JEE Main 2020 September 6 Question Paper Shift 2 | JEE Main & Advanced Previous Year Papers PDF Download

 Page 1


JEE Main 2020 Paper
         6
th
 September 2020 | (Shift-2), Maths     Page | 1
Date : 6
th
 September 2020
Time : 02 : 00 pm - 05 : 00 pm
Subject : Maths
Q.1 If the normal at an end of a latus rectum of an ellipse passes through an extremity of
the minor axis, then the eccentricity e of the ellipse satisfies:
(1) e
4
+2e
2
–1=0 (2) e
2
+2e–1=0 (3) e
4
+e
2
–1=0 (4) e
2
+e–1=0
Sol. (3)
(0, –b)
2
b
ae,
a
Equation of normal at 
2
,
b
ae
a
? ?
? ?
? ?
2 2
2 2
2
? ?
a x b y
a e
b ae
a
2 2 2
ax x
ay a e y ae
e e
? ? ? ? ?
It passes through (0,–b)
2 2 2 4
–b ae b a e ? ? ?
? ?
2 2 2 4
a 1 e a e ? ?
?
 e
4
 + e
2
 – 1 = 0
Q.2 The set of all real values of 
?
for which the function 
2
f x 1 x ( ) ( cos ) ? ? . x ( sin ) ? ? ,
x
2 2
,
? ? ? ?
? ?
? ?
? ?
, has exactly one maxima and exactly one minima, is:
(1) 
3 3
0
2 2
, { }
? ?
? ?
? ?
? ?
(2) 
1 1
0
2 2
, { }
? ?
? ?
? ?
? ?
(3) 
3 3
2 2
,
? ?
?
? ?
? ?
(4) 
1 1
2 2
,
? ?
?
? ?
? ?
Page 2


JEE Main 2020 Paper
         6
th
 September 2020 | (Shift-2), Maths     Page | 1
Date : 6
th
 September 2020
Time : 02 : 00 pm - 05 : 00 pm
Subject : Maths
Q.1 If the normal at an end of a latus rectum of an ellipse passes through an extremity of
the minor axis, then the eccentricity e of the ellipse satisfies:
(1) e
4
+2e
2
–1=0 (2) e
2
+2e–1=0 (3) e
4
+e
2
–1=0 (4) e
2
+e–1=0
Sol. (3)
(0, –b)
2
b
ae,
a
Equation of normal at 
2
,
b
ae
a
? ?
? ?
? ?
2 2
2 2
2
? ?
a x b y
a e
b ae
a
2 2 2
ax x
ay a e y ae
e e
? ? ? ? ?
It passes through (0,–b)
2 2 2 4
–b ae b a e ? ? ?
? ?
2 2 2 4
a 1 e a e ? ?
?
 e
4
 + e
2
 – 1 = 0
Q.2 The set of all real values of 
?
for which the function 
2
f x 1 x ( ) ( cos ) ? ? . x ( sin ) ? ? ,
x
2 2
,
? ? ? ?
? ?
? ?
? ?
, has exactly one maxima and exactly one minima, is:
(1) 
3 3
0
2 2
, { }
? ?
? ?
? ?
? ?
(2) 
1 1
0
2 2
, { }
? ?
? ?
? ?
? ?
(3) 
3 3
2 2
,
? ?
?
? ?
? ?
(4) 
1 1
2 2
,
? ?
?
? ?
? ?
JEE Main 2020 Paper
         6
th
 September 2020 | (Shift-2), Maths     Page | 2
Sol. (1)
f(x) = (1– cos
2
x) ( ? ? ? ?sinx)
f(x) = sin
2
x ( ? + sinx)
f’(x) =
 
2sinx cosx ( ? +sinx) + sin
2
x (cosx)
= sin2x 
sin
sin
2
x
x ?
? ?
? ?
? ?
? ?
= sin2x ? ? 2 3sin x ? ?
For extreme value f ?(x) = 0
Sin2x = 0 ? sinx = 0 ? x = 0 ? One point
2 ? + 3sinx ? = 0
? sinx = 
2
3
? ?
sinx
?
 (–1,1) – ? ? 0
–1 < 
2
3
? ?
 < 1 ? 
3 3
2 2
?
?
? ?
? 
?
 
3 3
,
2 2
? ? ?
? ?
? ?
 – ? ? 0
Q.3 The probabilities of three events A, B and C are given by P(A)=0.6, P(B)=0.4 and
P(C)=0.5. If P A B ? ( ) =0.8, P A C ( ) ? =0.3, P A B C ( ) ? ? =0.2, P B C ( ) ? ? ? and
P A B C ( ) ? ? ? ? , where 
0 85 0 95 . . ? ? ?
, then ? lies in the interval: al:
(1) [0.36,0.40] (2) [0.25,0.35] (3) [0.35,0.36] (4) [0.20,0.25]
Sol. (2)
P(A ? BUC) = P(A) + P(B) + P(C) – P(A ? B) – P(B ? C) –P(C ? A) + P(A ? B ? C)
? = 0.6 + 0.4 + 0.5 –P(A ? B) – ? –0.3 + 0.2
? = 1.4 – P(A ? B) – ? ? ? + ? = 1.4 – P(A ? B) ........(1)
again
P(A ? B) = P(A) + P(B) –P (A ? B)
0.8 = 0.6 + 0.4 – P(A ? B)
P(A ? B) = 0.2 ........(2)
Put the value P(A ? B) in equation (1)
? + ? = 1.2
? = 1.2 – ?
0.85 ? ? ?0.95  ? 0.85 
?
 1.2 – ? ?0.95
? ? [0.25, 0.35]
Page 3


JEE Main 2020 Paper
         6
th
 September 2020 | (Shift-2), Maths     Page | 1
Date : 6
th
 September 2020
Time : 02 : 00 pm - 05 : 00 pm
Subject : Maths
Q.1 If the normal at an end of a latus rectum of an ellipse passes through an extremity of
the minor axis, then the eccentricity e of the ellipse satisfies:
(1) e
4
+2e
2
–1=0 (2) e
2
+2e–1=0 (3) e
4
+e
2
–1=0 (4) e
2
+e–1=0
Sol. (3)
(0, –b)
2
b
ae,
a
Equation of normal at 
2
,
b
ae
a
? ?
? ?
? ?
2 2
2 2
2
? ?
a x b y
a e
b ae
a
2 2 2
ax x
ay a e y ae
e e
? ? ? ? ?
It passes through (0,–b)
2 2 2 4
–b ae b a e ? ? ?
? ?
2 2 2 4
a 1 e a e ? ?
?
 e
4
 + e
2
 – 1 = 0
Q.2 The set of all real values of 
?
for which the function 
2
f x 1 x ( ) ( cos ) ? ? . x ( sin ) ? ? ,
x
2 2
,
? ? ? ?
? ?
? ?
? ?
, has exactly one maxima and exactly one minima, is:
(1) 
3 3
0
2 2
, { }
? ?
? ?
? ?
? ?
(2) 
1 1
0
2 2
, { }
? ?
? ?
? ?
? ?
(3) 
3 3
2 2
,
? ?
?
? ?
? ?
(4) 
1 1
2 2
,
? ?
?
? ?
? ?
JEE Main 2020 Paper
         6
th
 September 2020 | (Shift-2), Maths     Page | 2
Sol. (1)
f(x) = (1– cos
2
x) ( ? ? ? ?sinx)
f(x) = sin
2
x ( ? + sinx)
f’(x) =
 
2sinx cosx ( ? +sinx) + sin
2
x (cosx)
= sin2x 
sin
sin
2
x
x ?
? ?
? ?
? ?
? ?
= sin2x ? ? 2 3sin x ? ?
For extreme value f ?(x) = 0
Sin2x = 0 ? sinx = 0 ? x = 0 ? One point
2 ? + 3sinx ? = 0
? sinx = 
2
3
? ?
sinx
?
 (–1,1) – ? ? 0
–1 < 
2
3
? ?
 < 1 ? 
3 3
2 2
?
?
? ?
? 
?
 
3 3
,
2 2
? ? ?
? ?
? ?
 – ? ? 0
Q.3 The probabilities of three events A, B and C are given by P(A)=0.6, P(B)=0.4 and
P(C)=0.5. If P A B ? ( ) =0.8, P A C ( ) ? =0.3, P A B C ( ) ? ? =0.2, P B C ( ) ? ? ? and
P A B C ( ) ? ? ? ? , where 
0 85 0 95 . . ? ? ?
, then ? lies in the interval: al:
(1) [0.36,0.40] (2) [0.25,0.35] (3) [0.35,0.36] (4) [0.20,0.25]
Sol. (2)
P(A ? BUC) = P(A) + P(B) + P(C) – P(A ? B) – P(B ? C) –P(C ? A) + P(A ? B ? C)
? = 0.6 + 0.4 + 0.5 –P(A ? B) – ? –0.3 + 0.2
? = 1.4 – P(A ? B) – ? ? ? + ? = 1.4 – P(A ? B) ........(1)
again
P(A ? B) = P(A) + P(B) –P (A ? B)
0.8 = 0.6 + 0.4 – P(A ? B)
P(A ? B) = 0.2 ........(2)
Put the value P(A ? B) in equation (1)
? + ? = 1.2
? = 1.2 – ?
0.85 ? ? ?0.95  ? 0.85 
?
 1.2 – ? ?0.95
? ? [0.25, 0.35]
JEE Main 2020 Paper
         6
th
 September 2020 | (Shift-2), Maths     Page | 3
Q.4 The common difference of the A.P. b
1
, b
2
,..... b
m
 is 2 more than the common difference
of A.P. a
1
, a
2
, ...a
n
. If a
40
 =–159, a
100
=–399 and b
100
=a
70
, then b
1
 is equal to:
(1) –127 (2) 81 (3) 127 (4) –81
Sol. (4)
A.P (a
1
, a
2
, a
3
 ..........a
n
)  (CD = D
a
)
(b
1
, b
2
, b
3
 .......... b
m
)     (CD = D
b
)
D
b
 = D
a
 + 2
a
40 
 
= –159
a
1
 + 39 D
a
 = –159 ––––(1)
a
100
 = –399
a
1
 + 99 D
a
 = –399 ––––(2)
Eqn (1) – (2)
–60 D
a
 = 240  ? D
a
 = –4
D
b
 = –4 + 2 = –2
a
1
 + 39(–4) =–159 ?
 
a
1 
=-3
b
100
 = a
70
b
1
 + 99 D
b
 = a
1
 + 69 D
a
b
1
 + 99 (-2) = (–3) + 69(–4)
b
1
 = –81
Q.5 The integral 
2
x x
e
1
e x 2 x dx . ( log ) ?
?
 equal :
(1) e(4e–1) (2) e(4e+1) (3) 4e
2
–1 (4) e(2e–1)
Sol. (1)
?
2
1
x x
x . e
 (2+lnx) dx
e
x
. x
x
 = t  
?
Upper Limt = e
2
.2
2
, Lower Limit = e
(e
x
.x
x
 + e
x 
x
x
(1+lnx)) dx = dt
e
x
 . x
x 
(2 + lnx) dx  = dt
2
4.e
e
dt
?
=  
2
4.
[t]
e
e
 = 4.e
2
 – e = e(4e–1)
Q.6 If the tangent to the curve, y=f(x)=xlog
e
x, (x>0) at a point (c,f(c)) is parallel to the
line-segment joining the points (1,0) and (e,e), then c is equal to:
(1) 
1
1 e
e
? ?
? ?
?
? ?
(2) 
e 1
e
?
(3) 
1
e 1 ?
(4) 
1
e 1
e
? ?
? ?
?
? ?
Sol. (4)
y = f(x) = x lnx
m
1
 = 
dy
dx
( , (c)) c f
 = (lnx+1) 
( , (c)) c f
 = lnc + 1
slope of the line joining (1, 0), (e, e)
m
2
 = 
e
e 1
? ?
? ?
?
? ?
m
2
 = m
1
  ? lnc + 1 = 
1 e
e
?
lnc = 
1
1 e
e
?
?
= 
1 e
1
?
? ?
? ?
? ? ?
?
1
e 1
c e
Page 4


JEE Main 2020 Paper
         6
th
 September 2020 | (Shift-2), Maths     Page | 1
Date : 6
th
 September 2020
Time : 02 : 00 pm - 05 : 00 pm
Subject : Maths
Q.1 If the normal at an end of a latus rectum of an ellipse passes through an extremity of
the minor axis, then the eccentricity e of the ellipse satisfies:
(1) e
4
+2e
2
–1=0 (2) e
2
+2e–1=0 (3) e
4
+e
2
–1=0 (4) e
2
+e–1=0
Sol. (3)
(0, –b)
2
b
ae,
a
Equation of normal at 
2
,
b
ae
a
? ?
? ?
? ?
2 2
2 2
2
? ?
a x b y
a e
b ae
a
2 2 2
ax x
ay a e y ae
e e
? ? ? ? ?
It passes through (0,–b)
2 2 2 4
–b ae b a e ? ? ?
? ?
2 2 2 4
a 1 e a e ? ?
?
 e
4
 + e
2
 – 1 = 0
Q.2 The set of all real values of 
?
for which the function 
2
f x 1 x ( ) ( cos ) ? ? . x ( sin ) ? ? ,
x
2 2
,
? ? ? ?
? ?
? ?
? ?
, has exactly one maxima and exactly one minima, is:
(1) 
3 3
0
2 2
, { }
? ?
? ?
? ?
? ?
(2) 
1 1
0
2 2
, { }
? ?
? ?
? ?
? ?
(3) 
3 3
2 2
,
? ?
?
? ?
? ?
(4) 
1 1
2 2
,
? ?
?
? ?
? ?
JEE Main 2020 Paper
         6
th
 September 2020 | (Shift-2), Maths     Page | 2
Sol. (1)
f(x) = (1– cos
2
x) ( ? ? ? ?sinx)
f(x) = sin
2
x ( ? + sinx)
f’(x) =
 
2sinx cosx ( ? +sinx) + sin
2
x (cosx)
= sin2x 
sin
sin
2
x
x ?
? ?
? ?
? ?
? ?
= sin2x ? ? 2 3sin x ? ?
For extreme value f ?(x) = 0
Sin2x = 0 ? sinx = 0 ? x = 0 ? One point
2 ? + 3sinx ? = 0
? sinx = 
2
3
? ?
sinx
?
 (–1,1) – ? ? 0
–1 < 
2
3
? ?
 < 1 ? 
3 3
2 2
?
?
? ?
? 
?
 
3 3
,
2 2
? ? ?
? ?
? ?
 – ? ? 0
Q.3 The probabilities of three events A, B and C are given by P(A)=0.6, P(B)=0.4 and
P(C)=0.5. If P A B ? ( ) =0.8, P A C ( ) ? =0.3, P A B C ( ) ? ? =0.2, P B C ( ) ? ? ? and
P A B C ( ) ? ? ? ? , where 
0 85 0 95 . . ? ? ?
, then ? lies in the interval: al:
(1) [0.36,0.40] (2) [0.25,0.35] (3) [0.35,0.36] (4) [0.20,0.25]
Sol. (2)
P(A ? BUC) = P(A) + P(B) + P(C) – P(A ? B) – P(B ? C) –P(C ? A) + P(A ? B ? C)
? = 0.6 + 0.4 + 0.5 –P(A ? B) – ? –0.3 + 0.2
? = 1.4 – P(A ? B) – ? ? ? + ? = 1.4 – P(A ? B) ........(1)
again
P(A ? B) = P(A) + P(B) –P (A ? B)
0.8 = 0.6 + 0.4 – P(A ? B)
P(A ? B) = 0.2 ........(2)
Put the value P(A ? B) in equation (1)
? + ? = 1.2
? = 1.2 – ?
0.85 ? ? ?0.95  ? 0.85 
?
 1.2 – ? ?0.95
? ? [0.25, 0.35]
JEE Main 2020 Paper
         6
th
 September 2020 | (Shift-2), Maths     Page | 3
Q.4 The common difference of the A.P. b
1
, b
2
,..... b
m
 is 2 more than the common difference
of A.P. a
1
, a
2
, ...a
n
. If a
40
 =–159, a
100
=–399 and b
100
=a
70
, then b
1
 is equal to:
(1) –127 (2) 81 (3) 127 (4) –81
Sol. (4)
A.P (a
1
, a
2
, a
3
 ..........a
n
)  (CD = D
a
)
(b
1
, b
2
, b
3
 .......... b
m
)     (CD = D
b
)
D
b
 = D
a
 + 2
a
40 
 
= –159
a
1
 + 39 D
a
 = –159 ––––(1)
a
100
 = –399
a
1
 + 99 D
a
 = –399 ––––(2)
Eqn (1) – (2)
–60 D
a
 = 240  ? D
a
 = –4
D
b
 = –4 + 2 = –2
a
1
 + 39(–4) =–159 ?
 
a
1 
=-3
b
100
 = a
70
b
1
 + 99 D
b
 = a
1
 + 69 D
a
b
1
 + 99 (-2) = (–3) + 69(–4)
b
1
 = –81
Q.5 The integral 
2
x x
e
1
e x 2 x dx . ( log ) ?
?
 equal :
(1) e(4e–1) (2) e(4e+1) (3) 4e
2
–1 (4) e(2e–1)
Sol. (1)
?
2
1
x x
x . e
 (2+lnx) dx
e
x
. x
x
 = t  
?
Upper Limt = e
2
.2
2
, Lower Limit = e
(e
x
.x
x
 + e
x 
x
x
(1+lnx)) dx = dt
e
x
 . x
x 
(2 + lnx) dx  = dt
2
4.e
e
dt
?
=  
2
4.
[t]
e
e
 = 4.e
2
 – e = e(4e–1)
Q.6 If the tangent to the curve, y=f(x)=xlog
e
x, (x>0) at a point (c,f(c)) is parallel to the
line-segment joining the points (1,0) and (e,e), then c is equal to:
(1) 
1
1 e
e
? ?
? ?
?
? ?
(2) 
e 1
e
?
(3) 
1
e 1 ?
(4) 
1
e 1
e
? ?
? ?
?
? ?
Sol. (4)
y = f(x) = x lnx
m
1
 = 
dy
dx
( , (c)) c f
 = (lnx+1) 
( , (c)) c f
 = lnc + 1
slope of the line joining (1, 0), (e, e)
m
2
 = 
e
e 1
? ?
? ?
?
? ?
m
2
 = m
1
  ? lnc + 1 = 
1 e
e
?
lnc = 
1
1 e
e
?
?
= 
1 e
1
?
? ?
? ?
? ? ?
?
1
e 1
c e
JEE Main 2020 Paper
         6
th
 September 2020 | (Shift-2), Maths     Page | 4
Q.7 If 
2
y x 1
? ?
? ?
? ?
?
? ?
 cosec x is the solution of the differential equation,
dy 2
p x y 0 x
dx 2
?
? ? ? ?
?
( ) cosec x,
, then the function p(x) is equal to:
(1) cosec x (2) cot x (3) tan x (4) sec x
Sol. 2
y = 
2
1 x
?
? ?
?
? ?
? ?
 cosec x
Differentiate w.r.t. x
2 dy
dx ?
?
 cosec x 
2
1
x
?
? ?
? ?
? ?
? ?
cosec x . cot x
2
1
dy x
dx ?
? ?
? ?
? ?
? ?
 cosec x . cot x = 
2
?
 cosec x
dy
y
dx
?
 cot x = 
2
?
 cosec x
Compare this differential equation with given differential equation
p(x) = cot x
Q.8 If ? and ? are the roots of the equation 2x(2x+1)=1, then ? is equal to:
(1) 2 1 ( ) ? ? ? (2) 2 1 ( ) ? ? ? ? (3) 
2
2 ?
(4) 2 1 ( ) ? ? ?
Sol. (2)
2x(2x+1) = 1
If ? & ? are the roots i.e ? & ? satisy this equation
2 ? (2 ? + 1) = 1 ?  ? (2 ? +1) = 
2
1
4x
2
 + 2x –1 = 0
? + ? = 
2
1 ?
= – ? (2 ? +1)
? = – ? (2 ? +1) – ? = – ? (2 ? +2) = –2 ? ( ? +1)
Q.9 For all twice differentiable functions f: R ? R, with f(0)=f(1)=f’(0)=0,
(1) f”(x)=0, at every point x ? (0,1) (2) f”(x)
?
0, at every point x ? (0,1)
(3) f”(x)=0, for some x ? (0,1) (4) f”(0)=0
Sol. (3)
Applying rolle’s theorem in [0,1] for function f(x)
f'(c) = 0, c ? (0,1)
again applying rolles theorem in [0,c] for function f'(x) s
f"(c
1
) = 0, c
1
?
(0,c)
Page 5


JEE Main 2020 Paper
         6
th
 September 2020 | (Shift-2), Maths     Page | 1
Date : 6
th
 September 2020
Time : 02 : 00 pm - 05 : 00 pm
Subject : Maths
Q.1 If the normal at an end of a latus rectum of an ellipse passes through an extremity of
the minor axis, then the eccentricity e of the ellipse satisfies:
(1) e
4
+2e
2
–1=0 (2) e
2
+2e–1=0 (3) e
4
+e
2
–1=0 (4) e
2
+e–1=0
Sol. (3)
(0, –b)
2
b
ae,
a
Equation of normal at 
2
,
b
ae
a
? ?
? ?
? ?
2 2
2 2
2
? ?
a x b y
a e
b ae
a
2 2 2
ax x
ay a e y ae
e e
? ? ? ? ?
It passes through (0,–b)
2 2 2 4
–b ae b a e ? ? ?
? ?
2 2 2 4
a 1 e a e ? ?
?
 e
4
 + e
2
 – 1 = 0
Q.2 The set of all real values of 
?
for which the function 
2
f x 1 x ( ) ( cos ) ? ? . x ( sin ) ? ? ,
x
2 2
,
? ? ? ?
? ?
? ?
? ?
, has exactly one maxima and exactly one minima, is:
(1) 
3 3
0
2 2
, { }
? ?
? ?
? ?
? ?
(2) 
1 1
0
2 2
, { }
? ?
? ?
? ?
? ?
(3) 
3 3
2 2
,
? ?
?
? ?
? ?
(4) 
1 1
2 2
,
? ?
?
? ?
? ?
JEE Main 2020 Paper
         6
th
 September 2020 | (Shift-2), Maths     Page | 2
Sol. (1)
f(x) = (1– cos
2
x) ( ? ? ? ?sinx)
f(x) = sin
2
x ( ? + sinx)
f’(x) =
 
2sinx cosx ( ? +sinx) + sin
2
x (cosx)
= sin2x 
sin
sin
2
x
x ?
? ?
? ?
? ?
? ?
= sin2x ? ? 2 3sin x ? ?
For extreme value f ?(x) = 0
Sin2x = 0 ? sinx = 0 ? x = 0 ? One point
2 ? + 3sinx ? = 0
? sinx = 
2
3
? ?
sinx
?
 (–1,1) – ? ? 0
–1 < 
2
3
? ?
 < 1 ? 
3 3
2 2
?
?
? ?
? 
?
 
3 3
,
2 2
? ? ?
? ?
? ?
 – ? ? 0
Q.3 The probabilities of three events A, B and C are given by P(A)=0.6, P(B)=0.4 and
P(C)=0.5. If P A B ? ( ) =0.8, P A C ( ) ? =0.3, P A B C ( ) ? ? =0.2, P B C ( ) ? ? ? and
P A B C ( ) ? ? ? ? , where 
0 85 0 95 . . ? ? ?
, then ? lies in the interval: al:
(1) [0.36,0.40] (2) [0.25,0.35] (3) [0.35,0.36] (4) [0.20,0.25]
Sol. (2)
P(A ? BUC) = P(A) + P(B) + P(C) – P(A ? B) – P(B ? C) –P(C ? A) + P(A ? B ? C)
? = 0.6 + 0.4 + 0.5 –P(A ? B) – ? –0.3 + 0.2
? = 1.4 – P(A ? B) – ? ? ? + ? = 1.4 – P(A ? B) ........(1)
again
P(A ? B) = P(A) + P(B) –P (A ? B)
0.8 = 0.6 + 0.4 – P(A ? B)
P(A ? B) = 0.2 ........(2)
Put the value P(A ? B) in equation (1)
? + ? = 1.2
? = 1.2 – ?
0.85 ? ? ?0.95  ? 0.85 
?
 1.2 – ? ?0.95
? ? [0.25, 0.35]
JEE Main 2020 Paper
         6
th
 September 2020 | (Shift-2), Maths     Page | 3
Q.4 The common difference of the A.P. b
1
, b
2
,..... b
m
 is 2 more than the common difference
of A.P. a
1
, a
2
, ...a
n
. If a
40
 =–159, a
100
=–399 and b
100
=a
70
, then b
1
 is equal to:
(1) –127 (2) 81 (3) 127 (4) –81
Sol. (4)
A.P (a
1
, a
2
, a
3
 ..........a
n
)  (CD = D
a
)
(b
1
, b
2
, b
3
 .......... b
m
)     (CD = D
b
)
D
b
 = D
a
 + 2
a
40 
 
= –159
a
1
 + 39 D
a
 = –159 ––––(1)
a
100
 = –399
a
1
 + 99 D
a
 = –399 ––––(2)
Eqn (1) – (2)
–60 D
a
 = 240  ? D
a
 = –4
D
b
 = –4 + 2 = –2
a
1
 + 39(–4) =–159 ?
 
a
1 
=-3
b
100
 = a
70
b
1
 + 99 D
b
 = a
1
 + 69 D
a
b
1
 + 99 (-2) = (–3) + 69(–4)
b
1
 = –81
Q.5 The integral 
2
x x
e
1
e x 2 x dx . ( log ) ?
?
 equal :
(1) e(4e–1) (2) e(4e+1) (3) 4e
2
–1 (4) e(2e–1)
Sol. (1)
?
2
1
x x
x . e
 (2+lnx) dx
e
x
. x
x
 = t  
?
Upper Limt = e
2
.2
2
, Lower Limit = e
(e
x
.x
x
 + e
x 
x
x
(1+lnx)) dx = dt
e
x
 . x
x 
(2 + lnx) dx  = dt
2
4.e
e
dt
?
=  
2
4.
[t]
e
e
 = 4.e
2
 – e = e(4e–1)
Q.6 If the tangent to the curve, y=f(x)=xlog
e
x, (x>0) at a point (c,f(c)) is parallel to the
line-segment joining the points (1,0) and (e,e), then c is equal to:
(1) 
1
1 e
e
? ?
? ?
?
? ?
(2) 
e 1
e
?
(3) 
1
e 1 ?
(4) 
1
e 1
e
? ?
? ?
?
? ?
Sol. (4)
y = f(x) = x lnx
m
1
 = 
dy
dx
( , (c)) c f
 = (lnx+1) 
( , (c)) c f
 = lnc + 1
slope of the line joining (1, 0), (e, e)
m
2
 = 
e
e 1
? ?
? ?
?
? ?
m
2
 = m
1
  ? lnc + 1 = 
1 e
e
?
lnc = 
1
1 e
e
?
?
= 
1 e
1
?
? ?
? ?
? ? ?
?
1
e 1
c e
JEE Main 2020 Paper
         6
th
 September 2020 | (Shift-2), Maths     Page | 4
Q.7 If 
2
y x 1
? ?
? ?
? ?
?
? ?
 cosec x is the solution of the differential equation,
dy 2
p x y 0 x
dx 2
?
? ? ? ?
?
( ) cosec x,
, then the function p(x) is equal to:
(1) cosec x (2) cot x (3) tan x (4) sec x
Sol. 2
y = 
2
1 x
?
? ?
?
? ?
? ?
 cosec x
Differentiate w.r.t. x
2 dy
dx ?
?
 cosec x 
2
1
x
?
? ?
? ?
? ?
? ?
cosec x . cot x
2
1
dy x
dx ?
? ?
? ?
? ?
? ?
 cosec x . cot x = 
2
?
 cosec x
dy
y
dx
?
 cot x = 
2
?
 cosec x
Compare this differential equation with given differential equation
p(x) = cot x
Q.8 If ? and ? are the roots of the equation 2x(2x+1)=1, then ? is equal to:
(1) 2 1 ( ) ? ? ? (2) 2 1 ( ) ? ? ? ? (3) 
2
2 ?
(4) 2 1 ( ) ? ? ?
Sol. (2)
2x(2x+1) = 1
If ? & ? are the roots i.e ? & ? satisy this equation
2 ? (2 ? + 1) = 1 ?  ? (2 ? +1) = 
2
1
4x
2
 + 2x –1 = 0
? + ? = 
2
1 ?
= – ? (2 ? +1)
? = – ? (2 ? +1) – ? = – ? (2 ? +2) = –2 ? ( ? +1)
Q.9 For all twice differentiable functions f: R ? R, with f(0)=f(1)=f’(0)=0,
(1) f”(x)=0, at every point x ? (0,1) (2) f”(x)
?
0, at every point x ? (0,1)
(3) f”(x)=0, for some x ? (0,1) (4) f”(0)=0
Sol. (3)
Applying rolle’s theorem in [0,1] for function f(x)
f'(c) = 0, c ? (0,1)
again applying rolles theorem in [0,c] for function f'(x) s
f"(c
1
) = 0, c
1
?
(0,c)
JEE Main 2020 Paper
         6
th
 September 2020 | (Shift-2), Maths     Page | 5
Q.10 The area (in sq.units) of the region enclosed by the curves y=x
2
–1 and y=1–x
2
 is equal
to :
(1) 
4
3
(2) 
7
2
(3) 
16
3
(4) 
8
3
Sol. (4)
(–1, 0) (1, 0)
y = 1–x
2
y = x –1
2
Total area = 4 
1
1 3
2
0 0
(1 ) 4
3
x
x dx x
? ?
? ? ?
? ?
? ?
?
= 4 
1 8
1 .
3 3
sq unit
? ?
? ?
? ?
? ?
Q.11 For a suitably chosen real constant a, let a function, f:R–{–a} ? R be defined by y
a x
f x
a x
( )
?
?
?
. Further suppose that for any real number x a ? ? and f(x)
?
–a, a,
(fof)(x)=x.Then 
1
f
2
? ?
?
? ?
? ?
 is equal to:
(1) –3 (2) 3 (3) 
1
3
(4) 
1
3
?
Sol. (2)
f(x) = 
a x
a x
?
?
f(f(x)) = 
( )
( )
a f x
x
a f x
?
?
?
( )
1
a ax a x
f x
x a x
? ?
? ?
? ?
1
1
x a x
a
x a x
? ? ? ?
?
? ?
? ?
? ?
? a = 1
So f(x) = 
1
1
x
x
?
?
1
3
2
f
? ? ?
?
? ?
? ?