1
WB JEE 2019
MCQ (Single Correct Answer)
+1
-0.25
Change Language
Consider the function f(x) = cos x2. Then,
A
f is of period 2$$\pi$$
B
f is of period $$\sqrt {2\pi } $$
C
f is not periodic
D
f is of period $$\pi$$
2
WB JEE 2019
MCQ (Single Correct Answer)
+2
-0.5
Change Language
Let a > b > 0 and I(n) = a1/n $$-$$ b1/n, J(n) = (a $$-$$ b)1/n for all n $$ \ge $$ 2, then
A
I(n) < J(n)
B
I(n) > J(n)
C
I(n) = J(n)
D
I(n) + J(n) = 0
3
WB JEE 2018
MCQ (Single Correct Answer)
+1
-0.25
Change Language
The domain of definition of $$f(x) = \sqrt {{{1 - |x|} \over {2 - |x|}}} $$ is
A
$$( - \infty , - 1) \cup (2,\infty )$$
B
$$[ - 1,1] \cup (2,\infty ) \cup ( - \infty , - 2)$$
C
$$( - \infty ,1) \cup (2,\infty )$$
D
$$[ - 1,1] \cup (2,\infty )$$

Here (a, b) $$ \equiv $$ {x : a < x < b} and [a, b] $$ \equiv $$ {x : a $$ \le $$ x $$ \le $$ b}
4
WB JEE 2018
MCQ (Single Correct Answer)
+1
-0.25
Change Language
If f : R $$ \to $$ R be defined by f (x) = ex and g : R $$ \to $$ R be defined by g(x) = x2. The mapping gof : R $$ \to $$ R be defined by (gof) (x) = g[f(x)] $$\forall $$x$$ \in $$R. Then,
A
gof is bijective but f is not injective.
B
gof is injective but g is injective
C
gof is injective but g is not bijective
D
gof is surjective and g is surjective
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