A) \[\frac{1}{2|b|}\sqrt{{{d}^{2}}-{{a}^{2}}}\]
B) \[\frac{1}{2|a|}\sqrt{{{d}^{2}}-{{a}^{2}}}\]
C) \[\frac{1}{2|d|}\sqrt{{{d}^{2}}-{{a}^{2}}}\]
D) None of these
Correct Answer: B
Solution :
\[a\sin x+b\cos (x+\theta )+b\cos (x-\theta )=d\] \[\Rightarrow \] \[a\sin x+2b\cdot \cos x\cdot \cos \theta =d\] \[\Rightarrow \] \[|d|\,\,\le \sqrt{{{a}^{2}}+4{{b}^{2}}+{{\cos }^{2}}\theta }\] \[\Rightarrow \] \[\frac{{{d}^{2}}-{{a}^{2}}}{4{{b}^{2}}}\le {{\cos }^{2}}\theta \] \[\Rightarrow \] \[|\cos \theta |\,\,\ge \frac{\sqrt{{{d}^{2}}-{{a}^{2}}}}{2|b|}\]
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