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

If $S_{r} = \left|\begin{array}{ccc} 2r & x & n(n+1) \\ 6r^{2}-1 & y & n^{2}(2n+3) \\ 4r^{3}-2nr & z & n^{3}(n+1) \end{array}\right|$, then the value of $\sum_{r=1}^{n} S_{r}$ is independent of

Let $f(x) = \left| \begin{array}{ccc} 2\cos^2 x & \sin(2x) & -\sin x \\ \sin(2x) & 2\sin^2 x & \cos x \\ \sin x & -\cos x & 0 \end{array} \right|$. Then,evaluate $\int_{0}^{\frac{\pi}{2}} [f(x) + f'(x)] dx$.

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 $f(x) = \begin{vmatrix} \sin x & \cos x & \tan x \\ x^3 & x^2 & x \\ 2x & 1 & 1 \end{vmatrix}$,then $\lim_{x \to 0} \frac{f(x)}{x^2}$ is

The rank of the matrix $\left[\begin{array}{cccc}3 & 2 & 1 & -4 \\ 2 & 3 & 0 & -1 \\ 1 & -6 & 3 & -8\end{array}\right]$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module