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 \\…

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 $\left(f(2019)^{f(2020)}-f(2020)^{f(2019)}=\right.$
  1. 1
  2. –1
  3. 0
  4. 2

Solution

$ \begin{aligned} & (c) f(x)=\left|\begin{array}{lll} \cos (x+a+b) & \sin (x+a+b) & 10 \\ \cos (x+b+c) & \sin (x+b+c) & 10 \\ \cos (x+a+c) & \sin (x+a+c) & 10 \end{array}\right| \\ & R_2 \rightarrow R_2-R_1 \\ & R_3 \rightarrow R_3-R_1 \end{aligned} $ $ \begin{aligned} & 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+a+c)-\cos (x+a+b) & \sin (x+a+c)-\sin (x+a+b) & 0 \end{array}\right| \\ & f(x)=\left|\begin{array}{ccc} \cos (x+a+b) & \sin (x+a+b) & 10 \\ -2 \sin \left(x+b+\frac{a+c}{2}\right) \sin \left(\frac{c-a}{2}\right) & 2 \cos \left(x+b+\frac{a+c}{2}\right) \sin \left(\frac{c-a}{2}\right) & 0 \\ -2 \sin \left(x+a+\frac{b+c}{2}\right) \sin \left(\frac{c-b}{2}\right) & 2 \cos \left(x+a+\frac{b+c}{2}\right) \sin \left(\frac{c+b}{2}\right) & 0 \end{array}\right| \\ & =4 \sin \left(\frac{c-b}{2}\right) \sin \left(\frac{c-a}{2}\right)\left|\begin{array}{ccc} \cos (x+a+b) & \sin (x+a+b) & 10 \\ -\sin \left(x+b+\frac{a+c}{2}\right) & \cos \left(x+b+\frac{a+c}{2}\right) & 0 \\ -\sin \left(x+a+\frac{b+c}{2}\right) & \cos \left(x+a+\frac{b+c}{2}\right) & 0 \end{array}\right| \\ & =4 \sin \left(\frac{c-b}{2}\right) \sin \left(\frac{c-a}{2}\right) \times 10\left[\sin \left(x+a+\frac{b+c}{2}\right) \cos \left(x+b+\frac{a+c}{2}\right)-\right. \\ & \left.\sin \left(x+b+\frac{a+c}{2}\right) \cos \left(x+a+\frac{b+c}{2}\right)\right] \\ & \end{aligned} $ $ \begin{aligned} = & 40 \sin \left(\frac{c-b}{2}\right) \sin \left(\frac{c-a}{2}\right) \sin \left(\frac{a-b}{2}\right)=\text { constant } \\ \therefore \quad f(x)= & \text { constant } \\ & f(2019)^{f(2020)}-f(2020)^{f(2019)}=0 \end{aligned} $ Hence, option (3) is correct

Asked in: AP EAMCET 2020 (22 Sep Shift 2)

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