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:
- A: Event that a person has a fever.
- B: Event that a person has a cough.
- C: Event that a person has a breathing problem.
We are given:
- P(A)=900190=9019
- P(B)=900220=9022
- P(C)=900220=9022
- P(A∪B)=900330=3011
- P(B∪C)=900350=187
- P(A∪C)=900340=4517
- P(A∩B∩C)=90030=301
We need to find the probability that a randomly chosen person has at most one symptom, which is equivalent to finding P(at most one symptom).
Using the principle of inclusion-exclusion, we have:
P(at most one symptom)=P(A)+P(B)+P(C)−2P(A∩B)−2P(B∩C)−2P(A∩C)+3P(A∩B∩C)
Substituting the given probabilities:
P(at most one symptom)=
9019+9022+9022−2×3011−2×187−2×4517+3×301
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 O, the point where the circle touches side AB as D, and the point where the circle touches side AC as E. Since the circle touches the circumcircle of triangle ABC internally, let's denote the center of the circumcircle as O′.
Given that AB=1, AC=3, and ∠BAC=2π, we know that triangle ABC is a right-angled triangle with BC=2 by Pythagoras' theorem.
Now, let's consider the following:
OD and OE are the radii of the circle. Since the circle touches AB and AC at right angles, OD and OE are the altitudes of triangle ABC from points B and C, respectively.
The incenter of triangle ABC coincides with the center of the circle that we're interested in.
The circumcenter of triangle ABC coincides with the center of the circumcircle O′
The circumradius of triangle ABC is the radius of the circumcircle, denoted as R.
Now, let's use the formula for the inradius of a right triangle:
r=2a+b−c
Where:
- a=AB=1
- b=AC=3
- c=BC=2
So, r=21+3−2=1.
This means that the radius of the circle is 1.
Therefore, the value of r 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
(C) 𝑇30 = 3454
(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 T1,T2,T3,… defined by T1=3 and Tn+1−Tn=an for n≥1, where an is the nth term of the arithmetic progression 7,15,23,….
First, let's find the nth term of the arithmetic progression 7,15,23,…:
an=a1+(n−1)d
where a1=7 and d=8.
So, an=7+(n−1)8=7+8n−8=8n−1.
Now, let's find Tn:
Tn+1−Tn=an
Tn+1=Tn+an
We are given T1=3, so we can find T2,T3,… using the formula Tn+1=Tn+8n−1.
Let's calculate T2,T3,…,T30:
T2=T1+a1=3+7=10
T3=T2+a2=10+15=25
Similarly, we can continue to find T4,T5,…,T30.
Now, we can calculate the sums as required.
(A) T20:
T20=T19+a19=(T18+a18)+a19
T20=((T17+a17)+a18)+a19
T20=…
T20=3+∑n=119an
T20=3+∑n=119(8n−1)
Now, we calculate the sum:
T20=3+∑n=119(8n−1)
T20=3+8∑n=119n−∑n=1191
We know that ∑n=119n=219(20)=190 and ∑n=1191=19.
So,
T20=3+8(190)−19
T20=3+1520−19
T20=1604
So, (A) is TRUE.
(B) ∑k=120Tk=3+∑k=120∑n=1kan
∑k=120∑n=1kan=∑k=120(k(8k−1))=∑k=120(8k2−k)
∑k=120Tk=3+∑k=120(8k2−k)
∑k=120Tk=3+8∑k=120k2−∑k=120k
We know ∑k=120k=220(21)=210 and ∑k=120k2=620(21)(41)=7×20×41=5740.
So,
∑k=120Tk=3+8(5740)−210
∑k=120Tk=3+45920−210
∑k=120Tk=10510
So, (B) is TRUE.
(C) We need to find T30:
T30=T29+a29=(T28+a28)+a29
T30=((T27+a27)+a28)+a29
T30=…
T30=3+∑n=129an
T30=3+∑n=129(8n−1)
Now, we calculate the sum:
T30=3+∑n=129(8n−1)
T30=3+8∑n=129n−∑n=1291
We know that ∑n=129n=229(30)=435 and ∑n=1291=29.
So,
T30=3+8(435)−29
T30=3+3480−29
T30=3454
So, (C) is TRUE.
(D) ∑k=130Tk=3+∑k=130∑n=1kan
∑k=130∑n=1kan=∑k=130(k(8k−1))=∑k=130(8k2−k)
∑k=130Tk=3+∑k=130(8k2−k)
∑k=130Tk=3+8(5740)−435
∑k=130Tk=3+45920−435
∑k=130Tk=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 P1 and P2, 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 P1 and P2:
(A) Line equation: 0x−1=0y−1=5z−1
For plane P1:
10(1)+15(1)+12(1)−60=10+15+12−60=37−60=−23=0
The line does not intersect P1.
For plane P2:
−2(1)+5(1)+4(1)−20=−2+5+4−20=7−20=−13=0
The line does not intersect P2.
(B) Line equation: −5x−6=2y=3z
For plane P1:
10(6)+15(0)+12(0)−60=60−60=0
The line intersects P1 at the point (6,0,0).
For plane P2:
−2(6)+5(0)+4(0)−20=−12−20=−32= (Does Not) =0
The line does not intersect P2.
C) Line equation: −2x=5y−4=4z
For plane P1:
10(0)+15(4)+12(0)−60=60−60=0
The line intersects P1 at the point (0,4,0).
For plane P2:
−2(0)+5(4)+4(0)−20=20−20=0
The line intersects P2 at the point (0,4,0).
(D) Line equation: 1x=−2y−4=3z
For plane P1:
10(1)+15(0)+12(3)−60=10+36−60=−14=0
The line does not intersect P1.
For plane P2:
−2(1)+5(0)+4(3)−20=−2+12−20=−10=0
The line does not intersect P2.
Based on the analysis, only option (C) has a line that intersects both planes P1 and P2 at distinct points. Therefore, option (C) is the correct answer.
