1
JEE Main 2020 (Online) 5th September Morning Slot
+4
-1
If the minimum and the maximum values of the function $$f:\left[ {{\pi \over 4},{\pi \over 2}} \right] \to R$$, defined by
$$f\left( \theta \right) = \left| {\matrix{ { - {{\sin }^2}\theta } & { - 1 - {{\sin }^2}\theta } & 1 \cr { - {{\cos }^2}\theta } & { - 1 - {{\cos }^2}\theta } & 1 \cr {12} & {10} & { - 2} \cr } } \right|$$ are m and M respectively, then the ordered pair (m,M) is equal to :
A
$$\left( {0,2\sqrt 2 } \right)$$
B
(-4, 0)
C
(-4, 4)
D
(0, 4)
2
JEE Main 2020 (Online) 5th September Morning Slot
+4
-1
Let $$\lambda \in$$ R . The system of linear equations
2x1 - 4x2 + $$\lambda$$x3 = 1
x1 - 6x2 + x3 = 2
$$\lambda$$x1 - 10x2 + 4x3 = 3
is inconsistent for:
A
exactly one positive value of $$\lambda$$
B
exactly one negative value of $$\lambda$$
C
exactly two values of $$\lambda$$
D
every value of $$\lambda$$
3
JEE Main 2020 (Online) 4th September Evening Slot
+4
-1
Suppose the vectors x1, x2 and x3 are the
solutions of the system of linear equations,
Ax = b when the vector b on the right side is equal to b1, b2 and b3 respectively. if

$${x_1} = \left[ {\matrix{ 1 \cr 1 \cr 1 \cr } } \right]$$, $${x_2} = \left[ {\matrix{ 0 \cr 2 \cr 1 \cr } } \right]$$, $${x_3} = \left[ {\matrix{ 0 \cr 0 \cr 1 \cr } } \right]$$

$${b_1} = \left[ {\matrix{ 1 \cr 0 \cr 0 \cr } } \right]$$, $${b_2} = \left[ {\matrix{ 0 \cr 2 \cr 0 \cr } } \right]$$ and $${b_3} = \left[ {\matrix{ 0 \cr 0 \cr 2 \cr } } \right]$$,
then the determinant of A is equal to :
A
$${3 \over 2}$$
B
4
C
2
D
$${1 \over 2}$$
4
JEE Main 2020 (Online) 4th September Evening Slot
+4
-1
If the system of equations
x+y+z=2
2x+4y–z=6
3x+2y+$$\lambda$$z=$$\mu$$
has infinitely many solutions, then
A
2$$\lambda$$ - $$\mu$$ = 5
B
$$\lambda$$ - 2$$\mu$$ = -5
C
2$$\lambda$$ + $$\mu$$ = 14
D
$$\lambda$$ + 2$$\mu$$ = 14
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