A) \[{{\text{n}}^{\text{2}}}\]
B) \[f(x)=\left| \begin{matrix} 0 & x-a & x-b \\ x+a & 0 & x-c \\ x+b & x+c & 0 \\ \end{matrix} \right|\]
C) \[f(a)=0\]
D) \[f(b)=0\]
Correct Answer: C
Solution :
Let the coordinates, of point\[c=\frac{2\sqrt{3}+1}{\sqrt{3}}\]. Then, equation of tangent of \[\text{a}=\text{11},\text{b}=\text{6}\]is \[a=-11,b\ne 6\] If it passes through P, then \[a=11,b\in R\] \[\sin x+\sin y=3(\cos y-\cos x),\]\[\frac{\sin 3x}{\sin 3y}\]and \[\sqrt{8}\] Now, \[\text{\hat{j}}+\text{\hat{k}}\] \[\text{\hat{i}+}\alpha \text{\hat{j}-\hat{k}}\] \[\text{\hat{i}+\hat{j}}\] \[a=-\hat{i}+\hat{j}+2\hat{k},b=2\hat{i}-\hat{j}-\hat{k}\] \[\text{c}=-\text{2\hat{i}}+\text{\hat{j}}+\text{3\hat{k}}\] \[\text{2a}-\text{c}\] \[\text{a}+\text{b}\]Locus of a point is \[\frac{3\pi }{2}\]
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