If f(x) = left| begin{array}{ccc} cos(x+a+b) | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given the determinant $f(x) = \left| \begin{array}{ccc} \cos(x+a+b) & \sin(x+a+b) & 10 \ \cos(x+b+c) & \sin(x+b+c) & 10 \ \cos(x+c+a) & \sin(x+c

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(C) Given the determinant $f(x) = \left| \begin{array}{ccc} \cos(x+a+b) & \sin(x+a+b) & 10 \ \cos(x+b+c) & \sin(x+b+c) & 10 \ \cos(x+c+a) & \sin(x+c+a) & 10 \end{array} ight|$.Applying row operations $R_2 ightarrow R_2 - R_1$ and $R_3 ightarrow R_3 - R_1$:$f(x) = \left| \begin{array}{ccc} \cos(x+a+b) & \sin(x+a+b) & 10 \ \cos(x+b+c) - \cos(x+a+b) & \sin(x+b+c) - \sin(x+a+b) & 0 \ \cos(x+c+a) - \cos(x+a+b) & \sin(x+c+a) - \sin(x+a+b) & 0 \end{array} ight|$.Expanding along the third column:$f(x) = 10 \cdot [(\cos(x+b+c) - \cos(x+a+b))(\sin(x+c+a) - \sin(x+a+b)) - (\sin(x+b+c) - \sin(x+a+b))(\cos(x+c+a) - \cos(x+a+b))]$.Using the identity $\sin(A-B) = \sin A \cos B - \cos A \sin B$,the expression inside the bracket simplifies to $\sin((x+c+a) - (x+b+c)) = \sin(a-b)$.Thus,$f(x) = 10 \sin(a-b)$,which is a constant independent of $x$.Since $f(x)$ is a constant,$f(2019) = f(2020) = k$.Therefore,$f(2019)^{f(2020)} - f(2020)^{f(2019)} = k^k - k^k = 0$.

If $f(x) = \left| \begin{array}{ccc} \cos(x+a+b) & \sin(x+a+b) & 10 \\ \cos(x+b+c) & \sin(x+b+c) & 10 \\ \cos(x+c+a) & \sin(x+c+a) & 10 \end{array} \right|$,then find the value of $f(2019)^{f(2020)} - f(2020)^{f(2019)}$.

Similar Questions

The determinant $\left| \begin{array}{ccc} ^x{C_1} & ^x{C_2} & ^x{C_3} \\ ^y{C_1} & ^y{C_2} & ^y{C_3} \\ ^z{C_1} & ^z{C_2} & ^z{C_3} \end{array} \right|$ equals:

If $y = \sin(mx)$,then the value of $\left| \begin{array}{ccc} y & y_1 & y_2 \\ y_3 & y_4 & y_5 \\ y_6 & y_7 & y_8 \end{array} \right|$ (where subscripts of $y$ denote the order of derivative) is:

If $A = \begin{vmatrix} x & 1 & 1 \\ 1 & x & 1 \\ 1 & 1 & x \end{vmatrix}$ and $B = \begin{vmatrix} x & 1 \\ 1 & x \end{vmatrix}$,then $\frac{dA}{dx}$ is equal to

If $f(x) = \left| \begin{array}{ccc} 2 \cos x & 1 & 0 \\ x - \frac{\pi}{2} & 2 \cos x & 1 \\ 0 & 1 & 2 \cos x \end{array} \right|$,then $f^{\prime}(\pi)$ is equal to

Which of the following matrices has rank $3$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module