1
JEE Main 2020 (Online) 5th September Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
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) 4th September Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
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
$${1 \over 2}$$
3
JEE Main 2020 (Online) 4th September Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
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
4
JEE Main 2020 (Online) 4th September Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
If $$A = \left[ {\matrix{ {\cos \theta } & {i\sin \theta } \cr {i\sin \theta } & {\cos \theta } \cr } } \right]$$, $$\left( {\theta = {\pi \over {24}}} \right)$$

and $${A^5} = \left[ {\matrix{ a & b \cr c & d \cr } } \right]$$, where $$i = \sqrt { - 1} $$ then which one of the following is not true?
A
$$a$$2 - $$c$$2 = 1
B
$$0 \le {a^2} + {b^2} \le 1$$
C
$$ a$$2 - $$d$$2 = 0
D
$${a^2} - {b^2} = {1 \over 2}$$

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