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If the equation `x^(2)+12+3sin(a+bx)+6x=0` has atleast one real solution, where `a, b in [0,2pi]`, then the value of a - 3b is `(n in Z)`A. `2n pi`B. `(2n+1)pi`C. `(4n-1)(pi)/(2)`D. `(4n+1)(pi)/(2)` |
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Answer» Correct Answer - C `x^(2)+12+3 sin (a+bx)+6x =0` `rArr (x+3)^(2)+3+3sin (a+bx)=0` `rArr (x+3)^(2)+3=-3 sin (a+ bx)` `L.H.S. ge 3` but `R.H.S. le 3` `L.H.S. = R.H.S. = 3` `therefore x=-3` and `sin (a + bx) =-1` `rArr sin (a-3b)=-1` or `a-3b=(4n-1)(pi)/(2), n in Z` |
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