Match the statements/expressions given in Column I with the values given in Column II.

(a) \(2 \sin ^2 \theta+\sin ^2 2 \theta=2\)
\(\Rightarrow \quad \sin ^2 2 \theta=2 \cos ^2 \theta\)
\(\Rightarrow \quad 4 \sin ^2 \theta \cos ^2 \theta=2 \cos ^2 \theta\)
\(\Rightarrow \quad \cos ^2 \theta=0\) or \(\sin ^2 \theta=\frac{1}{2}\)
\(\Rightarrow \quad \cos \theta=0\) or \(\sin \theta=\pm \frac{1}{\sqrt{2}}\)
\(\Rightarrow \quad \theta=\pm \frac{\pi}{4}\) or \(\frac{\pi}{2}\)
(b) \(f(x)=\left[\frac{6 x}{\pi}\right] \cos \left[\frac{3 x}{\pi}\right]\)
Possible points of discontinuity of \(\left[\frac{6 x}{\pi}\right]\) are
\[
\begin{gathered}
\frac{6 x}{\pi}=n, n \in I \\
\Rightarrow \quad x=\frac{n \pi}{6} \Rightarrow x=\frac{\pi}{6}, \frac{\pi}{3}, \frac{\pi}{2}, \pi \\
\lim _{x \rightarrow \pi^{-} / 6} f(x)=0 \cos 0=0 \\
\lim _{x \rightarrow \pi^{+} / 6} f(x)=1 \cos 0=1
\end{gathered}
\]
\(\therefore\) Discontinuous at \(x=\frac{\pi}{6}\).
Similarly, discontinuous at \(x=\frac{\pi}{3}, \frac{\pi}{2}, \pi\).
(c) Here, \(V=\left\|\begin{array}{lll}1 & 1 & 0 \\ 1 & 2 & 0 \\ 1 & 1 & \pi\end{array}\right\|=\pi\) cubic unit
(d) Given, \(\mathbf{a}+\mathbf{b}+\sqrt{3} \mathbf{c}=\mathbf{0}\)
\[
\begin{array}{ll}
\Rightarrow & \mathbf{a}+\mathbf{b}=-\sqrt{3} \mathbf{c} \\
\Rightarrow & |\mathbf{a}+\mathbf{b}|^2=\mid \sqrt{3} \mathbf{c}^2
\end{array}
\]

\[
\begin{array}{lrl}
\Rightarrow & a^2+b^2+2 \mathbf{a} \cdot \mathbf{b}=3 c^2 \\
\Rightarrow & 2+2 \cos \theta=3 \\
\Rightarrow & \cos \theta=\frac{1}{2} \Rightarrow \theta=\frac{\pi}{3}
\end{array}
\]