The general solution of the equation $\sin ^{100} \mathrm{x}-\cos ^{100} \mathrm{x}=1$ is
$\left\{2 n \pi+\frac{\pi}{3}: n \in I\right\}$
$\left\{n \pi+\frac{\pi}{4}: n \in I\right\}$
$\left\{n \pi \pm \frac{\pi}{2}: n \in I\right\}$
$\left\{2 \mathrm{n} \pi-\frac{\pi}{3}: \mathrm{n} \in \mathrm{I}\right\}$
If $\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{b}=\hat{i}-\hat{j}+\hat{k}, \vec{c}=\hat{i}+2 \hat{j}-\hat{k}$, then the value of $\left|\begin{array}{lll}\vec{a} \cdot \vec{a} & \vec{a} \cdot \vec{b} & \vec{a} \cdot \vec{c} \\ \vec{b} \cdot \vec{a} & \vec{b} \cdot \vec{b} & \vec{b} \cdot \vec{c} \\ \vec{c} \cdot \vec{a} & \vec{c} \cdot \vec{b} & \vec{c} \cdot \vec{c}\end{array}\right|$ is equal to
64
0
14
16
Number of elements in the range set of $f(x)=\left[\frac{x}{15}\right]\left[-\frac{15}{x}\right]$, for all $x \in(0,90$ ); (where [.] denotes the greatest integer function) is
8
7
6
5
Let 10 Bags $B_1, B_2, \ldots, B_{10}$ which contains $21,22, \ldots, 30$ different articles respectively. Then the total number of ways to bring out 10 articles from a Bag is
${ }^{31} \mathrm{C}_{20}+{ }^{21} \mathrm{C}_{10}$
${ }^{31} \mathrm{C}_{20}-{ }^{21} \mathrm{C}_{10}$
${ }^{30} \mathrm{C}_{20}-{ }^{20} \mathrm{C}_{10}$
${ }^{30} \mathrm{C}_{20}+{ }^{20} \mathrm{C}_{10}$
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