1
JEE Main 2020 (Online) 5th September Morning Slot
MCQ (Single Correct Answer)
+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 :
$$f\left( \theta \right) = \left| {\matrix{ { - {{\sin }^2}\theta } & { - 1 - {{\sin }^2}\theta } & 1 \cr { - {{\cos }^2}\theta } & { - 1 - {{\cos }^2}\theta } & 1 \cr {12} & {10} & { - 2} \cr } } \right|$$ are m and M respectively, then the ordered pair (m,M) is equal to :
2
JEE Main 2020 (Online) 4th September Evening Slot
MCQ (Single Correct Answer)
+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 :
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 :
3
JEE Main 2020 (Online) 4th September Evening Slot
MCQ (Single Correct Answer)
+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
x+y+z=2
2x+4y–z=6
3x+2y+$$\lambda $$z=$$\mu $$
has infinitely many solutions, then
4
JEE Main 2020 (Online) 4th September Morning Slot
MCQ (Single Correct Answer)
+4
-1
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?
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?
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