Determinants Matrices Question 58
Question: Let $ \Delta = $
$ \begin{vmatrix} \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{vmatrix} $ Then, $ \underset{h\to 0}{\mathop{\lim }}( \frac{\Delta }{h^{2}} ) $ is
Options:
A) $ 9sin^{2}x\cos x $
B) $ 3{{\cos }^{2}}x $
C) $ \sin x{{\cos }^{2}}x $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
- [b] Operating $ C_1\to C_1+C_2+C_3 $ and taking out the common factor from $ C_1, $ we have $ \Delta =\sin (x+h)(1+2cosh) $
$ \begin{vmatrix} 1 & \sin (x+h) & \sin (x+2h) \\ 1 & \sin x & \sin (x+h) \\ 1 & \sin (x+2h) & \sin x \\ \end{vmatrix} $
$ {Since,sin(x+h)+sin(x)+sin(x+2h) $
$ =\sin (x+h)(1+2cosh)} $ By operating $ R_2\to R_2-R_1 $ and $ R_3\to R_3-R_1 $ , We have $ \Delta =\sin (x+h)(1+2cosh) $
$ \begin{vmatrix} 1 & \sin (x+h) & \sin (x+2h) \\ 0 & \sin x-\sin (x+h) & \sin (x+h)-\sin (x+2h) \\ 0 & \sin (x+2h)-\sin (x+h) & \sin x-\sin (x+2h) \\ \end{vmatrix} $
$ ={ \begin{aligned} & ( 2\cos ( x+\frac{h}{2} )\sin +\frac{h}{2} )(2cos(x+h)sinh) \\ & +( 2\cos (x+\frac{3h}{2} )si{{. n\frac{h}{2} )}^{2}} \\ \end{aligned} } $ Therefore, $ \underset{h\to 0}{\mathop{\lim }}\frac{\Delta }{h^{2}}=3{{\cos }^{2}}x $ .
BETA
