InterviewSolution
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InterviewSolution
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Find the smallest positive number p for which the equation `cos (p sin x) = sin (p cos x)` has a solution `x in [0,2pi]`. |
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Answer» Correct Answer - Smallest positive value of `p =(pi)/(2sqrt(2))` Given, ` cos (p sinx ) = sin(p cos x ), AA x in [0, 2pi]` ` rArr cos (p sin x ) = cos ( (pi)/(2) - p cos x) ` ` rArr" " p sin x = 2 npi pm ((pi)/(2) - p cos x ) , ne in I ` ` " "[because cos theta = cos alpha rArr theta = 2n pi pm alpha, n in I ] ` ` rArr " " p sinx + p cos x = 2 npi + pi//2 ` or ` " " p sinx - p cos x = 2n pi - pi//2, n in I ` ` rArr " " p (sin x + cos x ) = 2 n pi + pi//2` or ` " " p (sinx - cos x ) = 2 n pi - pi //2, n in I ` ` rArr p sqrt 2 (cos ""(pi)/(4) sinx + sin ""(pi)/(4) cos x) = 2 n pi + (pi)/(2)` or ` psqrt2(cos ""(pi)/(4) sin x - sin ""(pi)/(4) cos x ) = 2n pi - (pi)/(4) , n in I ` `rArr " " psqrt 2[sin(x - pi//4)] = ((4n + 1)pi )/(2)` ` or " " psqrt2[sin(x - pi//4)] = ( 4n - 1 ) (pi)/(2), n in I` Now, `" " - 1 le sin (x + pi//4) le 1 ` ` rArr" " - psqrt2 le psqrt2 sin ( x pm pi//4) le p sqrt2` `rArr " " -psqrt2 le ((4n + 1 )*pi)/(2) le p sqrt2, n in I` or ` " "-psqrt2 le ((4n - 1 )pi)/(2) le psqrt2 , n in I ` Second inequality is always a subset of first, therefore we have to considor only first. It is sufficient to consier `n ge 0 `, because for ` n gt 0`, the solution will be same for `n ge 0.` If `n ge 0, " " - sqrt2p le ( 4n + 1) pi//2` ` rArr" " (4n + 1) pi//2 le sqrt2p ` For p to be least, n should be latest. ` rArr" " n = 0 ` ` rArr " " sqrt2 p ge pi//2 rArr p ge (pi)/(2sqrt2)` Therefore, least value of p = `(pi)/(2sqrt2)` |
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