" If "f(x)=min(x+1,sqrt(1-x))" ,then area bounded by "f(x)" and "x" -axis is: (in sq.units) "

Let f(x)=min (x+1,sqrt(1-x)) for all xlt=1. Then the area bounded by y=f(x) and the x-axis is (a) 7/3 sq units (b) 1/6 sq units (c) (11)/6 sq units (d) 7/6 sq units

Let f(x)=min (x+1,sqrt(1-x)) for all xlt=1. Then the area bounded by y=f(x) and the x-axis is (a) 7/3 sq units (b) 1/6 sq units (c) (11)/6 sq units (d) 7/6 sq units

" Area of the region bounded by "y=2+|x|-|x+1|-|x-1|" and "x" -axis is (in sq units "

Let f(x)= minimum (x+1,sqrt(1-x) for all x<=1 then the area bounded by y=f(x) and the x -axis,is

Let f(x)= minimum (x+1,sqrt(1-x)) for all xle1 . Then the area bounded by y=f(x) and the x-axis is a) (7)/(3) sq. units b) 1/6 sq. units c) 11/6 sq. units d) 7/6 sq. units

f(x) = min { x+1 , sqrt((1-x))} the area of the region bounded by the curve f(x) and X - axis is ….Sq. unit

Area of the region bounded by y=|x| and y=1-|x| in sq.units is

Let f(x) = minimum (x+1, sqrt(1-x))" for all "x le 1. Then the area bounded by y=f(x) and the x-axis is

Find area bounded by y = f^(-1)(x), x = 4, x = 10 and the x -axis, given that the area bounded by y = f(x) , x = 2, x = 6 and the x -axis is 30 sq. units, where f(2) = 4 and f(6) = 10 , and that f(x) is an invertible function.

Let f(x) = minimum (x+1, sqrt(1-x))" for all "x le 1. Then the area bounded by y=f(x) and the x-"axis" is