{1}{2}\\frac{\\Delta^{2}}{(s - a)^{2}}\\left( \\frac{c}{2R} + \\frac{b}{2R} - \\frac{a}{2R} \\right).$$\n$$\\Delta_{1} = \\frac{1}{2}\\frac{\\Delta^{2}}{(s - a)^{2}} \\cdot \\frac{(2s - 2a)}{2R} = \\frac{\\Delta^{2}}{2R(s - a)} = \\frac{r_{1}\\Delta}{2R}.$$\nTherefore,\n$$\\frac{\\Delta_{1}}{\\Delta} = \\frac{r_{1}}{2R}.$$\nSimilarly,\n$$\\frac{\\Delta_{2}}{\\Delta} = \\frac{r_{2}}{2R},\\quad\\frac{\\Delta_{3}}{\\Delta} = \\frac{r_{3}}{2R}\\quad\\text{and}\\quad\\frac{\\Delta_{0}}{\\Delta} = \\frac{r}{2R}.$$\nThus,\n$$\\frac{\\Delta_{1}}{\\Delta} + \\frac{\\Delta_{2}}{\\Delta} + \\frac{\\Delta_{3}}{\\Delta} - \\frac{\\Delta_{0}}{\\Delta} = \\frac{1}{2R}\\left( r_{1} + r_{2} + r_{3} - r \\right).$$\n$$\\frac{\\Delta_{1}}{\\Delta} + \\frac{\\Delta_{2}}{\\Delta} + \\frac{\\Delta_{3}}{\\Delta} - \\frac{\\Delta_{0}}{\\Delta} = \\frac{1}{2R}(4R) = 2.$$", "solution_images": ["images/image24.png"], "subject": "Mathematics", "topic": "Circles", "difficulty": "", "question_type": "numerical", "has_image": true, "exam": "JEE Advanced", "source_paper": "Mathematics_1-50_with_img.docx"}
+{"question_id": "ADV-M01-Q6", "question": "Let\n$M = \\left\\{ (x,y)\\mathbb{\\in R \\times R:}x^{2} + y^{2} \\leq r^{2} \\right\\}$\nwhere $r > 0$\nConsider the geometric progression\n$a_{n} = \\frac{1}{2^{n - 1}},\\ n = 1,2,3\\ldots.$ Let $S_{0} = 0$ and,\nfor $n \\geq 1,$let $S_{n}$ denote the sum of the first n terms of this\nprogressions. For $n \\geq 1,$let $C_{n}$ denote the circle with center\n$(S_{n - 1},0)$ and radius, $a_{n}$ and $D_{n}$ denote the circle with\ncenter $(S_{n - 1},S_{n - 1})$ and $a_{n}$.\nConsider $M$ with $r = \\frac{1025}{513}$. Let $k$ be the number of all\nthose circles $C_{n}$ that are inside $M$. Let $l$ be the maximum\npossible number of circles among these $k$ circles such that no two\ncircles intersect. Then", "question_images": [], "answer": "$3k + 2l = 40$", "solution": "$\\because a_{n} = \\frac{1}{2^{n - 1}}$\\\n$${S_{n} = 1 + \\frac{1}{2} + \\frac{1}{2^{2}} + \\ldots + \\frac{1}{2^{n - 1}}\n}{= 2\\left( 1 - \\frac{1}{2^{n}} \\right) = 2 - \\frac{1}{2^{n - 1}}\n}$$For circles $C_{n}$ to inside $M$\\\n$${S_{n - 1} + a_{n} < \\frac{1025}{513}\n}{\\Rightarrow \\ 2 - \\frac{1}{2^{n - 2}} + \\frac{1}{2^{n - 1}} < \\frac{1025}{513}\n}{\\Rightarrow \\ 1 - \\frac{1}{2^{n}} < \\frac{1025}{1026} = 1 - \\frac{1}{1026}\n}{\\Rightarrow 2^{n} < 2026 \\Rightarrow n \\leq 10 \\Rightarrow k = 10\n}$$Also $l = 5$\\\n$$3k + 2l = 30 + 10 = 40$$", "solution_images": [], "subject": "Mathematics", "topic": "Circles", "difficulty": "", "question_type": "subjective", "has_image": false, "exam": "JEE Advanced", "source_paper": "Mathematics_1-50_with_img.docx"}
+{"question_id": "ADV-M01-Q37", "question": "Tangents are parallel to the three sides are drawn to\nthe in-circle. If $x,y,z$ are the lengths of the parts of the tangents\nwithin the triangle then find the value of\n$$\\frac{x}{a} + \\frac{y}{b} + \\frac{z}{c}$$", "question_images": [], "answer": "1", "solution": "Let\n$$PQ = x,\\quad PQ \\parallel BC,$$\n$$SR = y,\\quad SR \\parallel AC,$$\n$$TU = z,\\quad TU \\parallel AB.$$\nLet $I$ be the in-centre of $\\bigtriangleup ABC$.\n[IMAGE] Consider $\\bigtriangleup APQ$.\nBy sine rule,\n$$\\frac{x}{\\sin A} = \\frac{AQ}{\\sin B} = \\frac{AP}{\\sin C}.$$\nTherefore,\n$$AQ = \\frac{\\sin B}{\\sin A}x = \\frac{b}{a}x,$$\n$$AP = \\frac{\\sin C}{\\sin A}x = \\frac{c}{a}x.$$\nAlso, $r$ will be ex-radius of $\\bigtriangleup APQ$.\n$$r = \\left( \\frac{x + PA + AQ}{2} \\right)\\tan\\frac{A}{2} = \\left\\lbrack \\frac{x + \\frac{bx}{a} + \\frac{cx}{a}}{2} \\right\\rbrack\\tan\\frac{A}{2}.$$\n$$r = \\frac{(a + b + c)}{2a}x\\tan\\frac{A}{2} = \\frac{sx}{a}\\tan\\frac{A}{2}.$$\nSimilarly, in $\\bigtriangleup BRS$,\n$$BR = \\frac{c}{b}y,\\quad BS = \\frac{a}{b}y\\quad\\text{and}\\quad r = \\frac{sy}{b}\\tan\\frac{B}{2}.$$\n$$\\left\\{ \\because\\ BR + BS + y = \\frac{c}{b}y + \\frac{a}{b}y + y = \\frac{sy}{b} \\right\\}.$$\nAnd in $\\bigtriangleup CTU$,\n$$CT = \\frac{a}{c}z,\\quad CU = \\frac{b}{c}z\\quad\\text{and}\\quad r = \\frac{sz}{c}\\tan\\frac{C}{2}.$$\nWe also know that,\n$$r = (s - a)\\tan\\frac{A}{2} = (s - b)\\tan\\frac{B}{2} = (s - c)\\tan\\frac{C}{2}.$$\nTherefore,\n$$\\frac{sx}{a}\\tan\\frac{A}{2} = (s - a)\\tan\\frac{A}{2} \\Rightarrow \\frac{sx}{a} = s - a,$$\n$$\\frac{sy}{b}\\tan\\frac{B}{2} = (s - b)\\tan\\frac{B}{2} \\Rightarrow \\frac{sy}{b} = s - b,$$\n$$\\frac{sz}{c}\\tan\\frac{C}{2} = (s - c)\\tan\\frac{C}{2} \\Rightarrow \\frac{sz}{c} = s - c.$$\nOn adding,\n$$s\\left( \\frac{x}{a} + \\frac{y}{b} + \\frac{z}{c} \\right) = 3s - (a + b + c) = s.$$\nTherefore,\n$$\\left( \\frac{x}{a} + \\frac{y}{b} + \\frac{z}{c} \\right) = 1.$$", "solution_images": ["images/image21.png"], "subject": "Mathematics", "topic": "Circles", "difficulty": "", "question_type": "numerical", "has_image": true, "exam": "JEE Advanced", "source_paper": "Mathematics_1-50_with_img.docx"}