`sin^(-1)(sin5)> x^2-4x` hold if (a) `x=2-sqrt(9-2pi)` (b) `x=2+sqrt(9-2pi)` (c) `x >2+sqrt(9-2pi)` (d) `x in (2-sqrt(9-2pi),2+sqrt(9-2pi))`

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)

lim_(xrarr1^(-)) (sqrtpi-sqrt(2sin^-1x))/sqrt(1-x)= (A) sqrt(2/pi) (B) sqrt(pi/2) (C) 1/pi (D) sqrt(1/pi)

If sin^(-1)x+sin^(-1)y=pi/2 and sin2x=cos2y , then (a) x=pi/8+sqrt(1/2-(pi^2)/(64)) (b) y= sqrt(1/2-(pi^2)/(64))-pi/(12) (c) x=pi/(12)+sqrt(1/2-(pi^2)/(64)) (d) y=sqrt(1/2-(pi^2)/(64))-pi/8

int_0^(4/pi) (3x^2sin(1/x)-xcos(1/x))dx= (A) (8sqrt(2))/pi^3 (B) (32sqrt(2))/pi^3 (C) (24sqrt(2))/pi^3 (D) sqrt(2048)/pi^3

int_0^3(3x+1)/(x^2+9)dx = (pi^)/(12)+log(2sqrt(2)) (b) (pi^)/2+log(2sqrt(2)) (c) (pi^)/6+log(2sqrt(2)) (d) (pi^)/3+log(2sqrt(2))

Derivative of sin(sin x^(2)) at x = sqrt((pi)/(2)) is

Prove that : int_0^(pi/2)sqrt(1-sin2x)dx=2(sqrt(2)-1)

sin^(-1)(x^2/4+y^2/9)+cos^(-1)(x/(2sqrt2)+y/(3sqrt2)-2)

If 4\ cos^(-1)x+sin^(-1)x=pi , then the value of x is (a) 3/2 (b) 1/(sqrt(2)) (c) (sqrt(3))/2 (d) 2/(sqrt(3))

If sin^(-1) x + sin^(-1) y = pi/2 , prove that x sqrt(1-y^2) + y sqrt(1-x^2) =1 .