The area of the region bounded by the lines `x=0, x=pi/2` and `f(x)=sinx, g(x)=cosx` is (A) `2(sqrt(2)+1)` (B) `sqrt(3)-1` (C) `2(sqrt(3)-1)` (D) `2(sqrt(2)-1)`

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)

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)

If f(x)=sin^2x and the composite function g(f(x))=|sinx| , then g(x) is equal to (a) sqrt(x-1) (b) sqrt(x) (c) sqrt(x+1) (d) -sqrt(x)

The area enclosed by the curves y=sinx+cosx and y=|cosx−sinx| over the interval [0,pi/2] is (a) 4(sqrt2-1) (b) 2sqrt2(sqrt2-1) (c) 2(sqrt2+1) (d) 2sqrt2(sqrt2+1)

If f(x)=sin^2x and the composite function g(f(x))=|sinx| , then g(x) is equal to sqrt(x-1) (b) sqrt(x) (c) sqrt(x+1) (d) -sqrt(x)

A solution of sin^-1 (1) -sin^-1 (sqrt(3)/x^2)- pi/6 =0 is (A) x=-sqrt(2) (B) x=sqrt(2) (C) x=2 (D) x= 1/sqrt(2)

A solution of sin^-1 (1) -sin^-1 (sqrt(3)/x^2)- pi/6 =0 is (A) x=-sqrt(2) (B) x=sqrt(2) (C) x=2 (D) x= 1/sqrt(2)

If tan^(-1)x+2cot^(-1)x=(2pi)/3, then x , is equal to (a) (sqrt(3)-1)/(sqrt(3)+1) (b) 3 (c) sqrt(3) (d) sqrt(2)

If tan^(-1)x+2cot^(-1)x=(2pi)/3, then x , is equal to (a) (sqrt(3)-1)/(sqrt(3)+1) (b) 3 (c) sqrt(3) (d) sqrt(2)

If tan^(-1)x+2cot^(-1)x=(2pi)/3, then x , is equal to (a) (sqrt(3)-1)/(sqrt(3)+1) (b) 3 (c) sqrt(3) (d) sqrt(2)