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InterviewSolution
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The least value of `a`for which the equation `4/(sinx)+1/(1-sinx)=a`has at least one solution in the interval `(0,pi/2)`9 (b)4 (c) 8(d) 1A. 9B. 4C. 8D. 1 |
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Answer» Correct Answer - 3 Since a =`((4)/(sinx)+(1)/(1-sinx))`,a is atleast `therefore (da)/(dx)=[(-4)/(sin^(2)x)+(1)/(1-sinx)^(2)]cosx=0` we have to find the value of x in the interval `(0,pi//2)` Thus cos `x ne 0` and the other factor when equated to zero gives sinx =`2//3` Now Put sin x=`2/3 and cos^(2) x=1-4/9=5/9` `therefore (d^(2)a)/(dx^(2)=0+[(8)/(8//27)+2xx27]5/9=81xx8/9=45gt0` Thus a is minimum and its value is `(4)/(2//3)+(1)/(1-2//3)=6+3=9` |
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