`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)

Solve : sqrt(9+2x)-sqrt(2x)=(5)/(sqrt(9+2x)) .

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

The set of values of a for which x^2+ax+sin^-1 (x^2-4x+5) +cos^-1 (x^2-4x+5)=0 has at least one solution is (A) (-oo, -sqrt(2pi)] cup[sqrt[(2pi, oo] (B) (-oo, -sqrt(2pi)] cup[sqrt((2pi, oo) (C) R (D) none of these

If int_ (0)^(oo) e^(-x^(2)) dx = (sqrt (pi))/(2), and int_ (0)^(oo) e^(-ax^(2) ) dx, a> 0 is (i) (sqrt (pi))/(2) (ii) (sqrt (pi))/(2a) (iii) 2 (sqrt (pi))/(a) (iv) (1)/(2) sqrt ((pi)/(a))

If (dy)/(dx) +y sec x=tan x," then (sqrt(2) +1) y((pi)/(4)) - y(0)=, (A) sqrt(2) - (pi)/(4) (B) sqrt(2) + (pi)/(4) (C) sqrt(2) - (pi)/(2) (D) sqrt(2) + (pi)/(2)

range of 3tan sqrt((pi^(2))/(9)-x^(2)) is

If f(x)=sqrt(1+cos^(2)(x^(2))), then f'((sqrt(pi))/(2)) is (sqrt(pi))/(6)(b)-sqrt(pi/6)1/sqrt(6)(d)pi/sqrt(6)

int_(0)^(pi//2)((pi)/(4)-x)/(sqrt(sin x)+ sqrt(cos x))dx=

lim_(x rarr1^(-))(sqrt(pi)-sqrt(2sin^(-1)x))/(sqrt(1-x))=(A)sqrt((2)/(pi))(B)sqrt((pi)/(2))(C)(1)/(pi)(D)sqrt((1)/(pi))