The area enclosed by the curve `y=sinx+cosxa n dy=|cosx-sinx|` over the interval `[0,pi/2]` is `4(sqrt(2)-2)` (b) `2sqrt(2)` (`sqrt(2)` -1) `2(sqrt(2)` +1) (d) `2sqrt(2)(sqrt(2)+1)`

The area enclosed by the curves y=sin x+cos x and y=|cos x-sin x| over the interval [0,(pi)/(2)] is (a) 4(sqrt(2)-1) (b) 2sqrt(2)(sqrt(2)-1)(c)2(sqrt(2)+1)(d)2sqrt(2)(sqrt(2)+1)

The area of the figure bounded by the curves y=cosx and y=sinx and the ordinates x=0 and x=pi/4 is (A) sqrt(2)-1 (B) sqrt(2)+1 (C) 1/sqrt(2)(sqrt(2)-1) (D) 1/sqrt(2)

(2+sqrt(2)+(1)/(2+sqrt(2))+(1)/(sqrt(2)-2)) simplifies to 2-sqrt(2)(b)2(c)2+sqrt(2)(d)2sqrt(2)

The value of sqrt(3-2sqrt(2)) is sqrt(2)-1(b)sqrt(2)+1(c)sqrt(3)-sqrt(2)(d)sqrt(3)+sqrt(2)

The greatest value of the function f(x)=(sin2x)/(sin(x+(pi)/(4))) on the interval (0,(pi)/(2)) is (1)/(sqrt(2))(b)sqrt(2)(c)1(d)-sqrt(2)

The area formed by triangular shaped region bounded by the curves y=sinx, y=cosx and x=0 is (A) sqrt(2)-1 (B) 1 (C) sqrt(2) (D) 1+sqrt(2)

(1)/(sqrt(9)-sqrt(8)) is equal to: 3+2sqrt(2)(b)(1)/(3+2sqrt(2)) (c) 3-2sqrt(2)(d)(3)/(2)-sqrt(2)

The area under the curve y=|cosx-sinx|, 0 le x le pi/2 , and above x-axis is: (A) 2sqrt(2)+2 (B) 0 (C) 2sqrt(2)-2 (D) 2sqrt(2)

Solve sinx +cosx=sqrt2 .

If (0,1),(1,1) and (1,0) be the middle points of the sides of a triangle,its incentre is (2+sqrt(2),2+sqrt(2))( b) [2+sqrt(2),-(2+sqrt(2))](2-sqrt(2),2-sqrt(2)) (d) [2-sqrt(2),(2+sqrt(2))]