A) On x
B) On n
C) Both on x and n
D) None of these
Correct Answer: B
Solution :
\[\Delta =\left| \,\begin{matrix} 1 & 1 & 1 \\ \cos nx & \cos (n+1)x & \cos (n+2)x \\ \sin nx & \sin (n+1)x & \sin (n+2)x \\ \end{matrix}\, \right|\] Applying \[{{C}_{1}}\to {{C}_{1}}+{{C}_{3}}-(2\cos x){{C}_{2}}\] \[\Delta =\left| \,\begin{matrix} 2(1-\cos x) & 1 & 1 \\ 0 & \cos (n+1)x & \cos (n+2)x \\ 0 & \sin (n+1)x & \sin (n+2)x \\ \end{matrix}\, \right|\] \[\Delta =2(1-\cos x)[\cos (n+1)x\sin (n+2)x\] \[-\cos (n+2)x\sin (n+1)x]\] \[\Delta =2(1-\cos x)\,[\sin (n+2-n-1)x]\]\[=2\sin x(1-\cos x)\] i.e., \[\Delta \] is independent of n.You need to login to perform this action.
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