1
JEE Main 2021 (Online) 31st August Morning Shift
+4
-1
If the following system of linear equations

2x + y + z = 5

x $$-$$ y + z = 3

x + y + az = b

has no solution, then :
A
$$a = - {1 \over 3},b \ne {7 \over 3}$$
B
$$a \ne {1 \over 3},b = {7 \over 3}$$
C
$$a \ne - {1 \over 3},b = {7 \over 3}$$
D
$$a = {1 \over 3},b \ne {7 \over 3}$$
2
JEE Main 2021 (Online) 31st August Morning Shift
+4
-1
If $${a_r} = \cos {{2r\pi } \over 9} + i\sin {{2r\pi } \over 9}$$, r = 1, 2, 3, ....., i = $$\sqrt { - 1}$$, then
the determinant $$\left| {\matrix{ {{a_1}} & {{a_2}} & {{a_3}} \cr {{a_4}} & {{a_5}} & {{a_6}} \cr {{a_7}} & {{a_8}} & {{a_9}} \cr } } \right|$$ is equal to :
A
a2a6 $$-$$ a4a8
B
a9
C
a1a9 $$-$$ a3a7
D
a5
3
JEE Main 2021 (Online) 27th August Evening Shift
+4
-1
Let $$A = \left( {\matrix{ {[x + 1]} & {[x + 2]} & {[x + 3]} \cr {[x]} & {[x + 3]} & {[x + 3]} \cr {[x]} & {[x + 2]} & {[x + 4]} \cr } } \right)$$, where [t] denotes the greatest integer less than or equal to t. If det(A) = 192, then the set of values of x is the interval :
A
[68, 69)
B
[62, 63)
C
[65, 66)
D
[60, 61)
4
JEE Main 2021 (Online) 27th August Evening Shift
+4
-1
Out of Syllabus
Let A(a, 0), B(b, 2b + 1) and C(0, b), b $$\ne$$ 0, |b| $$\ne$$ 1, be points such that the area of triangle ABC is 1 sq. unit, then the sum of all possible values of a is :
A
$${{ - 2b} \over {b + 1}}$$
B
$${{2b} \over {b + 1}}$$
C
$${{2{b^2}} \over {b + 1}}$$
D
$${{ - 2{b^2}} \over {b + 1}}$$
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