1
JEE Main 2020 (Online) 3rd September Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
The solution curve of the differential equation,

(1 + e-x)(1 + y2)$${{dy} \over {dx}}$$ = y2,

which passes through the point (0, 1), is :
A
y2 + 1 = y$$\left( {{{\log }_e}\left( {{{1 + {e^{ - x}}} \over 2}} \right) + 2} \right)$$
B
y2 + 1 = y$$\left( {{{\log }_e}\left( {{{1 + {e^{ x}}} \over 2}} \right) + 2} \right)$$
C
y2 = 1 + $${y{{\log }_e}\left( {{{1 + {e^{ - x}}} \over 2}} \right)}$$
D
y2 = 1 + $${y{{\log }_e}\left( {{{1 + {e^{ x}}} \over 2}} \right)}$$
2
JEE Main 2020 (Online) 3rd September Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
If $$\alpha $$ and $$\beta $$ are the roots of the equation
x2 + px + 2 = 0 and $${1 \over \alpha }$$ and $${1 \over \beta }$$ are the
roots of the equation 2x2 + 2qx + 1 = 0, then
$$\left( {\alpha - {1 \over \alpha }} \right)\left( {\beta - {1 \over \beta }} \right)\left( {\alpha + {1 \over \beta }} \right)\left( {\beta + {1 \over \alpha }} \right)$$ is equal to :
A
$${9 \over 4}\left( {9 - {q^2}} \right)$$
B
$${9 \over 4}\left( {9 + {q^2}} \right)$$
C
$${9 \over 4}\left( {9 - {p^2}} \right)$$
D
$${9 \over 4}\left( {9 + {p^2}} \right)$$
3
JEE Main 2020 (Online) 3rd September Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
If the number of integral terms in the expansion
of (31/2 + 51/8)n is exactly 33, then the least value of n is :
A
264
B
256
C
128
D
248
4
JEE Main 2020 (Online) 3rd September Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
If $$\Delta $$ = $$\left| {\matrix{ {x - 2} & {2x - 3} & {3x - 4} \cr {2x - 3} & {3x - 4} & {4x - 5} \cr {3x - 5} & {5x - 8} & {10x - 17} \cr } } \right|$$ =

Ax3 + Bx2 + Cx + D, then B + C is equal to :
A
-1
B
-3
C
9
D
1
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