Mathematics - Vector Question with Solution | TestHub

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