1
WB JEE 2017
MCQ (Single Correct Answer)
+1
-0.25
Change Language
Let $$A = \left( {\matrix{ {x + 2} & {3x} \cr 3 & {x + 2} \cr } } \right),\,B = \left( {\matrix{ x & 0 \cr 5 & {x + 2} \cr } } \right)$$. Then all solutions of the equation det (AB) = 0 is
A
1, $$-$$1, 0, 2
B
1, 4, 0, $$-$$2
C
1, $$-$$1, 4, 3
D
$$-$$1, 4, 0, 3
2
WB JEE 2017
MCQ (Single Correct Answer)
+1
-0.25
Change Language
The value of det A, where $$A\, = \left( {\matrix{ 1 & {\cos \theta } & 0 \cr { - \cos \theta } & 1 & {\cos \theta } \cr { - 1} & { - \cos \theta } & 1 \cr } } \right)$$, lies
A
in the closed interval [1, 2]
B
in the closed interval [0, 1]
C
in the open interval (0, 1)
D
in the open interval (1, 2)
3
WB JEE 2017
MCQ (Single Correct Answer)
+1
-0.25
Change Language
Let $$f:R \to R$$ be such that f is injective and $$f(x)f(y) = f(x + y)$$ for $$\forall x,y \in R$$. If f(x), f(y), f(z) are in G.P., then x, y, z are in
A
AP always
B
GP always
C
AP depending on the value of x, y, z
D
GP depending on the value of x, y, z
4
WB JEE 2017
MCQ (Single Correct Answer)
+1
-0.25
Change Language
On the set R of real numbers we define xPy if and only if xy $$ \ge $$ 0. Then, the relation P is
A
reflexive but not symmetric
B
symmetric but not reflexive
C
transitive but not reflexive
D
reflexive and symmetric but not transitive
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