Question

$\Delta =\left|\begin{array}{ccc}\sin x& \sin (x+h)& \sin (x+2h)\\ \sin (x+2h)& \sin x& \sin (x+h)\\ \sin (x+h)& \sin (x+2h)& \sin x\end{array}\right|$ find $\underset{h\to 0}{lim}\left(\frac{\Delta }{h^2}\right)$

• A. $9{\sin }^{3}x$
• B. $9\cos x{\sin }^{2}x$
• C. $9\sin x{\cos }^{2}x$
• D. $3{\sin }^{2}x{\cos }^{2}x$

Answer 1

Answer: (C)

$9\sin x{\cos }^{2}x$

$\because \Delta =\left|\begin{array}{ccc}\sin x& \sin (x+h)& \sin (x+2h)\\ \sin (x+2h)& \sin x& \sin (x+h)\\ \sin (x+h)& \sin (x+2h)& \sin x\end{array}\right|$

Applying $C_2\to C_2-C_1$ and $C_3\to C_3-C_2$
then
$\Delta =\left|\begin{array}{ccc}\sin x& \sin (x+h)-\sin x& \sin (x+2h)-\sin (x+h)\\ \sin (x+2h)& \sin x-\sin (x+2h)& \sin (x+h)-\sin x\\ \sin (x+h)& \sin (x+2h)-\sin (x+h)& \sin x-\sin (x+2h)\end{array}\right|$

$=\left|\begin{array}{ccc}\sin x& 2\cos \left(\frac{2x+h}{2}\right)sin\frac{h}{2}& 2cos(x+\frac{3h}{2})sin\frac{h}{2}\\ sin(x+2h)& -2cos(x+h)sinh& 2cos(x+\frac{h}{2})\sin \frac{h}{2}\\ \sin (x+h)& 2\cos (x+\frac{3h}{2})\sin \frac{h}{2}& -2\cos (x+h)\sin h\end{array}\right|$

$\therefore \frac{\Delta }{h^2}=\left|\begin{array}{ccc}\sin x& \frac{2\cos (x+\frac{h}{2})\sin \frac{h}{2}}{2\cdot \frac{h}{2}}& \frac{2\cos (x+\frac{3h}{2})\sin \frac{h}{2}}{2\cdot \frac{h}{2}}\\ \sin (x+2h)& \frac{-2\cos (x+h)\sin h}{h}& \frac{2\cos (x+\frac{h}{2})\sin \frac{h}{2}}{2\cdot \frac{h}{2}}\\ \sin (x+h)& \frac{2\cos (x+\frac{3h}{2})\sin \frac{h}{2}}{2\cdot \frac{h}{2}}& \frac{-2\cos (x+h)\sin h}{h}\end{array}\right|$

$\underset{h\to 0}{lim}\left(\frac{\Delta }{h^2}\right)=\left|\begin{array}{ccc}\sin x& \frac{2\cos (x+0)\cdot 1}{2}& \frac{2\cos (x+0)\cdot 1}{2}\\ \sin (x+0)& -2\cos (x+0)\cdot 1& \frac{2\cos (x+0)\cdot 1}{2}\\ \sin (x+0)& \frac{2\cos (x+0)\cdot 1}{2}& -2\cos (x+0)\cdot \end{array}\right|$

$=\left|\begin{array}{ccc}\sin x& \cos x& \cos x\\ \sin x& -2\cos x& \cos x\\ \sin x& \cos x& -2\cos x\end{array}\right|$

Applying $R_2\to R_2-R_1$ and $R_3\to R_3-R_1$

$\therefore \underset{h\to 0}{lim}\left(\frac{\Delta }{h^2}\right)=\left|\begin{array}{ccc}\sin x& \cos x& \cos x\\ 0& -3\cos x& 0\\ 0& 0& -3\cos x\end{array}\right|$

$=\sin x\cdot (-3\cos x)\cdot (-3\cos x)$ ....(All elements below leading diagonal are zero)
$=9\sin x{\cos }^{2}x$