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
$\frac{1}{2}$
The position vectors of two adjacent sides $\overrightarrow{O A}$ and $\overrightarrow{O B}$ of a rectangle $O A C B$ are $\vec{a}$ and $\vec{b}$ respectively, where $O$ is the origin. If $16|\vec{a} \times \vec{b}|=3(|\vec{a}|+|\vec{b}|)^2$ and $\theta$ be the acute angle between the diagonals $O C$ and $A B$, then the value of $\tan \left(\frac{\theta}{2}\right)$ is
$\frac{1}{3}$
$\frac{1}{\sqrt{3}}$
$\sqrt{3}$
1
The point of intersection of $\vec{r} \times \vec{a}=\vec{b} \times \vec{a}$ and $\vec{r} \times \vec{b}=\vec{a} \times \vec{b}$, where $\vec{a}=\hat{i}+\hat{j}$ and $\vec{b}=2 \hat{i}-\hat{k}$ is
$3 \hat{i}+2 \hat{j}+\hat{k}$
$\hat{i}-\hat{j}-\hat{k}$
$4 \hat{i}+2 \hat{j}-\hat{k}$
$3 \hat{i}+\hat{j}-\hat{k}$
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