Browse Questions

# The parameter on which the value of the determinant $\begin{vmatrix}1&x&x+1\\\cos(p-d)x&\cos px&\cos(p+d)x\\\sin(p-d)x&\sin px&\sin(P+d)x\end{vmatrix}$ does not defined upon is

$(a)\;a\qquad(b)\;p\qquad(c)\;d\qquad(d)\;x$

Let $A=\begin{vmatrix}1 &a&a^2\\\cos(p-d)x&\cos px&\cos(p+d)x\\\sin(p-d)x&\sin px&\sin(p+d)x\end{vmatrix}$
Apply $C_1\rightarrow C_1+C_3$
$\Delta=\begin{vmatrix}1+\alpha^2&a&a^2\\\cos(p-d)x&\cos px&\cos(p+d)x\\\sin(p-d)x&\sin px&\sin(p+d)x\end{vmatrix}$
$\Rightarrow \Delta=\begin{vmatrix}1+a^2&a&a^2\\2\cos px\cos dx&\cos px&\cos (p+d)x\\2\sin px\cos dx&\sin px&\sin(p+d)x\end{vmatrix}$
$C_1\rightarrow C_1-(2\cos dx)C_2$
$\Delta =\begin{vmatrix}1+a^2-2a\cos dx&a&a^2\\0&\cos px&\cos(p+d)x\\0&\sin px&\sin(p+d)x\end{vmatrix}$
Expanding along $C_1$ we get,
$\Delta=(1+a^2-2a\cos dx)[\sin (p+d)x\cos px-\sin px\cos(p+d)x]$
$\Rightarrow \Delta=(1+a^2-2a\cos dx)[\sin \{(p+d)x-px\}]$
$\Rightarrow \Delta=(1+a^2-2a\cos dx)[\sin dx]$
Which is independent of p.
Hence (b) is the correct answer.