JEE Main 2020 Answer Key Mathematics - Evening Shift (09-01-2020) JEE Notes | EduRev

JEE : JEE Main 2020 Answer Key Mathematics - Evening Shift (09-01-2020) JEE Notes | EduRev

 Page 1


   
 
    
   
 
   
 
                       
 
 
 
PART : MATHEMATICS 
 
SECTION – 1  
Straight Objective Type   (lh/ks oLrqfu"B izdkj) 
This section contains 19 multiple choice questions. Each question has 4 choices (1), (2), (3) and (4) for its 
answer, out of which Only One is correct. 
bl [k.M esa 19 cgq&fodYih iz'u gSaA izR;sd iz'u ds 4 fodYi (1), (2), (3) rFkk (4) gSa] ftuesa ls flQZ ,d lgh gSA 
 
1. If f(x) =  
?
?
?
?
?
?
?
?
?
? ?
?
? ?
1 x
2
1
x – 1
2
1
x
2
1
2
1
x 0 x
 
 g(x) = 
2
2
1
– x ?
?
?
?
?
?
 then find the area bounded by f(x) and g(x) from x = 
2
1
 to x = 
2
3
.  
;fn f(x) =  
?
?
?
?
?
?
?
?
?
? ?
?
? ?
1 x
2
1
x – 1
2
1
x
2
1
2
1
x 0 x
 
 g(x) = 
2
2
1
– x ?
?
?
?
?
?
 rc f(x) rFkk g(x) ds }kjk x = 
2
1
 ls x = 
2
3
rd ifjc} {ks=kQy Kkr dhft;sA  .  
 (1) 
3
1
–
4
3
    (2) 
3
1
4
3
?    (3) 2 3    (4) 3 3   
Ans. (1) 
Sol. 
C( 2 / 3 ,1– 2 / 3 ) 
 ?
?
?
?
?
?
2
1
,
2
1
 
D 
B 
?
?
?
?
?
?
0 ,
2
1
A 
x = 
2
3
 
( 2 / 3 ,0) 
 
Required area = Area of trapezium ABCD – 
?
?
?
?
?
?
?
2 / 3
2 / 1
2
2
1
– x dx 
vHkh"V {ks=kQy = leyEc prqHkZt ABCD dk {ks=kQy – 
?
?
?
?
?
?
?
2 / 3
2 / 1
2
2
1
– x dx 
 
Page 2


   
 
    
   
 
   
 
                       
 
 
 
PART : MATHEMATICS 
 
SECTION – 1  
Straight Objective Type   (lh/ks oLrqfu"B izdkj) 
This section contains 19 multiple choice questions. Each question has 4 choices (1), (2), (3) and (4) for its 
answer, out of which Only One is correct. 
bl [k.M esa 19 cgq&fodYih iz'u gSaA izR;sd iz'u ds 4 fodYi (1), (2), (3) rFkk (4) gSa] ftuesa ls flQZ ,d lgh gSA 
 
1. If f(x) =  
?
?
?
?
?
?
?
?
?
? ?
?
? ?
1 x
2
1
x – 1
2
1
x
2
1
2
1
x 0 x
 
 g(x) = 
2
2
1
– x ?
?
?
?
?
?
 then find the area bounded by f(x) and g(x) from x = 
2
1
 to x = 
2
3
.  
;fn f(x) =  
?
?
?
?
?
?
?
?
?
? ?
?
? ?
1 x
2
1
x – 1
2
1
x
2
1
2
1
x 0 x
 
 g(x) = 
2
2
1
– x ?
?
?
?
?
?
 rc f(x) rFkk g(x) ds }kjk x = 
2
1
 ls x = 
2
3
rd ifjc} {ks=kQy Kkr dhft;sA  .  
 (1) 
3
1
–
4
3
    (2) 
3
1
4
3
?    (3) 2 3    (4) 3 3   
Ans. (1) 
Sol. 
C( 2 / 3 ,1– 2 / 3 ) 
 ?
?
?
?
?
?
2
1
,
2
1
 
D 
B 
?
?
?
?
?
?
0 ,
2
1
A 
x = 
2
3
 
( 2 / 3 ,0) 
 
Required area = Area of trapezium ABCD – 
?
?
?
?
?
?
?
2 / 3
2 / 1
2
2
1
– x dx 
vHkh"V {ks=kQy = leyEc prqHkZt ABCD dk {ks=kQy – 
?
?
?
?
?
?
?
2 / 3
2 / 1
2
2
1
– x dx 
 
   
 
    
   
 
   
 
                       
 
 
= 
2
1
 
?
?
?
?
?
?
?
?
2
1 – 3
 
?
?
?
?
?
?
?
?
?
2
3
– 1
2
1
 – 
2
3
2
1
3
2
1
– x
3
1
?
?
?
?
?
?
?
?
?
?
?
?
?
?
 
= 
4
3
 – 
3
1
 
 
 
2. z is a complex number such that |Re(z)| + |Im (z)| = 4 then |z| can't be 
z ,d lfeJ la[;k bl izdkj gS fd |Re(z)| + |Im (z)| = 4 rc |z| ugh gks ldrk 
 (1) 7    (2) 10   (3) 
2
17
   (4) 8   
Ans. (1) 
Sol. z = x + iy 
 |x| + |y| = 4 
 Minimum value of |z| = 2 2 
 Maximum value of |z| = 4 
 |z| ? ? ? 16 , 8 
 So |z| can't be 7 
 
 
(0, 4) 
(0, –4) 
(4, 0) (–4, 0) 
 
Sol. z = x + iy 
 |x| + |y| = 4 
 |z| dk U;wure eku = 2 2 
 |z| dk vf/kdre eku = 4  
 |z| ? ? ? 16 , 8 
 vr% |z|= 7 ugh gks ldrk 
 
 
(0, 4) 
(0, –4) 
(4, 0) (–4, 0) 
 
 
 
Page 3


   
 
    
   
 
   
 
                       
 
 
 
PART : MATHEMATICS 
 
SECTION – 1  
Straight Objective Type   (lh/ks oLrqfu"B izdkj) 
This section contains 19 multiple choice questions. Each question has 4 choices (1), (2), (3) and (4) for its 
answer, out of which Only One is correct. 
bl [k.M esa 19 cgq&fodYih iz'u gSaA izR;sd iz'u ds 4 fodYi (1), (2), (3) rFkk (4) gSa] ftuesa ls flQZ ,d lgh gSA 
 
1. If f(x) =  
?
?
?
?
?
?
?
?
?
? ?
?
? ?
1 x
2
1
x – 1
2
1
x
2
1
2
1
x 0 x
 
 g(x) = 
2
2
1
– x ?
?
?
?
?
?
 then find the area bounded by f(x) and g(x) from x = 
2
1
 to x = 
2
3
.  
;fn f(x) =  
?
?
?
?
?
?
?
?
?
? ?
?
? ?
1 x
2
1
x – 1
2
1
x
2
1
2
1
x 0 x
 
 g(x) = 
2
2
1
– x ?
?
?
?
?
?
 rc f(x) rFkk g(x) ds }kjk x = 
2
1
 ls x = 
2
3
rd ifjc} {ks=kQy Kkr dhft;sA  .  
 (1) 
3
1
–
4
3
    (2) 
3
1
4
3
?    (3) 2 3    (4) 3 3   
Ans. (1) 
Sol. 
C( 2 / 3 ,1– 2 / 3 ) 
 ?
?
?
?
?
?
2
1
,
2
1
 
D 
B 
?
?
?
?
?
?
0 ,
2
1
A 
x = 
2
3
 
( 2 / 3 ,0) 
 
Required area = Area of trapezium ABCD – 
?
?
?
?
?
?
?
2 / 3
2 / 1
2
2
1
– x dx 
vHkh"V {ks=kQy = leyEc prqHkZt ABCD dk {ks=kQy – 
?
?
?
?
?
?
?
2 / 3
2 / 1
2
2
1
– x dx 
 
   
 
    
   
 
   
 
                       
 
 
= 
2
1
 
?
?
?
?
?
?
?
?
2
1 – 3
 
?
?
?
?
?
?
?
?
?
2
3
– 1
2
1
 – 
2
3
2
1
3
2
1
– x
3
1
?
?
?
?
?
?
?
?
?
?
?
?
?
?
 
= 
4
3
 – 
3
1
 
 
 
2. z is a complex number such that |Re(z)| + |Im (z)| = 4 then |z| can't be 
z ,d lfeJ la[;k bl izdkj gS fd |Re(z)| + |Im (z)| = 4 rc |z| ugh gks ldrk 
 (1) 7    (2) 10   (3) 
2
17
   (4) 8   
Ans. (1) 
Sol. z = x + iy 
 |x| + |y| = 4 
 Minimum value of |z| = 2 2 
 Maximum value of |z| = 4 
 |z| ? ? ? 16 , 8 
 So |z| can't be 7 
 
 
(0, 4) 
(0, –4) 
(4, 0) (–4, 0) 
 
Sol. z = x + iy 
 |x| + |y| = 4 
 |z| dk U;wure eku = 2 2 
 |z| dk vf/kdre eku = 4  
 |z| ? ? ? 16 , 8 
 vr% |z|= 7 ugh gks ldrk 
 
 
(0, 4) 
(0, –4) 
(4, 0) (–4, 0) 
 
 
 
   
 
    
   
 
   
 
                       
 
 
3. If ;fn f(x) = 
3 x 4 x c x
2 x 3 x b x
1 x 2 x a x
? ? ?
? ? ?
? ? ?
 and rFkk a – 2b + c = 1 then rc 
 (1)  f(50) = 1     (2) f(–50) = – 1 
 (3) f(50) = 501     (4) f(50) = – 501 
Ans. (1) 
Sol. Apply R1 = R1 + R3 – 2R2 iz;ksx djus ij 
 ? f(x) = 
3 x 4 x c x
2 x 3 x b x
0 0 1
? ? ?
? ? ? ? f(x) = 1 ? f(50) = 1 
 
4. Let an is a positive term of a GP and 
?
?
?
?
100
1 n
1 n 2
200 a , 
?
?
?
100
1 n
n 2
100 a find 
?
?
200
1 n
n
a 
ekuk an xq.kksÙkj Js<+h dk /kukRed in gS rFkk 
?
?
?
?
100
1 n
1 n 2
200 a , 
?
?
?
100
1 n
n 2
100 a , 
?
?
200
1 n
n
a dk eku gS& 
 (1) 300    (2) 150    (3) 175    (4) 225  
Ans. (2) 
Sol. Let GP is a, ar, ar
2
 ……..  
 
?
?
?
100
1 n
1 n 2
a = a3 + a5 + ….. a201 = 200  ? 
1 – r
) 1 – r ( ar
2
200 2
 = 200 ….(1) 
?
?
100
1 n
n 2
a = a2 + a4 + ……. a200 = 100 = 
1 – r
) 1 – r ( ar
2
200
 = 100  ….(2) 
 Form (1) and (2) r = 2 
 add both ?
? a2 + a3 + ……….. a200 + a201 = 300 ? ? ? r(a1 + ………… a200) = 300 
?
?
200
1 n
n
a = 
r
300
 = 150 
Sol. ekuk a, ar, ar
2
 …….. xq.kksÙkj Js<+h esa gS 
 
?
?
?
100
1 n
1 n 2
a = a3 + a5 + ….. a201 = 200  ? 
1 – r
) 1 – r ( ar
2
200 2
 = 200 ….(1) 
?
?
100
1 n
n 2
a = a2 + a4 + ……. a200 = 100 = 
1 – r
) 1 – r ( ar
2
200
 = 100  ….(2) 
 lehdj.k (1) rFkk (2) ls r = 2 
 nksuksa dk ;ksx djus ij ?
? a2 + a3 + ……….. a200 + a201 = 300 ? ? ? r(a1 + ………… a200) = 300 
?
?
200
1 n
n
a = 
r
300
 = 150 
 
 
Page 4


   
 
    
   
 
   
 
                       
 
 
 
PART : MATHEMATICS 
 
SECTION – 1  
Straight Objective Type   (lh/ks oLrqfu"B izdkj) 
This section contains 19 multiple choice questions. Each question has 4 choices (1), (2), (3) and (4) for its 
answer, out of which Only One is correct. 
bl [k.M esa 19 cgq&fodYih iz'u gSaA izR;sd iz'u ds 4 fodYi (1), (2), (3) rFkk (4) gSa] ftuesa ls flQZ ,d lgh gSA 
 
1. If f(x) =  
?
?
?
?
?
?
?
?
?
? ?
?
? ?
1 x
2
1
x – 1
2
1
x
2
1
2
1
x 0 x
 
 g(x) = 
2
2
1
– x ?
?
?
?
?
?
 then find the area bounded by f(x) and g(x) from x = 
2
1
 to x = 
2
3
.  
;fn f(x) =  
?
?
?
?
?
?
?
?
?
? ?
?
? ?
1 x
2
1
x – 1
2
1
x
2
1
2
1
x 0 x
 
 g(x) = 
2
2
1
– x ?
?
?
?
?
?
 rc f(x) rFkk g(x) ds }kjk x = 
2
1
 ls x = 
2
3
rd ifjc} {ks=kQy Kkr dhft;sA  .  
 (1) 
3
1
–
4
3
    (2) 
3
1
4
3
?    (3) 2 3    (4) 3 3   
Ans. (1) 
Sol. 
C( 2 / 3 ,1– 2 / 3 ) 
 ?
?
?
?
?
?
2
1
,
2
1
 
D 
B 
?
?
?
?
?
?
0 ,
2
1
A 
x = 
2
3
 
( 2 / 3 ,0) 
 
Required area = Area of trapezium ABCD – 
?
?
?
?
?
?
?
2 / 3
2 / 1
2
2
1
– x dx 
vHkh"V {ks=kQy = leyEc prqHkZt ABCD dk {ks=kQy – 
?
?
?
?
?
?
?
2 / 3
2 / 1
2
2
1
– x dx 
 
   
 
    
   
 
   
 
                       
 
 
= 
2
1
 
?
?
?
?
?
?
?
?
2
1 – 3
 
?
?
?
?
?
?
?
?
?
2
3
– 1
2
1
 – 
2
3
2
1
3
2
1
– x
3
1
?
?
?
?
?
?
?
?
?
?
?
?
?
?
 
= 
4
3
 – 
3
1
 
 
 
2. z is a complex number such that |Re(z)| + |Im (z)| = 4 then |z| can't be 
z ,d lfeJ la[;k bl izdkj gS fd |Re(z)| + |Im (z)| = 4 rc |z| ugh gks ldrk 
 (1) 7    (2) 10   (3) 
2
17
   (4) 8   
Ans. (1) 
Sol. z = x + iy 
 |x| + |y| = 4 
 Minimum value of |z| = 2 2 
 Maximum value of |z| = 4 
 |z| ? ? ? 16 , 8 
 So |z| can't be 7 
 
 
(0, 4) 
(0, –4) 
(4, 0) (–4, 0) 
 
Sol. z = x + iy 
 |x| + |y| = 4 
 |z| dk U;wure eku = 2 2 
 |z| dk vf/kdre eku = 4  
 |z| ? ? ? 16 , 8 
 vr% |z|= 7 ugh gks ldrk 
 
 
(0, 4) 
(0, –4) 
(4, 0) (–4, 0) 
 
 
 
   
 
    
   
 
   
 
                       
 
 
3. If ;fn f(x) = 
3 x 4 x c x
2 x 3 x b x
1 x 2 x a x
? ? ?
? ? ?
? ? ?
 and rFkk a – 2b + c = 1 then rc 
 (1)  f(50) = 1     (2) f(–50) = – 1 
 (3) f(50) = 501     (4) f(50) = – 501 
Ans. (1) 
Sol. Apply R1 = R1 + R3 – 2R2 iz;ksx djus ij 
 ? f(x) = 
3 x 4 x c x
2 x 3 x b x
0 0 1
? ? ?
? ? ? ? f(x) = 1 ? f(50) = 1 
 
4. Let an is a positive term of a GP and 
?
?
?
?
100
1 n
1 n 2
200 a , 
?
?
?
100
1 n
n 2
100 a find 
?
?
200
1 n
n
a 
ekuk an xq.kksÙkj Js<+h dk /kukRed in gS rFkk 
?
?
?
?
100
1 n
1 n 2
200 a , 
?
?
?
100
1 n
n 2
100 a , 
?
?
200
1 n
n
a dk eku gS& 
 (1) 300    (2) 150    (3) 175    (4) 225  
Ans. (2) 
Sol. Let GP is a, ar, ar
2
 ……..  
 
?
?
?
100
1 n
1 n 2
a = a3 + a5 + ….. a201 = 200  ? 
1 – r
) 1 – r ( ar
2
200 2
 = 200 ….(1) 
?
?
100
1 n
n 2
a = a2 + a4 + ……. a200 = 100 = 
1 – r
) 1 – r ( ar
2
200
 = 100  ….(2) 
 Form (1) and (2) r = 2 
 add both ?
? a2 + a3 + ……….. a200 + a201 = 300 ? ? ? r(a1 + ………… a200) = 300 
?
?
200
1 n
n
a = 
r
300
 = 150 
Sol. ekuk a, ar, ar
2
 …….. xq.kksÙkj Js<+h esa gS 
 
?
?
?
100
1 n
1 n 2
a = a3 + a5 + ….. a201 = 200  ? 
1 – r
) 1 – r ( ar
2
200 2
 = 200 ….(1) 
?
?
100
1 n
n 2
a = a2 + a4 + ……. a200 = 100 = 
1 – r
) 1 – r ( ar
2
200
 = 100  ….(2) 
 lehdj.k (1) rFkk (2) ls r = 2 
 nksuksa dk ;ksx djus ij ?
? a2 + a3 + ……….. a200 + a201 = 300 ? ? ? r(a1 + ………… a200) = 300 
?
?
200
1 n
n
a = 
r
300
 = 150 
 
 
   
 
    
   
 
   
 
                       
 
 
5. If ;fn  
dx
dy
 = 
2 2
y x
xy
?
, y(1) = 1 and rFkk y(x) = e then rc x = ? 
 (1) 
2
3
e   (2) 3 e  (3) 2 e  (4) 
2
e
 
Ans. (2) 
Sol.  Put y = vx  j[kus ij 
 
dx
dy
= v + x 
dx
dv
 
 v + x 
dx
dv
= 
2 2 2
2
x v x
vx
?
 
 ? 
3
2
v
v 1 ?
dv = –
x
1
dx 
? ?
?
?
?
?
?
?
?
?
v
1
v
1
3
dv = 
?
dx
x
1 –
 
? 
2
1 –
2
v
1
 + ?nv = – ?nx + c 
? –
2
2
y 2
x
 = – ?ny + c 
 When tc x = 1, y = 1 then rc  
 –
2
1
 = c 
 ? x
2
 = y
2
(1 + 2 ?ny) 
 ?  x
2
 = e
2
(3) 
 ?  x = ± 3 e 
 So blfy;s x = e 3 
 
 
6. Let probability distribution is  
ekuk izkf;drk forj.k bl izdkj gS 
 xi  : 1 2 3 4 5 
 Pi : k
2
  2k k 2k 5k
2
 
 then value of p(x > 2) is  
rc p(x > 2) dk eku gS 
 (1) 
12
7
   (2) 
36
1
   (3) 
6
1
   (4) 
36
23
 
Ans. (4) 
Sol. 
?
? 1 p
i
 ? 6k
2
 + 5k = 1 
 6k
2
 + 5k – 1 = 0 
 6k
2
 + 6k – k – 1 = 0 
Page 5


   
 
    
   
 
   
 
                       
 
 
 
PART : MATHEMATICS 
 
SECTION – 1  
Straight Objective Type   (lh/ks oLrqfu"B izdkj) 
This section contains 19 multiple choice questions. Each question has 4 choices (1), (2), (3) and (4) for its 
answer, out of which Only One is correct. 
bl [k.M esa 19 cgq&fodYih iz'u gSaA izR;sd iz'u ds 4 fodYi (1), (2), (3) rFkk (4) gSa] ftuesa ls flQZ ,d lgh gSA 
 
1. If f(x) =  
?
?
?
?
?
?
?
?
?
? ?
?
? ?
1 x
2
1
x – 1
2
1
x
2
1
2
1
x 0 x
 
 g(x) = 
2
2
1
– x ?
?
?
?
?
?
 then find the area bounded by f(x) and g(x) from x = 
2
1
 to x = 
2
3
.  
;fn f(x) =  
?
?
?
?
?
?
?
?
?
? ?
?
? ?
1 x
2
1
x – 1
2
1
x
2
1
2
1
x 0 x
 
 g(x) = 
2
2
1
– x ?
?
?
?
?
?
 rc f(x) rFkk g(x) ds }kjk x = 
2
1
 ls x = 
2
3
rd ifjc} {ks=kQy Kkr dhft;sA  .  
 (1) 
3
1
–
4
3
    (2) 
3
1
4
3
?    (3) 2 3    (4) 3 3   
Ans. (1) 
Sol. 
C( 2 / 3 ,1– 2 / 3 ) 
 ?
?
?
?
?
?
2
1
,
2
1
 
D 
B 
?
?
?
?
?
?
0 ,
2
1
A 
x = 
2
3
 
( 2 / 3 ,0) 
 
Required area = Area of trapezium ABCD – 
?
?
?
?
?
?
?
2 / 3
2 / 1
2
2
1
– x dx 
vHkh"V {ks=kQy = leyEc prqHkZt ABCD dk {ks=kQy – 
?
?
?
?
?
?
?
2 / 3
2 / 1
2
2
1
– x dx 
 
   
 
    
   
 
   
 
                       
 
 
= 
2
1
 
?
?
?
?
?
?
?
?
2
1 – 3
 
?
?
?
?
?
?
?
?
?
2
3
– 1
2
1
 – 
2
3
2
1
3
2
1
– x
3
1
?
?
?
?
?
?
?
?
?
?
?
?
?
?
 
= 
4
3
 – 
3
1
 
 
 
2. z is a complex number such that |Re(z)| + |Im (z)| = 4 then |z| can't be 
z ,d lfeJ la[;k bl izdkj gS fd |Re(z)| + |Im (z)| = 4 rc |z| ugh gks ldrk 
 (1) 7    (2) 10   (3) 
2
17
   (4) 8   
Ans. (1) 
Sol. z = x + iy 
 |x| + |y| = 4 
 Minimum value of |z| = 2 2 
 Maximum value of |z| = 4 
 |z| ? ? ? 16 , 8 
 So |z| can't be 7 
 
 
(0, 4) 
(0, –4) 
(4, 0) (–4, 0) 
 
Sol. z = x + iy 
 |x| + |y| = 4 
 |z| dk U;wure eku = 2 2 
 |z| dk vf/kdre eku = 4  
 |z| ? ? ? 16 , 8 
 vr% |z|= 7 ugh gks ldrk 
 
 
(0, 4) 
(0, –4) 
(4, 0) (–4, 0) 
 
 
 
   
 
    
   
 
   
 
                       
 
 
3. If ;fn f(x) = 
3 x 4 x c x
2 x 3 x b x
1 x 2 x a x
? ? ?
? ? ?
? ? ?
 and rFkk a – 2b + c = 1 then rc 
 (1)  f(50) = 1     (2) f(–50) = – 1 
 (3) f(50) = 501     (4) f(50) = – 501 
Ans. (1) 
Sol. Apply R1 = R1 + R3 – 2R2 iz;ksx djus ij 
 ? f(x) = 
3 x 4 x c x
2 x 3 x b x
0 0 1
? ? ?
? ? ? ? f(x) = 1 ? f(50) = 1 
 
4. Let an is a positive term of a GP and 
?
?
?
?
100
1 n
1 n 2
200 a , 
?
?
?
100
1 n
n 2
100 a find 
?
?
200
1 n
n
a 
ekuk an xq.kksÙkj Js<+h dk /kukRed in gS rFkk 
?
?
?
?
100
1 n
1 n 2
200 a , 
?
?
?
100
1 n
n 2
100 a , 
?
?
200
1 n
n
a dk eku gS& 
 (1) 300    (2) 150    (3) 175    (4) 225  
Ans. (2) 
Sol. Let GP is a, ar, ar
2
 ……..  
 
?
?
?
100
1 n
1 n 2
a = a3 + a5 + ….. a201 = 200  ? 
1 – r
) 1 – r ( ar
2
200 2
 = 200 ….(1) 
?
?
100
1 n
n 2
a = a2 + a4 + ……. a200 = 100 = 
1 – r
) 1 – r ( ar
2
200
 = 100  ….(2) 
 Form (1) and (2) r = 2 
 add both ?
? a2 + a3 + ……….. a200 + a201 = 300 ? ? ? r(a1 + ………… a200) = 300 
?
?
200
1 n
n
a = 
r
300
 = 150 
Sol. ekuk a, ar, ar
2
 …….. xq.kksÙkj Js<+h esa gS 
 
?
?
?
100
1 n
1 n 2
a = a3 + a5 + ….. a201 = 200  ? 
1 – r
) 1 – r ( ar
2
200 2
 = 200 ….(1) 
?
?
100
1 n
n 2
a = a2 + a4 + ……. a200 = 100 = 
1 – r
) 1 – r ( ar
2
200
 = 100  ….(2) 
 lehdj.k (1) rFkk (2) ls r = 2 
 nksuksa dk ;ksx djus ij ?
? a2 + a3 + ……….. a200 + a201 = 300 ? ? ? r(a1 + ………… a200) = 300 
?
?
200
1 n
n
a = 
r
300
 = 150 
 
 
   
 
    
   
 
   
 
                       
 
 
5. If ;fn  
dx
dy
 = 
2 2
y x
xy
?
, y(1) = 1 and rFkk y(x) = e then rc x = ? 
 (1) 
2
3
e   (2) 3 e  (3) 2 e  (4) 
2
e
 
Ans. (2) 
Sol.  Put y = vx  j[kus ij 
 
dx
dy
= v + x 
dx
dv
 
 v + x 
dx
dv
= 
2 2 2
2
x v x
vx
?
 
 ? 
3
2
v
v 1 ?
dv = –
x
1
dx 
? ?
?
?
?
?
?
?
?
?
v
1
v
1
3
dv = 
?
dx
x
1 –
 
? 
2
1 –
2
v
1
 + ?nv = – ?nx + c 
? –
2
2
y 2
x
 = – ?ny + c 
 When tc x = 1, y = 1 then rc  
 –
2
1
 = c 
 ? x
2
 = y
2
(1 + 2 ?ny) 
 ?  x
2
 = e
2
(3) 
 ?  x = ± 3 e 
 So blfy;s x = e 3 
 
 
6. Let probability distribution is  
ekuk izkf;drk forj.k bl izdkj gS 
 xi  : 1 2 3 4 5 
 Pi : k
2
  2k k 2k 5k
2
 
 then value of p(x > 2) is  
rc p(x > 2) dk eku gS 
 (1) 
12
7
   (2) 
36
1
   (3) 
6
1
   (4) 
36
23
 
Ans. (4) 
Sol. 
?
? 1 p
i
 ? 6k
2
 + 5k = 1 
 6k
2
 + 5k – 1 = 0 
 6k
2
 + 6k – k – 1 = 0 
   
 
    
   
 
   
 
                       
 
 
 (6k – 1) (k + 1) = 0 ? k = – 1 (rejected vekU; ) ; k = 
6
1
  
 P(x > 2) = k + 2k + 5k
2
  
 = 
36
5
6
2
6
1
? ? = 
36
5 12 6 ? ?
 = 
36
23
 
 
7. 
? ?
?
? ? ? ?
?
2 tan 2 sec cos
d
2
= ?tan ? + 2log f(x) + c then ordered pair ( ?,f(x)) is 
? ?
?
? ? ? ?
?
2 tan 2 sec cos
d
2
= ?tan ? + 2log f(x) + c rc Øfer ;qXe  ( ?,f(x)) gS& 
 (1) (1, 1 + tan ?)  (2) (1, 1 – tan ?)  (3) (–1, 1 + tan ?) (4) (–1, 1 – tan ?) 
Ans. (3) 
Sol. 
?
?
?
?
?
? ?
?
2 2
2
2
tan – 1
tan 2
tan – 1
tan 1
sec
 d ? 
 = 
? ?
? ?
?
? ?
? ?
2
2 2
tan 1
tan – 1 sec
d ? ?
= 
? ?
?
? ?
? ?
tan 1
tan – 1 sec
2
d ? ?
tan ? ?= t ? sec
2
? d ? = dt  
= 
?
?
?
?
?
?
?
? t 1
t – 1
dt = 
?
?
?
?
?
?
?
?
? dt
t 1
2
1 – 
= – t + 2 log (1+t) + C 
= –tan ? + 2 log (1 + tan ?) + C   
? ? = –1 and f(x) = 1 + tan ?  
 
8. If p ? ? (p ? ? q) is false. Truth value of p & q will be 
;fn p ? ? (p ? ? q) vlR; gS] rc p & q dk lR;rrk eku gksxk 
(1) TT   (2) TF   (3) F T   (4) F F 
Ans. (1) 
 
Sol.  
p q ?q p ? ?q p ?(p ? ?q) 
T T F F F 
T F T T T 
F T F F T 
F F T F T 
 
 
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