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Consider the system of equations sin x cos 2y=(a^(2)-1)^(2)+1, cos x sin 2y = a+1 The number of values of a for which the system has a solution is |
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Answer» 1 `sin X COS 2Y=(a^(@)-1)^(2)+1`, and `cos x sin 2y=a+1` ...(i) Since the L.H.S. of both the equations does not exceed 1, the given system may have solutions only for a's such that `(a^(2)-1)^(2)+1 le 1 and -1 le a +1 le 1` ...(II) `(a^(2)-1)^(2)+1 le 1` or `(a^(2)-1)^(2) le 0` or `(a^(2)-1)^(2)=0` or `a=1` For `a=1`, equation `cos x sin 2y=a+1` does not hold. Thus, `a=-1` only and we get `sin x cos 2y=1` `cos x sin 2y =0` ...(iii) `sin x cos 2y =1` `rArr sin x=1, cos 2y =1` or `sin x=-1, cos 2y=-1` for which `cos x sin 2y=0`. |
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