A) a unique solution
B) infinite number of solutions
C) no solution
D) None of the above
Correct Answer: C
Solution :
\[-\sqrt{{{\alpha }^{2}}+{{b}^{2}}}\le \alpha \sin x+b\cos x\le \sqrt{{{\alpha }^{2}}+{{b}^{2}}}\]. Given that, \[a\sin x+b\cos x=c\] We know, \[-\sqrt{{{a}^{2}}+{{b}^{2}}}\le a\sin x+b\cos x\le \sqrt{{{a}^{2}}+{{b}^{2}}}\] \[-\sqrt{{{a}^{2}}+{{b}^{2}}}\le c\le \sqrt{{{a}^{2}}+{{b}^{2}}}\] \[\left| c \right|\le \sqrt{{{a}^{2}}+{{b}^{2}}}\] But it is given \[\left| c \right|>\sqrt{{{a}^{2}}+{{b}^{2}}}\] Hence, no solution exists for a\[\sin x+b\cos x=c\].
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