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1
JEE Main 2020 (Online) 3rd September Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
For the frequency distribution :
Variate (x) :      x1   x2   x3 ....  x15
Frequency (f) : f1    f2   f3 ...... f15
where 0 < x1 < x2 < x3 < ... < x15 = 10 and
$$\sum\limits_{i = 1}^{15} {{f_i}} $$ > 0, the standard deviation cannot be :
A
6
B
1
C
4
D
2
2
JEE Main 2020 (Online) 3rd September Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
2$$\pi $$ - $$\left( {{{\sin }^{ - 1}}{4 \over 5} + {{\sin }^{ - 1}}{5 \over {13}} + {{\sin }^{ - 1}}{{16} \over {65}}} \right)$$ is equal to :
A
$${{7\pi } \over 4}$$
B
$${{5\pi } \over 4}$$
C
$${{3\pi } \over 2}$$
D
$${\pi \over 2}$$
3
JEE Main 2020 (Online) 3rd September Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
Let [t] denote the greatest integer $$ \le $$ t. If for some
$$\lambda $$ $$ \in $$ R - {1, 0}, $$\mathop {\lim }\limits_{x \to 0} \left| {{{1 - x + \left| x \right|} \over {\lambda - x + \left[ x \right]}}} \right|$$ = L, then L is equal to :
A
1
B
2
C
0
D
$${1 \over 2}$$
4
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 + $${y{{\log }_e}\left( {{{1 + {e^{ x}}} \over 2}} \right)}$$

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