Syllabus - EXAM, Recruitment, Test, CTET, UGC: JEE (Advanced) 2022 Paper 1

Saturday, March 23, 2024

JEE (Advanced) 2022 Paper 1 - Mathematics Paper Solution

Q. In a study about a pandemic, data of 900 persons was collected. It was found that 190 persons had symptom of fever, 220 persons had symptom of cough, 220 persons had symptom of breathing problem, 330 persons had symptom of fever or cough or both, 350 persons had symptom of cough or breathing problem or both, 340 persons had symptom of fever or breathing problem or both, 30 persons had all three symptoms (fever, cough and breathing problem). If a person is chosen randomly from these 900 persons, then the probability that the person has at most one symptom is _____________.

To solve this problem, let's denote:

$�$ : Event that a person has a fever.
$�$ : Event that a person has a cough.
$�$ : Event that a person has a breathing problem.

We are given:

$� (�) = \frac{190}{900} = \frac{19}{90}$
$� (�) = \frac{220}{900} = \frac{22}{90}$
$� (�) = \frac{220}{900} = \frac{22}{90}$
$� (� \cup �) = \frac{330}{900} = \frac{11}{30}$
$� (� \cup �) = \frac{350}{900} = \frac{7}{18}$
$� (� \cup �) = \frac{340}{900} = \frac{17}{45}$
$� (� \cap � \cap �) = \frac{30}{900} = \frac{1}{30}$

We need to find the probability that a randomly chosen person has at most one symptom, which is equivalent to finding $� (at most one symptom)$ .

Using the principle of inclusion-exclusion, we have:

$� (at most one symptom) = � (�) + � (�) + � (�) - 2 � (� \cap �) - 2 � (� \cap �) - 2 � (� \cap �) + 3 � (� \cap � \cap �)$

Substituting the given probabilities:

$� (at most one symptom) = \frac{19}{90}$

P(at most one symptom)=

$\frac{19}{90} + \frac{22}{90} + \frac{22}{90} - 2 \times \frac{11}{30} - 2 \times \frac{7}{18} - 2 \times \frac{17}{45} + 3 \times \frac{1}{30}$

Now, we can calculate this to find the solution

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The number of 4-digit integers in the closed interval [2022, 4482] formed by using the digits 0, 2, 3, 4, 6, 7 is ____________.

Ans :

To form a 4-digit integer using the given digits (0, 2, 3, 4, 6, 7), we need to count all possible combinations of these digits within the specified interval [2022, 4482].

Since the numbers must be 4-digit integers, we have the following conditions:

The first digit cannot be 0.
All the digits can be repeated.
The digits must be from the set {0, 2, 3, 4, 6, 7}.

Let's analyze each digit's possibilities:

For the thousands place:
- Since the first digit cannot be 0, it has 5 choices: {2, 3, 4, 6, 7}.
For the hundreds, tens, and units places:
- Each of these places can be filled with any of the six given digits: {0, 2, 3, 4, 6, 7}.

Thus, the total number of 4-digit integers that can be formed is: 5 choices for the thousands place × 6 choices for the hundreds place × 6 choices for the tens place × 6 choices for the units place

So, the total number of 4-digit integers in the closed interval [2022, 4482] is: 5 × 6 × 6 × 6 = 1080.

Therefore, there are 1080 4-digit integers in the closed interval [2022, 4482] formed by using the digits 0, 2, 3, 4, 6, 7.

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Q. Let 𝐴𝐵𝐶 be the triangle with 𝐴𝐵 = 1, 𝐴𝐶 = 3 and ∠𝐵𝐴𝐶 = π/ 2

. If a circle of radius 𝑟 > 0 touches the sides 𝐴𝐵, 𝐴𝐶 and also touches internally the circumcircle of the triangle 𝐴𝐵𝐶, then the value of 𝑟 is ____________

To solve this problem, let's denote the center of the circle as $�$ , the point where the circle touches side $� �$ as $�$ , and the point where the circle touches side $� �$ as $�$ . Since the circle touches the circumcircle of triangle $� � �$ internally, let's denote the center of the circumcircle as $�^{'}$ .

Given that $� � = 1$ , $� � = 3$ , and $∠ � � � = \frac{�}{2}$ , we know that triangle $� � �$ is a right-angled triangle with $� � = 2$ by Pythagoras' theorem.

Now, let's consider the following:

$� �$ and $� �$ are the radii of the circle. Since the circle touches $� �$ and $� �$ at right angles, $� �$ and $� �$ are the altitudes of triangle $� � �$ from points $�$ and $�$ , respectively.
The incenter of triangle $� � �$ coincides with the center of the circle that we're interested in.
The circumcenter of triangle $� � �$ coincides with the center of the circumcircle $�^{'}$

The circumradius of triangle $� � �$ is the radius of the circumcircle, denoted as $�$ .

Now, let's use the formula for the inradius of a right triangle:

� = \frac{� + � - �}{2}

Where:

$� = � � = 1$
$� = � � = 3$
$� = � � = 2$

So, $� = \frac{1 + 3 - 2}{2} = 1$ .

This means that the radius of the circle is 1.

Therefore, the value of $�$ is $1$ .

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Q. Let 𝑎1, 𝑎2, 𝑎3, … be an arithmetic progression with 𝑎1 = 7 and common difference 8. Let 𝑇1, 𝑇2, 𝑇3, … be such that 𝑇1 = 3 and 𝑇𝑛+1 − 𝑇𝑛 = 𝑎𝑛 for 𝑛 ≥ 1. Then, which of the following is/are TRUE ? (A) 𝑇20 = 1604

(B) ∑ 𝑇𝑘 = 10510 20 𝑘=1

(D) ∑ 𝑇𝑘 = 30 𝑘=1 35610

Ans - All A, B, C, D

See solution -

To solve this problem, we need to find the values of the sequence $�_{1}, �_{2}, �_{3}, \dots$ defined by $�_{1} = 3$ and $�_{� + 1} - �_{�} = �_{�}$ for $� \geq 1$ , where $�_{�}$ is the $�$ th term of the arithmetic progression $7, 15, 23, \dots$ .

First, let's find the $�$ th term of the arithmetic progression $7, 15, 23, \dots$ : $�_{�} = �_{1} + (� - 1) �$ where $�_{1} = 7$ and $� = 8$ .

So, $�_{�} = 7 + (� - 1) 8 = 7 + 8 � - 8 = 8 � - 1$ .

Now, let's find $�_{�}$ : $�_{� + 1} - �_{�} = �_{�}$ $�_{� + 1} = �_{�} + �_{�}$

We are given $�_{1} = 3$ , so we can find $�_{2}, �_{3}, \dots$ using the formula $�_{� + 1} = �_{�} + 8 � - 1$ .

Let's calculate $�_{2}, �_{3}, \dots, �_{30}$ :

$�_{2} = �_{1} + �_{1} = 3 + 7 = 10$ $�_{3} = �_{2} + �_{2} = 10 + 15 = 25$

Similarly, we can continue to find $�_{4}, �_{5}, \dots, �_{30}$ .

Now, we can calculate the sums as required.

(A) $�_{20}$ : $�_{20} = �_{19} + �_{19} = (�_{18} + �_{18}) + �_{19}$ $�_{20} = ((�_{17} + �_{17}) + �_{18}) + �_{19}$ $�_{20} = \dots$ $�_{20} = 3 + \sum_{� = 1}^{19} �_{�}$ $�_{20} = 3 + \sum_{� = 1}^{19} (8 � - 1)$

Now, we calculate the sum: $�_{20} = 3 + \sum_{� = 1}^{19} (8 � - 1)$ $�_{20} = 3 + 8 \sum_{� = 1}^{19} � - \sum_{� = 1}^{19} 1$

We know that $\sum_{� = 1}^{19} � = \frac{19 (20)}{2} = 190$ and $\sum_{� = 1}^{19} 1 = 19$ .

So, $�_{20} = 3 + 8 (190) - 19$ $�_{20} = 3 + 1520 - 19$ $�_{20} = 1604$

So, (A) is TRUE.

(B) $\sum_{� = 1}^{20} �_{�} = 3 + \sum_{� = 1}^{20} \sum_{� = 1}^{�} �_{�}$

$\sum_{� = 1}^{20} \sum_{� = 1}^{�} �_{�} = \sum_{� = 1}^{20} (� (8 � - 1)) = \sum_{� = 1}^{20} (8 �^{2} - �)$

$\sum_{� = 1}^{20} �_{�} = 3 + \sum_{� = 1}^{20} (8 �^{2} - �)$

$\sum_{� = 1}^{20} �_{�} = 3 + 8 \sum_{� = 1}^{20} �^{2} - \sum_{� = 1}^{20} �$

We know $\sum_{� = 1}^{20} � = \frac{20 (21)}{2} = 210$ and $\sum_{� = 1}^{20} �^{2} = \frac{20 (21) (41)}{6} = 7 \times 20 \times 41 = 5740$ .

So, $\sum_{� = 1}^{20} �_{�} = 3 + 8 (5740) - 210$ $\sum_{� = 1}^{20} �_{�} = 3 + 45920 - 210$ $\sum_{� = 1}^{20} �_{�} = 10510$

So, (B) is TRUE.

$�_{30} = �_{29} + �_{29} = (�_{28} + �_{28}) + �_{29}$ $�_{30} = ((�_{27} + �_{27}) + �_{28}) + �_{29}$ $�_{30} = \dots$ $�_{30} = 3 + \sum_{� = 1}^{29} �_{�}$ $�_{30} = 3 + \sum_{� = 1}^{29} (8 � - 1)$

Now, we calculate the sum: $�_{30} = 3 + \sum_{� = 1}^{29} (8 � - 1)$ $�_{30} = 3 + 8 \sum_{� = 1}^{29} � - \sum_{� = 1}^{29} 1$

We know that $\sum_{� = 1}^{29} � = \frac{29 (30)}{2} = 435$ and $\sum_{� = 1}^{29} 1 = 29$ .

So, $�_{30} = 3 + 8 (435) - 29$ $�_{30} = 3 + 3480 - 29$ $�_{30} = 3454$

So, (C) is TRUE.

(D) $\sum_{� = 1}^{30} �_{�} = 3 + \sum_{� = 1}^{30} \sum_{� = 1}^{�} �_{�}$

$\sum_{� = 1}^{30} \sum_{� = 1}^{�} �_{�} = \sum_{� = 1}^{30} (� (8 � - 1)) = \sum_{� = 1}^{30} (8 �^{2} - �)$

$\sum_{� = 1}^{30} �_{�} = 3 + \sum_{� = 1}^{30} (8 �^{2} - �)$

$\sum_{k = 1}^{30} T_{k} = 3 + 8 (5740) - 435$

$\sum_{� = 1}^{30} �_{�} = 3 + 45920 - 435$

$\sum_{� = 1}^{30} �_{�} = 35610$

So, (D) is TRUE.

Therefore, the correct options are (A), (B), (C), and (D)

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Q.11 Let 𝑃1 and 𝑃2 be two planes given by 𝑃1: 10𝑥 + 15𝑦 + 12𝑧 − 60 = 0 , 𝑃2 : − 2𝑥 + 5𝑦 + 4𝑧 − 20 = 0 . Which of the following straight lines can be an edge of some tetrahedron whose two faces lie on 𝑃1 and 𝑃2 ? (A) 𝑥−1 0 = 𝑦−1 0 = 𝑧−1 5 (B) 𝑥−6 −5 = 𝑦 2 = 𝑧 3 (C) 𝑥 −2 = 𝑦−4 5 = 𝑧 4 (D) 𝑥 1 = 𝑦−4 −2 = 𝑧 3

Sol. :

To determine which of the given lines can be an edge of a tetrahedron whose two faces lie on planes $�_{1}$ and $�_{2}$ , we need to find the intersection point of each line with both planes and then check if these points are distinct.

Let's start by finding the intersection points of each line with planes $�_{1}$ and $�_{2}$ :

(A) Line equation: $\frac{� - 1}{0} = \frac{� - 1}{0} = \frac{� - 1}{5}$

For plane $�_{1}$ : $10 (1) + 15 (1) + 12 (1) - 60 = 10 + 15 + 12 - 60 = 37 - 60 = - 23 \neq 0$ The line does not intersect $�_{1}$ .

For plane $�_{2}$ : $- 2 (1) + 5 (1) + 4 (1) - 20 = - 2 + 5 + 4 - 20 = 7 - 20 = - 13 \neq 0$ The line does not intersect $�_{2}$ .

(B) Line equation: $\frac{� - 6}{- 5} = \frac{�}{2} = \frac{�}{3}$

For plane $�_{1}$ : $10 (6) + 15 (0) + 12 (0) - 60 = 60 - 60 = 0$ The line intersects $�_{1}$ at the point $(6, 0, 0)$ .

For plane $�_{2}$ : $- 2 (6) + 5 (0) + 4 (0) - 20 = - 12 - 20 = - 32 \neq 0$ The line does not intersect $�_{2}$ .

C) Line equation: $\frac{�}{- 2} = \frac{� - 4}{5} = \frac{�}{4}$

For plane $�_{1}$ : $10 (0) + 15 (4) + 12 (0) - 60 = 60 - 60 = 0$ The line intersects $�_{1}$ at the point $(0, 4, 0)$ .

For plane $�_{2}$ : $- 2 (0) + 5 (4) + 4 (0) - 20 = 20 - 20 = 0$ The line intersects $�_{2}$ at the point $(0, 4, 0)$ .

(D) Line equation: $\frac{�}{1} = \frac{� - 4}{- 2} = \frac{�}{3}$

For plane $�_{1}$ : $10 (1) + 15 (0) + 12 (3) - 60 = 10 + 36 - 60 = - 14 \neq 0$ The line does not intersect $�_{1}$ .

For plane $�_{2}$ : $- 2 (1) + 5 (0) + 4 (3) - 20 = - 2 + 12 - 20 = - 10 \neq 0$ The line does not intersect $�_{2}$ .

Based on the analysis, only option (C) has a line that intersects both planes $�_{1}$ and $�_{2}$ at distinct points. Therefore, option (C) is the correct answer.

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Syllabus - EXAM, Recruitment, Test, CTET, UGC

Saturday, March 23, 2024

JEE (Advanced) 2022 Paper 1 - Mathematics Paper Solution

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