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) 1

Options

A. 9
B. 4
C. 8
D. 1

Answer

3

Explanation

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`