ALPS Mathematics Advanced Problem Solving – Questions & Solutions (JEE 2022)

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ALPS Mathematics 2208| JEE 2022
Syllabus: Vectors, Three Dimensional Geometry, Probability
Solutions
Day – 1
1.(C) Let an be the number of strings of H and T of length n with no two adjacent H’s, then a1  2, a2  3.
Also, an  2  an 1  an ( the string must begin with T or HT)
13 13
So, a3  5, a4  8, a5  8  5  13  the required probability =  .
25 32
2
2.(ABCD) We have P (AB)  max {P (A), P (B)} = ,
3
Next, P(A  B) = P(A) + P(B)  P(A  B)  P(A) + P(B) –1 = 1/6
1 1 1
And P(A  B)  min {P(A), P(B)} =   P(A  B) 
2 6 2
1 1 1
Also P(A  B) = P(A)  P(A  B)   
2 6 3
Lastly P(A  B) = P(B)  P(A  B)
2 1 1
2/3  1/2  P(A  B)   1/6  P(A  B) 
3 6 2
3.(AD) Clearly P  A  P  C 
3 3
1 1 1 3 1
Event A is no boy or exactly one boy in family  P  A     3C1     
2 2 8 8 2
3 3
1 1 3
Event B is 2 boys, 1 girl or 1 boy, 2 girls P  B   3C1    3C2   
 
2  
2 4
3 3
1 1 1
Event C is no girl or exactly one girl P  C      3C1   
2 2 2
3
1 3
Event A  B is one boy and two girls P  A  B   3C1   
 
2 8
3
Event B  C is one girl and two boys P  B  C  =
8
AC    A  B  C is also 
3
P  A  B    P  A   P  B  , so, A and B are independent neither P  A  B  C   P  A  P  B   P  B  nor
8
P  A  C   P  A  P  C  So, ABC is not independent

4.(C) We have pr  (r  b )a  c  pr  b  (r  b )a  b  c  b
c b c b c c .b
 pr  b  0  c  b  r .b   pr  a c r   2 a .
p p p p

VMC | JEE-2022 | Mathematics 1 ALPS | Week 8

, Vidyamandir Classes: Innovating for Your Success

*5.(AB) aˆ  bˆ  cos 2θ and | aˆ  bˆ |  1  1  1  2aˆ  bˆ  1 ; 2aˆ  bˆ  1  cos 2θ  1/ 2  1/2 < cos 2  1
x 1 y 1 z 1
6.(C) Let any point on the given line    r is (2r + 1, r + 1, r + 1)
2 1 1
On putting this point on the plane x + y + z = 1
1
we get r  
2

1 1
 Given line cuts the plane x + y + z = 1 at the point Q   0,, 

2 2
x 1 y 1 z 1
Now the line from point (1, 1, 1) and perpendicular to the given plane is     … (1)
1 1 1
 Any point on the line (1) is (+1, +1, +1).
1 1 1 1
Therefore line (1) cuts the given plane at point R when    point R is  , , 
3  3 3 3
1 1 1
x
y z
Hence the equation of required line QR is 3 3 3
1 1 1 1 1
0  
3 3 2 3 2
1 1 1
x y z
or 3 3 3  3x  1  3 y  1  3z  1 .
1 1 1 2 1 1
 
3 6 6
x  2 y  1 0 1
7.(CD) For the points where the line intersects the curve, we have z = 0. Therefore  
3 2 1
x2 y 1
  1 and = 1 x = 5 and y = 1
3 2
Put these values in xy = c2, we get c =  5 .
8.(8) The queens will be facing each other in case both are placed on the same column, the same row or the same
diagonal of the chess board. There are eight rows, eight columns, two principal diagonals and four sets each
containing diagonal lines of 7, 6, 5, 4, 3, 2, 1 squares. Hence the probability of the queens taking on each
other is

P ( A) 
18 8C2  4  7
C2  6C2  5C2  4C2  3C2  2C2 
64
C2
18  8  7  4  7  6  6  5  5  4  4  3  3  2  2  1

64  63
18  7   7  3  3  5  5  2  2  3  3  1 18  7  7  3  3  7  14  18  6  2 26 13
     .
8  63 8  63 8 9 72 36
13 23
Hence the probability that the two queen to not take on each other is P( A)  1   .
36 36
9.(0.20) Let C = the event of the selected number being composite,
E = the event of the there being no remainder.
n  C  15 3
 , P C  
2
P C   
n  S  25 5 5

, PE / C 
4 1
PE / C 
15 10

P  E   P C  . P  E / C   P C  .P  E / C  =
3 4 2 1 4 1 1
.  .     0.2
5 15 5 10 25 25 5




VMC | JEE-2022 | Mathematics 2 ALPS | Week 8