The true set of values of ' $K$ ' for which $\sin ^{-1}\left(\frac{1}{1+\sin ^2 x}\right)=\frac{K \pi}{6}$ may have a solution is
$\left[\frac{1}{6}, \frac{1}{2}\right]$
$\left[\frac{1}{4}, \frac{1}{2}\right]$
$[2,4]$
$[1,3]$
A mapping is selected at random from all mappings $f: A \rightarrow A$, where set $A=\{1,2,3 \ldots, n\}$. If the probability that the mapping is injective is $\frac{3}{32}$, then the value of $n$ is
8
14
3
4
Let $A=[a, \infty)$ denotes the domain, then $f:(a, \infty) \rightarrow B$, which is defined by $f(x)=2 x^3-3 x^2+6$ will have an inverse for the smallest real value of ' $a$ ' if
$\mathrm{a}=0, \mathrm{~B}=[6, \infty)$
$\mathrm{a}=2, \mathrm{~B}=[10, \infty)$
$\mathrm{a}=1, \mathrm{~B}=[5, \infty)$
$\mathrm{a}=-1, \mathrm{~B}=[5, \infty)$
If $a=\mathop {\lim }\limits_{n \to \infty } \cos ^{2 n} x,(x=n \pi)$ and $b=\mathop {\lim }\limits_{n \to \infty } \cos ^{2 n} x,(x \neq n \pi)$, then numerical value of the area of the triangle whose vertices are (a, b), (-2, 1) and (2, 1) is
2
4
1
$\frac{1}{2}$
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