Question

Match the expression on the left with the appropriate value on the right

Answer 1

A) $f\left(x\right)=\left|\begin{array}{ccc}\sin (x+\alpha )& \sin (x+\beta )& \sin (x+\gamma )\\ \cos (x+\alpha )& \cos (x+\beta )& \cos (x+\gamma )\\ \sin (\beta +\gamma )& \sin (\gamma +\alpha )& \sin (\alpha +\beta )\end{array}\right|$
$\Rightarrow f^{\prime }\left(x\right)=\left|\begin{array}{ccc}\cos (x+\alpha )& \cos (x+\beta )& \cos (x+\gamma )\\ \cos (x+\alpha )& \cos (x+\beta )& \cos (x+\gamma )\\ \sin (\beta +\gamma )& \sin (\gamma +\alpha )& \sin (\alpha +\beta )\end{array}\right|+\left|\begin{array}{ccc}\sin (x+\alpha )& \sin (x+\beta )& \sin (x+\gamma )\\ -\sin (x+\alpha )& -\sin (x+\beta )& -\sin (x+\gamma )\\ \sin (\beta +\gamma )& \sin (\gamma +\alpha )& \sin (\alpha +\beta )\end{array}\right|+\left|\begin{array}{ccc}\sin (x+\alpha )& \sin (x+\beta )& \sin (x+\gamma )\\ \cos (x+\alpha )& \cos (x+\beta )& \cos (x+\gamma )\\ 0& 0& 0\end{array}\right|=0+0+0=0$
$\Rightarrow f\left(x\right)$ is constant function
$\Rightarrow {\sum }_{j=-5}^{5}f\left(j\right)=0$
B) $\sin 2x=1\Rightarrow x=\frac{\pi }{4}$
$f\left(x\right)=\left|\begin{array}{ccc}0& \cos x& -\sin x\\ \sin x& 0& \cos x\\ \cos x& \sin x& 0\end{array}\right|=\left|\begin{array}{ccc}0& \cos \frac{\pi }{4}& -\sin \frac{\pi }{4}\\ \sin \frac{\pi }{4}& 0& \cos \frac{\pi }{4}\\ \cos \frac{\pi }{4}& \sin \frac{\pi }{4}& 0\end{array}\right|=\left|\begin{array}{ccc}0& \frac{1}{\sqrt{2}}& -\frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}}& 0& \frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}& 0\end{array}\right|=-\frac{1}{\sqrt{2}}(0-\frac{1}{2})-\frac{1}{\sqrt{2}}(\frac{1}{2}-0)=0$
C) $f\left(x\right)=\left|\begin{array}{ccc}\sin x& \sec x& x^2-1\\ cosecx& x\sin x& \cos x\\ \tan x& x\tan x& x^2+1\end{array}\right|$
$f(-x)=\left|\begin{array}{ccc}\sin (-x)& \sec (-x)& {(-x)}^2-1\\ cosec(-x)& (-x)\sin (-x)& \cos (-x)\\ \tan (-x)& (-x)\tan (-x)& {(-x)}^2+1\end{array}\right|=\left|\begin{array}{ccc}-\sin x& \sec x& x^2-1\\ -cosecx& x\sin x& \cos x\\ -\tan x& x\tan x& x^2+1\end{array}\right|=-\left|\begin{array}{ccc}\sin x& \sec x& x^2-1\\ cosecx& x\sin x& \cos x\\ \tan x& x\tan x& x^2+1\end{array}\right|$
As $f\left(x\right)=f(-x)\Rightarrow f\left(x\right)$ is odd function
${\int }_{-\pi /3}^{\pi /3}f\left(x\right)=0$
D) $D=\left|\begin{array}{ccc}1+{\sin }^{2}x& {\cos }^{2}x& \sin 2x\\ {\sin }^{2}x& 1+{\cos }^{2}x& \sin 2x\\ {\sin }^{2}x& {\cos }^{2}x& 1+\sin 2x\end{array}\right|$
Applying $C_1\to C_1+C_2+C_3$, we get
$D=\left|\begin{array}{ccc}1+{\sin }^{2}x+{\cos }^{2}x+\sin 2x& {\cos }^{2}x& \sin 2x\\ 1+{\sin }^{2}x+{\cos }^{2}x+\sin 2x& 1+{\cos }^{2}x& \sin 2x\\ 1+{\sin }^{2}x+{\cos }^{2}x+\sin 2x& {\cos }^{2}x& 1+\sin 2x\end{array}\right|=\left|\begin{array}{ccc}2+\sin 2x& {\cos }^{2}x& \sin 2x\\ 2+\sin 2x& 1+{\cos }^{2}x& \sin 2x\\ 2+\sin 2x& {\cos }^{2}x& 1+\sin 2x\end{array}\right|=(2+\sin 2x)\left|\begin{array}{ccc}1& {\cos }^{2}x& \sin 2x\\ 1& 1+{\cos }^{2}x& \sin 2x\\ 1& {\cos }^{2}x& 1+\sin 2x\end{array}\right|$
Again applying $R_2\to R_2-R_1,R_3\to R_3-R_1$, we get
$D=(2+\sin 2x)\left|\begin{array}{ccc}1& {\cos }^{2}x& \sin 2x\\ 0& 1& 0\\ 0& 0& 1\end{array}\right|=(2+\sin 2x)$
Hence $Max\{2+\sin 2x\}=3$