Saved Bookmarks
| 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» 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`
|
|