1.

Let y=f(x) be the solution of the differential equation (dy)/(dx)+k/7 x tan x=1+xtanx-sinx, where f(0)=1 and let k be the minimum value of g(x) where g(x)="max"|(sqrt(193)-1)/2cosy+cos(y+(pi)/3)-x| where yepsilonR then Area bounded by y=f(x) and its inverse between x=(pi)/2 and x=(7pi)/2 is

Answer»

12
6
9
8

Solution :`-7 LE ((sqrt(193)-1))/2 cosy+cos(y+(pi)/3) le 7`
So, `g(x)={(|x+7|, ,, xge0),(|x-7|, ,, XLT0):}`
So, `g(x)_("min")=7`
So, `f(x)=x+cosx`s
`A = 3xx2 int_(pi//2)^(3pi//2) (x-(x+cos x)dx)`
`=-3xx2(SIN x)_(pi//2)^(3pi//2)`
`= -6(-1-1)=12`


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