JEE Main July 2022

Question Paper With Text Solution

28 July | Shift-1

MATHEMATICS

MATRIX

JEE Main & Advanced | XI-XII Foundation| VI-X Pre-Foundation

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 Question Paper With Text Solution (Mathematics)
JEE Main July 2022 | 28 July Shift-1
JEE MAIN JULY 2022 | 28TH JULY SHIFT-1
SECTION – A
Question ID : 100601
Differential Equation

1. Let the solution curve of the differential equation xdy   

x 2  y 2  y dx , x > 0, intersect the line
x = 1 at y = 0 and the line x = 2 at y = . Then the value of  is :

ekuk vody lehdj.k xdy   

x 2  y 2  y dx , x > 0 dk gy oØ] js[kk
x = 1 dksy = 0 ij rFkk js[kk

x = 2 dksy = ij dkVrk gSA rks

 dk eku gS:
1 3 3 5
(1) (2) (3)  (4)
2 2 2 2
Ans. Official Answer NTA (2)

Sol. xdy   x 2  y 2  y dx

xdy  ydx  x 2  y 2 dx

xdy  ydx y 2 dx
 1  .
x2 x2 x

dy / x dx

y
2 x
1  
x

y 2 
 y
ln     1   ln x  R
x x 
 

y  y2  x 2
 cx
x

y  y 2  x 2  cx 2
x =1, y = 0  0 +1 = C  C = 1
Curve is y  x 2  y 2  x 2
x = 2, y = 
2  4  2  4

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 Question Paper With Text Solution (Mathematics)
JEE Main July 2022 | 28 July Shift-1
4   2  16   2  8
3

2
Question ID : 100602
ITF
2. Considering only the principal values of the inverse trigonometric functions, the domain of the function

 x 2  4x  2 
f  x   cos 1  2  is :
 x 3 

 x 2  4x  2 
izfrykse f=kdks.kferh; Qyu ds dsoy eq[; eku ysrs gq,]
f  Qyu
x   cos 1  2  dk izkar:gS
 x 3 

 1  1  1
(1)  ,  (2)   ,   (3)   1 ,   (4)  , 
 4  4   3   3
Ans. Official Answer NTA (2)

x 2  4x  2
Sol. 1
x2  3
2 2
  x 2  4x  2    x 2  3
2 2
  x 2  4x  2    x 2  3  0

  2x 2  4x  5   4x  1  0

1
 4x  1  0  x  
4
Question ID : 100603
Vectors
  
3. Let the vectors a  1  t  ˆi  1  t  ˆj  kˆ , b  1  t  ˆi  1  t  ˆj  2kˆ and c  tiˆ  tjˆ  kˆ , t  R be such that
   
for , ,   R, a   b  c  0        0 . Then, the set of all values of t is :
(1) a non-empty finite set (2) equal to N
(3) equal to R – {0} (4) equal to R
 
ekuk lfn'k a  1  t  ˆi  1  t  ˆj  kˆ , b  1  t  ˆi  1  t  ˆj  2kˆ rFkkc  tiˆ  tjˆ  kˆ , t  R bl izdkj gS fd
   
, ,   R, ds fy, a   b  c  0        0 rkst ds lHkh ekuksa dk leqPp; : gS
(1) ,d vfjDr ifjfer leqPp; gS (2) N ds cjkcj gS

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 Question Paper With Text Solution (Mathematics)
JEE Main July 2022 | 28 July Shift-1
(3) R – {0} ds cjkcj gS (4) R ds cjkcj gS
Ans. Official Answer NTA (3)
Sol. By its given condition
  
: a, b, c are linearly independent vectors

 a b c   0 ....(i)

Now, a b c 

1 t 1 t 1
 1 t 1 t 2
t t 1

C2  C1  C2

1 t 2 1
1 t 2 2
t 0 1

1 t 2 1
 2 1 t 2 2
t 0 1

 2 1  t   1  t   t 

 2 3t   6t

a b c   0  t  0
 
Question ID : 100604
ITF
4. Considering the principal values of the inverse trigonometric functions, the sum of all the solutions of the equation
cos–1(x) – 2sin–1(x) = cos–1(2x) is equal to :
izfrykse f=kdks.kferh; Qyu ds eq[; eku ysrs gq, lehdj.k
cos–1(x) – 2sin–1(x) = cos–1(2x) ds lHkh gyksa dk
;ksx gS
:
1 1
(1) 0 (2) 1 (3) (4) 
2 2
Ans. Official Answer NTA (1)

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 Question Paper With Text Solution (Mathematics)
JEE Main July 2022 | 28 July Shift-1
Sol. 1 1
cos x  2 sin x  cos 2x 1

 
cos 1 x  2   cos 1 x   cos 1 2x
2 
cos 1 x    2 cos 1 x  cos 1 2x
3cos 2 x    cos 1 2x ....(1)
cos  3cos 1 x   cos    cos 1 2x 

4x 3  3x  2x
1
4x 3  x  x  0, 
2
All satisfy the original equation
1 1
sum   to   0
2 2
Question ID : 100605
Mathematical Reasoning
5. Let the operations *,  {, }. If (p * q) (p ~q) is a tautology, then the ordered pair (*, ) is :
ekuk *,  {, } gSA ;fn(p * q) (p ~q) ,d iqu#fDr gS] rks Øfer ;qXe
(*, ) gS:
(1) (, ) (2) (, ) (3) (, ) (4) (, )
Ans. Official Answer NTA (2)
Sol. Well check each option
For A  = v of 0 = A
 pvq    pv  q 
 pv  q   q 

 pv  c   p
For B : * = v, O = v
 pvq    pv  q   t using Venn Diagrams

Question ID : 100606
Vectors

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 Question Paper With Text Solution (Mathematics)
JEE Main July 2022 | 28 July Shift-1
 
6. Let a vector a has magnitude 9. Let a vector b be such that for every (x, y)  R × R – {(0, 0)}, the vector
     
   
xa  yb is perpendicular to the vector 6ya  18xb . Then the value of a  b is equal to :

ekuk ,d lfn'k a dk ifjek.k9 gSA ekuk ,d lfn'kb bl izdkj gS fd izR;sd(x, y)  R × R – {(0, 0)} ds
     
  
fy,] lfn'k xa  yb , lfn'k 6ya  18xb ds yacor gSA rks  a  b dk eku cjkcj gS:

(1) 9 3 (2) 27 3 (3) 9 (4) 81

Ans. Official Answer NTA (2)
    
Sol.  
| a | 9 & xa  yb . 6ya  18xb  0 
 6xy | a | 18x  a.b   6y  a.b   18xy | b |  0
2 2 2 2

 6xy | a | 3 | b |    a.b   y  3x   0
2 2 2 2

This should hold x, y  R  R

 
 | a |2  3 | b |2 & a.b  0
2 2
Now a  b | a |2 | b |2  a.b  
| a |2
 a.
3
| a |2 81
 | a  b |   27 3
3 3
Question ID : 100607
Circle
7. For t  (0, 2), if ABC is an equilateral triangle with vertices A(sin t, – cot t), B(cos t, sin t) and C(a, b) such
1
that its orthocentre lies on a circle with centre 1,  , then (a2 – b2) is equal to :
 3
t  (0, 2) ds fy,] ;fn 'kh"kksZa
A(sin t, – cot t), B(cos t, sin t) rFkk
C(a, b) ds ,d leckgq f=kHkqt
ABC dk yac
 1  gS
dsUnz] ,d o`Ùk ftldk dsUnz
1,  , ij fLFkr gS] rks
(a2 – b2) cjkcj gS:
 3
3 77 80
(1) (2) 8 (3) (4)
8 9 9
Ans. Official Answer NTA (2)
Sol. s  sin t,c  cos t
Let orthocentre be (h,k)
Since it if an equilateral triangle hence orthocentre coincides with centroid.
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 Question Paper With Text Solution (Mathematics)
JEE Main July 2022 | 28 July Shift-1
 a + s + c = 3h, b + s – c = 3k
2 2 2 2
  3h  a    3k  b    s  c    s  c   2  s 2  c 2   2
2 2
 a  b 2
h   K    ,
 3  3 9

a b
circle centre at  , 
3 3
a b 1
Gioves,  1,   a  3, b  1
3 3 3
 a 2  b2  8
Question ID : 100608
Set & Relations
8. For  N, consider a relation R on N given by R = {(x, y) : 3x + y is a multiple of 7}. The relation R is an
equivalence relation if and only if :
(1)  = 14
(2)  is a multiple of 4
(3) 4 is the remainder when  is divided by 10
(4) 4 is the remainder wehn  is devided by 7
 N ds fy, , N ij ,d laca/k R,R = {(x, y) : 3x + y, 7 dk ,d xq.kt gS} }kjk fn;k x;k gSA laca/k
R ,d
rqY;rk laca/k gS ;fn vkSj dsoy: ;fn
(1)  = 14 gS
(2) , 4 dk ,d xq.kt gS
(3)  dks10 ls foHkkftr djus ij 'ks"kQy
4 gS
(4)  dks7 ls foHkkftr djus ij 'ks"kQy4gS
gS
Ans. Official Answer NTA (4)
Sol. For R to be reflexive  x R x
 3x +  x = 7x  (3 + ) x = 7K
 3 +  = 7  = 7 – 3 = 7N + 4, K,,N  I
 when  divided by 7, remainder is 4.
R to be symmetric xRy  yRx
3x + y = 7N1, 3y + x = 7N2
 (3 + ) (x + y) = 7 (N1 + N2) = 7N3
Which holds when 3 +  is multiple of 7

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 Question Paper With Text Solution (Mathematics)
JEE Main July 2022 | 28 July Shift-1
  = 7N + 4 (as did earlier)
R to be transitive
xRy&yRz  xRz .
3x + y = 7N1 & 3y + z = 7N2 and
3x + z = 7N3
 3x + 7N2 – 3y = 7N3
 7N1 - y + 7N2 –3y = 7N3
 7 (N1 + N2) – (3 + y = 7N3
 (3 + ) y = 7N
Which is true again when 3 +  divisible by 7, i.e.
when  divided by 7, remainder is 4.
Question ID : 100609
Probability
9. Out of 60% female and 40% male candidates appearing in an exam, 60% candidates qualify it. The number of
females qualifying the exam is twice the number of males qualifying it. A candidate is randomly chosen from the
qualified candidates. The probability, that the chosen candidate is a female, is :
60% efgyk rFkk
40% iq#"k vH;kfFkZ;kas }kjk nh xbZ ,d60%
ijh{kk
vH;FkhZ
esa lQy gksrs gSaA ijh{kk esa l
gksus okyh efgykvksa dh la[;k] ijh{kk esa lQy gksus okys iq:"kksa dh la[;k dh nks xquk gSA lQ
,d vH;kFkhZ ;kn`PN;k pquk tkrk gSA pqus x, vH;kFkhZ ds efgyk gksus : dh izkf;drk gS
3 11 23 13
(1) (2) (3) (4)
4 16 32 16
Ans. Official Answer NTA (1)

Sol.

40 2
Probability that chosen candidate is female  
60 3
Question ID : 100610
Differential Equation

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 Question Paper With Text Solution (Mathematics)
JEE Main July 2022 | 28 July Shift-1
10. If y = y(x), x  (0, /2) be the solution curve of the differential equation
dy  
 sin 2
2x 
dx
  8sin 2 2x  2sin 4x  y  2e 4x  2 sin 2x  cos 2x  , with y    e  , then y   is equal
4 6
to :
dy
;fn vody lehdj.k  sin 2
2x    8sin 2 2x  2sin 4x  y  2e 4x  2 sin 2x  cos 2x  , x  (0, /2)
dx
 
y    e  dk gy oØ y = y(x) gS
, rksy   cjkjcj gS:
4 6
2 2  /3 2 2  /3 1 2  /3 1 2  /3
(1) e (2) e (3) e (4) e
3 3 3 3
Ans. Official Answer NTA (1)
Sol. Given differential equation can be re-written as
dy 2e 4x
  8  4 cot 2x  y   2 sin x  cos 2x 
dx sin 2 2x
which is a linear diff. equation.
I.f .  e 8 4cot 2x dx  e8x  2Cu sin 2x 
 e8x .sin 2 2x
 solution is
y  e8x .sin 2 2x    2e 4x  2sin 2x  cos 2x  dx  C

 e 4x .sin 2x  C
   
Given y    e  C  0
4
e4x
y 
sin 2x

4.
 e 6 2  23
y    e
 6  sin  2.   3
 
 6
Question ID : 100611
Parabola
11. If the tangents drawn at the points P and Q on the parabola y2 = 2x – 3 intersect at the point R(0, 1), then the
orthocentre of the triangle PQR is :

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 Question Paper With Text Solution (Mathematics)
JEE Main July 2022 | 28 July Shift-1
;fn ijoy; y2 = 2x – 3 ds fcUnqvksa
P rFkk
Q ij [khaph xbZ Li'kZ js[kk,sa
R(0,fcUnq
1), ij feyrh gSaA rks f=kHkqt
PQR dk yac dsUnz: gS
(1) (0, 1) (2) (2, –1) (3) (6, 3) (4) (2, 1)
Ans. Official Answer NTA (2)
Sol. y2 = 2x – 3 …(1)
Equation of chord of contact
PQ : r = 0
yx1 = (x + 0) – 3
y=x–3 ....(2)

from (1) and (2)

(x . 3)2 = 2x – 3
x2 – 8x +12 = 0
(x – 2)(x – 6) = 0
x = 2or 6
y = –1or 3

1
MPQ  1
4
2 1
MQR  
6 3
2 1
MPR  
6 3
2
MPR   1
2

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 Question Paper With Text Solution (Mathematics)
JEE Main July 2022 | 28 July Shift-1
MPQ  MPR    PQ  PR
Orthocentre = P (2, –1)
Question ID : 100612
Circle
11
12. Let C be the centre of the circle x2 + y2 – x + 2y = and P be a point on the circle. A line passes through the
4

point C, makes an angle of with the line CP and intersects the circle at the points Q and R. Then the area of
4
the triangle PQR (in unit2) is :
11
ekuk o`Ùk
x2 + y2 – x + 2y = dks dsUnz
C gS rFkk o`Ùk ij ,d fcUnq
P gSA ,d js[kk] fcUnq
C ls gksdj tkrh gS]
4

js[kkCP ls  dks dks.k cukrh gS rFkk o`Ùk dks

Q rFkk
fcUnqvksa
R ij dkVrh gSA rks f=kHkqt
PQR dk {ks=kQy
(oxZ
4
bdkbZ)esa
gS:
 
(1) 2 (2) 2 2 (3) 8sin   (4) 8 cos  
8 8
Ans. Official Answer NTA (2)
11
Sol. x 2  y 2  x  2y 
4
2
 1 2 2
 x     y  1   2 
 2
Or  PQR
1
PR  QK sin 2 
3


 4.6 sin
8

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 Question Paper With Text Solution (Mathematics)
JEE Main July 2022 | 28 July Shift-1
1
PQ  QR cos 22
2

 4 cos
8
1
As PQR  PR  PQ
2
1   
  42 sin   4 cos 
2 6  8
 4
 4sin  2 2
4 2
Question ID : 100613
Binomial Theorem
13. The remainder when 72022 + 32022 is divided by 5 is :
72022 + 32022 dks6 ls foHkkftr djus ij 'ks"kQy :gS
(1) 0 (2) 2 (3) 3 (4) 4
Ans. Official Answer NTA (3)
Sol. 72022 + 32022
= (49)1011 + (9)1011
= (50 – 1)1011 + (10 – 1)1011
= 5 –1+ 5K –1
= 5m – 2
Remainder = 5 – 2 = 3
Question ID : 100614
Matrices

0 1 0 
14. Let the matrix A  0 0 1  and the matrix B0 = A49 + 2A98. If Bn = Adj(Bn–1) for all n  1, then det(B4) is
 
1 0 0 
equal to :

0 1 0 
A  0 0 1  rFkk vkO;wg
ekuk vkO;wg B0 = A49 + 2A98 gSaA ;fn lHkh
n  1 ds fy, Bn = Adj(Bn–1) gS] rks
1 0 0 

det(B4) cjkcj gS:

(1) 328 (2) 330 (3) 332 (4) 336

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Ans. Official Answer NTA (3)

0 1 0  0 1 0
A  0 0 1 0 0 1
2
Sol.
1 0 0  1 0 0

0 0 1
 1 0 0
0 1 0

a  R2

1 0 0
 0 0 1
0 1 0

R2  R3

1 0 0 
0 1 0   I
 
0 0 1
B0 = A49 + 2A98
= A+ 2I
Bn = Adj (Bn – 1)
B4 = Adj(Adj(Adj(AdjB0))
4
| B0 |
n 1

| B0 |16

0 1 0  2 0 0
B0  0 0 1   0 2 0
1 0 0 0 0 2

2 1 0
 0 2 1 
1 0 2 
= 2(4 – 0) –1(0 – 1)
=9
B4 (9)16 = (3)32

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 Question Paper With Text Solution (Mathematics)
JEE Main July 2022 | 28 July Shift-1
Question ID : 100615
Complex number
1
15. Let S1   z1  C : z1  3   and S2  z 2  C : z 2  z 2  1  z 2  z 2  1 . Then, for z1  S 1 and
 
 2
z2  S2, the least value of |z2 – z1| is :

ekukS1  z1  C : z1  3  1  rFkkS2  z 2  C : z 2  z 2  1  z 2  z 2  1  gSaA rks

z 1  S 1 rFkk
 2
z2  S2 ds fy, |z2 – z1| dk fuEure eku gS
:
1 3 5
(1) 0 (2) (3) (4)
2 2 2
Ans. Official Answer NTA (3)
2 2
Sol. z2  z2  1  z2  z2  1

 
 z 2  z 2  1 z 2  z 2  1   z 2  z 2  1  z 2   z 2  1  
 
 z 2 z 2  122  1  z 2  z 2  1  z 2  z 2  1  z 2  1 
2 2
 z2  1  z2  1

 z
 z2  z2
 2 
 1    z 2  1  2 z 2  z 2 
 z  z  z
2 2 2 1  z2  1  2  0

 z 2  z 2  0 or z 2  1  z 2  1  2  0
 z2 lie on imaginary axis. Or on real axis within [–1, 1]
1 1
Also z1  3  lie on circle having centre 3 and radius .
2 2

5 3
Clearly |z1 – z2| min  1 
2 2
Question ID : 100616

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 Question Paper With Text Solution (Mathematics)
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3D Geometry
16. The foot of the perpendicular from a point on the circle x2 + y2 = 1, z = 0 to the plane 2x + 3y + z = 6 lies on
which one of the following curves?
o`Ùk
x2 + y2 = 1, z = 0 ds ,d fcUnq ls lery 2x + 3y + z = 6 ij Mkys x, yac dk ikn fuEu esa ls fdl oØ
ij gS ?
(1) (6x + 5y – 12)2 + 4(3x + 7y – 8)2 = 1, z = 6 – 2x – 3y
(2) (5x + 6y – 12)2 + 4(3x + 5y – 9)2 = 1, z = 6 – 2x – 3y
(3) (6x + 5y – 14)2 + 9(3x + 5y – 7)2 = 1, z = 6 – 2x – 3y
(4) (5x + 6y – 14)2 + 9(3x + 7y – 8)2 = 1, z = 6 – 2x – 3y
Ans. Official Answer NTA (2)

Sol.

h  cos  k  sin  w  0
 
2 3 1
1 2 cos   3sin   6 

14
2  2 cos   3sin   6 
h  cos
14
10 cos   6sin   12

14
3
k  sin    2 cos   3sin   6 
14
5sin   6 cos   18
k
14
Elementaary sin  and cos 
2 2
 5h  6k  12   4  3h  5k  9   1
Question ID : 100617
Sequence & progression
5x 2 
17. If the minimum value of f  x    5 , x > 0, is 14, then the value of  is equal to :
2 x

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 Question Paper With Text Solution (Mathematics)
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5x 2 
;fn f  x    5 , x > 0, dk fuEure eku14 gS] rks
 dk eku cjkcj gS:
2 x
(1) 32 (2) 64 (3) 128 (4) 256
Ans. Official Answer NTA (3)
x2 x2 x2 x2 x2  
Sol.      5 5
2 2 2 2 2 2x 2x
1
 2  7
 7 7 
2 
2/7
7.   
 14
2
(2)1/7 = 22
(22)7/2 = 27
 = 128
Question ID : 100618
Sequence & progression
18. Let ,  and  be three positive real numbers. Let f(x) = x5 + x3 + x, x  R and g : R  R be such that
g(f(x)) = x for all x  R. If a1, a2, a3, ......, an be in arithmetic progression with mean zero, then the value of

 1 n 
f  g   f  a i    is equal to :
  n i 1 
ekuk,  rFkkrhu /kukRed okLrfod la[;k,sa gSA ekuk
f(x) = x5 + x3 + x, x  R rFkkg : R  R bl izdkj
gSa fd lHkh
x  R ds fy, g(f(x)) = x gSA ;fna1, a2, a3, ......, an ,d lekarj Js<+h esa esa gS] ftudk

 1 n 
ek/; 'kwU; gS]frks
g  f  a i    dk eku cjkcj gS:
  n i 1 
(1) 0 (2) 3 (3) 9 (4) 27
Ans. Official Answer NTA (1)
Sol. Consider a case when  =  = 0 then
f(x) = yx
x
gx 
y
1 n y
 f  a i    a1  a 2  .....  a n 
n i 1 n
=0

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 Question Paper With Text Solution (Mathematics)
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 f  g  0   f  0
0
Question ID : 100619
Binomial Theorem
2
19. Consider the sequence a1, a2, a3, ..... such that a1 = 1, a2 = 2 and a n  2   a n for n = 1, 2, 3, ....... . If
a n 1

 1  1  1   1 
 a1  a   a2  a   a3  a   a 30  a 
 2
 . 3
 . 4
 ....  31
  2  61 C31  , then  is equal to :
 a3   a4   a5   a 32 
     
     

2
vuqØea1, a2, a3, ..... dk fopkj dhft, ftlds fy, a1 = 1, a2 = 2 rFkka n  2   a n , n = 1, 2, 3, .......
a n 1

 1  1  1   1 
 a1  a   a2  a   a3  a   a 30  a 
gSaA ;fn 2
 . 3
 . 4
 ....  31
  2  61 C31  gS
, rks cjkcj gS:
 a3   a4   a5   a 32 
     
     

(1) –30 (2) –31 (3) –60 (4) –61

Ans. Official Answer NTA (3)
Sol. an+2 an+1 – an+1 – an = 2
Series will satisfy
a1a2, a2a3, a3a4, a4a5
1.2 2.2 2.3 2.4
1 1
an  a n 2 
a n 1 a n 1

a n2 a n2

1
 1
a n 1a n  2

1
 1
2  r  1

2r  1

2  r  1
now proof is given by

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 Question Paper With Text Solution (Mathematics)
JEE Main July 2022 | 28 July Shift-1


30  2r  1
r 1 2  r  1


1.3.5.......61
230.  2.3......31


1.3.5.......61  230  30
31.230 230  30
61
 60
2 31. 30
 = –60
Question ID : 100620
Differential Equation
x
20. The minimum value of the twice differentiable function f  x    e x  t f '  t  dt   x 2  x  1 e x , x  R, is:
0

x
nks ckj vodyuh; Qyu f  x    e x  t f '  t  dt   x 2  x  1 e x , x  R dk fuEure eku gS
:
0

2 2
(1)  (2) 2 e (3)  e (4)
e e
Ans. Official Answer NTA (1)
x
f ' t 
Sol. f  x   ex .  t
dt
0 e

x
f ' t  f ' x 
f '  x   ex .  t
dt  e x . x
0 e e
  2x  1 .e x   x 2  x  1 .e x 
x
f ' t 
 t
dt  x 2  x
0 e

f ' x 
 2x  1
ex
f '  x    2x  1 .e x
1
f ' x  0  x  
2
f  x    2x  1 .e x  2ex  C

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 Question Paper With Text Solution (Mathematics)
JEE Main July 2022 | 28 July Shift-1
f  0   1
–1 = 1 – 2 + C
C=0
f(x) = ex (2x – 1)
 1  2
f   
 2 e
SECTION – B
Question ID : 100621
P&C
21. Let S be the set of all passwords which are six to eight characters long, where each character is either an
alphabet from {A, B, C, D, E} or a number from {1, 2, 3, 4, 5} with the repetition of characters allowed. If the
number of passwords in S whose at least one character is number from {1, 2, 3, 4, 5} is  × 56, then  is equal
to ___________.
ekuk N% ls vkB fpUg yacs lHkh ladsr&'kCnksa
{A, B, C, D, E} ls ,d v{kj ;k {1, 2, 3, 4, 5} ls ,d vad gS
rFkk ftuesa fpUgksa dh iqujko`fÙk dh vuqefr gS]
S gSS
dk leqPPk;
;fn
S esa mu ladsr&'kCnksa] ftudk de ls de
,d fpUg {1, 2, 3, 4, 5} esa ls ,d vad gS] fd la[;k
 × 56 gS
, rks cjkcj gS_________A
Ans. Official Answer NTA (7073)
Sol. Required no. = Total – no character from {1, 2, 3, 4, 5}
= (106 – 56) + (107 – 57) + (108 – 58)
= 106 (1 + 10 + 100) – 56 (1 + 5 + 25)
= 106 × 111 – 56 × 31
= 26 × 56 × 111 – 56 × 31
= 56 (26 × 111 – 31)
 56  7073



 = 7073
Question ID : 100622
3D Geometry
 56 43 111 
22. Let P(–2, –1, 1) and Q  , ,  be the vertices of the rhombus PRQS. If the direction ratios of the
 17 17 17 
diagnonal RS are , –1, , where both  and  are integers of minimum absolute values, then 2 + 2 is equal
to ___________.
56 43 111 
ekukP(–2, –1, 1) rFkkQ  , ,  ,d leprqHkqZt
PRQS ds 'kh"kZ gSaA ;fnRS
fod.kZ
dsfnd~~&vuqikr
 17 17 17 

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 Question Paper With Text Solution (Mathematics)
JEE Main July 2022 | 28 July Shift-1
, –1, gSa] tgk¡
 rFkk nksuksa fuEure fujis{k eku ds iw.kkZad
2 + 2gSa]
cjkcjrks
gS_________A
Ans. Official Answer NTA (450)
Sol. RS   , 1,  

56 43 111 
DR of PQ    2,  1,  1
 17 17 17 
 90 60 94 
 , , 
 17 17 17 
90 60 94
   1    0
17 17 17
90 + 94 = 60
60  90

94
30  2  3 

94

  30
 3  2 
94
 3  2
 
15 47
   15,   15
 2  2  225  225
= 450
Question ID : 100623
Monotonocity
23. Let f : [0, 1]  R be a twice differentiable function in (0, 1) such that f(0) = 3 and f(1) = 5. If the line
y = 2x + 3 intersects the graph of f at only two distinct points in (0, 1), then the least number of points x  (0,
1), at which f ''  x   0 , is ___________.
ekukf : [0, 1]  R, varjky (0, 1) esa nks ckj vodyuh; gS rFkk
f(0) = 3 gSaA ;fn js[kk
f(1) = 5 gSaA ;fn js[kk
y = 2x + 3, f ds xzkQ dks
(0, 1) esa dsoy nks fHkUUk fcUnqvksa ij dkVrh
xgS]
(0,rks
1) dh
fcUnqvksa
U;wure la[;k]
ftu ij f ''  x   0 gS
, gS
__________A
Ans. Official Answer NTA (2)

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 Question Paper With Text Solution (Mathematics)
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Sol.

f ' (a) = f ' (b) = f ' (c) = 2

 f '' (x) is zero
for atleast x1  (a, b) & x2  (b, c)
Question ID : 100624
Definite Integration
3
15x 3
24. If  dx   2   3 , where ,  are integers, then  +  is equal to ___________.
2 3
0 2
1 x  1  x 
3
15x 3
;fn  dx   2   3 gS
, tgk¡ ,  iw.kkZad gS]
 + rks
 cjkcj________A
2 3
0
1 x  2
1  x 
Ans. Official Answer NTA (10)
Sol. Put 1 + x2 = t2
2x dx = 2t dt
X dx = t dt
2 15  t 2  1 t dt

1
t 2  t3
2 t  t 2  1
15  dt
t 1 t
1

Put 1 + t= u2
dt = 2u du
3
15   u 4  2u 2  du
2

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 Question Paper With Text Solution (Mathematics)
JEE Main July 2022 | 28 July Shift-1
3
 u 5 2u 3 
30   
 5 3  2

1
30 
5
 5
3  2
5
  52  3

3 
3  2 

1 2 

30  9 3  4 2  3 3  2 2 
5 3 
  
 1 8 
30    3  2
 5 15 
6 3  16 2   2   3
  16,   6
    10
Question ID : 100625
Matrices
1 1  1 
25. Let A    and B    , ,   R. Let  1 be the value of  which satisfies
2    1 0 
2 2 2
 A  B  A2    and 2 be the value of  which satisfies (A + B)2 = B2. Then |1 – 2| is equal to
2 2
___________.

1 1  1  2 2 2
ekukA    rFkkB  1 0  , ,  R gSaA ekuk
 A  B  A2    dks larq"V djus okyk
2     2 2
 dk eku  gS rFkk
(A + B)2 = B2 dks larq"V djus okyk
 dk eku gSaA |rks
1
– 2| cjkcj gS_______A
Ans. Official Answer NTA (2)
  1 0 
Sol. AB  
 3 

2   1 0    1 0 
 A  B   
 3   3 

   1 2 0 
 
3    1  3  2 

1  1  1  1 
A2    
2   2 

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 Question Paper With Text Solution (Mathematics)
JEE Main July 2022 | 28 July Shift-1

 1 1   
 2 
 2  2   2 
2
 1    1     1 0 
  
 2  4  2  3      1  2 
 = 1 = 1
 1  1
B2    
1 0  1 0 
2
2  1     1 0 
    
 1  3    1  3  2 

   0,   1   2
1   2  1   1  2
Question ID : 100626
Sequence & progression
26. For p, q  R, consider the real valued function f(x) = (x – p)2 – q, x  R and q > 0. Let a1, a2, a3 and a4 be in
an arithmetic progression with mean p and positive common difference. If |f(ai)| = 500 for all
i = 1, 2, 3, 4 then the absolute difference between the roots of f(x) = 0 is ___________.
p, q  R, q > 0, ds fy, okLrfod eku Qyu f(x) = (x – p)2 – q, x  R dk fopkj dhft,A ekuka1, a2, a3 rFkk
a4 ,d /kukRed lkoZ varj dh lekarj Js<+h esa gS rFkk budk
p gSA
ek/;;fni = 1, 2, 3, 4 ds fy,
|f(ai)| = 500 gS] rks
f(x) = 0 ds ewyksa dk fujis{k varj gS A
___________
Ans. Official Answer NTA (50)
Sol. f(x) = 0  (x – p)2 – q = 0.
Roots are p  q, p  q absolute difference between roots 2 q .
Now, |f (ai)| = 500
Let a1, a2, a3, a4 are a1 a + d, a + 2d, a + 3d
|f(a4)| = 500
|(a1 – p)2 – q| = 500
 (a1 – p)2 – q = 500
9 2
 d  q  500 ........(1)
4
and |f(a1)|2 = |f(a2)|2
((a1 - p)2 - q)2 = ((a2 - p)2 – q)2

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 Question Paper With Text Solution (Mathematics)
JEE Main July 2022 | 28 July Shift-1
 ((a1 - p) – (a2 - p) ) ((a1 - p) – q + (a2 - p)2 - q) = 0
2 2 2

9 d2
 d2  q   q  0
4 4
10d 2 5d 2
2q  q
4 4
4q
 d2 
5
9 4.q
From equation (1) .  q  500
4 5
4q
 500
5
50
and 2 q  2   50
2
Question ID : 100627
Hyperbola
x 2 y2
27. For the hyperbola H : x2 – y2 = 1 and the ellipse E :   1 , a > b > 0, let the
a 2 b2
x 2 y2
vfrijoy; H : x2 – y2 = 1 rFkk nh?kZo`Ùk E : 2  2  1 , a > b > 0 ds fy,] ekuk
a b
(1) eccentricity of E be reciprocal of the eccentricity of H, and
E dh mRdsUnzrk] H dh mRdsUnzrk dh O;qRØe.kh; gSa] rFkk

5
(2) the line y  x  K be a common tangent of E and H.
2

5
js[kky  x  K , E rFkkH dh ,d mHk;fu"B Li'kZ js[kk gSA
2
Then 4(a2 + b2) is equal to ___________.
rks4(a2 + b2) cjkcj gS___________.
Ans. Official Answer NTA (3)

b2
Sol. eE  1  ,e H  2
a2
1
If  eE 
eH

a 2  b2 1
 
a2 2
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 Question Paper With Text Solution (Mathematics)
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2a2–2b = a2
a2 = 2b2
5
and y  x  k is tangent to ellipse then
2
5 3
K 2  a 2   b2 
2 2
3 1 1
6b 2   b 2  and a 2 
2 4 2
 4.  a 2  b 2   3
Question ID : 100628
Statistics
1
28. Let x1, x2, x3, ......., x20 be in geometric progression with x1 = 3 and the common ratio . A new data is
2
constructed replacing each xi by (xi – i)2. If x is the mean of new data, then the greatest integer less than or
equal x to is ___________.
1
ekukx1 = 3, x2, x3, ......., x20 ,d xq.kksÙkj Js<+h esa gSa] ftldk lkoZgSA
vuqikr
izR;sd
xi dh txg (xi – i)2.
2
ysdj u;s vk¡dM+s cuk, tkrs gSaA ;fn u;s vk¡dM+s
x gS
dk rks
ek/;egÙke iw.kkZad
 x gS
_________A
Ans. Official Answer NTA (142)
20
  1 
3 1    
 2   1 
Sol. x10    6 1  20 
1  2 
1
2
20
2
   x i 1 
i 1

20
2
   x i    i   2x i i
i 1

20
  1 
9 1    
20
 4   1 
 
2
Now    x i   12 1  40 
i 1 1  2 
1
4
20 1
 i 2   20  21  41  2870
i 1 6

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 Question Paper With Text Solution (Mathematics)
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20 1 1 1
 x i i  s  3  2.3  3.3 2  4.3 3  ......AGP
i 1 2 2 2
 22 
 6   20 
 2 

12  22 
12  40
 2870  12  2  20 
2  2 
x
20
2858  12 22  1
x   
20  240 220  20

 x   142
 
Question ID : 100629
Limits
100
  x  2 cos x 3  2  x  2 cos x  2  3sin  x  2 cos x   x
29. lim  3 2
 is equal to ___________.
x 0  
  x  2   2  x  2   3sin  x  2  
100
  x  2 cos x 3  2  x  2 cos x  2  3sin  x  2 cos x   x
cjkcj gS
__________A
lim  3 2

x 0  
  x  2   2  x  2   3sin  x  2  
Ans. Official Answer NTA (1)
x
  x  2 cos x 3  2  x  2 cos x  2  3sin  x  2 cos x  
Sol. lim  3 2

x 10  
  x  2   2  x  2   3sin  x  2  
Form 1
  x  2cos x 3  2 x  2cos x 2  3sin  x  2cos x    100
lim   1
x 0 
  x  23  2 x  22 3sin  x  2   x
 
e


 x  2cos x 3  2 x  2cos x 2  3sin  x  2cos x    x  2 3  2 x  2 2  3sin  x  2 
100  
lim 
 
 
x 0 x
   x  2 3  2 x  22  3sin  x  2  
  
e
3 3 2 2
100   x  2cos x    x  2   2 x  2cos x   2 x  2   3sin  x  2cos x  3sin  x  2   
lim
x 0 x  8 8  3sin 2 
 
e
2 2
100 3 x  2cos x  1 2sin x  3 x  2   4 x  2cos x 
lim
16  3sin 2 x0 x 1 2sin x  4 x  2 3cos x  2cos x 1 2sin x 3cos x  2 
e
100  12 3 4  818  3cos 2 3cos 2 
 
16  3sin 2  1 
e
Using L'H rule.
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 Question Paper With Text Solution (Mathematics)
JEE Main July 2022 | 28 July Shift-1
= eº = 1
Question ID : 100630
Quadratic Equation
3x 2  9x  17 5x 2  7x  19
30. The sum of all real value of x for which  is equal to ___________.
x 2  3x  10 3x 2  5x  12
3x 2  9x  17 5x 2  7x  19
x ds lHkh ekuksa] ftlds fy,2  gS] dk ;ksx cjkcj __________
gS A
x  3x  10 3x 2  5x  12
Ans. Official Answer NTA (6)
3x 2  9x  17 5x 2  7x  19
Sol. 
x 2  3x  10 3x 2  5x  12
x 2  3x  10  2x 2  12x  7 3x 2  5x  12  2x 2  12x  7

x 2  3x  10 3x 2  5x  12
2x 2  12x  7 2x 2  12x  7
1  1 
x 2  3x  10 3x 2  5x  12
 1 1 
 2x  12x  7   2
2
 2 0
 x  3x  10 3x  5x  12 
2x2 – 12x + 7 = 0 OR 3x2 + 5x + 12 = x2 + 3x +10
12  D
x 2x2 + 2x + 2 = 0
4
x2 + x +1= 0
Sum or Roots = 6 No solution.

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