$ \begin{array}{l|l} \hline \text { Column I } & \text { Column II } \\ \hline \text { A. Root(s) of the…

$ \begin{array}{l|l} \hline \text { Column I } & \text { Column II } \\ \hline \text { A. Root(s) of the equation } 2 \sin ^2 \theta+\sin ^2 2 \theta=2 & \text { p. } \frac{\pi}{6} \\ \hline \begin{array}{l} \text { B. Points of discontinuity of the function } \\ f(x)=\left[\frac{6 x}{\pi}\right] \cos \left[\frac{3 x}{\pi}\right] \text {, where }[y] \text { denotes the largest } \\ \text { integer less than or equal to } y \end{array} & \text { q. } \frac{\pi}{4} \\ \hline \begin{array}{l} \text { C. Volume of the parallelopiped with its edges } \\ \text { represented by the vectors } \hat{\mathbf{i}}+\hat{\mathbf{j}}, \hat{\mathbf{i}}+2 \hat{\mathbf{j}} \text { and } \hat{\mathbf{i}}+\hat{\mathbf{j}}+\pi \hat{\mathbf{k}} \end{array} & \text { r. } \frac{\pi}{3} \\ \hline \begin{array}{l} \text { D. Angle between vectors } \mathbf{a} \text { and } \mathbf{b}, \text { where } \mathbf{a}, \mathbf{b} \text { and } \mathbf{c} \text { are } \\ \text { unit vectors, satisfying } \vec{\mathbf{a}}+\vec{\mathbf{b}}+\sqrt{3} \vec{\mathbf{c}}=\overrightarrow{0} \end{array} & \text { s. } \frac{\pi}{2} \\ \hline \end{array} $
  1. (A) q,s, (B) p,r,s,t, (C) t, (D) r
  2. (A) p,s, (B) q,r,s, (C) s, (D) q
  3. (A) q,s, (B) q,r,s, (C) t, (D) r
  4. (A) p,s, (B) p,r,s,t, (C) t, (D) q

Solution

(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} $

Asked in: JEE Advanced 2009 (Paper 2)

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